اﻟﻔﺻل اﻟﺧﺎﻣس اﻟﺗﺣﻠﯾل اﻟﺑﯾﯾزى ﻟﺑﻌض اﻟﺗوزﯾﻌﺎت ﺗﺣت اﺧﺗﺑﺎرات اﻟﺣﯾﺎه
٣٠١
) (١-٥اﻟﺗوزﯾﻊ اﻻﺳﻰ ﺑﻣﻌﻠﻣﺔ واﺣدة ) (١-١-٥ﻓﺗرة ﺛﻘﺔ ﻟﻣﺗوﺳط اﻟﺣﯾﺎة وﺗﻘدﯾر داﻟﺔ اﻟﺻﻼﺣﯾﺔ ﺑﺎﻻﻋﺗﻣﺎد ﻋﻠﻰ داﻟﺔ اﻻﻣﻛ ﺎن ﻓ ﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺔ اﻟﻛﺎﻣﻠﺔ )ا( ﻓﺘﺮة ﺛﻘﺔ ﻟﻤﺘﻮﺳﻂ اﻟﺤﻴﺎﻩ)اﻟﻤﻌﻠﻤﺔ (
ﺑﻔﺮض أن nﻣﻦ اﻟﻮﺣﺪات وﺿﻌﺖ ﻟﻼﺧﺘﺒﺎر وان اﻻﺧﺘﺒﺎر ﻳﻨﺘﻬﻲ ﺑﻌﺪ ﻓﺸﻞ ﻛﻞ اﻟﻮﺣﺪات .ﺑﻔﺮض أن X1 , X 2 , , X nﻋﻴﻨﺔ
ﻋﺸﻮاﺋﻴﺔ ﲤﺜﻞ أزﻣﻨﺔ اﻟﻔﺸﻞ وان أزﻣﻨﺔ اﻟﻔﺸﻞ ﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ ﺑﺪاﻟﺔ ﻛﺜﺎﻓﺔ اﺣﺘﻤﺎﻟﻴﺔ: x,>0.
1 x e
,
= )f(x;
وﺑﺎﻋﺘﺒﺎر أن ﻣﺸﺎﻫﺪات اﻟﻌﻴﻨﺔ x1 , x 2 , , x nﰎ اﳊﺼﻮل ﻋﻠﻴﻬﺎ واﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة . داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : xi
n
1 L(x1 , x 2 ,..., x n | ) L n e i1 .
ﻣﻘﺪر اﻹﻣﻜﺎن اﻻﻛﺒﺮ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ اﻟﺤﻞ ﻟﻠﻤﻌﺎدﻟﺔ :
ln L 0 . then : n i
.
x i 1
ln L = - n ln -
n
xi ln L n i 1 2 .
ln L ﺑﻮﺿﻊ 0 ˆ
ﻓﺎن :
n xi 2 0. ˆ ˆ n
x
n i 1 i 2 , ˆ ˆ n i
=x .
x i 1
n
ˆ
٣٠٢
ﻹﻳﺠﺎد ﺗﻮزﻳﻊ ˆ ) اﻟﻤﻀﺒﻮط ( exactوﺗﻮﻗﻌﻪ وﺗﺒﺎﻳﻨﻪ ﻳﺘﺒﻊ اﻟﺘﺎﻟﻰ : n i
X
)
i 1
n
(E(ˆ ) E(X) E
1 n 1 E(X i ) n . n i1 n أي ان ˆ ﻣﻘﺪر ﻏﻴﺮ ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ . اﻟﺘﺒﺎﻳﻦ ﻟﻠﻤﻘﺪر ˆ ﻳﺤﺴﺐ ﻛﺎﻟﺘﺎﻟﻲ :
1 n ) Xi n i 1
(Var(ˆ ) = Var(X) = Var n
) Var(X i
i 1
1 n2
1 2 2 n = . n2 n
= =
ﺑﻤـﺎ ان X1 , X 2 , , X nﻣﺘﻐﻴــﺮات ﻋﺸـﻮاﺋﻴﺔ ﻣﺴــﺘﻘﻠﺔ وﻟﻬــﺎ ﻧﻔـﺲ اﻟﺘﻮزﻳــﻊ ﻓـﻴﻤﻜﻦ إﻳﺠـﺎد ﺗﻮزﻳــﻊ ˆ ﻣــﻦ اﻟﺪاﻟـﺔ اﻟﻤﻮﻟــﺪة ﻟﻠﻌــﺰوم ﻛﺎﻟﺘﺎﻟﻲ : M Xi (t) (1 t) 1 ,i 1, 2,...n, M n (t) (1 t) n . Xi i1
t n ) , n
(t) (1
n
M
Xi i1 n
n
n
واﻟﺘ ــﻲ ﺗﻤﺜ ــﻞ اﻟﺪاﻟ ــﺔ اﻟﻤﻮﻟ ــﺪة ﻟﻠﻌ ــﺰوم ﻟﻤﺘﻐﻴ ــﺮ ﻋﺸ ــﻮاﺋﻲ ﻳﺘﺒ ــﻊ ﺟﺎﻣ ــﺎ ﺑﻤﻌﻠﻤﺘ ــﻴﻦ ) ( , nاى ) . G( , nﺳ ــﻮف ﻧﺜﺒ ــﺖ ان اﻟﻤﻘﺪر ˆ ﻫﻮ ﻣﻘﺪر ﻏﻴﺮ ﻣﺘﺤﻴـﺰ ﺑﺄﻗـﻞ ﺗﺒـﺎﻳﻦ أي MVUEوذﻟـﻚ ﺑﺎﺗﺒـﺎع اﻟﺨﻄـﻮات اﻟﺘﺎﻟﻴـﺔ .داﻟـﺔ ﻛﺜﺎﻓـﺔ اﻻﺣﺘﻤـﺎل ﻟﻠﻤﺘﻐﻴـﺮ ˆ ﺗﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﻟﻲ :
, ˆ > 0.
n ( )n ˆn n 1 ˆ ˆ f() e )(n
2 ln L n xi 2 2 3 , 2 2 ln L n 2n n (E ) 2 3 2 2 2 2 1 ˆ) . (Var 2 ln L n n ( E ) 2
ﻹﺛﺒﺎت أن ˆ إﺣﺼﺎء ﻛﺎﻓﻲ ﻧﺘﺒﻊ اﻵﺗﻲ :
٣٠٣
xi 1 L= n e ,
xi
n
u= x i ,
,
)
L = (1).(-n e
i 1
N(x) . K(u,).
ﺣﻴﺚ N(x) =1و
xi
. K(u,)=-n e n
U sufficient for .
i
X
i 1
ˆ sufficient for . n U ˆ =U وﺑﻤﺎ ان Xi إﺣﺼﺎء ﻛﺎﻓﻲ ﻟﻠﻤﻌﻠﻤﺔ وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن إﺣﺼﺎء ﻛﺎﻓﻲ أﻳﻀﺎ )ﻻن اى داﻟﺔ ﻓﻰ اﺣﺼﺎء ﻛﺎﻓﻰ n i 1
ﻫﻰ اﺣﺼﺎء ﻛﺎﻓﻰ( .وﻳﻤﻜﻦ اﺛﺒﺎت ذﻟﻚ ﺑﻄﺮﻳﻘﺔ اﺧﺮى وذﻟﻚ ﺑﺈﺳﺘﺨﺪام ﺑﺎﺳﺘﺨﺪام اﻻﺣﺘﻤﺎل اﻟﺸﺮﻃﻲ ﻛﺎﻟﺘﺎﻟﻰ : )L(x, u , )h(u
g(x | u)
ﺣﻴﺚ ) h(uداﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻺﺣﺼﺎء . Uﻳﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﻰ ﺗﻮزﻳﻊ Uﻛﺎﻟﺘﺎﻟﻰ : n
Xi E xp() U= Xi G(, n), i 1
u > 0.
1 ( )n u h(u) = u n-1 e ) (n
,
)L(x, u )h(u
=) g(x|u
u
)(n , u n 1
1 e n u
1 u n 1e ) (n
=
n
U ˆ إﺣﺼــﺎء ﻛــﺎﻓﻲ ﺣﻴــﺚ ان ) g(x|uﻻ ﻳﻌﺘﻤــﺪ ﻋﻠــﻰ أي أن U= Xiإﺣﺼــﺎء ﻛــﺎﻓﻲ ﻟﻠﻤﻌﻠﻤــﺔ وﺑﺎﻟﺘــﺎﻟﻲ ﻓــﺎن n
ﻟﻠﻤﻌﻠﻤﺔ .
ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﻓﺘﺮات ﺛﻘـﺔ أو اﺧﺘﺒـﺎرات ﻓـﺮوض ﻧﺤﺘـﺎج ﻹﻳﺠـﺎد ﻗﻴﻤـﺔ ﻣﺤﻮرﻳـﺔ )داﻟـﺔ ﻓـﻰ اﻟﻤﻌﻠﻤـﺔ واﻟﻤﻘـﺪر ﺗﻮزﻳﻌﻬـﺎ ﻻ ﻳﻌﺘﻤـﺪ ﻋﻠﻰ اﻟﻤﻌﻠﻤﺔ ( ﻟﺬﻟﻚ ﺳﻨﺴﺘﺨﺪم اﻟﻌﻼﻗﺔ ﺑﻴﻦ ﺗﻮزﻳﻊ اﻟﺠﺎﻣﺎ وﺗﻮزﻳﻊ ﻣﺮﺑﻊ ﻛﺎي ﺣﻴﺚ :
٣٠٤
1 u ( )n h(u) = u n-1 ee (n)
,
u > 0.
2u z
let
z u 2
du =
dz . 2
1 ( )n z z g1 (z) ( ) n 1 e 2 (n) 2 2 1 ( )n () n z z z n 1 n 1 2 2 z e e (n) 2 n (n) 2n Z 22n .
:وﻣﻨﻬﺎ ﻳﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﻰ ﻓﺘﺮة ﺛﻘﺔ ﻛﺎﻟﺘﺎﻟﻰ P( 2
1 2
2nˆ 2 ) 1 . 2
1
2
2
1
2
2
2
2
2
٣٠٥
1 1 2 ) 1 . 2 2nˆ 2
1
( P
2
ˆ2n ˆ2n 2 ) 1 . 2 2
1
( P
2
ﺣﻴﺚ :
ˆ 2nˆ 2n , 2 2 ﻳﻤﺜﻼن اﻟﺤﺪﻳﻦ اﻻدﻧﻰ واﻻﻋﻠﻰ ﻟﻔﺘﺮة اﻟﺜﻘﺔ ﻟـ ﺣﻴﺚ 1 2 2
2
2 , 2ﺗﺴﺘﺨﺮﺟﺎن ﻣﻦ ﺟﺪول ﺗﻮزﻳﻊ ﻣﺮﺑﻊ
1
2
ﻛﺎى ﻣﻦ اﻟﻤﻠﺤﻖ ) (٥ﻋﻨﺪ درﺟﺎت ﺣﺮ ﻳﺔ . 2n
)ا(ﺗﻘﺪﻳﺮ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﺻﻼﺣﻴﺔ اﻟﻮﺣﺪة )أو اﻟﻨﻈﺎم( ﺗﻌﺮف ﺑﺄ ﺎ اﺣﺘﻤﺎل أن اﻟﻮﺣﺪة ﺻﺎﳊﺔ ﻟﻠﻌﻤﻞ ﻋﻠﻰ اﻷﻗﻞ ﻟﻔﱰة ﻣﻌﻴﻨﺔ ﻣﻦ اﻟﺰﻣﻦ. R t0 P X t0 1 F t 0 .
ﻣﺜﺎل ﺑﻔﺮض اﻧﻪ ﰎ اﺧﺘﻴﺎر ﻋﻴﻨﻪ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ 10ﻣﺼﺎﺑﻴﺢ وﺣﺴﺒﺖ أزﻣﻨﺔ ﻓﺸﻠﻬﺎ وﻛﺎﻧﺖ: 125,189,210,351,465,580,630,760,810,870 إذا ﰎ ﺷﺮاء ﻣﺼﺒﺎح ﻣﻦ ﻧﻔﺲ اﻟﻨﻮﻋﻴﺔ ،ﻓﻤﺎ اﺣﺘﻤﺎل اﻧﻪ ﺳﻴﻈﻞ ﻳﻌﻤﻞ ﺣﱴ ﻋﻠﻰ اﻷﻗﻞ 600ﺳﺎﻋﺔ ؟ اﻟﺣــل:
اوﻻ :ﻋﺪم ﺗﻌﻴﲔ ﺗﻮزﻳﻊ ﻷزﻣﻨﺔ اﻟﻔﺸﻞ ﳝﻜﻦ ﺗﻘﺪﻳﺮ اﻻﺣﺘﻤﺎل ﻛﻤﺎ ﻳﻠﻲ: ﻋﺪد اﳌﺼﺎﺑﻴﺢ اﻟﱵ ﺗﻌﻤﻞ ﺣﱴ اﻟﺰﻣﻦ 600 ____________________________________ = R * t 0 ﻋﺪد اﳌﺼﺎﺑﻴﺢ اﳌﺴﺘﺨﺪﻣﺔ ﰲ اﻻﺧﺘﺒﺎر ٣٠٦
=
4 . 10 x 499
: ﺣﻴﺚ : ﺗﻌﻴﲔ اﻟﺘﻮزﻳﻊ اﻵﺳﻲ ﻛﺘﻮزﻳﻊ ﻷزﻣﻨﺔ اﻟﻔﺸﻞ ﻟﻠﻤﺼﺎﺑﻴﺢ: ﺛﺎﻧﻴﺎ x
1 R t 0 e dx t0 t0
e
. t0 x
ˆ x Rˆ t 0 e
.
600 Rˆ 600 e 499 0.3006.
: ﻧﺘﺒﻊ اﻟﺘﺎﱃ
ˆ ) R(t 0
ﻻﳚﺎد ﺗﻮﻗﻊ
X G ,n n
ˆ ) R(t R(t 0 ˆ 0 ) g x dx 0
e 0
t 0 x
nx
n
1 n n 1 x e dx n
1 n n
n
x
n 1 t 0 nx x
e
dx.
0
: وﻟﻜﻦ داﻟﺔ ﺑﺴﻞ ﻫﻲ
٣٠٧
r 1
b ax r x
a 2 x e dx 2 k r 1 2 ab 0 b nt k r k r , a 1, b r n 1 r n 1 r 1 n
n nt 1 2 ˆ R(t) 2 k n 2 n nt
n
2 nt 2 nt k n 2 . n
. ﻣﻘﺪر ﻟﻴﺲ ﻟﻪ أﻗﻞ ﺗﺒﺎﻳﻦRˆ t 0 اى ان ﻣﻘﺪر داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻣﺘﺤﻴﺰ اى ان ˆ ) ﻻﳚﺎد ﺗﺒﺎﻳﻦ : ﻧﺘﺒﻊ اﻟﺘﺎﱃR(t 0
ˆ ) R(t 0
2
e
2t 0 x
n
0
n
n 1 n x e
nx
dx.
nx y. y x dy dx. n n then : let
ˆ ) R(t 0
2
1 n n
2nt
n
n 1
2nt 0 n y n
n 1 y y e 0 n n
y 1 y n 1e y dy. n 0
2nt , n 1 r. n r 1
a 1, b
n r 1.
٣٠٨
n 2 2nt 0 2nt 0 ˆ R(t 0 ) 2 k 2 n n 1 2nt 0 2 2nt 0 2 k n 2 . n
2
2
ˆ ) R(t ˆ ˆ V ar R(t 0 0 ) R t 0
2
n
n 2 2nt 0 2 2nt 0 4 nt 0 2 nt 0 k n 2 kn 2 n n 2 n n 2 2nt 0 2 2nt 0 2 n2 nt 0 2 2 nt 0 k 2 2 k 2 n n n n 2 n n n 2 2nt 0 2 2nt 0 2 2 nt 0 2 2 nt 0 kn 2 kn 2 . n n
: ﻧﺘﺒﻊ اﻟﺘﺎﱃMVUE ﻹﳚﺎد اﳌﻘﺪر اﻟﻐﲑ ﻣﺘﺤﻴﺰ ﺑﺄﻗﻞ ﺗﺒﺎﻳﻦ أزﻣﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻟﻠﻔﺸﻞ وﻛﻞ ﻣﻨﻬﺎ ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻷﺳﻰ ﻋﻠﻰX1 ,, X n إذا ﻛﺎﻧﺖ : اﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ x
1 f x e
, x 0, 0.
MVUE ﻟﺪاﻟﻪ اﻟﺼﻼﺣﻴﺔ ﲝﻴﺚ ﻳﻜﻮنR t 0 اﳌﻄﻠﻮب إﳚﺎد اﳌﻘﺪر : ﻣﻌﺮﻓﺔ ﻛﺎﻷﰐG x1 , x 2 ,..., x n ﺑﻔﺮض أن اﻟﺪاﻟﺔ G x1 1 x1 t 0
e.w.
t G X1 1 P X1 t (0)P X1 t R 0 .
. R t 0 ﻣﻘﺪر ﻏﲑ ﻣﺘﺤﻴﺰ ﻟﺪاﻟﻪ اﻟﺼﻼﺣﻴﺔG X1 أي أن ﻓﺈﻧﻨﺎ ﻧﻌﺮف اﻟﺘﻘﺪﻳﺮ اﻟﺘﺎﱃ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ واﻟﺬى ﻟﻪ ﻟﻠﻤﻌﻠﻤﺔMVUE إﺣﺼﺎء ﻛﺎﰲ وX ˆ وﺣﻴﺚ أن :MVUE اﻟﺼﻔﺔ
٣٠٩
t G(X ) | x) R 0 1
G x1 g1 (x1 | x)dx1 t0
g1 (x1 | x)dx1. t0
. X1 Y1 و (, n 1) ﲟﻌﺘﲔ
X 2 , X3 ,..., X n Y2 , Y3 ,..., Yn
ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ
n
ﺣﻴﺚ
X1
و
X 2 , X 3 ,..., X n
ﻳﺘﻢ ﺗﻘﺴﻴﻢ اﻟﻌﻴﻨﺔ إﱃ ﻗﺴﻤﲔ
ﺳﻮف ﻧﺜﺒﺖ ﺑﺎﺳﺘﺨﺪام اﻟﺪاﻟﺔ اﳌﻮﻟﺪة ﻟﻠﻌﺰوم ان اﳌﺘﻐﲑ ﻋﺸﻮاﺋﻲ
Y Yi i 2
: ﻛﺎﻟﺘﺎﱃ 1
n
n
Yi i 2
t 1 t i2
1 t
(n 1)
.
n
Y
i
Y
i2
n 1
.
Y t n t 1 Yi n 1 i 2
n 1
.
n 1
.
, n 1 n 1
إذن اﻟﺘﻮزﻳﻊ اﳌﺸﱰك ﻟﻠﻤﺘﻐﲑﻳﻦ. ﻣﺴﺘﻘﻠﲔ g 2 x1 , y f1 x1 f 2 y
Y, X1
Y, X1
n 1 y
n 1
x
ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ
1 1 n 1 n2 e y e n 1n 1
n 1
n 1
n n 1
y
n 2
e
x n 1 y 1
.
G x1 | x G x1 g1 x1 | x dx1 t0
g1 x1 | x dx1 t0
g 3 x1 , x dx1. g4 x t0
g 2 x1 , y
n 1
n 1
n 1n
y
٣١٠
n2
e
1 n 1 y x1
.
ﻣﺴﺘﻘﻠﲔ ﻓﺈن
Y
أي أن
Y, X1
ﲟﺎ أن
nx x1 n 1 y nx x1 . n 1 1 0 n . 1 n n 1 n 1 n 1
y
J=
g 3 x1 , x
n 1
n 1
nx x 1 n n 1 n 1 n 1
n2
n n 1 nx x 1 n n 1 n 1 n 1
n 2
e
n x
: ﺣﻴﺚ
n 2
e
nx
n n 1
.
,n n
ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ
X
وﻟﻜﻦ
nx
g4 x
g1 x1 | x
nn n 1 x e . n n
: وﻋﻠﻰ ذﻟﻚ
g 3 x1 , x g4 x
n2
n 2
nx
n n 1 nx x1 e n n 1 n 1 n 1 nx nn n 1 x e n n n2 n n2 n nx nx n n 1 nx x1 n 1 n x e n 1 n 1 e n 1 n
n n 1
n2
n n n 1 n 1 x n 1
n 1
nx x 1 n 1 n 1
n2
r
n n 1 1 n n 2 x nx x1 . n n
٣١١
: وﻋﻠﻰ ذﻟﻚ
n2 n n 1 x1 n2 1 n g1 x1 | x x nx 1 nn nx n2
nx.
1
x1 1 nx
n n 1 n n 2 nnx
n 1 1 x1 n 2 , 0 x nx
nx
t E G x | x G x g x | x dx R 0 1 1 1 1 t0
g1 x1 | x dx1. t0
وﻋﻠﻰ ذﻟﻚ : n 2
dx1 n 1
|nx t0 .
nx
t n 1 1 x1 R 0 nx t0 nx
n 1
1 x1 1 nx nx n 1 nx n 1
|nx t0
x 1 1 nx
n 1
t 0 nx.
n 1
, t 0 nx.
t 1 0 nx
t 1 t 0 R 0 nx
وﻋﻠﻰ ذﻟﻚ R t 0 ﻣﻘﺪر ﻟﻪ ﺻﻔﺔ MVUEﺣﻴﺚ : .
t0
t R t e R 0 0
وﺗﺒﺎﻳﻨﻪ اﻗﻞ ﻣﻦ اى ﺗﺒﺎﻳﻦ ﳌﻘﺪر اﺧﺮ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ وﺳﻮف ﻳﱰك اﳚﺎد ﺗﺒﺎﻳﻦ R t 0 ﻛﺘﻤﺮﻳﻦ . ) (٢-١-٥ﺗﻘدﯾر اﻻﻣﻛﺎن اﻻﻛﺑر ﻟﻣﺗوﺳط اﻟﺣﯾﺎة ﻓﻰ ﺣﺎﻟﺔ اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻻول ﰲ اﺧﺘﺒﺎرات اﳊﻴﺎة ﻏﺎﻟﺒﺎً ﻳﻜﻮن ﻣﻦ اﻟﻀﺮوري ﺗﻘـﺪﻳﺮ ﻣﺘﻮﺳـﻂ ﺗﻮزﻳـﻊ أﺳـﻲ ﻣـﻦ ﺑﻴﺎﻧـﺎت ﻣﺮاﻗﺒـﺔ .ﺳـﻮف ﻳـﺘﻢ ﻣﻨﺎﻗﺸـﺔ اﳌﺸـﻜﻠﺔ ﻋﻨـﺪﻣﺎ ﺗﻼﺣــﻆ ﻛــﻞ وﺣــﺪة ﰲ ﻓــﱰة ﻣﻌﻄــﺎة ﻣــﻦ اﻟــﺰﻣﻦ .إذا ﻓﺸــﻠﺖ اﻟﻮﺣــﺪة ﰲ ﻫــﺬﻩ اﻟﻔــﱰة ﻓــﺈن زﻣــﻦ اﻟﻔﺸــﻞ )ﺑﺎﻟﻀــﺒﻂ( ﻟﻠﻮﺣــﺪة ﻳﻌــﺮف .وﻏــﲑ ذﻟ ـ ـ ـ ــﻚ ﻓ ـ ـ ـ ــﺈن اﳌﻌ ـ ـ ـ ــﺮوف أن ﻋﻤ ـ ـ ـ ــﺮ اﻟﻮﺣ ـ ـ ـ ــﺪة ﻳﺰﻳ ـ ـ ـ ــﺪ ﻋ ـ ـ ـ ــﻦ ﻫ ـ ـ ـ ــﺬﻩ اﻟﻔ ـ ـ ـ ــﱰة .وﻋﻠ ـ ـ ـ ــﻰ ذﻟ ـ ـ ـ ــﻚ ﺑﻔ ـ ـ ـ ــﺮض أن ﻟ ـ ـ ـ ــﺪﻧﻴﺎ ﻋ ـ ـ ـ ــﺪد ﺛﺎﺑ ـ ـ ـ ــﺖ ﻣـ ـ ـ ـ ــﻦ اﳌﺸﺎﻫﺪات x1 , x 2 ,, x nﻣﺄﺧﻮذﻳﻦ ﻣﻦ ﻣﺘﻐﲑ ﻋﺸﻮاﺋﻲ ﻟﻪ ﺗﻮزﻳﻊ أﺳﻲ ﺑﺪاﻟﺔ ﻛﺜﺎﻓﺔ اﺣﺘﻤﺎل: ٣١٢
1 x f (x) e x i Ti
, x 0 , 0. ﲢﺖ اﻟﻘﻴﺪ أن x iﺗﻌﺮف ﻓﻘﻂ إذا ﻛﺎﻧﺖ ),(i 1,2,...,n
T1 ,T2 , ,Tn
ﺣﻴﺚ : T1 ,T2 , ,Tn ﻗﻴﻢ ﻣﻌﻄﺎة و اﻟﱵ ﰲ ﺗﻄﺒﻴﻘﺎت ﻛﺜﲑة ﺗﻜﻮن ﻣﺘﺴﺎوﻳﺔ أي أن T1 T2 Tn let
x i Ti x i Ti
1 ai 0
if if n
r ai , i 1
ﺣﻴــﺚ rﻋــﺪد اﻟﻮﺣــﺪات اﻟــﱵ ﺗﻔﺸــﻞ وﻳﺴــﺠﻞ ﻋﻤﺮﻫــﺎ .و rﻣﺘﻐــﲑ ﻋﺸ ـﻮاﺋﻲ .اﳌﻄﻠــﻮب إﳚــﺎد ﺗﻘــﺪﻳﺮ ﺑﻨﻘﻄــﺔ ﻟﻠﻤﻌﻠﻤــﺔ ، ﺑﺎﺳــﺘﺨﺪام ﻃﺮﻳﻘﺔ اﻹﻣﻜﺎن اﻷﻛﱪ. n
L (f (x i ))a i (1 F(Ti ))1a i i 1
) 1 a ixi Ti (1a i e e i 1 n
a i x i (1a i )Ti
n
n
ai
i 1
e
s
i 1
r
e . n
where s a i x i (1 a i )Ti i 1
s ln L r ln , ln L r s 2 r s 2 then : s ˆ ,r 0 r ﻟﺪراﺳﺔ ﺧﺼﺎﺋﺺ اﳌﻘﺪر ˆ ﺳﻮف ﻧﻮﺟﺪ أوﻻً ﺗﻮﻗﻊ اﻻﺣﺼﺎء Sﺣﻴﺚ : ٣١٣
n
E(S) E{ a i X i (1 a i )Ti } i 1
since : E[a i X i (1 a i )Ti ] E(a i X i ) E(1 a i )Ti , E(a i X i ) E(a i )E(X i | a i ) (1)P(a i 1)E(X i | a i 1) 0P(a i 0)E(X i | a i 0)
P(a i 1)E(X i | a i 1) P(a i 1) P(X i Ti ) [1 e
Ti
Ti
x
Ti
x
] Pi
1 x 0 e dx E(X i | a i 1) P(X i Ti )
1 x 0 e dx (1 e
e
Ti
T i
.
)
x
1 0 x e dx . Pi Ti
1 x x e dx 0 T
x x i 1 Ti [e x 0 e dx] 0 x x 1 Ti T 2 [e x 0 e (1) 0i ]
Ti e
Ti
(1 e
Ti
)
Pi TQ i then : E(X i | a i 1)
(1 e
Ti
) Te i
Ti
Ti
(1 e ) Pi Ti Qi ٣١٤ ,Q i 1 Pi . Pi
,
then : E(X i | a i 1)P(a i 1)
Pi TQ i i Pi Pi
Pi Ti Qi
. (١-٥)
E(1 ai )Ti E(Ti ) TE(a i i) Ti T[(1)P(a i i 1) (0)P(a i 0)] Ti T[P(a i i 1)] T[1 i P(a i 1)]
(٢-٥)
TP(a i i 0) TP(X i i Ti )
Te i
Ti
TQ i i. : ( ﻓﺈن٢-٥)(و١-٥) وﻣﻦ
E[a i X i (1 a i )Ti ] Pi TQ i i TQ i i Pi (1 e n
n
Ti
).
n
E{ [a i X i (1 a i )Ti ]} Pi (1 e i 1
i 1
Ti
).
i 1
: ﻫﻨﺎك ﻃﺮﻳﻘﺔ أﺧﺮى ﻹﳚﺎد n
E{ a i X i (1 a i )Ti }. i 1
Ti
x x 1 x [ e dx Ti e dx] i 1 0 Ti n
(1)
(2) Ti
x x (1) e dx, 0
: ﺑﺎﻟﺘﻜﺎﻣﻞ ﺑﺎﻟﺘﺠﺰيء ﺣﻴﺚ T
x x i 1 Ti [x e ( 1) 0 e dx 0
Ti e
x
e
x
Ti
0
٣١٥
Ti e
Ti e
Ti
Ti
[e
[1 e
Ti
x
Ti
].
x
1 e e dx Ti 1 Ti
(2) Ti
e 0 ]
Te i
Ti
.
Ti
: ( ﻓﺈن2) (1) وﻣﻦ n
E[[a i x i Ti (1 a i )]] i 1
Ti e
Ti
[1 e
Ti
] Tie
Ti
Ti
[1 e ] Pi . where :
Pi [1 e
Ti
].
: ﰲ ﺣﺎﻟﺔ اﻟﻌﻴﻨﺔ اﳌﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻻول ﻣﻘﺪر ﻣﺘﺤﻴﺰˆ اﻵن ﺳﻮف ﻧﺜﺒﺖ أن Cov(r, ˆ ) E(rˆ ) E(r)E(ˆ )
sin ce : n
Ti
n
E(rˆ ) E(S) Pi (1 e ) , i 1
i 1
n
n
E(r) E( a i ) [E(a i )] i 1
i 1
n
[(1)P(a i 1) (0)P(a i 0)] i 1 n
n
P(a i 1) P(X i Ti ) i 1
i 1
n
(1 e i 1 n
Pi . i 1
٣١٦
Ti
)
n
n
i 1
i 1
Cov(r, ˆ ) Pi E(ˆ ) Pi n
Pi [ E(ˆ )]. i 1
then : E(ˆ ).
) ˆCov(r, n
Pi
i 1
then : .
) ˆCov(r, n
E(ˆ )
Pi
i 1
اﻟﺘﻮزﻳــﻊ اﻟﺘﻘ ـﺮﻳﱯ asymptotic distrihutionﻟﻠﻤﻘــﺪر ˆ ﳝﻜــﻦ اﳊﺼــﻮل ﻋﻠﻴــﻪ ﻣــﻦ ﺧﺼــﺎﺋﺺ ﻣﻘــﺪﻳﺮات اﻹﻣﻜــﺎن اﻻﻋﻈــﻢ ﺣﻴﺚ ﳝﻜﻦ إﺛﺒﺎت أن : 2 ˆ ~ N(, ). n Pi i 1
ﺣﻴﺚ اﻧﻪ ﻋﻨﺪﻣﺎ ﺗﻜﻮن ﺣﺠﻢ ةاﻟﻌﻴﻨﺔ ﻛﺒﲑا ﻓﺈن ﻣﻦ ﺧﻮاص ﻣﻘﺪرات اﻻﻣﻜﺎن اﻻﻛﱪ ان :
1 ˆ N(, ), )nI( 2 ln L (nI() E ) 2 2 . n اى ان ˆ ﺗﻘﺮﻳﺒﺎ ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻟﻄﺒﻴﻌﻰ ﲟﺘﻮﺳﻂ وﺗﺒﺎﻳﻦ Pi i 1
اﻟﱪﻫﺎن :
,
s
ln L r ln n
s [a i x i (1 a i )Ti ], i 1
ln L r s 2 2 ln L r 2s 2 3. 2 ٣١٧
2 ln L )E(S r E E 2 2 3 2 P P 2i 2 3 i
1 ] [2Pi Pi 2 1 2 Pi T i 1 n 2 (1 e ). i1
) (٣-١-٥ﺗﻘدﯾر اﻻﻣﻛﺎن اﻻﻛﺑر ﻟﻣﺗوﺳط اﻟﺣﯾﺎة ﻓﻰ ﺣﺎﻟﺔ اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ ﺣﻴﺚ r nو y y1 , y 2 , , y rواﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة . داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : r !n f (yi )[1 F(yr )]n r (n r)! i 1 y
L(y1 , y 2 ,..., y n | ) L y
r !n ) 1 ( i ) ( r [ e ][e ]n r (n r)! i 1 r
1
] yi (n r) y r n! 1 ( [ i 1 e . (n r)! r
ﻣﻘﺪر اﻹﻣﻜﺎن اﻷﻛﱪ MLEﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ اﳊﻞ ﻟﻠﻤﻌﺎدﻟﺔ : ln L 0 ˆ
وﻳﺘﻢ ﺑﺎﳋﻄﻮات اﻟﺘﺎﻟﻴﺔ: r
] [ y i (n r)y r i 1
ln L r ln
r
] [ yi (n r)y r .
i 1
2
ln L r
ﺑﻮﺿﻊ : ٣١٨
ln L 0. ˆ
r
yi (n r)y r r i 1 ˆ ˆ 2 r u ˆ ,u= y i (n r)y r . r i 1
اﻻن ﻳﺘﻢ دراﺳﺔ اﻟﺘﻮزﻳﻊ اﳌﻀﺒﻮط ﻟﻠﻤﻘﺪر ˆ. داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻻﺣﺘﻤﺎﻟﻴﺔ اﳌﺸﱰﻛﺔ ﻟﻺﺣﺼﺎءات اﻟﱰﺗﻴﺒﻴﺔ ﰲ اﻟﻌﻴﻨﺔ ﺗﻌﻄﻰ ﻛﺎﻟﺘﺎﱄ: r
] [ y i (n r)y r i 1
)
1 g(y1 , y 2 ,..., y r ) r exp(
r
اﻻن اﳌﻄﻠﻮب اﳚﺎد ﺗﻮزﻳﻊ ] . U [ Yi (n r)Yr i 1
ﺑﻔﺮض اﻟﺘﺤﻮﻳﻠﺔ اﻷﺣﺎدﻳﺔ:
Z1 n (Y1 Y0 ), ) Z2 (n 1)(Y2 Y1 ) Z3 (n 2)(Y3 Y2 Zi (n i 1)(Yi Yi1 ) , i 1,2,...,r , Y0 0 واﻟﺘﺤﻮﻳﻠﺔ اﻟﻌﻜﺴﻴﺔ ﳍﺎ ﻫﻲ: Z Z Zi Yi 1 2 ... , i 1,2,,r, n n 1 n i 1 وﻫﺬا ﻳﻌﲎ ان : r
r
Yi (n r)Yr Zi U . i 1
i 1
وﻣﻨﻬﺎ ﻳﺘﻢ اﳚﺎد ﺟﺎﻛﻮﺑﻴﺎن اﻟﺘﺤﻮﻳﻞ ﻛﺎﻟﺘﺎﱄ:
٣١٩
y1 z r
y1 z 2
y1 z1
y 2 y 2 !n z 2 z r . !)(n r y r z r
y 2 J z1
y r z 2
y r z1
وﻣﻨﻬﺎ : r
zi ) 1 ( i 1 g(z1 ,z 2 ,...,z r ) r e . ﳑﺎ ﻳﻌﲏ إن اﳌﺘﻐﲑات Z iﻣﺘﻐﲑات ﻋﺸﻮاﺋﻴﺔ ﻣﺴﺘﻘﻠﺔ وﻣﺘﻄﺎﺑﻘﺔ وﻛﻞ ﻣﻨﻬﺎ ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻲ ﺑﺎﳌﻌﻠﻤﺔ . Zi E xp() U G(, r).
ﺣﻴﺚ ان ﺗﻮزﻳﻊ Uﻫﻮ : u > 0.
1 ( )r u r-1 = )h(u u e )(r
,
r
ﻛﻤﺎ ﳝﻜﻦ ﳝﻜﻦ إﳚﺎد ﺗﻮزﻳﻊ U Ziﻣﻦ اﻟﺪاﻟﺔ اﳌﻮﻟﺪة ﻟﻠﻌﺰوم ﻛﺎﻟﺘﺎﱄ : i 1
1
M Zi (t) (1 t) ,i 1, 2,...r (t) (1 t) r .
r
M
Zi i1
واﻟــﱵ ﲤﺜــﻞ اﻟﺪاﻟــﺔ اﳌﻮﻟــﺪة ﻟﻠﻌــﺰوم ﳌﺘﻐــﲑ ﻋﺸ ـﻮاﺋﻲ ﻳﺘﺒــﻊ ﺟﺎﻣــﺎ ﲟﻌﻠﻤﺘــﲔ ). (, rاﻟﺘﻮزﻳــﻊ اﳌﻀــﺒﻮط ﻟﻠﻤﻘــﺪر ˆ ﳝﻜــﻦ اﳚــﺎدﻩ ﻣــﻦ اﻟﺪاﻟــﺔ اﳌﻮﻟﺪة ﻟﻠﻌﺰوم ﻛﺎﻟﺘﺎﱃ : t r ) . r
(t) (1
r
Zi i1
Mˆ (t) M U (t) M r
r
r
واﻟﱵ ﲤﺜﻞ اﻟﺪاﻟﺔ اﳌﻮﻟﺪة ﻟﻠﻌﺰوم ﳌﺘﻐﲑ ﻋﺸﻮاﺋﻲ ﻳﺘﺒﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ ). ( , r ﻹﳚﺎد اﻟﺘﻮﻗﻊ واﻟﺘﺒﺎﻳﻦ ﻟﺘﻮزﻳﻊ ˆ ) اﳌﻀﺒﻮط ( exactﻧﺘﺒﻊ اﻟﺘﺎﱃ: 1 1 E(ˆ ) E(U) (r) . r r . ٣٢٠
أي ان ˆ ﻣﻘﺪر ﻏﲑ ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ . اﻟﺘﺒﺎﻳﻦ ﻟﻠﻤﻘﺪر ˆ ﳛﺴﺐ ﻛﺎﻟﺘﺎﱄ : 1 r ) Zi r i 1
(Var(ˆ ) = Var n
) Var(Z i
i 1
1 n2
1 2 2 (r =) . r2 r
= =
وداﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻠﻤﺘﻐﲑ ˆ ﺗﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱄ : ˆ > 0.
,
r ( )n ˆr r 1 ˆ ˆ f() e )(r
ﻻﺛﺒﺎت أن اﳌﻘﺪر ˆ ﻏﲑ ﻣﺘﺤﻴﺰ ﺑﺄﻗﻞ ﺗﺒﺎﻳﻦ أي MVUEﻧﺘﺒﻊ اﻟﺘﺎﱃ .: 2 ln L r u 2 2 3. 2 2 ln L r 2r r (E ) 2 3 2 2 2 2 1 ˆ) . (Var 2 ln L r r ( E ) 2
أي أن اﳌﻘﺪر ˆ ﻫﻮ ﻣﻘﺪر ﻏﲑ ﻣﺘﺤﻴﺰ ﺑﺄﻗﻞ ﺗﺒﺎﻳﻦ أي . MVUE ﻹﺛﺒﺎت أن ˆ إﺣﺼﺎء ﻛﺎﰲ sufficientﻧﺘﺒﻊ اﻵﰐ : u
1 L= r e
u
L = (1).(-r e ) , N(x) . K(u,). u
ﺣﻴﺚ N(x) =1و . K(u,)=-r e U sufficient for . ˆ sufficient for . r U U= Ziإﺣﺼﺎء ﻛﺎﰲ ﻟﻠﻤﻌﻠﻤﺔ وﺑﺎﻟﺘﺎﱄ ﻓﺈن n i=1
ˆ إﺣﺼﺎء ﻛﺎﰲ أﻳﻀﺎ .
) (٤-١-٥ﺗﻧﺑؤ اﻟﻌﯾﻧﺔ اﻟواﺣدة ٣٢١
إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات
اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ . ﰱ ﻫﺬﻩ اﳊﺎﻟﺔ ﻓﺈن اﻟﺘﻨﺒﺆ ﺳﻴﻜﻮن ﻻﺣﺪ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ اﳌﺘﺒﻘﻴﺔ ﻣﻦ اﳊﺠﻢ ) (n rوﻫﻰ . y r1, yr 2 ,, ynﻟﻠﻤﺸﺎﻫﺪات اﳌﺘﺒﻘﻴﺔ ذات اﳊﺠﻢ ) (n rﺣﻴﺚ r<nإذا وﺿﻌﻨﺎ ys yr sﻟﻠﺘﻌﺒﲑ ﻋﻦ اﳌﺸﺎﻫﺪة ذات اﻟﱰﺗﻴﺐ sﺣﻴﺚ 1 s n r ﻓﺈن ﻓﱰة ﺛﻘﺔ ﻟﻠﻤﺸﺎﻫﺪة ysﺑﺸﺮط وﺟﻮد اﳌﺸﺎﻫﺪات y1 , y2 ,, yrﺗﻌﻄﻰ ﻣﻦ اﻟﻨﻈﺮﻳﺔ اﻟﺘﺎﻟﻴﺔ : ﻧظرﯾﺔ ﺑﻔﺮض أن:
Ys Yr , U وﻋﻠﻰ ذﻟﻚ ﻓﱰة اﻟﺜﻘﺔ ﻣﻦ ﺟﺎﻧﺐ واﺣﺪ ﻟﻠﻤﺸﺎﻫﺪة ysﺑﺸﺮط وﺟﻮد اﳌﺸﺎﻫﺪات y1 , y2 ,, yrﳝﻜﻦ اﳚﺎدﻫﺎ V
ﻛﺎﻟﺘﺎﱃ :
P(V v 0 ) 1 . Ys Yr v 0 ) 1 . U P(Ys uv 0 Yr ) 1 . ( P
ﻣﻦ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ v0وﲟﻌﻠﻮﻣﻴﺔ ﻛﻞ ﻣﻦ v 0 , u, y rﻓﺈن : P(Ys v 0 u y r ) 1 .
ﺣﻴﺚ اﳊﺪ اﻻدﱏ واﻻﻋﻠﻰ ﻟﻠﻔﱰة ﻫﻮ ) (0, v0u yrواﻟﱴ ﲤﺜﻞ ﻓﱰة ﺛﻘﺔ ذات ﺟﺎﻧﺐ واﺣﺪ وﺑﻨﻔﺲ اﻟﺸﻜﻞ ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﻓﱰة ﺛﻘﺔ ذات ﺟﺎﻧﺒﲔ. ﻹﳚﺎد
v0
v0
ﻣﻦ اﳌﻌﺎدﻟﺔ اﻟﺘﻜﺎﻣﻠﻴﺔ g(v)dv 1 ﻳﻜﻔﻲ أن ﳒﺪ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل 0
Z Ys Yr ,
و Uﺗﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ ) ، (,nاى ان ). U ~ G(, n اﳌﻄﻠﻮب ﺗﻮزﻳﻊ : وﻟﺬﻟﻚ ﳓﺘﺎج إﱃ اﻟﺘﻮزﻳﻊ اﳌﺸﱰك ﻟـ Zو U وﻟﺬﻟﻚ ﻧﻮﺟﺪ أوﻻً ﺗﻮزﻳﻊ Zوﻟﻨﺄﺧﺬ اﻟﺘﺤﻮﻳﻠﺔ :
Z , U
V
٣٢٢
)g(v
.ﺑﻔﺮض ان :
z yx x yr
wx
ﺣﻴﺚ
, y ys
. J 1 وﻣﻨﻪ ﺟﺎﻛﻮﺑﻴﺎن اﻟﺘﺤﻮﻳﻞ f (x, y)
x w, y z w
واﻟﺘﺤﻮﻳﻠﺔ اﻟﻌﻜﺴﻴﺔ
n! f (x)f (y)[F(x)]r 1[F(y) F(x)]s r 1[1 F(y)]n s , (r 1)!(s r 1)!(n s)! 0x y n! let c (r 1)!(s r 1)!(n s)!
:ﺑﻔﺮض أن X1 , X 2 ,, X n
: ﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻷﺳﻲ ﺣﻴﺚi,i,d ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﳍﺎ ﻧﻔﺲ اﻟﺘﻮزﻳﻊ وﻣﺴﺘﻘﻠﺔ 1 x f (x) e , x 0, 0, : ﻫﻮYr ,Ys وﲢﺖ ﻓﺮض اﻟﺘﻮزﻳﻊ اﻷﺳﻲ ﻓﺈن اﻟﺘﻮزﻳﻊ اﳌﺸﱰك ﻻﺣﺼﺎﺋﻴﲔ 2
x
x
y
x
y
(n s 1) 1 f (x, y) c (1 e )r 1 (e e )s r 1 e e
let z y x wx
yzw
0xy
0z ,
xw 0 w , 1 1 | J | 1. 0 1
w (n r 1)w z (n s1)z 1 r 1 s r 1 (1 e ) e (1 e ) e 0 z ,0 w , 2 z z w w (n s 1) (n r 1) 1 s r 1 r 1 f z (z) c(1 e ) e 0 2 (1 e ) e dw.
f1 (z, w) c
let
y1 e
w
1 w dy1 e dw
w
dw e dy1
٣٢٣
1 z z (n s 1) s r 1
f 2 (z) c(1 e )
e
1
(1 y ) 1
r 1
y1n r dy1
0
z z (n s 1) s r 1
c (1 e )
e
(n r 1,r)
n! (r 1)!(n r)! (r 1)!(s r 1)!(n s)! n!
z z (n s1) (n r)! 1 s r 1 (1 e ) e 0z (s r 1)!(n s)! u 1 r 1 f3 (u) u e ,u 0, 0; (r) r
U ~ G(,r)
: ﻣﺴﺘﻘﻠﲔZ, U ﲟﺎ أن f 4 (z,u) f 2 (z)f 3 (u).
f (z,u)
z z u (n s1) (n r)! 1 r 1 s r 1 u (1 e ) e e . r 1 (s r 1)!(n s)!(r 1)! z let v u w 1 u , z vw 1 w v | J | 1 w1 0 1
wu wu w 1 1 (n s 1) 1 c r1 s r 1 f 5 (v, w1 ) r 1 w1 (1 e ) e e wv w 1 1 (1 v(n s 1)) c r s r 1 f 6 (v) r1 w1 (1 e ) e dw1 0 w s r 1 s r 1 1 (1 u(n s 1 j) c r j r 1 w1 (1) e j 0 j0
w 1 (1 u(n s1) j) c s r 1 s r 1 j r r 1 (1) w1e dw1. j j0 0 w 1 (1 v(n s 1) j) let y. y w1 (1 v(n s 1) j)
٣٢٤
dy dw1 , 1 u(n s 1) j
c s r 1 s r 1 y j f 6 (v) r 1 (1) j j0 1 v(n s 1) 0
r
y dy e j 1 v(n s 1) j
cr 1 s r 1 s r 1 1 j r 1 ( 1) y re ydy r 1 j (1 v(n s 1) j) 0 j0 s r 1 (r 1) c (1) j . j [1 v(n s 1) j]r 1 j 0 s r 1
then : f6 (v)
s r 1 s r 1 (n r)!r 1 j ( 1) j (s r 1)!(n s)! j0 [1 v(n s 1) j]r 1
s r 1 s r 1 r 1 j (1) j (s r,n s 1) j0 (1 v(n s 1) j) r 1
0 u .
. ﻫﺬا اﻟﺘﻮزﻳﻊ ﻻ ﻳﻌﺘﻤﺪ ﻋﻠﻰ اﻟﻤﻌﻠﻤﺔ then : P(V v 0 ) 1 v0
f (v)dv 1 . 6
0
: ﻓﺈنv 0 , u, y r وﲟﻌﻠﻮﻣﻴﺔ ﻛﻞ ﻣﻦv0 ﻣﻦ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰv0 P(Ys v 0 u y r ) 1 .
. ( واﻟﱴ ﲤﺜﻞ ﻓﱰة ﺛﻘﺔ ذات ﺟﺎﻧﺐ واﺣﺪ0, v0u yr ) ﺣﻴﺚ اﳊﺪ اﻻدﱏ واﻻﻋﻠﻰ ﻟﻠﻔﱰة ﻫﻮ :ﻋﻨﺪ اﻟﺮﻏﺒﺔ ﻓﻲ إﻳﺠﺎد ﻓﺘﺮة ﺛﻘﺔ ذات ﺟﺎﻧﺒﻴﻦ أي ان P(a V b) 1 . b
f (v)dv 1 . 6
a
: ﻓﺈنb,a وﺑﻌﺪ إﻳﺠﺎد
: ﺗﺼﺒﺢ ٣٢٥
Ys Yr b) 1 . U ﻣﻦ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ a,bوﲟﻌﻠﻮﻣﻴﺔ ﻛﻞ ﻣﻦ a,b,u, y rﻓﺈن : P(a
P(au y r Ys bu Yr ) 1 .
ﺣﻴﺚ اﳊﺪ اﻻدﱏ واﻻﻋﻠﻰ ﻟﻠﻔﱰة ﻫﻮ ) (au yr ,bu yrواﻟﱴ ﲤﺜﻞ ﻓﱰة ﺛﻘﺔ ذات ﺟﺎﻧﺒﲔ. ﻋﻨﺪﻣﺎ r 1ﻓﺈن : s 2 s 2 1 1 j . )( 1 (s 1, n s 1) j0 j (1 v(n s 1) j) 2
f6 (v)
ﻋﻨﺪﻣﺎ r 1,s nﻓﺈن : n 2 n 2 1 1 j f6 (v) . )( 1 (n 1,1) j0 j (1 v(1) j) 2
) (٥-١-٥ﺗﻘدﯾرات ﺑﯾﯾزﯾﺔ ﺗﺣت ﻓرض ﺗوزﯾﻊ ﺟﺎﻣﺎ اﻟﻌﻛﺳﻰ ﻛﺗوزﯾﻊ ﻗﺑﻠﻰ ﻓﻰ ﺣﺎﻟﺔ اﻟﻣﻌﺎﯾﻧ ﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ ) Shalaby (1990وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃ : إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ ﺣﻴﺚ r nو y y1 , y 2 , , y rواﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة . داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : r !n f (yi )[1 F(y r )]n r (n r)! i 1 y
L(y | ) L
y
r !n ) 1 ( i ) ( r [ e ][e ]n r (n r)! i 1 r
1
] yi (n r ) yr n! 1 ( [ i1 e . (n r)! r
ﻟﻴﻜﻦ : r
u [ yi (n r)y r ]. i 1
٣٢٦
: ﻫﻮ اﳊﻞ ﻟﻠﻤﻌﺎدﻟﺔ ﻟﻠﻤﻌﻠﻤﺔMLE ﻣﻘﺪر اﻹﻣﻜﺎن اﻷﻛﱪ :وﻳﺘﻢ ﺑﺎﳋﻄﻮات اﻟﺘﺎﻟﻴﺔ
ln L 0 ˆ
ln L u r ln , ln L r u 2.
: ﺑﻮﺿﻊ ln L 0 ˆ
r u ˆ ˆ 2 r
u ˆ r
y
i
(n r)y r
i 1
r
,
: ﻫﻮ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ اﻟﻌﻜﺴﻲ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃ واذا ﻛﺎن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ ﻟﻠﻤﻌﻠﻤﺔ
()
g 1 –g / e (g 1)
; g 1 , , 0.
: ﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻧﺘﺒﻊ اﻟﺘﺎﱃ
n! –(rg-1) 1(u) g1 e d 0 ()L(y | )d (g 1) (n r)! 0
n! g1 . (r g 1) (u )(r g-1) . (g 1) (n r)!
( y)
()L(y | )
( u)
()L(y | )d 0
(u )(r g1) (rg) e (r g 1)
(u )
٣٢٧
,r g 1 ; , 0.
اﻟﻌﺰوم ﻏﻴﺮ اﻟﻤﺮﻛﺰﻳﺔ ﻟﻠﻤﻌﻠﻤﺔ
ﻫﻲ: S
E( u) S ( | u) d 0
d
) (u
e
) (r g s 1
0
)(u ) (r g 1 ) (r g 1
)(u ) (r g 1 )(u ) (r g s 1) (r g s 1 ) (r g 1 ) (r g s 1 (u )s ; r g 1, s 1, 2,.... ) (r g 1 ﺑﺎﺳﺘﺨﺪام اﻟﻌﺰوم ﻏﲑ اﳌﺮﻛﺰﻳﺔ ﳝﻜﻦ اﻟﺘﻮﺻﻞ ﻟﻠﻤﻘﺪر اﻟﺒﻴﻴﺰي ﻟﻠﻤﻌﻠﻤﺔ وﺗﺒﺎﻳﻨﻪ ،ﻓﺒﻮﺿﻊ ) (s=1ﻳﺘﻢ اﻟﺘﻮﺻﻞ ﻟﻠﻤﻘﺪر اﻟﺒﻴﻴﺰي ﻟﻠﻤﻌﻠﻤﺔ ﲢﺖ ﻓﺮض داﻟﺔ ﻣﺮﺑﻊ اﳋﺴﺎرة ﻛﺎﻟﺘﺎﱄ :
) (r g 2 ) (u )(r g 1 ) (u rg2
1* E( u)
; r g 2.
وﺑﺎﺳﺘﺨﺪام اﻟﻌﺰم اﻟﺜﺎﱐ ﳝﻜﻦ اﻟﺘﻮﺻﻞ إﱃ ﺗﺒﺎﻳﻦ اﳌﻘﺪر ﻛﻤﺎ ﻳﻠﻲ : 2
Var(1* ) E( 2 u) E( u) (u ) 2 (u ) 2 (r g 3) ) (r g 2 (r g 2) 2 1 1 ( (u ) 2 ) )(r g 2)(r g 3 (r g 2) 2 (u ) 2 . )(r g 2) 2 (r g 3
اﳌﻨﻮال اﻟﺒﻴﻴﺰي ﻟﻠﻤﻌﻠﻤﺔ وﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ *2وﳝﻜﻦ اﺳﺘﻨﺘﺎﺟﻪ ﺑﺎﺗﺒﺎع اﳋﻄﻮات اﻟﺘﺎﻟﻴﺔ : ٣٢٨
أوﻻً :إﳚﺎد اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻄﺒﻴﻌﻲ ﻟﺪاﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي :
(u )(r g1) )(u ln [( u)] ln . (r g)ln (r g 1) ﺛﺎﻧﻴﺎ :ﺑﺎﺷﺘﻘﺎق ﻃﺮﰱ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻤﺔ ﻛﻤﺎ ﻳﻠﻲ : ) ln [f( u)] (r g) (u . 2
ﺛﺎﻟﺜﺎً :ﲟﺴﺎواة اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﺑﺎﻟﺼﻔﺮ وﺣﻞ اﳌﻌﺎدﻟﺔ ﻳﻜﻮن اﳌﻨﻮال اﻟﺒﻴﻴﺰي :
) (u rg
; r g 0.
داﻟﺔ اﻟﻜﺜﺎﻓﺔ ﻟﻠﻤﻨﻮال اﻟﺒﻌﺪى
*2
*2
ﻳﺘﻢ اﺳﺘﻨﺘﺎﺟﻪ ﺑﺎﺳﺘﺨﺪام اﳋﻄﻮات اﻟﺘﺎﻟﻴﺔ :
أوﻻً :ﻳﺼﺎغ اﳌﻨﻮال اﻟﺒﻴﻴﺰي ﺑﺎﺳﺘﺨﺪام اﻟﻮﺳﻂ اﳊﺴﺎﰊ ﻛﻤﺎ ﻳﻠﻲ :
) (u u rg rg rg u , . rg rg
*2
أي أن :
u (r g) ( *2 ). ﺛﺎﻧﻴﺎَ :ﲟﺎ أن اﻹﺣﺼﺎء Uﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﺑﺎﳌﻌﻠﻤﺘﲔ ) ( r,θﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃ : , u 0.
u
r h (u | ) u r 1 e ) (r
ﺣﻴﺚ : ٣٢٩
du (r g) d *2 . : ﻛﻤﺎ ﻳﻠﻲU ﺑﺎﺳﺘﺨﺪام داﻟﺔ اﻟﻜﺜﺎﻓﺔ ﻟﻺﺣﺼﺎء*2 ﻓﺈﻧﻪ ﳝﻜﻦ إﳚﺎد داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻠﻤﻨﻮال اﻟﺒﻴﻴﺰي r * h1 (2 ) (r g ) r 1 ( *2 ) r 1 e (r ) 1 rg r * ( ) ( 2 ) r 1 e (r)
( r g ) ( *2 )
( r g ) ( *2 )
: ﻳﺘﻢ إﳚﺎدﻩ ﻛﺎﻟﺘﺎﱄ 1 rg E ( *2 ) (r )
r
* 2
(r g )
( *2 ) r 1 e
( r g ) ( *2 )
; *2
*2 0.
اﻟﻘﻴﻤﺔ اﳌﺘﻮﻗﻌﺔ ﻟﻠﻤﻨﻮال اﻟﺒﻌﺪى
d *2 .
* : ﻳﺘﻢ ﺣﻞ اﻟﺘﻜﺎﻣﻞ اﻟﺴﺎﺑﻖ ﻛﻤﺎ ﻳﻠﻲz ( 2 ) ﺑﺎﺳﺘﺨﺪام اﻟﺘﻌﻮﻳﺾ
r (rg )z 1 rg r 1 E ( ) (z )z e dz (r) 0 * 2
r (rg)z (rg)z 1 rg r r 1 z e dz z e dz 0 (r ) 0 r ( r 1) r 1 rg rg rg (r) (r 1) (r ) r rg r . rg
. ﻣﻘﺪر ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ*2 وﻳﺘﻀﺢ ﳑﺎ ﺳﺒﻖ ان اﳌﻨﻮال اﻟﺒﻌﺪى
٣٣٠
: *2 ﻣﺘﻮﺳﻂ ﻣﺮﺑﻊ اﳋﻄﺄ ﻟﻠﻤﻨﻮال اﻟﺒﻌﺪى : ﻓﺈنθ ﻣﻘﺪر ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ*2 ﲟﺎ أن اﳌﻨﻮال اﻟﺒﻌﺪى M S E ( *2 ) V ar ( *2 ) .
: *2 وﺑﺎﻟﺘﺎﱄ ﻣﺘﻮﺳﻂ ﻣﺮﺑﻊ اﳋﻄﺄ ﻟﻠﻤﻨﻮال اﻟﺒﻌﺪى M S E ( *2 ) E ( *2 ) 2 E ( *2 ) 2 2 E ( *2 ) 2 . * 2 : ﻛﻤﺎ ﻳﻠﻲE[ ( 2 ) ] ًوﳊﻞ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﻳﺘﻢ إﳚﺎد أوﻻ
1 rg E[( ) ] (r) * 2
2
r
* 2
2
* 2
( )
r 1
e
( r g ) ( *2 )
d *2
: ﻳﺘﻢ ﺣﻞ اﻟﺘﻜﺎﻣﻞ اﻟﺴﺎﺑﻖ ﻛﻤﺎ ﻳﻠﻲz ( *2 ) ﺑﺎﺳﺘﺨﺪام اﻟﺘﻌﻮﻳﺾ r (r g)z 1 rg 2 r 1 E[ ( ) ] (z ) z e dz (r) 0 r (r g)z (r g)z (r g)z 1 r g r 1 r 2 r 1 z e dz 2 z e dz z e dz 0 0 (r) 0 r (r 2) (r 1) r 1 rg rg rg rg 2 2 (r 1) (r) (r 2) (r) * 2
2
2
r r(r 1) 2 . 2 rg rg
: ﳓﺼﻞ ﻋﻠﻰ*2 ﰲ ﻣﻌﺎدﻟﺔ ﻣﺘﻮﺳﻂ ﻣﺮﺑﻊ اﳋﻄﺎ ﻟﻠﻤﻨﻮال اﻟﺒﻌﺪىE[ ( *2 ) 2 ] وﺑﺎﻟﺘﻌﻮﻳﺾ ﺑﻘﻴﻤﺔ M SE ( *2 ) E[ ( *2 ) 2 ] 2 E ( *2 ) 2 2
r r 2 r(r 1) 2 2 2 rg rg rg r 2 ( g ) 2 . 2 r g ﻓﱰات اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰﻳﺔ ﻟﻠﻤﻌﻠﻤﺔ
: ﲝﻞ اﳌﻌﺎدﻟﺘﲔ اﳌﺘﻤﺎﺛﻠﺔ ﻟﻠﻤﻌﻠﻤﺔ100(1- )% ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﻓﱰات اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰﻳﺔ ٣٣١
. 2
( u ) d
( u ) d 2
,
t2
t1
0
ﻟﻠﺤﺪ اﻷدﱏ t1واﳊﺪ اﻷﻋﻠﻰ ، t 2أي أن: P(t1 t 2 ) 1 .
اﻟﻔﱰة ) ( t1 , t 2ﻫﻲ ﻓﱰة اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰﻳﺔ 100(1- )%اﳌﺘﻤﺎﺛﻠﺔ ﻟﻠﻤﻌﻠﻤﺔ ، وﲟﺎ ان ﺗﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ )( u, r g 1
ﻓﺈن : )2( u ) 22(r g 1 *2 ) 2 (r g ) 22(r g 1
أي أن P(t1 t 2 ) 1 ﳓﺼﻞ ﻋﻠﻴﻬﺎ ﻛﺎﻟﺘﺎﱃ : 2(r g ) *2 P 2 1 2
2 1 2
1 1 P 2 2 * 1 (2 r g ) 2 1 2 2 2(r g ) *2 2( r g ) *2 P 2 2 1 . 1 2 2
اى ان :
2(r g ) *2 2(r g ) *2 , 2 2 1 2 2
ﻳﻤﺜﻼن اﻟﺤﺪﻳﻦ اﻻدﻧﻰ واﻻﻋﻠﻰ ﻟﻔﺘﺮة اﻟﺜﻘﺔ ﻟـ ﺣﻴﺚ
2
2 , 2ﺗﺴﺘﺨﺮﺟﺎن ﻣﻦ ﺟﺪول ﺗﻮزﻳﻊ ﻣﺮﺑﻊ ﻛﺎى ﻣﻦ اﻟﻤﻠﺤﻖ
1
2
) (٥ﻋﻨﺪ درﺟﺎت ﺣﺮ ﻳﺔ ). 2(r g 1 ٣٣٢
ﳐﺎﻃﺮة ﺑﻴﻴﺰ : أوﻻً :ﳐﺎﻃﺮة ﺑﻴﻴﺰ ﳌﻘﺪر اﳌﻜﺎن اﻻﻛﱪ ˆ) اﳌﻘﺪر ﻏﲑ اﳌﺘﺤﻴﺰ ﺑﺄﻗﻞ ﺗﺒﺎﻳﻦ ﻟﻠﻤﻌﻠﻤﺔ : ( ﲟﺎ أن ˆ ﻣﻘﺪﻳﺮ ﻏﲑ ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ ﻓﺈن : 2 M S E ( ˆ ) V ar ( ˆ ) . r ﳐﺎﻃﺮة ﺑﻴﻴﺰ ﳌﻘﺪر اﳌﻜﺎن اﻻﻛﱪ ˆ ﻫﻮ :
r ( ˆ )
M SE ( ˆ ) ( ) d 0
2 g 1 – g e / d )r (g 1
0
g 1 (g 3) -(g 3 ) )r (g 1 g 3 , r 0.
2 )r (g 2)(g 3
;
ﺛﺎﻧﻴﺎً :ﳐﺎﻃﺮة ﺑﻴﻴﺰ ﻟﻠﻤﻘﺪر اﻟﺒﻴﻴﺰي *: 1 ﲟﺎ أن * 1ﻣﻘﺪﻳﺮ ﻏﲑ ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ ﻓﺈن : M S E ( 1* ) V ar ( 1* ).
) (u وﺣﻴﺚ أن rg2
1*
داﻟﺔ ﰲ اﻹﺣﺼﺎء ، Uﻓﻴﺠﺐ إﳚﺎد داﻟﺔ اﻟﺘﻮزﻳﻊ ﻟﻺﺣﺼﺎء Uﻛﺎﻟﺘﺎﱃ:
) ( ) d
h (u
g (u )
0
u r g r 1 g 1 u e e d ) (r 0 ) (g 1
u r 1 g 1 ) ( r g 1 ) (r g 1) (u (r) (g 1) u r 1 g 1 ) ( r g 1 . ) (u (r, g )1
وﻣﻨﻪ ﳝﻜﻦ اﺳﺘﻨﺘﺎج ﳐﺎﻃﺮة ﺑﻴﻴﺰ ﻟﻠﻤﻘﺪر اﻟﺒﻴﻴﺰي * 1ﻛﻤﺎ ﻳﻠﻲ : ٣٣٣
* 1
r ( )
V ar (
* 1
) g (u ) du
0
0
u r 1 g 1 (u ) ( r g 1) (u ) 2 2 (r, g 1) (r g 2 ) (r g 3)
( r g 3) g 1 2 (r g 2) (r g 3) (r, g 1)
(1
0
du
u ( r g 3) ) du.
: ﻛﻤﺎ ﻳﻠﻲ1* ﻳﺘﻢ ﺣﻞ اﻟﺘﻜﺎﻣﻞ اﻟﺴﺎﺑﻖ وإﳚﺎد ﳐﺎﻃﺮة ﺑﻴﻴﺰ ﻟﻠﻤﻘﺪر اﻟﺒﻴﻴﺰيz
u , ﺑﺎﺳﺘﺨﺪام اﻟﺘﻌﻮﻳﺾ
(r 2) r (r g 3) r 1 r(1* ) z dz (1 z) 2 (r g 2) (r g 3) (r, g 1) 0 (r, g 3) 2 2 (r g 2) (r g 3) (r, g 1)
2 (r g 2)(g 2)(g 3)
;
rg 3 .
: *2 ﳐﺎﻃﺮة ﺑﻴﻴﺰ ﻟﻠﻤﻘﺪر اﻟﺒﻴﻴﺰي ﻟﻠﻤﻨﻮال: ًﺛﺎﻟﺜﺎ * 2
r( ) MSE(*2 ) () d 0
r2 (g )2 g1 –g / e d 2 (g 1) (r g) 0
g1 r 2–g e / d g22–g e/ d 2g1–g e/ d 2–g e/ d 2 (r g) (g 1) 0 0 0 0 g1 r(g 3)-(g3) g2(g 3)-(g3) 2g(g 2)-(g2) 2(g 1)-(g1) 2 (r g) (g 1) 2 (r g 6) (r g)2 (g 2)(g 3)
;
(r g) 3.
: ﻛﺎﻟﺘﺎﱃ*2 ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻘﺪر1* اﻻن ﻳﺘﻢ ﺣﺴﺎب ﻧﺴﺒﺔ ﳐﺎﻃﺮة ﺑﻴﻴﻴﺰ ﻟﻠﻤﻘﺪر ٣٣٤
) *r( 1 2 )r(g 2)(g 3 r r( 1* , ˆ ) , * 2 )r( 2 ) (g 2)(g 3)(r g 2 )(r g 2 ) r(*2 ) 2 (r g 6 )r(g 2)(g 3) r(r g 6 , 2 ˆ 2 (r g) 2 )r() (r g) (g 3)(g 3
r( *2 , ˆ )
اى ان : r(1* , ˆ ) r(*2 , ˆ ) r(1* ) r( ˆ ) r( *2 ).
وﻫﺬا ﻳﻌﲎ ان اﻓﻀﻞ ﻣﻘﺪر ﰱ ﻫﺬﻩ اﳊﺎﻟﺔ ﻫﻮ *. 1 ﻟﻠﻨﺘﺎﺋﺞ اﻟﺴﺎﺑﻘﺔ وﺑﻮﺿﻊ r nﻓﺈﻧﻨﺎ ﳓﺼﻞ ﻋﻠﻰ ﻧﺘﺎﺋﺞ ﲣﺺ اﻟﻌﻴﻨﺔ اﻟﻜﺎﻣﻠﺔ .
) (٦-١-٥ﺗﻘدﯾرات ﺑﯾﯾزﯾﺔ ﺗﺣت ﻓرض اﻟﺗوزﯾﻊ اﻟﻣﻧﺗظم ﻓﻰ ﺣﺎﻟﺔ اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ
اﻟﻔﺗ رة )(,
ﻛﺗوزﯾ ﻊ ﻗﺑﻠ ﻰ ﻓ ﻰ
ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ ) Bhattacharya (1967وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃ : إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ ﺣﻴﺚ r nو y y1 , y 2 , , y rواﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة . داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : r !n f (yi )[1 F(y r )]n r (n r)! i 1 y
L(y | )
y
r !n ) 1 ( i ) ( r [ e ][e ]n r (n r)! i 1 r
1
] yi (n r ) yr n! 1 ( [ i 1 e . (n r)! r
ﻟﻴﻜﻦ : r
u [ yi (n r)y r ], i 1
وإذاﻛﺎن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ اﻟﺘﻮزﻳﻊ اﻟﻤﻨﺘﻈﻢ ﻓﻰ اﻟﻔﺘﺮة ) (, اﻟﺘﺎﻟﻰ : ; 0 .
1 a
(a 1)() a 1 () a 1 a 1
٣٣٥
: ﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻧﺘﺒﻊ اﻟﺘﺎﱃ
( y)
()L(y | )
( u)
()L(y | )d 0 u
(a 1)()a 1 –a n! –r e a1 a 1 (n r)! u (a 1)()a1 –a n! –r e d a 1 a1 (n r)!
u
–(a r) e – r
–(a r)
u
, .
–r
e d
: اﻻن
–(a r )
u
–r
–(a r)
e d 0
let w=
u
–r
e d
–(a r)
0
u
e – r d
u u u = d 2 dz z z
–(a r )
u
u 0
–r
e d
u
(a r 1)
u w
(a r )
e
w
u u u d 2 0 w w
–(a r )
u w e d w2
u u (a r 1) 1 w e dw w (a r 1) 1e w dw . 0 w 0
u u (r 1, ) (r 1, ) x
where (n,x)= e-t t (n 1)dt. 0
( u ) u
– ( a r 1)
e
u
, . u u (r 1, ) (r 1, ) : ﲢﺖ ﻓﺮض داﻟﺔ ﻣﺮﺑﻊ اﳋﺴﺎرة ﻛﺎﻟﺘﺎﱄ ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻰ اﳌﻘﺪر اﻟﺒﻴﻴﺰي ﻟﻠﻤﻌﻠﻤﺔ
( a r 1)
٣٣٦
*
u
–(a r 2)
E( u)
(a r 1)
–(a r 2)
u
e d
u u (r 1, ) (r 1, )
, .
u
u u e d u (a r 2) (r 1, ) (r 1, )
u u u (r 2, ) (r 2, ) * . u u (r 1, ) (r 1, ) : ﻧﺘﺒﻊ اﻻﺗﻰ* ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺗﺒﺎﻳﻦ
E(2 u)
u
(a r 1)
u
–(a r 3) e d
u u (r 1, ) (r 1, )
u 2 * (r 3, u) * y y * , (n, y) (n, ) (n, ). (r 1, u)
: اذن * u 2 * (r 3, u) 2 (r 2, u) Var( ) * u * (r 1, u) (r 2, u)
2
*
u 2 * (r 3, u) * (r 2, u) *
2
[ (r 1, u)]
2
.
n
ﲤﺜﻞ زﻣﻦt 0 ﰱ ﺣﺎﻟﺔ اﳌﻌﺎﻳﻨﺔ ﻣﻦ اﻟﻨﻮع اﻻول ﻓﺈن. u x i وr n ﰱ ﺣﺎﻟﺔ اﻟﻌﻴﻨﺔ اﻟﻜﺎﻣﻠﺔ ﳓﺼﻞ ﻋﻠﻰ اﻟﻨﺘﺎﺋﺞ ﺑﻮﺿﻊ i 1
وﻋﻠﻰ ذﻟﻚ داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ. ﺗﺼﺒﺢ ﻣﺘﻐﲑا ﻋﺸﻮاﺋﻴﺎr اﻧﺘﻬﺎء اﻟﺘﺠﺮﺑﺔ واﶈﺪد ﻣﺴﺒﻘﺎ ﻣﻦ ﻗﺒﻞ اﻟﺒﺎﺣﺚ وﰱ ﻫﺬﻩ اﳊﺎﻟﺔ ﻓﺈن : ﻛﺎﻵﰐ r
L(x1 , x 2 ,..., x n | ) f (x i )[1 F(t 0 )]n r i 1
r
1 [ e i 1
x ( i )
t [( 0 )]nr
][e
]
u
1 ( ) e r r
where u
x
i
(n r)t 0 .
i 1
٣٣٧
) (٧-١-٥ﻣﻘﺎرﻧ ﺔ ﺑﯾﯾزﯾ ﺔ ﺗﺣ ت ﻓ رض اﻟﺗوزﯾ ﻊ اﻻﺳ ﻰ ﺑﻣﻌﻠﻣ ﺔ اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ
ﻛﺗوزﯾ ﻊ ﻗﺑﻠ ﻰ ﻓ ﻰ ﺣﺎﻟ ﺔ
ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ ) Bhattacharya (1967وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃ : إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ ﺣﻴﺚ r nو y y1 , y 2 , , y rواﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة . داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : r !n f (yi )[1 F(y r )]n r (n r)! i 1
L(y | )
y
r ) ( r !n 1 y [ exp( i )][e ]n r (n r)! i 1 r
1
] yi (n r) yr n! 1 [ i 1 e . (n r)! r
ﻟﻴﻜﻦ :
r
u [ yi (n r)y r ]. i 1
وإذاﻛﺎن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ اﻟﺘﻮزﻳﻊ اﻟﺘﺎﱃ :
, 0 ; 0 .
ﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻧﺘﺒﻊ اﻟﺘﺎﱃ :
٣٣٨
1 () e
( y)
()L(y | )
()L(y | )d 0
u
u
1 n! –r e e (n r)!
0
1 e
( u) u
n! –r e d (n r)!
u
– re
0
u – r
e
. d
: اﻻن
٣٣٩
0
b ax x
x e
a dx 2 b
1 2
K 1 2 ab
1 let a= , r
–r
e e
( u)
2 (u)
r 1
u
r 1 2
u
u K r 1 2
u – r
(u) e u 2K r 1 2
.
: ﲢﺖ ﻓﺮض داﻟﺔ ﻣﺮﺑﻊ اﳋﺴﺎرة ﻛﺎﻟﺘﺎﱄ ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻰ اﳌﻘﺪر اﻟﺒﻴﻴﺰي ﻟﻠﻤﻌﻠﻤﺔ u
(u) r 1 1–r E( u) e d u 0 2K r 1 2 *
(u) r 1 u 2K r 1 2
1 2 u
K r2 2 u K r 1 2
(r 2) 2
u K r 2 2
u . u : ﻧﺘﺒﻊ اﻻﺗﻰ* ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺗﺒﺎﻳﻦ
٣٤٠
u
(u) r 1 2–r E( u) e d. u 0 2K r 1 2 1 let a= , r, b u, 2
r 1
E( 2 u)
(u) 1 2 u u 2K r 1 2
K r 3 2 u K r 1 2
(r 3) 2
u K r 3 2
u . u : اذن
u K r 3 2 K r 2 2 u Var(* ) u K r 1 2 u K r 1 2
u u
2
2 u u u K r 3 2 2 K r 1 2 K r 2 2 . u K r 1 2
u
R(t) ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ
اﳌﻘﺪر اﻟﺒﻴﻴﺰى : ﲟﺎ ان
-
t
R(t) e , : وﺑﺈﺳﺘﺨﺪام داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺎ ﳝﻜﻦ اﳚﺎد اﳌﻘﺪر اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻛﺎﻟﺘﺎﱃ ٣٤١
*
R (t ) E[R(t)] R(t) ( | u) d 0
u t
(u) r 1 – r e u 0 2K r 1 2 1 let a= , r, b u t, r 1
d.
(u) 1 2 u (u t) 2K r 1 2 ut r 1 K r 1 2 2 u u ut K r 1 2 ut (r 1) K r 1 2 t 2 . 1 u u K r 1 2
R * (t)
(r 2) 2
ut K r 2 2
: اﻳﻀﺎ ﳝﻜﻦ اﳚﺎد ﺗﺒﺎﻳﻦ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﰱ ﻫﺬﻩ اﳊﺎﻟﺔ ﻛﺎﻟﺘﺎﱃ
٣٤٢
2
Var R * (t) E R 2 (t) u E(R(t ) u) ,
E[R 2 (t )] R 2 (t ) ( | u) d 0
u 2 t
( u) r 1 – r e u 0 2K r 1 2 1 let a= , r, b u 2t,
( u) r 1 E[R (t)] u 2K r 1 2 2
2t 1 u
( r 1) 2
d.
u 2t 2K r 1 2
u 2t K r 1 2 u K r 1 2
( r 1) (u t) 2
.
Var R * (t) E R 2 (t ) u E(R (t) u)
2
2
2t 1 u
1
(r 1) 2
u 2t u t (r 1) K r 1 2 K r 1 2 t 2 1 . u u u K r 1 2 K r 1 2
2t Var R * (t ) 1 2 u u K 2 r 1
( r 1) 2
u 2t K r 1 2
2
u u t K r 1 2 K r 1 2 .
ﻛﺗوزﯾ ﻊ ﻗﺑﻠ ﻰ ﻓ ﻰ ﺣﺎﻟ ﺔprior quasi-density ( ﺗﻘدﯾرات ﺑﯾﯾزﯾ ﺔ ﺗﺣ ت ﻓ رض٨-١-٥) اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ : وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃBhattacharya (1967) ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ اﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪاتr ﻫﻰ الy1 y 2 y r إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان واﳌﻄﻠﻮبy y1 , y 2 , , y r وr n ﺣﻴﺚ ﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔn اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ : داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ. ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة ٣٤٣
r n! f (yi )[1 F(yr )]n r (n r)! i 1
L(y | )
y
r ( r ) n! 1 y [ exp( i )][e ]n r (n r)! i 1 r
1
yi (n r ) yr ] n! 1 ( [ i 1 e . (n r)! r
: ﻟﻴﻜﻦ r
u [ yi (n r)y r ] i 1
: ﻫﻮ ﲢﺖ ﻓﺮض ان اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ ل a ,0 . : ﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻧﺘﺒﻊ اﻟﺘﺎﱃ.Bhattacharya (1967 )واﳌﺎﺧﻮذ ﻣﻦ ﻗﺒﻞ
( x)
() L(y | )
( u)
()L(y | )d 0
u
n! –r e (n r)! u n! –r –a 0 (n r)! e d –a
0
–(a r)
e
–( r)
u
e
u
d u
1 u ( )(a r ) e u (a r 1)
: ﲢﺖ ﻓﺮض داﻟﺔ ﻣﺮﺑﻊ اﳋﺴﺎرة ﻛﺎﻟﺘﺎﱄ ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻰ اﳌﻘﺪر اﻟﺒﻴﻴﺰي ﻟﻠﻤﻌﻠﻤﺔ
٣٤٤
u (u) (a r ) –(a r) 1 E( u) e d . 0 (r) (a r 1)) u u u let w= = d 2 dw. w w (a r ) 1 (u) u * u (a r ) 1 w (a r) 1e w 2 dw (a r 1) 0 w *
*
( a r 1)1 u w e w dw 0 (a r 1)
(u) u (a r 2) , r 2. (a r 1) ar2 : ﻧﺘﺒﻊ اﻻﺗﻰ* ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺗﺒﺎﻳﻦ ( a r 1)1
( a r ) 2 (a r ) 2 u u E( u) u w e w 2 dw (a r 1) 0 w 2
( a r 3)1 u2 w e w dw (a r 1) 0
u2 u2 (a r 3) . (a r 1) (a r 2)(a r 3) : اذن 2
u u2 Var( ) (a r 2)(a r 3) a r 2 u2 u2 . a r 2 a r 3 (r 2)(r 3) (r 2) 2 (r 3) *
( ﺗﻘدﯾرات ﺑﯾﯾزﯾﺔ ﻓﻰ ﺣﺎﻟﺔ اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ﺗﺣت ﻓرض اﻟﺗوزﯾﻊ اﻟﻘﺑﻠ ﻰ٩-١-٥) اﻟﻣراﻓق : وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃBhattacharya (1967 ) ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ
٣٤٥
إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ ﺣﻴﺚ r nو y y1 , y 2 , , y rواﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة . داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : r !n f (yi )[1 F(y r )]n r (n r)! i 1
L(y1 , y 2 ,..., y n | )
r y y !n 1 [ exp( i )][exp( r )]n r (n r)! i 1
n! 1 1 r exp( [ yi (n r)y r ]. (n r)! r i 1
ﻟﻴﻜﻦ : r
] u [ yi (n r)y r i 1
ﻻﺧﺘﻴﺎر اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ اﳌﺮاﻓﻖ ﻣﻦ اﳌﻌﻠﻮم ان داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻻﺑﺪ ان ﻳﻜﺘﺐ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃ : .
a ( ) b( x ) c( )d (x )]
f (x | ) e
وﻋﻠﻰ ذﻟﻚ ﻓﺈن : n
.
d ( x i )] ) na ( )) c( i 1
L(y | ) L e
وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ اﳌﺮاﻓﻖ ﻳﻜﺘﺐ ﻋﻠﻰ اﻟﺸﻜﻞ اﻻﺗﻰ : ,
1a ( ) 1c( )]
( ) e
واﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻋﻠﻰ اﻟﺸﻜﻞ اﻻﺗﻰ : n
, 2 1 n, 2 1 d(x i ).
2 a ( ) 2 c( )]
( | x) e
i 1
ﲢﺖ ﻓﺮض اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﻓﺈن Lﳝﻜﻦ ﻛﺘﺎﺑﺘﻬﺎ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃ : n
x i 1
a() ln , c( )
i1
r ln
e
n x i i 1
1 L re
وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ اﳌﺮاﻓﻖ ﻫﻮ : ,
1 1 ln ]
( ) e
اى ان :
.
1 ] 1
e
( )
1 1 ln ]
اى ان اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ اﳌﺮاﻓﻖ ﻫﻮ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ اﻟﻌﻜﺴﻰ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃ : ٣٤٦
( ) e
1 ( ) ( )
1
e
]
,0< , , 0.
: ﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻧﺘﺒﻊ اﻟﺘﺎﱃ ()L(y | )
( y)
( | u)
()L(y | )d 0 –r
0
e
–( r 1)
let w=
e
u –r –(1)
0
u –(1)
e
e
–( r 1)
e d
u
e
u
. d
u u u d 2 dw. w w
: اذن
0
–( r 1) e
u w
u
d
u w
0
–( r)
0
w
( r ) 1 w
e
–( r 1)
e
w
u dw w
u dw w2
–( r )
( r).
: اذن ) (u )(r ) –(r1) (u ( u) e (r ) (r 1)
u
(u )
e . (u )(r )
: ﲢﺖ ﻓﺮض داﻟﺔ ﻣﺮﺑﻊ اﳋﺴﺎرة ﻛﺎﻟﺘﺎﱄ ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻰ اﳌﻘﺪر اﻟﺒﻴﻴﺰي ﻟﻠﻤﻌﻠﻤﺔ
٣٤٧
(u ) (r ) E( u) (r ) *
(u ) (r ) (r ) *
*
0
0
u w
–(r )
(r )
e
u
d
u e w 2 dw w
(T ) ( r1)1 w w e dw (r ) 0
(u ) (u ) (r 1) , r 1. (r ) (r 1) : ﻧﺘﺒﻊ اﻻﺗﻰ* ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺗﺒﺎﻳﻦ
(u ) (r ) E( u) (r ) 2
(u ) (r ) (r )
0
0
u w
–(a r ) 1
(r ) 1
e
u
d
u e w 2 dw w
(u ) 2 ( r2 )1 w w e dw (r ) 0
(u ) 2 (u ) 2 (r 2) . (r ) (r 1)(r 2) : اذن 2
(u ) (u )2 (r 1)(r 2) (r 1) (u )2 (u ) 2 . r 1 r 2 (r 1)2 (r 2) (r 1) 2 (r 2) Var(* )
اﻟﻤﻘﺪر اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ
:ﲟﺎ ان ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻫﻰ
-
t
R(t ) e , : ﺑﺈﺳﺘﺨﺪام داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺎ ﳝﻜﻦ اﳚﺎد اﳌﻘﺪر اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻛﺎﻟﺘﺎﱃ
٣٤٨
*
R (t ) E[R(t)] R(t ) ( | u) d 0
(u )(r ) (r ) (u )(r ) (r )
0
0
–(r 1)
e
( t u )
t u w
(u )(r ) (t u ) (r )
d
(r 1)
t u e w dw 2 w
( r )
w
0
( r ) 1
e w dw
( r )
( r ) (u ) (r )(t u ) (r )
t u u
(r )
t 1 u
(r )
.
: اﻳﻀﺎ ﳝﻜﻦ اﳚﺎد ﺗﺒﺎﻳﻦ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﰱ ﻫﺬﻩ اﳊﺎﻟﺔ ﻛﺎﻟﺘﺎﱃ 2
Var R * (t) E R 2 (t) u E(R (t) | u) 2
E[R (t)] R 2 (t ) ( | u) d 0
(u )(r ) (r ) (u )(r ) (r )
0
0
–(r 1)
e
(2t u )
2t u w
(u )(r ) (2t u ) (r )
d
(r 1)
2t u e w dw 2 w
( r )
0
w
( r ) 1
( r )
( r ) (u ) (r )(2t u ) (r )
2t u u
(r )
2t 1 u ٣٤٩
(r )
.
e w dw
2
Var R * (t) E R 2 (t u) E(R (t | u) ) 2(r
.
t 1 u
) (r
2t 1 u
) (١٠-١-٥اﻟﺗﻘدﯾر اﻟﺑﯾﯾزى ﻓﻰ ﺣﺎﻟﺔ ﻧﻘص اﻟﻣﻌﻠوﻣﺎت ﻋن ﻓﻰ ﺣﺎﻟﺔ اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ ﺣﻴﺚ r nو y y1 , y 2 , , y rواﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻣﺘﻮﺳﻂ زﻣﻦ اﳊﻴﺎة . داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : r
1
r
n! 1 y n r y L y1 , y 2 ,..., y r . e n r ! r
i
i1
:اﯾﺠﺎد ﻣﻘﺪرات ﺑﯿﯿﺰ ﺗﺤﺖ ﻓﺮض اﻟﺘﻮزﯾﻊ ﻷﺳﻲ ﺑﻤﻌﻠﻤﺘﯿﻦ: r
u yi n r y r .
let
i 1
وﺑﻔﺮض ان : n! r 1 u k e d n r ! 0 !n u r r. ! n r
ﲢﺖ ﻓﺮض ان اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ ل ﻫﻮ : 1 g .
ﻓﺈن اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل ﻫﻮ : u r r
ﲟﺎ ان داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻟﻠﺘﻮزﻳﻊ اﻻﺳﻰ ﻫﻰ :
u
e
r 1
u
u 1 r u e . r
.
t
R(t) e
٣٥٠
ﻓﺈن اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻫﻮ : R * (t) E R(t) x
R(t) x d 0 r
1 u u e d r
t
e 0
u r r 1 1u t e d r 0 u t ut d d 2 ur r * R (t) u t r 1e d r 0 r u r t u r r r t * R (t) 1 . u
ut
let
) (١١-١-٥اﻟﺗﻘدﯾر اﻟﺑﯾﯾ زى ﻟﻣﺗوﺳ ط اﻟﺣﯾ ﺎة وداﻟ ﺔ اﻟﺻ ﻼﺣﯾﺔ ﻟﻠﺗوزﯾ ﻊ اﻻﺳ ﻰ ذو اﻟﺑﺗ ر اﻟﻣزدوج ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺔ اﻟﻛﺎﻣﻠﺔ ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ ) Shalaby (1994وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃ :
إذا ﻛﺎﻧﺖ Xﳍﺎ اﻟﺘﻮزﻳﻊ اﻵﺳﻲ ﺑﺎﳌﻌﻠﻤﺔ و ﻗﺪ ﰎ ﻋﻠﻴﻬﺎ ﺑﱰ ﻣﺰدوج ﺗﻜﻮن داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﳍﺎ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱄ : t2 1 x t1 f(x| t1 <X< t 2 ) = e (e e ) 1 > 0 , 0 <t1 t 2 ; t1 x t 2 .
داﻟﺔ اﻹﻣﻜﺎن اﻷﻋﻈﻢ ﻟﻌﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﺣﺠﻤﻬﺎ nﻣﻦ اﻟﺘﻮزﻳﻊ اﻵﺳﻲ ذو اﻟﺒﱰ اﳌﺰدوج ﺗﻜﻮن ﻛﺎﻟﺘﺎﱃ:
٣٥١
n
L(x1 , x 2 ,..., x n ; , t1 , t 2 ) L f (x i ) i 1
n
xi i 1
n
e 1 = . n t t 1 2 e e t 2 t1 n = e e e
u
n
n
t 2 t1 t1 n = e e 1 e
= n e = n e
u
(T nt1 )
G
n
n
(t t ) 2 1 1 e .
R 1 e
n
n
where G = u-nt1
, R= t 2 t1
,u = x i i=1
: ﻫﻮ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ اﻟﻌﻜﺴﻲ ﻋﻠﻰ اﻟﺸﻜﻞ ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ h h 1 () e ( 1)
; h, > 0 .
> 1 ﻫﻲ داﻟﺔ ﺟﺎﻣﺎ و( 1) ﺣﻴﺚ أن . ﺳﻮف ﻧﺴﺘﻔﻴﺪ ﻣﻦ داﻟﺔ ﺟﺎﻣﺎ اﻟﻨﺎﻗﺼﺔ اﻟﺘﺎﻟﻴﺔ ﰱ اﳊﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻟﻠﻤﻌﻠﻤﺔ
(m 1, a) y m e y dy a
aj =(m 1)e , m = 0,1,2, ... j0 j! m
a
٣٥٢
: ﻳﻜﻮن اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻟﻠﻤﻌﻠﻤﺔ ( | x)
L(x | )()
.
L(x | )() d 0
f (x) L(x | ) () d 0
(h+G) R h -1 -( +n) = e (1-e ) -n d ( -1) 0
(h+G) jR h -1 n j1 - -( +n) = e j e d ( -1) 0 j=0
(h+G+jR) -1 n j1 h -( +n) = e d. j ( -1) 0 j=0 h+G+jR h+G+jR (h+G+jR) let w = = d = dw, w w2 if =0 w= , if = w=0,
٣٥٣
0
-1 h+G+jR n j1 h f (x) = w j (-1) j=0
-( +n)
(h+G+jR) e-w dw 2 w
-1 1-( +n) n j1 h n 2 -w = w h+G+jR e dw j ( -1) j=0 0
-1 1-( +n) n j1 h = h+G+jR ( n 1) j ( -1) j=0
1-( +n)
-1 n j1 h ( n 1) ( n )1 h+jR = G 1+ j (-1) h j=0
n 1 h -1 n j1 A j = j n 1 ( n 1) ( -1) j=0 G
h -1 = K 1 , ( -1) -1
where
n 1
h+jR n j1 A j -1 A j = 1+ , K j n 1 ( n 1) h G j=0
G n 1 K n j1 n 1 ( n 1) Aj j j=0 R h -1 -( +n) - (h+G) e (1-e )-n (-1) ( | u) h -1 K 1 , ( -1)
K
-( +n)
e
-
(h+G)
-
R -n
(1-e )
; >0 .
: ﻫﻮU=u ﺑﺸﺮط ﻟﻠﻤﻌﻠﻤﺔm اﻟﻌﺰم اﻟﻼﻣﺮﻛﺰي ذو اﻟﺮﺗﺒﺔ ` m
m
( | u) E( ) m ( | u) d, m 1, 2,... 0
٣٥٤
=K
-( +n-m)
e
-
(h+G)
-
(h+G)
-
R -n
(1-e ) d
0
=K
-( +n-m)
e
0
Rj
n j1 - j e d j=0
n j1 -( +n-m) =K e j 0 j=0
(h+G+Rj)
d
n j1 =K ( +n-m-1) (G+h+Rj)1-( +n-m) j j=0
h+Rj -( +n-m-1) n j1 1-( +n-m) =K G ( +n-m-1) (1+ ) j G j=0
n j1 1-( +n-m) =K G (+n-m-1) A j +n-m-1 j j=0
G =
n 1
n j1 +n-m-1 1-( +n-m) G ( +n-m-1) j Aj j=0 n j1 +n-1 ( +n-1) Aj j j=0
n j1 +n-m-1 A j G m ( +n-m-1) j=0 j = . n j1 where , m = 1,2,... ( +n-1) +n-1 j Aj j=0 . m و ﳝﻜﻦ إﳚﺎد اﳌﺘﻮﺳﻂ و اﻟﺘﺒﺎﻳﻦ ﻣﻦ اﻟﻌﺰوم اﻟﻼﻣﺮﻛﺰﻳﺔ ذو اﻟﺮﺗﺒﺔ
R(t ) وداﻟﺔ اﻟﺼﻼﺣﻴﺔ اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى ﻟﻜﻞ ﻣﻦ اﳌﻌﻠﻤﺔ : ﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة ﳌﺮﺑﻊ اﳋﻄﺄ ﻫﻮ* اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى
٣٥٥
n j1 +n-1-1 j Aj G( +n-1-1) j=0 * E() 1` ( | u) . n j1 ( +n-1) +n-1 j Aj j=0
n j1 +n-2 j Aj G( +n-2) = . j=0 , n j1 ( +n-2)( +n-2) +n-1 j Aj j=0
*
n j1 +n-2 j Aj G = . j=0 . n j1 ( +n-2) +n-1 j Aj j=0
n j1 +n-3 j Aj 2 G ( +n-3) j=0 E(2 ) `2 ( | u) . n j1 ( +n-1) +n-1 j Aj j=0
n j1 +n-3 j Aj G2 j=0 . n j1 . ( +n-2)( +n-3) +n-1 j Aj j=0 : ( و ﻣﻨﻬﺎ ﻳﻜﻮن اﻟﺘﺒﺎﻳﻦ اﻟﺒﻌﺪي ) اﳌﺨﺎﻃﺮة اﻟﺒﻌﺪﻳﻪ
Var(* ) E(2 ) E
2
2 n j1 +n-3 n j1 +n-2 Aj j Aj j G2 ( +n-3) j=0 j=0 . = n j 1 n j 1 ( +n-2)( +n-3) +n-1 ( +n-2) +n-1 A j j j A j j=0 j=0
٣٥٦
اﳌﻨﻮال اﻟﺒﻌﺪى : ﳝﻜﻦ إﳚﺎدﻩ ﺑﺎﺷﺘﻘﺎق داﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻛﻤﺎ ﻳﻠﻲ اﳌﻨﻮال اﻟﺒﻌﺪي -( +n)
-
(h+G)
-
R -n
( | u) K e (1-e ) d( | u) d R R - (h+G) -n -( +n)-1 -n -( +n) e (1-e ) -( +n) (1-e ) R K (h+G) (h+G) R (h+G) n( R)e -n-1 2 e -( +n) e (1-e ) 2 (h+G) R R -( +n+1) -n -n -( +n+2) -( +n) e (1-e ) (h+G)(1-e ) K (h+G) (h+G+R) R e nR-( +n-2) e (1-e )-(n+1) R (h+G) R R - (h+G) nR -( +n+1) -n -1 =K(1-e ) e (1-e ) e -( +n)+
d( | u) 0 d R R (h+G) nR -1 -( +n)+ (1-e ) e 0 -
R R -1
-( +n)+(h+G) nR(1-e ) e 0. R R - - 1 -1 (h+G) nR(1-e ) e , ( +n)
R - 1 (h+G) nRe . ( +n)
. و ﻳﺘﻢ إﳚﺎد اﳌﻨﻮال اﻟﺒﻌﺪي ﲝﻞ اﳌﻌﺎدﻟﺔ ﻏﲑ اﳋﻄﻴﺔ اﻟﺴﺎﺑﻘﺔ ٣٥٧
ˆ ﺗﻘﺪﻳﺮ اﻻﻣﻜﺎن اﻻﻛﱪ
L(x | ) n e
G
n
R 1 e ,
G R n G R n G n n 1e 1 e n 2 e 1 e dL R R n 1 d n G R e n( e ) 1 e 2
n e
G
R 1 e
n
R 1 n G nR R 2 2 e 1 e 1
R dL n G nR R 0 2 2 e 1 e 0 d
n G nRe
R
R 1 e
1
1
R R ˆ 1 (G nRe ˆ 1 e ˆ ) n 1 Rˆ 1 (G nR e 1 ) n R ˆ 1 (G nRe ˆ ) . n وةةة
ااﻻﻟﻠﻜﻤﻢ
٣٥٨
ﺗﻘﺪﻳﺮ اﻻﻣﻜﺎن اﻻﻛﱪ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ : ﺣﻴﺚ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻫﻰ R(t)
t
t2
t1
t2
e e
.
e e وﻋﻠﻰ ذﻟﻚ ﻣﻘﺪر اﻻﻣﻜﺎن اﻻﻛﱪ ﻟﺪاﻟﺔ. وان اى داﻟﺔ ﰱ ﻣﻘﺪر اﻻﻣﻜﺎن اﻻﻛﱪ ﻫﻰ ﻣﻘﺪر اﳌﻜﺎن اﻻﻛﱪ ﳍﺬﻩ اﻟﺪاﻟﺔ
: اﻟﺼﻼﺣﻴﺔ ﻫﻮ
t ˆ
t2 ˆ
ˆ ) e e . R(t t1 t2 ˆ e e ˆ
اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ t1 t 0 t 2 ﺣﯾثR(t 0 ) ر اﯾﺎﻟﺔ اﻟﺻﻼﺣﯾﺔ : ﺣﻴﺚR * (t) ﺑﺎﺳﺘﺨﺪام داﻟﺔ اﳋﺴﺎرة ﳌﺮﺑﻊ اﳋﻄﺄ ﳝﻜﻦ إﳚﺎد
R * (t) E R(t )
= R(t ) (|u) d 0
= 0
t
t2
t1
t2
(e e (e
e
=K
-( +n)
e
t t1
)
K
(1 e
e
(t t1 ) -( +n)
e
(1 e
( t 2 t1 )
-( +n)
e
R
`
(h+G)
-
R -n
(1-e ) d
(1 e
(R R ` )
(1 e
)
e
-
(h+G)
R
-
R -n
(1-e ) d
)
(t 2 t1 (t t1 ))
(1 e
0
(t 2 t )
(1 e
0
=K
e
-
)
0
=K
-( +n)
R
)
)
e
-
(h+G)
-
R -n
(1-e ) d
) e
-
(h+G)
)
R t 2 t1 , R ` t t1 ٣٥٩
-
R -n
(1-e ) d
=K
R` -( +n)
e
(1 e
(R R ` )
) e
-
(h+G)
-
(h+G)
-
R -(n+1)
(1-e )
d
0
=K
R` -( +n)
e
(1 e
(R R ` )
) e
0
=K -( +n) 0
=K 0
j
j=0
-( +n)
n+j
j=0
n+j j
e
e
-
-
(jR+R ` h G)
(jR+R ` h G )
(1 e
jR
n+j - j e d j=0
(R R ` )
) d
d K
-( +n)
0
n+j je j=0
`
((1+j)R h G )
(jR+R h G ) ((1+j)R h G ) n+j -( +n) -( +n) =K e d e d j 0 j=0 0
n+j =K ( n 1) j j=0 (jR+R ` h G) ( n 1) ((1+j)R h G) ( n 1)
n+j =K ( n 1)G ( n 1) j j=0 jR+R ` h (1+j)R h ( n 1) ( n 1) ( 1) ( 1) G G
n+j ( n 1) A j1( n 1) j Bj j=0 = n+j-1 ( n 1) j Aj j=0 jR+R ` h (1+j)R h where B j ( 1) 1 , A j1 ( 1) 1 G G
٣٦٠
d
ﻓﱰة ﺛﻘﺔ ﻟﻠﻤﻌﻠﻤﺔ
ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﻓﱰة ﺛﻘﺔ (t1 , t 2 ) HBDﻟﻠﻤﻌﻠﻤﺔ واﻟﱴ ﳚﺐ ان ﲢﻘﻖ اﻟﺸﺮﻃﲔ اﻟﺘﺎﻟﻴﲔ : : ) (1) ( t 1 u ) ( t 2 u t2
f ( u ) d 1 .
( 2)
t1
وﳓﺼﻞ ﻋﻠﻰ ﻫﺬﻩ اﻟﻔﱰة ﻛﺎﻟﺘﺎﱃ :
) -n
R t2
-
(1 -e
) (h+ G t2
-
e
) -( +n n
n
.
Kt2
-n
)
R t 1 -e 1 R 1-e t 2
R t 1 -e 1 R 1 -e t 2
R t1
-
(1 -e 1
) (h +G t1
-
e
)-( + n
(1) K t 1
1
]) t 1 ( + n) [-(h +G )( t1 - t 2 ) ( e t2 1
]) t 1 ( + n) (h+ G )[ t1 t 2 )(t 2 t 1 ) ( e t2
ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﺣﺪود اﻟﺜﻘﺔ وذﻟﻚ ﲝﻞ اﳌﻌﺎدﻟﺘﲔ اﻟﺴﺎﺑﻘﺘﲔ ﺑﺎﺳﺘﺨﺪام اﳊﺎﺳﺐ اﻻﱃ . ﻓﻴﻤﺎ ﻳﻠﻰ اﳊﺎﻻت اﳋﺎﺻﺔ ﻣﻦ اﻟﻨﺘﺎﺋﺞ اﻟﺴﺎﺑﻘﺔ : )ا( ﻋﻨﺪﻣﺎ t 2 ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ اﻟﻨﺘﺎﺋﺞ اﳋﺎﺻﺔ ﺑﺎﻟﺒﱰ ﻣﻦ اﻟﻴﺴﺎر . )ب( ﻋﻨﺪﻣﺎ t1 0ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ اﻟﻨﺘﺎﺋﺞ اﳋﺎﺻﺔ ﺑﺎﻟﺒﱰ ﻣﻦ اﻟﻴﻤﲔ . )ج( ﻋﻨﺪﻣﺎ t1 0, t 2 ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ اﻟﻨﺘﺎﺋﺞ اﳋﺎﺻﺔ ﺑﺎﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ واﺣﺪة ﰱ اﻟﻌﻴﻨﺔ اﻟﻜﺎﻣﻠﺔ .
) (١٢-١-٥اﻟﺗﻘدﯾر اﻟﺑﯾﯾزى ﻟﻣﺗوﺳ ط اﻟﺣﯾ ﺎة وداﻟ ﺔ اﻟﺻ ﻼﺣﯾﺔ ﻟﻠﺗوزﯾ ﻊ اﻻﺳ ﻰ ﻓ ﻰ ﺣﺎﻟ ﺔ اﻟﻌﯾﻧﺔ اﻟﻣراﻗﺑﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ﻣن ﺟﮭﺗﯾن ٤ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ ) Shalaby (1990وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃ : ٣٦١
ﺑﻔﺮض ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nوﺿﻌﺖ ﻟﻼﺧﺘﺒﺎر ﲤﺜﻞ أزﻣﻨﺔ اﻟﻔﺸﻞ وان أزﻣﻨﺔ اﻟﻔﺸﻞ ﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻷﺳﻲ ﲟﻌﻠﻤﺔ وﻛﺎﻧﺖ اﻻﺣﺼﺎءات اﻟﱰﺗﻴﺒﻴﺔ ﻟﻠﻌﻴﻨﻪ اﻟﻌﺸﻮاﺋﻴﺔ ﻫﻰ Yn1 1 ,Yn1 2 ,...,Yn n 2وﺑﺎﻋﺘﺒﺎر أن y n1 1 yn1 2 ... y n n 2ﰎ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﺣﻴﺚ n1ﰎ ﲢﺪﻳﺪﻫﺎ ﻗﺒﻞ اﻟﺘﺠﺮﺑﺔ وﺑﺎﻟﺘﺎﱃ ﻓﺈن n1ﻣﻦ اﻟﻮﺣﺪات زﻣﻦ ﻓﺸﻠﻬﻤﺎ اﻗﻞ ﻣﻦ . y n1 1اﻳﻀﺎ n 2ﰎ ﲢﺪﻳﺪﻫﺎ ﻗﺒﻞ اﻟﺘﺠﺮﺑﺔ وﺑﺎﻟﺘﺎﱃ ﻓﺈن n 2ﻣﻦ اﻟﻮﺣﺪات ﺻﺎﳊﺔ ﻟﻠﻌﻤﻞ ﺑﻌﺪ اﻟﺰﻣﻦ . y n n 2اى ان ) (n n1 n 2ﺣﺴﺒﺖ
ازﻣﻨﺔ ﻓﺸﻠﻬﺎ .اﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ ﻟﻠﻤﻌﻠﻤﺔ . ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ داﻟﺔ اﻹﻣﻜﺎن ﻛﺎﻟﺘﺎﱄ: n! n n 2 L f (yi ) P(X y n11 ) n1 P(X y n n 2 ) n 2 n1 n 2 in11 n2
y n n2
n2
n1
e
yn1 1
y - i
- 1- e n n2 n y t1* 1 - i -ne in11 1- e
*
- t2 e
n1
n n 2
n! 1 e n1 n 2 in11
!n n1 n 2
- t1 -n -s e 1- e
!n n1 n 2
*
n
=
=
t 2 t1 n = e e e ,
ﺣﻴﺚ n 2 t *2 , n =n-n 1 -n 2 . :
s
=
n n2
i
y
* 2
* 1
t y n1 1 , t y n n 2 ,s
i n1 1
اﻟﻌﻴﻨﺔ اﳌﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱏ ﻣﻦ ﺟﺎﻧﺐ واﺣﺪ ﳓﺼﻞ ﻋﻠﻴﻬﺎ ﺑﻮﺿﻊ n1 0 or n 2 0
ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ اﻟﻌﻜﺴﻲ :
; h, > 0 .
h h 1 () e )( 1
اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل ﻫﻮ :
٣٦٢
( | x)
L(x | )()
( | s).
L(x | )() d 0
f (x) L(x | ) () d 0
*
(s+h) t -1 n! h -1 -( +n) n1 = e (1-e ) d n1 n 2 (-1) 0 *
(s+h+jt1 ) n1 n! h -1 j -( +n) = (1) e d n1 n 2 (-1) j=0 j 0 n1 h+jt1* n! h -1 j -(n-1) = (1) (n-1)s (1 ),n = +n . j n1 n 2 (-1) j=0 s
: ﻫﻮ وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل ( | s) k
-n
e
-
(s+h)
-
t1*
(1-e ) n1
n1 h+jt1* ( n-1 -(n-1) (k ) ( 1) (n -1)s (1 ) ) j s j=0 1
n1
n1
j
n1 ( 1) (n -1)s -(n -1) D (nj -1) , j j=0 j
h+jt1* 1 where D j (1 ) . s
: ﻫﻮs ﺑﺸﺮط ﻟﻠﻤﻌﻠﻤﺔm اﻟﻌﺰم اﻟﻼﻣﺮﻛﺰي ذو اﻟﺮﺗﺒﺔ
٣٦٣
` m
m
( | s) E( ) m ( | s) d 0
k
-(n-m)
e
-
(s+h)
-
t1* n1
(1-e ) d
0
n1
n1 =k (1) j (n-m-1)s -(n-m-1) D(nj -m-1) j j=0 n1
n1 (n-m-1) (1) j D j (n -m-1) m j=0 s n1 , n1 (n -1) j (n-1) (1) D j j j=0 m 1,2,... . m و ﳝﻜﻦ إﳚﺎد اﳌﺘﻮﺳﻂ و اﻟﺘﺒﺎﻳﻦ ﻣﻦ اﻟﻌﺰوم اﻟﻼﻣﺮﻛﺰﻳﺔ ذو اﻟﺮﺗﺒﺔ R(t ) وداﻟﺔ اﻟﺼﻼﺣﻴﺔ اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى ﻟﻜﻞ ﻣﻦ اﳌﻌﻠﻤﺔ j
: ﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة ﳌﺮﺑﻊ اﳋﻄﺄ ﻫﻮ* اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى n1
n1 (n-2) (1) D j s (n -2) j j=0 * ` E( | s) 1 ( | s) , n1 (n-1) n1 j (n-1) (1) D j j j=0 n1
j
n1 (n-3) (1) D j 2 s (n -3) j j=0 E(2 ) `2 ( | s) . n1 (n -1) n1 (1) j D(nj -1) j j=0
٣٦٤
j
Var() n1
n1 n1 n1 j (n-3) j (n-2) (1) j D j (1) j D j s 2 (n -3) j=0 s2 j=0 . n1 n1 n1 (n-1) ((n -1))2 n1 j n-1) j (n-1) (1) D j (1) j D j j j=0 j=0
2
2 n1 n1 n1 n1 j (n-3) j (n-2) (1) D j (1) j D j j s2 (n -3) j=0 j=0 . n1 n1 n1 (n -2)(n-3) n1 (n -2) (1) j D(nj -1) (1) j D(nj -1) j=0 j j j=0
اﳌﻨﻮال اﻟﺒﻌﺪى : ﳝﻜﻦ إﳚﺎدﻩ ﺑﺎﺷﺘﻘﺎق داﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻛﻤﺎ ﻳﻠﻲ اﳌﻨﻮال اﻟﺒﻌﺪي ( | s) k-n e ( | s) k
-n
e
-
(s+h)
-
-
(s+h)
-
t1* n1
(1-e )
t1*
(1-e )n1 .
d( | s) d t* (h+s) -1 - (h+s) -(n+1) -n (h+s) n1 (-n ) e (1-e ) e 2 k (h+s) t1* t1* * t1 -n n1 -1 e (n (1-e ) e ) . 1 2
٣٦٥
*
*
*
(s h ) t t n s h n1 t1* t1 1 1 d( | s) n n1 1 0 e (1 e ) 2 2 e (1 e ) 0 d * 1 1
n s h n t (e
t1*
1) 1 0
*
t 1 1 * 1 s h n t (e 1) 1 1 n *
t 1 1 * s h n 1 t1 e . n
اﳌﻨﻮال اﻟﺒﻌﺪي . ﳝﻜﻦ إﳚﺎدﻩ ﲝﻞ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﺑﺎﺳﺘﺨﺪام اﳊﺎﺳﺐ اﻻﱃ R(t) اﳌﻘﺪر اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ
: ﲟﺎ ان
-
t
R(t) e , t1* < t<t *2
: ﺑﺈﺳﺘﺨﺪام داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺎ ﳝﻜﻦ اﳚﺎد اﳌﻘﺪر اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻛﺎﻟﺘﺎﱃ
٣٦٦
*
R (t 0 ) E[R (t)] R(t )( | s) d 0
k
-n
e
-
(t+s+h)
-
t1* n1
(1-e ) d
0
(t+s+h+jt * )
n1
n1 1 j -n k (1) e d j 0 j=0 n1 n1 k j = (1) (t +s+h+jt1* )-(n-1) (n -1) j=0 j -(n -1)
n1
* 1
k t+h+jt j -(n -1) ( 1) s 1 j (n-1) j=0 s n1
n1
n1 n-1 (1) E j 1 t+h+jt1* j j=0 n1 ,E j 1 . n1 s (1) j D nj -1 j j=0 j
: اﻳﻀﺎ ﳝﻜﻦ اﳚﺎد ﺗﺒﺎﻳﻦ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﰱ ﻫﺬﻩ اﳊﺎﻟﺔ ﻛﺎﻟﺘﺎﱃ
Var R(t) E(R(t) s) (E(R (t) s)) 2 .
٣٦٧
2
E[R (t)] R 2 (t)( | s) d 0
k -n e
-
(2t+s+h)
-
t1*
(1-e ) n1 d
0
n1
n1
-n k ( 1) j e j 0 j=0
-
(2t+s+h+jt1* )
-
t1*
(1-e ) n1 d
n1 n1 k (2t+h+jt 1* ) j -(n-1) ( 1) j s 1 s (n -1) j=0
n1
n1 2t+h+jt 1* j -(n -1) ( 1) s 1 s k j j=0 n1 n1 (n -1) j ( 1) D (nj -1) j j=0 n1
-(n -1)
-(n -1)
n1 n-1 ( 1) H j * 1 * 1 j t+h+jt 2t +h+jt j=0 1 1 n1 ,E 1 , H 1 j j . n1 s s j n -1 ( 1) D j j j=0 j
Var R(t ) E(R (t) s) (E(R (t) s)) 2 2
n1
n1 n1 n1 j n-1 j n-1 (1) H j (1) j E j j j=0 . j=0 n1 n n1 n1 1 j n-1 j n-1 ( 1) D ( 1) j j j Dj j=0 j=0
٣٦٨
ﻟﻠﻣﻌﻠﻣﺗ ﯾن ,
) (١٣-١-٥ﺗﻘ دﯾرات اﻻﻣﻛ ﺎن اﻻﻛﺑ ر اﻟﻌﯾﻧﺔ اﻟﻛﺎﻣﻠﺔ
ﻟﻠﺗوزﯾ ﻊ اﻻﺳ ﻰ ﺑﻣﻌﻠﻣﺗ ﯾن ﻓ ﻰ ﺣﺎﻟ ﺔ
ﺑﻔـﺮض ان X X1 ,X 2 ,..., X n ﻋﻴﻨــﺔ ﻋﺸـﻮاﺋﻴﺔ ﻣــﻦ اﳊﺠــﻢ nﻣــﻦ اﻟﻮﺣــﺪات ﳐﺘــﺎرة ﻣــﻦ ﺗﻮزﻳــﻊ ﻟــﻪ داﻟــﺔ ﻛﺜﺎﻓــﺔ اﻻﺣﺘﻤــﺎل اﻻﺳﻰ ﲟﻌﻠﻤﺘﲔ , واﻟﱴ ﺗﺎﺧﺬ اﻟﺸﻜﻞ اﻟﺘﺎﱃ :
, x ,
x
1 e
)f (x; ,
0 , e.w. ﺑﺎﻋﺘﺒﺎر أن y1 y 2 y nﰎ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﺣﻴﺚ y y1 , y 2 , , y nواﳌﻄﻠﻮب ﺗﻘﺪﻳﺮ اﳌﻌﺎﱂ . , ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ داﻟﺔ اﻹﻣﻜﺎن ﻛﺎﻟﺘﺎﱄ:
yi ,
1 n ( y i ) i 1
,
1 e n
L
ﺑﻔــﺮض ان اﳌﻌﻠﻤــﺔ ﻣﻌﻠﻮﻣــﺔ ﲝﻴــﺚ ان 0واﳌ ـﺮاد ﺗﻌﻈــﻴﻢ داﻟــﺔ اﻻﻣﻜــﺎن .ﻳــﺘﻢ ذﻟــﻚ ﺑﺘﺼــﻐﲑ )
n i
(y i 1
ﻋﻨﺪﻣﺎ ﻳﻜﻮن ﻟـ أﻛﱪ ﻗﻴﻤﺔ وﺣﻴﺚ إن: y1 , y 2 , , y n .
وﻟﺬﻟﻚ ﻓﺈن اﳌﻘﺪر ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ: ˆ M in(X1 ,X 2 , ,X n ) Y1 . وﻫﺬا ﻳﻌﲎ ان ˆ ﻣﺴﺘﻘﻞ ﻋﻦ 0وﺑﺎﻟﺘﺎﱃ ﻋﻦ .
اﻻن داﻟﺔ اﻻﻣﻜﺎن ﺗﺼﺒﺢ :
n
1
1 ( yi y1 ) L n e i 1 .
ﻣﻘﺪر اﻹﻣﻜﺎن ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ اﳊﻞ ﻟﻠﻤﻌﺎدﻟﺔ : ln L 0 ˆ
٣٦٩
وذﻟ ــﻚ
then : n
) y1
i
(y i 1
.
ln L = - n ln -
n
) (yi y1 ln L n i 1 . 2
ln L ﺑﻮﺿﻊ 0 ˆ
ﻓﺎن : ) n (yi y1 0 ˆ ˆ 2 n
) y1
i
(y i 1
.
ˆ 2
n ˆ n
) y1 .
i
(y i 1
n
ˆ
وﺑﺎﻟﺘﺎﱃ ﻓﺈن ﻣﻘﺪرات اﻻﻣﻜﺎن ﻟﻠﻤﻌﻠﻤﺘﲔ , ﳘﺎ : 1 n ˆ Y1 , ˆ (Yi Y1 ) X Y1. n i 1
اﳌﻘﺪرﻳﻦ ˆ ˆ , ﻣﺴﺘﻘﻠﲔ ﻻن ˆ , ﻣﺴﺘﻘﻠﲔ اى ان اﻟﺘﻐﺎﻳﺮ ﺑﻴﻨﻬﻤﺎ ﻳﺴﺎوى ﺻﻔﺮ اى ان . Cov(ˆ , ˆ ) 0 اﻻن ﻳﺘﻢ دراﺳﺔ اﻟﺘﻮزﻳﻊ اﳌﻀﺒﻮط ﻟﻜﻞ ﻣﻦ اﳌﻘﺪرﻳﻦ. داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻻﺣﺘﻤﺎﻟﻴﺔ اﳌﺸﱰﻛﺔ ﻟﻺﺣﺼﺎءات اﻟﱰﺗﻴﺒﻴﺔ ﰲ اﻟﻌﻴﻨﺔ ﺗﻌﻄﻰ ﻛﺎﻟﺘﺎﱄ: ) g(y1 , y 2 ,..., y n ) n!f (y1 )f (y 2 )f (y n
n
1
( yi ) n! i 1 ne . ; y1 y 2 y n . ﻧﻌﺘﱪ اﻟﺘﺤﻮﻳﻠﺔ اﻷﺣﺎدﻳﺔ: Z1 n (Y1 Y0 ),Y0 ,
) Z2 (n 1)(Y2 Y1 ) Z3 (n 2)(Y3 Y2 Zi (n i 1)(Yi Yi1 ) , i 1,2,...,n , Y0 0 واﻟﺘﺤﻮﻳﻠﺔ اﻟﻌﻜﺴﻴﺔ ﳍﺎ ﻫﻲ: ٣٧٠
Yi
Z1 Z Zi 2 ... , i 1,2,,n, n n 1 n i 1 :وﻣﻨﻬﺎ ﻧﻮﺟﺪ ﺟﺎﻛﻮﺑﻴﺎن اﻟﺘﺤﻮﻳﻞ ﻛﺎﻟﺘﺎﱄ
y1 z1
y1 z 2
y1 z n
y 2 J z1
y 2 z 2
y 2 z n
y n z1
y n z 2
1 n 1 n 1 n
y n z n
0
0
1 0 1 . n 1 n! 1 1 n 1 : ﺳﻮف ﻧﺜﺒﺖ ان
n
n
Z (Y ). i
i
i 1
i 1
: ﲟﺎ ان Z1 n(Y1 ) nY1 n
(٣-٥) : اﻻن n
n
(Y ) Y n. i
i
i 1
i 1
: ( ﻓﺈن٣-٥) ﺑﻘﻴﻤﺘﻬﺎ ﰱn وﺑﺎﻟﺘﻌﻮﺋﺾ ﻋﻦ n
n
n
(Y ) Y Z i
i 1
i
1
nY1 Yi nY1 Z1
i 1
i 1
n
n
n
n
(Yi Y1 ) Z1 (Yi Y1 ) Z1 Zi Z1 Zi . i 1
i2
i 2
i 1
:( ﺗﻌﻄﻰ ﻛﺎﻟﺘﺎﱄi 1,2,,n) وZ i وﻋﻠﻰ ذﻟﻚ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل اﳌﺸﱰﻛﺔ ﻟـ n
zi
1 [ i 1 ] e n n 1 [ zi ] e i 1
h(z1 ,z 2 ,...,z n )
, 0 z i ,
0 , e.w. . ﻣﺘﻐﲑات ﻋﺸﻮاﺋﻴﺔ ﻣﺴﺘﻘﻠﺔ وﻣﺘﻄﺎﺑﻘﺔ وﻛﻞ ﻣﻨﻬﺎ ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻲ ﺑﺎﳌﻌﻠﻤﺔZ i ﳑﺎ ﻳﻌﲏ إن اﳌﺘﻐﲑات
٣٧١
ﲟﺎ إن : n
n
n
Z Y (n 1)Y (Y Y ), 1
1
i
i
i
i 2
i2
i2
اى ان :
1 n 1 n 1 n ˆ (Yi Y1 ) (Yi Y1 ) Zi . n i1 n i2 n i2 اﻟﺪاﻟﺔ اﳌﻮﻟﺪة ﻟﻠﻌﺰوم ﻟﻼﺣﺼﺎء ˆ ﻫﻰ : )t (n 1 ) . n
(t) (1
n
Zi i=2
Mˆ (t) M
n
n
واﻟﱴ ﲤﺜﻞ اﻟﺪاﻟﺔ اﳌﻮﻟﺪة ﻟﻠﻌﺰوم ﳌﺘﻐﲑ ﻋﺸﻮاﺋﻰ ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﺎﱂ ) . ( , n 1وذﻟﻚ ﻻن : اﳌﺘﻐﲑات Zi ,i 1,2,...n 1ﻣﺘﻐﲑات ﻋﺸﻮاﺋﻴﺔ ﻣﺴﺘﻘﻠﺔ وﻣﺘﻄﺎﺑﻘﺔ وﻛﻞ ﻣﻨﻬﺎ ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻲ ﺑﺎﳌﻌﻠﻤﺔ . n
اﻟﺪاﻟﺔ اﳌﻮﻟﺪة ﻟﻠﻌﺰوم ﻟﻠﻤﻘﺪر Ziﻫﻰ : i2
.
اى ان ﺗﻮزﻳﻊ
n i
Z i2
) (n 1
)M n (t) (1 t Zi i=2
ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ ). ,(n 1
وذﻟﻚ ﻻن اﳌﺘﻐﲑات Zi ,i 1,2,...n 1ﻣﺘﻐﲑات ﻋﺸﻮاﺋﻴﺔ ﻣﺴﺘﻘﻠﺔ وﻣﺘﻄﺎﺑﻘﺔ وﻛﻞ ﻣﻨﻬﺎ ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻲ ﺑﺎﳌﻌﻠﻤﺔ . اﻻن : ˆ Y1
ﺣﻴﺚ Y1اﻟﱰﺗﻴﺐ اﻻﺣﺼﺎء اﻻﺻﻐﺮوﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﺗﻮزﻳﻌﺔ ﻣﻦ اﻟﺼﻴﻐﺔ اﻟﺘﺎﻟﻴﺔ :
n f (y1 ) 1 F(y1 )n 1 , 0 y1 , g1 (y1 ) , e.w. 0 n
اى ان Y1ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺘﲔ . , اﻻن : Z1 1 ) E(Z1 ) . n n n
٣٧٢
(E(ˆ ) E
: اﻳﻀﺎ. ﻣﻘﺪر ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔˆ اى ان Z1 1 2 ) 2 Var(Z1 ) 2 . n n n n n 1 1 n 1 E(ˆ ) E( Zi ) E(Zi ) . n i 2 n i 2 n
Var(ˆ ) Var(
: اﻳﻀﺎ. ﻣﻘﺪر ﻣﺘﺤﻴﺰ ﻟﻠﻤﻌﻠﻤﺔˆ اى ان Var(ˆ ) Var(
1 n 1 Zi ) 2 n i 2 n
n
Var(Z ) i
i2
n 1 2 . n2
: ﻣﻘﺪرﻳﻦ ﻣﺘﺤﻴﺰﻳﻦ ﻓﻴﻤﻜﻦ اﳊﺼﻮل ﻣﻨﻬﻤﺎ ﻋﻠﻰ ﻣﻘﺪرﻳﻦ ﻏﲑ ﻣﺘﺤﻴﺰﻳﻦ ﻛﺎﻟﺘﺎﱃˆ , ˆ ﲟﺎ ان اﳌﻘﺪران n ˆ n 1 n (Y Y1 ), ˆ Y1 { [ (Y Y1 )]} n 1 n 1 n 1 n n 1 1 nY Y Y1 [ (Y Y1 )] 1 . n 1 n 1
: اى ان
n nY Y (Y Y1 ), 1 . n 1 n 1 E( )
n n n 1 n2 n 1 E( ) . , Var( ) . 2 2 2 n 1 n 1 n (n 1) n
2 , n 1
1 E( ) E(ˆ ) E( ) , n n n 1 2 1 2 2 1 Var( ) Var(ˆ ) 2 Var() 2 2 2 (1 ). n n n n 1 n n 1
: اﻻن ﻻﺛﺒﺎت ﻫﻞ اﳌﻘﺪرﻳﻦ ﻣﺴﺘﻘﻠﲔ ام ﻻ ﻧﺘﺒﻊ اﻻﺗﻰ
ˆ nˆ ˆ nˆ Cov( , ) Cov(ˆ , ) Cov( , ) n 1 n 1 n 1 n 1 n n n (n 1)2 ˆ ˆ ˆ Cov(, ) Var() . (n 1)2 (n 1)2 (n 1) 2 n2 2 . n(n 1)
. ﻏﲑ ﻣﺴﺘﻘﻠﲔˆ , ˆ اى ان اﳌﻘﺪران
٣٧٣
n
اﻻن ﺳﻮف ﻧﺜﺒﺖ ان } { (Yi Y1 ), Y1اﺣﺼﺎءات ﻛﺎﻓﻴﺔ ﻟﻠﻤﻌﻠﻤﺘﲔ ) ˆ. (ˆ , ﻧﻔﺮض ان : i 2
n
) . U Y1 , V (Yi Y1وﲟﺎ ان ) ˆ (ˆ , ﻣﺴﺘﻘﻠﲔ ﻓﺈن اى دوال ﻓﻴﻬﻤﺎ ﻣﺴﺘﻘﻠﲔ .اى ان ) (U, Vﻣﺴﺘﻘﻠﲔ اﻳﻀﺎ .ﲟﺎ i 2
ان U Y1ﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺘﲔ , n
ﺣﻴﺚ داﻟﺔ ﻛﺜﺎﻓﺘﻪ اﻻﺣﺘﻤﺎﻟﻴﺔ ﺗﺎﺧﺬ اﻟﺸﻜﻞ اﻟﺘﺎﱃ :
n u
, u ,
n e
)g1 (u; , 0
, e.w. وﲟﺎ ان : n
n
V (Yi Y1 ) Zi i 2
i 2
اى ان Vﺗﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ ، , n 1ﺣﻴﺚ داﻟﺔ ﻛﺜﺎﻓﺘﻪ اﻻﺣﺘﻤﺎﻟﻴﺔ ﺗﺎﺧﺬ اﻟﺸﻜﻞ اﻟﺘﺎﱃ :
v 1 n 2 g 2 (v ;)= n-1 v e ) (n 1
, v > 0.
اى ان اﻟﺘﻮزﻳﻊ اﳌﺸﱰك ﻟﻠﻤﺘﻐﲑﻳﻦ ) (V, Uﻫﻮ :
, v > 0,u>.
n u
v 1 n2 v e )n-1(n 1
n
g (u,v) e
وﺑﺎﺳﺘﺨﺪام اﻻﺣﺘﻤﺎل اﻟﺸﺮﻃﻲ:
n
1
( yi ) 1 i1 e v
1 v n 2e n-1 ) (n 1
n u
٣٧٤
e
n
)L(y; )g(u, v
h(y, u, v)
n
e
1 n ( yi ) i 1
v n u
n n v n 2 e (n 1)
(n 1) e n vn2
(
yi v n nu n)
n ( yi v nu) (n 1) exp i 1 n vn2 N ( yi v ny1 ) ( n 2) (n 1)v exp i 1 n
n [ (yi y1 ) v] ( n 2) (n 1)v exp i 2 n n n [ (y y ) (yi y1 )] ( n 2) i 1 (n 1)v i2 exp i 1 n (n 1)v ( n 2) . n n
{ اﺣﺼﺎءات ﻛﺎﻓﻴﺔ ﻣﺸﱰﻛﺔ (Yi Y1 ), Y1} اى ان. ﺑﻞ داﻟﺔ ﻓﻘﻂ ﰱ اﳌﺸﺎﻫﺪات, وﻫﺬﻩ اﻟﻨﺴﺒﺔ ﻻ ﺗﻌﺘﻤﺪ ﻋﻠﻰ i 2
. , ﻟﻠﻤﻌﻠﻤﺘﲔ : ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﻣﻘﺪر ﳌﺘﻮﺳﻂ اﻟﺘﻮزﻳﻊ ﻧﺘﺒﻊ اﻻﺗﻰ : ﲟﺎ ان nˆ ˆ ˆ , ˆ , Y Y1 , ˆ Y1 , n 1 n 1
: وﲟﺎ ان E(X) ,
: وﻋﻠﻰ ذﻟﻚ ﻣﻘﺪر اﻻﻣﻜﺎن اﻻﻋﻈﻢ ﳌﺘﻮﺳﻂ اﻟﺘﻮزﻳﻊ ﻫﻮ ˆ ˆ Y1 Y Y1 Y. ٣٧٥
اﳌﻘﺪر ﳌﺘﻮﺳﻂ اﻟﺘﻮزﻳﻊ اﻟﻐﲑ ﻣﺘﺤﻴﺰ ﻫﻮ : ˆ ˆn ˆ ˆ ˆ ˆ (n 1) ˆ ˆ Y. n 1 n 1 n 1 1 ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ R(t) exp (t ) , t ﻓﺈن اﳌﻘﺪر اﻟﺬى ﻳﻌﺘﻤﺪ ﻋﻠﻰ ﻣﻘﺪرات اﳌﻜﺎن اﻻﻋﻈﻢ ﻫﻮ :
t Y1 R(t) exp , t Y1. Y Y1
) (١٤-١-٥ﺑﻌ ض اﻟﻧظرﯾ ﺎت اﻟﺗ ﻰ ﺗﺧ ص اﻟﺗﻘ دﯾرات اﻟﺑﯾﯾزﯾ ﺔ ﻟﻠﻣﻌﻠﻣ ﺔ ﺗﺣ ت ﻓ رض اﻟﺗوزﯾﻊ اﻻﺳﻰ ﺑﻣﻌﻠﻣﺗﯾن و ﻣﻌﻠوﻣﺔ ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺔ اﻟﻛﺎﻣﻠﺔ ٤ ﻟﺘﻜﻦ X1 , X 2 ,...,X nﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ ﺗﻮزﻳﻊ ﻟﻪ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل اﻻﺳﻰ ﲟﻌﻠﻤﺘﲔ , واﻟﱴ ﺗﺎﺧﺬ اﻟﺸﻜﻞ اﻟﺘﺎﱃ : x
, x ,
1 e
)f (x; , 0
, e.w. n
وﻛﺎﻧﺖ ﻣﻌﻠﻮﻣﺔ واﳌﻄﻠﻮب اﳚﺎد ﺗﻮزﻳﻊ
i
X
. U اﻫﺘﻤﺎﻣﻨﺎ ﻫﻨـﺎ ﰱ اﳊﺼـﻮل ﻋﻠـﻰ ﺗﻮزﻳـﻊ Uوذﻟـﻚ ﻻﺳـﺘﺨﺪاﻣﻪ ﰱ
i 1
اﳊﺼﻮل ﻋﻠﻰ ﳐﺎﻃﺮة ﺑﻴﻴﺰ ﻟﺒﻌﺾ اﻟﺘﻮزﻳﻌﺎت اﻟﻘﺒﻠﻴﺔ اﻟﱴ ﳝﻜﻦ اﺳﺘﺨﺪاﻣﻬﺎ وﻫﻰ :
)أ( ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ اﻟﻌﻜﺴﻲ
h h 1 1 () e )( 1
); h, > 0 (٤-٥ ﺣﻴﺚ أن ) ( 1ﻫﻲ داﻟﺔ ﺟﺎﻣﺎ و > 1 ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﺗﻮزﻳﻊ Uﻣﻦ اﻟﻌﻼﻗﺔ اﻟﺘﺎﻟﻴﺔ :
)(٥-٥
h(u) g(u | ) ( )d . 0
ﺣﻴﺚ ﺗﻮزﻳﻊ ) g(u | اوﺟﺪﻩ ) Shalaby and Abdelmoneim (1990ﺑﺎﺳﺘﺨﺪام اﻟﺪاﻟﺔ اﳌﻤﻴﺰة وﻫﻮ : )(٦-٥
, u n .
( u n )
n g(u | ) (u n) n 1 e ) (n
٣٧٦
ﻧﻈﺮﻳﺔ :
ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ ) (٤-٥وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﺸﺮﻃﻰ ﻟﻠﻤﺘﻐﲑ Uﲢﺖ ﺷﺮط ﻣﻌﻠﻮﻣﻴﺔ ﻫﻮ اﳌﻌﻄﻰ ﰱ ) (٦-٥اذا واذا ﻓﻘﻂ ﻛﺎن ﺗﻮزﻳﻊ Uﻫﻮ : h n ) ( n 1 f1 (u) (u n) n 1 1 h , u n. )(n, 1
)(٧-٥ ﺣﻴﺚ ) (n, 1داﻟﺔ ﺑﻴﺘﺎ و . 1 اﻟﱪﻫﺎن : اﻟﺸﺮط اﻟﻀﺮورى واﻟﻜﺎﰱ ﻟﻠﺤﺎﻟﺔ اﻟﺴﺎﺑﻘﺔ ﻳﻌﻄﻰ ﻛﺎﻟﺘﺎﱃ : اوﻻ :اﻟﺸﺮط اﻟﻀﺮورى :
إذا ﻛﺎن اﻟﺘﻮزﻳﻊ اﻟﺸﺮﻃﻰ ﻟﻠﻤﺘﻐﲑ Uﲢﺖ ﺷﺮط ﻣﻌﻠﻮﻣﻴﺔ ﻫﻮ اﳌﻌﻄﻰ ﰱ ) (٦-٥و ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ ) (٤-٥وﻋﻠﻰ ذﻟﻚ ﺗﺒﻌﺎ ل ) (٥-٥ﻓﺈن ﺗﻮزﻳﻊ Uﻫﻮ ) (٧-٥ﻛﻤﺎ ﺳﻮف ﻳﺜﺒﺖ اﻻن:
f1 (u) g(u | ) 1 ( )d. 0
) (u n h -1 - - h n 1 (u n) e e d ) (n ) ( -1 0 -n
) h -1 (u n) n 1 -(n+ ) - (u n h e d ( -1) (n) 0 )(u n h )(u n h let ==z z )(u n h = d dz. z2 )(u n h dz z2
) ( n
e -z
h 1 (u n) n 1 u n h f1 (u) ( -1) (n) 0 z
٣٧٧
h 1 (u n) n 1 (u n h) -(n+ ) (u n h) z n+ -2 e -z dz 0 ( -1) (n) h 1 u n (u n) n 1 ( 1) -(n+1) h (n 1) (n+ 1). ( -1) (n) h h n u n f1 (u)= (u n) n 1 ( 1) -(n+1) . (n, -1) h
: اﻟﺸﺮط اﻟﻜﺎﰱ: ﺛﺎﻧﻴﺎ hn u n f1 (u)= (u n) n 1 ( 1) -(n+1) (n, -1) h
h -1 - - h g(u | ) e d ( -1) 0 h n (n -1) (u n) n 1[h (u n)] ( n -1) (n) h (n -1) -
g(u | ) e
-
h
h 1d.
0
h 1 (n -1) (u n) n 1[h (u n)] (n -1) (n) -
g(u | ) e
-
h
h 1d
0
h (n -1)(u n ) n 1 - (n) g(u | ) e d. [h (u n)] (n 1) 0
1 1 1 = d 2 dz. z z (n -1)(u n) n 1 -hz 1 (n) g(u | z) z e dz. 0 [h (u n )] (n 1) z2 let z=
(u n) (n 1) (n -1) (n)L h z 2 g(u | z) ( n 1) [h (u n)] ٣٧٨
: ﻣﻌﻠﻤﺔ اﻟﺘﺤﻮﻳﻞ وh ﻫﻰ ﲢﻮﻳﻠﺔ ﻻﺑﻼس ﺣﻴﺚL h ﺣﻴﺚ (k)
h a
k
z k 1e za .
let k=n+-1,a=(u-n),h=h, (u-n) n-1z n2 e z(u-n) (n)z 2g(u | z) 1 g (u | z) (u-n) n-1e z(u-n) z n . (n) 1 let = , z (u-n) 1 n-1 g(u | ) (u-n) e n ,u n. (n)
: ﻧﻈﺮﻳﺔ : ﻫﻮ )ب( ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ
h 1 2 () 2 e h ( 1)
; h, > 0, >1. (٨-٥)
. ﳓﺼﻞ ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ =2 وﻋﻨﺪﻣﺎ > 1 ﻫﻲ داﻟﺔ ﺟﺎﻣﺎ و( 1) ﺣﻴﺚ أن ( اذا واذا ﻓﻘﻂ ﻛﺎن٦-٥) ﻫﻮ اﳌﻌﻄﻰ ﰱ ﲢﺖ ﺷﺮط ﻣﻌﻠﻮﻣﻴﺔU وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﺸﺮﻃﻰ ﻟﻠﻤﺘﻐﲑ : ﻫﻮU ﺗﻮزﻳﻊ 1 (n 1) 2
1 2h ( n 3) f 2 (u) K ( n 1) 2 h(u n) , u n. (u n) 2 (n) ( 1) (٩-٥)
٣٧٩
Johnson and Kotz(1970) واﻟﱴ ﲤﺜﻞ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ اﳌﺮﻛﺐ واﻟﺬى ﻟﻪ ارﺑﻌﺔ ﻣﻌﺎﱂ واﳌﻌﺮف ﻣﻦ ﻗﺒﻞ : اﻟﱪﻫﺎن : اﻟﺸﺮط اﻟﻀﺮورى ﻳﻌﻄﻰ ﻛﺎﻟﺘﺎﱃ: اوﻻ ( وﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ٦-٥) ﻫﻮ اﳌﻌﻄﻰ ﰱ ﲢﺖ ﺷﺮط ﻣﻌﻠﻮﻣﻴﺔU إذا ﻛﺎن اﻟﺘﻮزﻳﻊ اﻟﺸﺮﻃﻰ ﻟﻠﻤﺘﻐﲑ
( ﻛﻤﺎ ﺳﻮف ﻳﺜﺒﺖ اﻻن٩-٥) ﻫﻮU ( ﻓﺈن ﺗﻮزﻳﻊ٥-٥) ( وﻋﻠﻰ ذﻟﻚ ﺗﺒﻌﺎ ل٨-٥) ﻫﻮ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ :
f 2 (u) g(u | ) 2 ( )d 0
(u n ) h -1 2 -h n 1 (u n) e e d (n) ( -1) 0 -n
h 1 (u n) n 1 n 2 - (u n ) h e d. ( -1) (n) 0 z let h =z = . h dz d = , h
h 1 (u n) n 1 z f 2 (u) ( -1) (n) 0 h 1
(n-1)
h (u n) z ( -1) (n)h -n-2 0
h
n 1
(n-1)
(u n) ( -1) (n)
z
-n-2
-n-2
0
٣٨٠
( n 2)
e e
e
-
-
-
(u n ) h z z
(u n ) z z
(u n )h z z
1 dz h 1 dz h .
1 dz h
ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﻟﺚ ﻣﻦ اﻟﺪرﺟﺔK a (b) modified Bessel function ﺑﺎﺳﺘﺨﺪام داﻟﺔ ﺑﻴﺴﻞ اﳌﻌﺪﻟﺔ ﺣﻴﺚ ؛a ﻣﻦ
z
1 2 1
0
e
-z
y z
dz 2K 1 2 (2 y ) . 1 ( 2 1 ) y2
: ﻓﺈن
2h n 1 f 2 (u) ( -1) (n)
(u n) (n-1) h h(u n)
1 (n+1- ) 2
K (n 1) 2 h(u n) ,
where y=h(u n), 1 , 2 n 1 1 (n+ 1) 2
1 (n+ 3) 2h f 2 (u) (u n) 2 K (n 1) 2 h(u n) , u n. ( -1) (n)
: اﻟﺸﺮط اﻟﻜﺎﰱ: ﺛﺎﻧﻴﺎ
f 2 (u)= g(u | ) 2 ( )d 0
1 (n+1) 2
1 (n+ 3) 2h 2 (u n ) K (n 1) 2 h(u n) ( -1) (n)
h 1 2 e-h g(u | )d. ( -1) 0
٣٨١
1 (n+1) 2
2h (n)
1 (n+3) n-1 2 (u n ) (u n) K ( n 1) 2 h(u n) (u n) n-1
h
1
2 e -h g(u | )d
0
(u n) 2 h
1 ( n 1) 2
K ( n 1) 2 h(u n)
(n) 2 e -h g(u | )d. n-1 (u n) 0
(n) 2 L ( g(u | )) h (u n) n-1
(u n) 2 h
.Doetsch (1971)
n 2
e
-
1 ( n 1) 2
(u n )
K ( n 1) 2 h(u n)
ﺣﻴﺚ اﳉﺎﻧﺐ اﻻﳝﻦ ﻣﻦ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﻫﻮ ﲢﻮﻳﻠﺔ ﻻﺑﻼس ﻟﻠﺪاﻟﺔ
(u n ) (n) 2 n 2 g(u | ) e . n-1 (u n)
(u n )
n g(u | ) (u n) n 1 e (n)
, u n.
: ﻧﻈﺮﻳﺔ ٣٨٢
)ج( ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ :
1pq p1 3 () ( )q1 )(p,q
0< < ,p,q>0. )(١٠-٥
وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﺸﺮﻃﻰ ﻟﻠﻤﺘﻐﲑ Uﲢﺖ ﺷﺮط ﻣﻌﻠﻮﻣﻴﺔ ﻫﻮ اﳌﻌﻄﻰ ﰱ ) (٦-٥اذا واذا ﻓﻘﻂ ﻛﺎن ﺗﻮزﻳﻊ Uﻫﻮ : (u n )
1 f 3 (u) n (u n) n 1 e ) (n) (p,q x pn 1 x q 1 ux ( LU ) (1 ) e . 1 x 1 x
, u n.
(u n ) x
1 e 2 )(1 x
)(١١-٥ ﺣﻴﺚ L Uﲤﺜﻞ ﲢﻮﻳﻠﺔ ﻻﺑﻼس ﲟﻌﻠﻤﺔ اﻟﺘﺤﻮﻳﻠﺔ U
اوﻻ :اﻟﺸﺮط اﻟﻀﺮورى ﻳﻌﻄﻰ ﻛﺎﻟﺘﺎﱃ : إذا ﻛﺎن اﻟﺘﻮزﻳﻊ اﻟﺸﺮﻃﻰ ﻟﻠﻤﺘﻐﲑ Uﲢﺖ ﺷﺮط ﻣﻌﻠﻮﻣﻴﺔ ﻫﻮ اﳌﻌﻄﻰ ﰱ )(٦-٥وﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ ﻫﻮ ) (١٠-٥وﻋﻠﻰ ذﻟﻚ ﺗﺒﻌﺎ ل ) (٥-٥ﻓﺈن ﺗﻮزﻳﻊ Uﻫﻮ ) (١١-٥ﻛﻤﺎ ﺳﻮف ﻳﺜﺒﺖ اﻻن :
f 3 (u) g(u | ) 3 ( )d 0
٣٨٣
: اذن (u n )
1pq f 3 (u) (u n)n 1 pn 1 ( )q1 e 0 (n) (p,q)
d .
: ﺑﻮﺿﻊ z
d z dz . (u n) z
1 1 n 1 p n 1 q 1 f 3 (u) n (u n) z (1 z) e 0 (n) (p,q)
dz
. : ﺑﻮﺿﻊ
z
x z dx x dz , 1 x 1 z (1 x) 2
: اذن
f3(u)
1 x pn1 x q1 n1 (u n ) ( ) (1 ) 0 1 x n(n) (p,q) 1 x
(un) (x1) x
.e
1 ux -ux e e dx (1 x)2 (un)
1 n1 n (u n) e (n) (p,q)
(un) x
x pn1 x q1 LU( ) (1 ) e 1 x 1 x
1 ux e ),u n (1 x)2
: اﻟﺸﺮط اﻟﻜﺎﰱ: ﺛﺎﻧﻴﺎ ٣٨٤
f3 (u)=g(u| )3()d 0
1 1 p1 z (1 z)q1g (u| z) dz (p,q) 0
=
1 x p1 x q1 1 x ux -ux ( ) (1 ) g(u| )e e dx 2 (p,q) 0 1 x 1 x (1 x) 1 x
x p1 1 x q1 1 x ux LU ( ) (1 ) g (u|)( )e 2 (p,q) 1 x 1 x (1 x) 1 x (un)
1 n (u n)n1e (n) (p,q) 1 ux e ). (1 x)2
(un) x
x pn1 x q1 LU ( ) (1 ) e 1 x 1 x
: وﺑﺄﺧﺬ ﻣﻌﻜﻮس ﻻﺑﻼس ﻟﻜﻞ ﻣﻦ اﻟﻄﺮﻓﲔ ﻓﺈن x p1 x q1 1 x ( ) (1 ) g (u| ) 1 x 1 x (1 x) 2 1 x (u n)
1 = n (u n) n 1 e (n)
(u n) x
x pn 1 x q 1 1 ( ) (1 ) e 2 1 x 1 x (1 x)
.
: ﺑﻮﺿﻊ z
x , x 1
: اذن (u n ) (u n) (1 z) z
z -n g(u | z) n (u n) n 1 e (n)
٣٨٥
e
.
ﺑﻮﺿﻊ : ,
z
(u n ) (u n) (u n) ( )n n 1 g(u | ) n (u n) e e e . ) (n
اذن : ,u n.
(u n)
n n 1 g(u | ) (u n) e )(n
) (١٥-١-٥ﺗﻘ دﯾرات ﺑﯾﯾزﯾ ﺔ ﻟﻣﻌ ﺎﻟم اﻟﺗوزﯾ ﻊ اﻻﺳ ﻰ ﺑﻣﻌﻠﻣﺗ ﯾن وداﻟ ﺔ اﻟﺻ ﻼﺣﯾﺔ ﻓ ﻰ ﺣﺎﻟ ﺔ اﻟﻣﻌﺎﯾﻧﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ٤ إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﳊﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱐ وﺑﻔﺮض ان y1 y 2 y rﻫﻰ ال rاﻻوﱃ ﻣﻦ اﳌﺸﺎﻫﺪات اﳌﺮﺗﺒﺔ واﳌﺎﺧﻮذة ﻣﻦ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ nﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺘﲔ , ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃ:
, x ,
x
1
e
, e.w.
)f (x; , 0
وﺑﺎﻋﺘﺒﺎر ان y y1 , y 2 , , y rﰎ اﳊﺼﻮل ﻋﻠﻴﻬﺎ واﳌﻄﻠﻮب اﳚﺎد ﺗﻘﺪﻳﺮ ﻟﻠﻤﻌﺎﱂ . , داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ :
:اﯾﺠﺎد ﻣﻘﺪرات ﺑﯿﯿﺰ ﺗﺤﺖ ﻓﺮض اﻟﺘﻮزﯾﻊ اﻷﺳﻲ ﺑﻤﻌﻠﻤﺘﯿﻦ:
٣٨٦
r
1
r
n! 1 y n r y L y , e . n r !
i
r
i 1
: اﻻن r
r
i 1
i 1
yi n r y r yi r n r yr n r r
yi n n r y r [ry1 ry1 ny1 i 1
ny1 ] r y i ry1 n r y r n r y1 i 1 ny1 n r
y i y1 n r y r y1 n y1 . i 1
r
let s yi y1 n r y r y1
;k
i 1
n! n r !
k 1s n y L y , r e . : , ﲢﺖ ﻓﺮض اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ اﻟﺘﺎﱃ ل
g ,
1 a
1
;a, 0 , y1 .
: ﻫﻮ, وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل
, y
k y1
k
a r
a r
e
e
1 s n y1
1 s n y1
. dd
0
: ﺣﻴﺚ
٣٨٧
y1
k
1
1 a r exp 0 s n y1 dd y1
s n y
1
a r 1
a r 1 d
y1
a r 1 s n y1
a r 1
d
y a r 2 a r 1 s n y . 1 n a r 2 1
a r 1 a r 2 a r 2 k
1
a r2 ns a r 2
, x k
ns a r 2 k ar2
a r
e
1 s n y1
.
اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل The marginal posterior of : ﻟـ
1 y , y d 0
k
a r
e
1 s n y1
d
0
k s n y1
a r 1
a r 1
a r 1 ns a r 2 s n y1 a r 2 a r 2 ar2 n a r 2 s a r 2 a r 1 . s n y1
y1
E y d
n a r 2s
a r 2
y1
s n y
٣٨٨
1
a r 1
d.
let u du d
dv s n y1 v
a r 1
a r 2 1 s n y1 n a r 2
y a r 2 n a r 2 sa r 2 s n y1 n a r 2 y a r 2 1 s n y1 n a r 2 1
1
1 sa r 2 s a r 2 y1 s a r 3 n a r 3 s y1 . n a r 3
The Marginal y1
y , y d
k
a r
y1
e
1 s n y1
d
1
k
a r
s n y e n
y1
1
k a r 1 ns e n s a r 2 a r 1 ns e ar2
s a r 2 a r 2 s e d ar20 s a r 2 a r 3 s a r3 ar2 a r3 s s . ar2 a r 2
٣٨٩
, 0.
دى ﻟـ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل :
s , n a r 3 s . a r3
y1
) (١٦-١-٥ﻣﻘﺎرﻧﺔ ﺑﯾن زﻣن اﻟﺗﺟرﺑﺔ ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺔ اﻟﻛﺎﻣﻠﺔ واﻟﻣﻌﺎﯾﻧﺔ اﻟﻣراﻗﺑﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ٤ﺑﻔﺮض اﻧﻨﺎ اﺟﺮﻳﻨﺎ اﺧﺘﺒﺎر ﺣﻴﺎة ﺑﺎﺳﺘﺨﺪام ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﱏ وﻗﺪ ﰎ ا ﺎء اﻟﺘﺠﺮﺑﺔ ﺑﻌﺪ ﻓﺸﻞ r
ﻣﻦ اﻟﻮﺣﺪات ﻣﻦ ﻋﻴﻨﺔ ﺣﺠﻤﻬﺎ nواﳌﻄﻠﻮب ﻣﻘﺎرﻧﺔ ﻫﺬﻩ اﻟﺘﺠﺮﺑﺔ ﺑﺘﺠﺮﺑﺔ اﺳﺘﺨﺪﻣﺖ ﻓﻴﻬﺎ ﻋﻴﻨﺔ ﻛﺎﻣﻠﺔ ﺣﺠﻤﻬﺎ r
وا ﻴﺖ اﻟﺘﺠﺮﺑﺔ ﺑﻌﺪ ﻓﺸﻞ ﻛﻞ اﻟﻮﺣﺪات اﻟﱴ ﻋﺪدﻫﺎ . r
) X (1,n) ,X (2,n ) , ,X (r,n
let
n r
و: ) X (1,r ) X (2,r ) , ,X (r,r
وﻟﺤﺴﺎب اﻟﻨﺴـﺒﺔ ﺑـﻴﻦ اﻻﺳـﻠﻮﺑﻴﻴﻦ ﻓـﻰ اﻟﻤﻌﺎﻳﻨـﺔ وﻛﻴـﻒ ﺗـﺆدى اﻟﻤﻌﺎﻳﻨـﺔ ﻓـﻰ ﺣﺎﻟـﺔ اﻟﻤﺮاﻗﺒـﺔ ﻣـﻦ اﻟﻨـﻮع اﻟﺜـﺎﻧﻰ اﻟـﻰ اﺧﺘﺼــﺎر زﻣــﻦ اﻟﺘﺠﺮﺑــﺔ ﻋــﻦ اﻟﻤﻌﺎﻳﻨــﺔ اﻟﻜﺎﻣﻠــﺔ ﻓﺈﻧﻨــﺎ ﻧﺤﺘــﺎج اﻟــﻰ ﺣﺴــﺎب اﻟﻘﻴﻤــﺔ اﻟﻤﺘﻮﻗﻌــﺔ ﻟﻠﻤﺘﻐﻴــﺮ ) X (r,rأي: ) ) E(X (r,rو اﻟﻘﻴﻤﺔ اﻟﻤﺘﻮﻗﻌﺔ ﻟﻠﻤﺘﻐﻴﺮ ) X (r ,nأي E(X (r,n) ) :اﻟﻤﻄﻠﻮب اوﻻ إﻳﺠﺎد : .
) ) E(X (r,n ) ) E(X (r,r
واﺛﺒﺎت ان ﻫﺬﻩ اﻟﻨﺴﺒﺔ أﻗﻞ ﻣﻦ 1وﺑﺬﻟﻚ ﻧﺜﺒﺖ اﻧﻪ اذا اﺟﺮﻳﻨﺎ اﺧﺘﺒﺎر ﺣﻴﺎة ﺑﺎﺳﺘﺨﺪام ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع
اﻟﺜﺎﻧﻰ و ﺗﻢ اﻧﻬﺎء اﻟﺘﺠﺮﺑﺔ ﺑﻌﺪ ﻓﺸﻞ rاﻓﻀﻞ ﻣﻦ اﺟﺮاء اﻟﺘﺠﺮﺑﺔ ﺑﺎﺳﺘﺨﺪام ﻋﻴﻨﺔ ﻛﺎﻣﻠﺔ ﺣﺠﻤﻬﺎ r
واﻧﻬﺎء اﻟﺘﺠﺮﺑﺔ ﺑﻌﺪ ﻓﺸﻞ ﻛﻞ اﻟﻮﺣﺪات اﻟﺘﻰ ﻋﺪدﻫﺎ . r ﺳﻮف ﻧﺜﺒﺖ ذﻟﻚ ﺗﺤﺖ ﻓﺮض أن:
) 1 ( x f (x) e
x n, 0. ٣٩٠
: ﺳﻮف ﻧﺘﺒﻊ اﻟﺨﻄﻮات اﻟﺘﺎﻟﻴﺔ Z1 n(Y 1 ) Z2 (n 1)(Y2 Y1 ) Zr (n r 1)(Yr Yr 1 ) Zis ~ exp() W1 Y1
,i,i,d
W2 Y2 Y1 Wr Yr Yr 1 r
W Y i
r
i 1 r
E( Wi ) E(Yr ) i 1
Wis ~ exp(
Z ), Wi . n j 1 n j 1
r
E(W ) E(Y ) i
r
i 1
r
j1
r 1 . n j 1 j1 n j 1
٣٩١
r
1 j1 (n j 1)
E(Yr )
r
1 n j 1
E[X (r,n) ] j1 r
1 r j1
E[X (r,r) ] j1 r
1 j1 n j 1 E[X (r,n) ] 1 E[X (r,r ) ] r j1 r j 1
r
(n j 1)1 j1 r
(r j 1)1 j1
r (n j 1)1 j1 r (r j 1)1 j1 r
r 1
(n j 1) (r j 1) j1
1
.
j1
: اى ان. اﻹﺷﺎرة ﺳﺎﻟﺒﺔ دﻟﻴﻞ ﻋﻠﻰ ﺣﺪوث ﺗﻮﻓﻴﺮ ﻓﻰ اﻟﺰﻣﻦ ﻋﻨﺪ اﺳﺘﺨﺪم ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ E(X r,n ) E[X (r,r ) ].
. ﻻ ﺗﻌﺘﻤﺪ ﻋﻠﻰ ﻣﻌﺎﱂ اﻟﺘﻮزﻳﻊ : ﻓﺈن 0 ﻋﻨﺪﻣﺎ r
E[X (r,r ) ] i 1
r
1 . r i 1
1 E[X (r,n) ] i1 n i 1) . E[X (r,r) ] r ( 1 ) i 1 r i 1 (
٣٩٢
:اﻟﻤﻄﻠﻮب اﻳﻀﺎ ﺣﺴﺎب P(X( r,n) X (r,r) )
.و إﺛﺒﺎت ان ﻫﺬا اﻻﺣﺘﻤﺎل ﻻ ﺗﻌﺘﻤﺪ ﻋﻠﻰ اﻟﺘﻮزﻳﻊ let
x (r,n ) x x (r,r ) y.
r cut of n واﻷﺧﺮىr cut of r ﺗﺠﺮﺑﺘﻴﻦ ﻣﺴﺘﻘﻠﻴﻦ أﺣﺪﻫﻤﺎ اﻷول
g r,n (x)
n! (F(x))r1 (1 F(x))n r f (x) (r 1)!(n r)! g r,r (y) r(F(y)) r1 f (y) , y x (r,r )
x x (r,n )
, x
, y .
: ﻣﺴﺘﻘﻠﲔ ﻓﺈنx, y ﲟﺎ أن g r,n (x)g r,r (y) f1 (x, y). let
h r,n (u)
u F(x) , v F(y).
n! u r 1 (1 u)n r (r 1)!(n r)! h r,r (v) rv r 1
f 2 (u, v)
0 v 1
n!r v r 1u r 1 (1 u) n r (r 1)!(n r)! 1
P(X Y) rv
r 1
0
P(X Y) P(U V)
0 u 1,
0 v 1
v n!u r 1(1 u) n r du dv, 0 (r 1)!(n r)!
let w 1 u 1 v w 1. w u 1 0 1 v v ٣٩٣
1
P(X Y) rv 0
1
r 1
n! (1 w)r 1 w n r dwdv (r 1)!(n r)!1v
n 1 1 r 1 1 r 1 r 1 n rv (1) jw n jr dwdv j r 1 0 1 v j0 1
n 1 1 r 1 r 1 r 1 w n jr 1 j n rv j (1) n j r 1 r 1 j 0 0 1v r 1 (1) j 1 1 r 1 r 1 n 1 j r 1 n r j1 n dv v dv v (1 v) r 1 j0 (n r j 1) 0 0 r 1 (1) j 1 1 r 1 j r 1 r 1 n r j1 c (1 y) dy (1 y) y dy , (n r j 1) j 0 0 0 n 1 c rn , r 1 r 1 (1) j j0 j c I1 I2 , n j j 1 1 r 1 r 1 i I1 (1) yidy i 0 i 0 r 1
r 1 (1)i r 1 i , i 1 i 0 1 r 1 r 1 i I2 (1) y n r ji1dy i0 i 0 r 1 (1)i r 1 i . i 0 n r j i 2
٣٩٤
r 1 r 1 r 1 i i ( )1 ( )1 r 1 i n 1 r 1 j r 1 i P(X Y) nr n r j 1 r 1 i 1 n r j i 2 i 0 i 0 i 0 r 1 r 1 j ( )1 (1)i r 1 r n 1 n r j i 2 i 1 j i nr (i 1)(n r j i 2) n r j1 r 1 i 0 i 0
r 1 r 1 (1)i j r 1 r n 1 j i nr . r 1 (i 1)(n r j i )2 i 0 i 0
اى ان ) P(X Yﻻ ﻳﻌﺘﻤﺪ ﻋﻠﻰ ﺷﻜﻞ اﻟﺘﻮزﻳﻊ. اﻳﻀﺎ ﻳﻤﻜﻦ ﺣﺴﺎب اﻟﻨﺴﺒﺔ اﻟﻤﺌﻮﻳﺔ ﻟﻼﺧﺘﺰال ﻓﻲ زﻣﻦ اﻟﺤﻴﺎة ﻟﻠﺘﻮزﻳﻊ اﻻﺳﻰ ﲟﻌﻠﻤﺔ وﲢﺴﺐ ﻛﺎﻟﺘﺎﱃ : اوﻻ ﳓﺴ ــﺐ اﻟﻘ ــﻴﻢ اﳌﺘﻮﻗﻌ ــﺔ ﻟﻺﺣﺼ ــﺎءات اﻟﱰﺗﻴﺒﻴ ــﺔ ،ﰲ ﻋﻴﻨ ــﺎت ﻣ ــﻦ اﳊﺠ ــﻢ nﻣ ــﻦ ﺗﻮزﻳ ــﻊ أﺳ ــﻲ ﻗﻴﺎﺳ ــﻲ )( 1
واﳌﻌﻄﺎة ﰲ اﳉﺪول اﻟﺘﺎﱃ ﺣﻴﺚ ﻣﻌﻄﺎة ﻟﻘﻴﻢ r=1,2,…,10,n=2,3,…,10,وﺗﺴﻤﻰ ﻫﺬﻩ اﻟﻘﻴﻢ درﺟﺎت أﺳﻴﺔ score exponentialوﳍﺎ ﺗﻄﺒﻴﻘﺎت ﻣﻔﻴﺪة ﰲ اﻻﺳﺘﺪﻻل اﻟﻼﻣﻌﻠﻤﻲ. 10
9
8
7
6
5
4
3
2
r n
0.100 0.211 0.336 0.479 0.646 0.846 1.096 1.429 1.929 2.929
0.111 0.236 0.379 0.546 0.746 0.996 1.329 1.829 2.829
0.125 0.268 0.435 0.635 0.885 1.218 1.718 2.718
0.143 0.310 0.510 0.760 1.093 1.593 2.593
0.167 0.367 0.617 0.950 1.450 2.450
0.200 0.450 0.783 1.283 2.283
0.250 0.583 1.083 2.083
0.333 0.833 1.833
0.500 1.500
1 2 3 4 5 6 7 8 9 10
ﻣﺜﺎل ٣٩٥
ﰲ اﺧﺘﺒﺎرات اﳊﻴﺎة ،وﺿﻌﺖ ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ n 9ﰲ اﻻﺧﺘﺒﺎر .أزﻣﻨﺔ اﳊﻴﺎة ﺑﺎﻟﺴﺎﻋﺎت ﻟﻮﺣﺪات اﻟﺘﺠﺮﺑﺔ ﳝﻜﻦ اﻋﺘﺒﺎرﻫﻢ ﻣﺘﻐﲑات ﻋﺸﻮاﺋﻴﺔ ﻣﺴﺘﻘﻠﺔ ﻣﻦ ﺗﻮزﻳﻊ أﺳﻲ ﲟﺘﻮﺳﻂ ﺣﻴﺎة 100ﺳﺎﻋﺔ .ﳊﺴﺎب اﻟﺘﻮﻓﲑ ﰲ زﻣﻦ اﳊﻴﺎة اﻟﻜﻠﻲ ﻋﻨﺪ إ ﺎء اﻟﺘﺠﺮﺑﺔ ﺑﻌﺪ ﻓﺸﻞ اﻟﻮﺣﺪة رﻗﻢ . (r 5)5وإذا ﻛﺎﻧﺖ X5,9ﲤﺜﻞ زﻣﻦ اﻧﺘﻬﺎء اﻟﺘﺠﺮﺑﺔ ﺑﻌﺪ ﻓﺸﻞ اﻟﻮﺣﺪات اﻟﱵ ﻋﺪدﻫﺎ r 5و X5,5ﳝﺜﻞ زﻣﻦ اﻧﺘﻬﺎء اﻟﺘﺠﺮﺑﺔ إذا أوﻗﻔﻨﺎ اﻟﺘﺠﺮﺑﺔ ﺑﻌﺪ ﻓﺸﻞ اﻟﻮﺣﺪة رﻗﻢ 5ﻣﻦ ﻋﻴﻨﺔ ﺣﺠﻤﻬﺎ . n 5ﻣﻦ اﳉﺪول اﻟﺴﺎﺑﻖ ﳒﺪ أﻧﻪ ﻟﻌﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﳊﺠﻢ n 9ﻣﻦ ﺗﻮزﻳﻊ أﺳﻲ ﻗﻴﺎﺳﻲ ،ﻓﺈن اﻟﻘﻴﻤﺔ اﳌﺘﻮﻗﻌﺔ ل Y5,9ﻫﻰ 0.749واﻟﻘﻴﻢ اﳌﺘﻮﻗﻌﺔ ل Y5,5ﻫﻰ . 2.283وﻋﻠﻰ ذﻟﻚ : E(X 5,5 ) 100(2.283) 228.3, E(X 5,9 ) 100(0.746) 74.6.
وﻋﻠﻰ ذﻟﻚ اﻟﻨﺴﺒﺔ اﳌﺌﻮﻳﺔ ﻟﻼﺧﺘﺰال ﰲ زﻣﻦ اﳊﻴﺎة ﻫﻮ: 228.3 74.6 100 67.32%. 228.3
) (٢-٥ﺗوزﯾﻊ واﯾﺑل ﺑﻣﻌﻠﻣﺗﯾن ) (١-٢-٥ﺗﻘدﯾرات ﺑﯾﯾز ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺔ اﻟﻛﺎﻣﻠﺔ ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ ) Sinha and Gutman (1988وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃ :
ﻳﻌﺘﱪ ﺗﻮزﻳﻊ واﺑﻴﻞ ﳌﻌﻠﻤﺘﲔ ﳕﻮذج ﻣﻔﻴﺪ ﰲ ﳎﺘﻤﻌﺎت اﳊﻴﺎة اﳌﺮﺗﺒﻄﺔ ﺑﺪراﺳﺎت اﻟﺼﻼﺣﻴﺔ و اﻟﻜﻔﺎءة و ﲡﺎرب اﺧﺘﺒﺎرات اﳊﻴﺎة ...و ﻷﳘﻴﺔ ﻫﺬا اﻟﺘﻮزﻳﻊ ﺳﻮف ﻳﻜﻮن اﻫﺘﻤﺎﻣﻨﺎ ﰱ ﻫﺬا اﻟﺒﻨﺪ ﺑﻄﺮق ﺗﻘﺪﻳﺮ ﻣﻌﺎﱂ واﺑﻴﻞ و pو إﳚﺎد اﻟﺘﻮزﻳﻌﺎت اﳍﺎﻣﺸﻴﺔ اﻟﺒﻌﺪﻳﺔ ﻟﻜﻞ ﻣﻦ اﳌﻌﺎﳌﺘﲔ و . p ﺑﻔﺮض أن nﻣﻦ اﻟﻮﺣﺪات وﺿﻌﺖ ﻟﻼﺧﺘﺒﺎر وان اﻻﺧﺘﺒﺎر ﻳﻨﺘﻬﻲ ﺑﻌﺪ ﻓﺸﻞ ﻛﻞ اﻟﻮﺣﺪات .ﺑﻔﺮض أن X1 , X 2 , , X nﻋﻴﻨﺔ
ﻋﺸﻮاﺋﻴﺔ ﲤﺜﻞ أزﻣﻨﺔ اﻟﻔﺸﻞ وان أزﻣﻨﺔ اﻟﻔﺸﻞ ﺗﺘﺒﻊ ﺗﻮزﻳﻊ واﻳﺒﻞ ﲟﻌﻠﻤﺘﲔ ﺑﺪاﻟﺔ ﻛﺜﺎﻓﺔ اﺣﺘﻤﺎﻟﻴﺔ: xp
, x 0.
داﻟﺔ اﻻﻣﻜﺎن ﺗﺄﺧﺬ اﻟﺼﻮرة:
, p, 0
p f x , p x p1e
٣٩٦
n
L x ,p f x i ,p i 1
x pi
p 1
n p x i e . i 1 Jeffreys' 1961 و ﻋﻠﻰ اﻋﺘﺒﺎر اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ اﳌﻘﱰح ﻣﻦ ﻗﺒﻞ n
n
1 p : ﻳﻌﻄﻰ ﺑﺎﻟﺼﻴﻐﺔx ﲢﺖ ﺷﺮط , p ﻓﺈن اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻟﻠﻤﻌﺎﱂ
,p
p 1
, p x
n 1
p
n 1
p 1
n 1
p
n 1
e
e
x pi
x pi
n
, xi. i 1
dpd
0 0
واﻗﺻر ﻓﺗرة ﺛﻘﺔp )أ( ﺗﻘدﯾر ﺑﯾﯾز ﻟﻣﻌﻠﻣﺔ اﻟﺷﻛل p ﺗﻘﺪﻳﺮ ﺑﻴﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ: اوﻻ
. p و ﻫﻲ داﻟﺔ ﰲ اﳌﻌﻠﻤﺔ1 p x ﻣﻘﻠﻘﻠﺔ و ﻟﺬﻟﻚ ﳒﺮي اﻟﺘﻜﺎﻣﻞ ﺑﺎﻟﻨﺴﺒﺔ ﳍﺎ ﻟﻨﺤﺼﻞ ﻋﻠﻰ ﻫﻨﺎ ﺗﻌﺘﱪ
p 1
n 1
p
n 1
e
1 p x
x pi
d
0
p 1
n 1
p
n 1
e
x pi
. ddp
0 0
n
n p x ip n i 1 n n p 1 n 1 p x i n dp 0 p i 1 p 1p n 1 n ,p 0. n k1 xp i i 1 p 1
n 1
: ﺣﻴﺚ n
n k p x ip dp. i 1 0 1 1
p 1
n 1
: ﺗﻌﻄﻰ ﺑﺎﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔp و ﻛﺬﻟﻚ ﳒﺪ أن اﻟﻘﻴﻤﺔ اﳌﺘﻮﻗﻌﺔ ﻟـ ٣٩٧
n
n E p x p1 p x dp p x ip k1dp. i 1 0 n
p 1
واﻟﱴ ﰎ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﺑﺎﺳﺘﺨﺪام اﳊﺎﺳﺐ اﻻﱃ ﺑﺎﻟﻄﺮق اﻟﻌﺪدﻳﺔ وذﻟﻚ ﻣﻦ ﻗﺒﻞ Sinha and Gutman
).(1988 ﺛﺎﻧﻴﺎ :اﻗﺼﺮ ﻓﱰة ﺛﻘﺔ
ﻟﻘﺪ اوﺟﺪ ) Sinha and Gutman (1988ﻓﺗرﻩ اﺣﺗﻣﺎل ﺑﯾﯾز اﻟﻣﺗﻣﺎﺛﻠﻪ ذات اﻟﺟﺎﻧﺑﯾن Symmetric 100(1-)% Two Side Bayes Probability Interval ﻟﻠﻣﻌﻠﻣﻪ ) pﻟﻼﺧﺗﺻﺎر ﺗﻛﺗب (100 1 %TBPIوﯾﻣﻛن اﻟﺣﺻول ﻋﻠﯾﻬﺎ ﺑﺣل اﻟﻣﻌﺎدﻟﺗﯾن:
, 2
p x dp
t1
0
p x dp 2 .
t2
واﻟﻔﺗرة اﻟﺳﺎﺑﻘﺔ ﺗﺳﻣﻰ ﻣﺗﻣﺎﺛﻠﺔ ﻻن ﻗﺳﻣت ﺑﺎﻟﺗﺳﺎوى ﺑﯾن طرﻓﻰ اﻟﺗوزﯾﻊ اﻟﺑﻌدى ﺣﯾث t1
ﻫو اﻟﺣد اﻻدﻧﻰ و t 2ﻫو اﻟﺣد اﻻﻋﻠﻰ ﺑﺣﯾث أن :
P(t1 p t 2 ) 1 .
ﻛﻤﺎ اوﺟﺪ أﻗﺻر ﻓﺗرة ﺛﻘﻪ )اﻟﻣرﻏوﺑﻪ او اﻟﻣﻌﺗﻣدة( ﻟﻠﻣﻌﻠﻣﺔ pوذﻟك ﺑﺈﺗﺑﺎع اﻟﺷروط اﻟﺗﺎﻟﯾﺔ : -
P(t1 p t 2 ) 1
. 1 t1 x 1 t 2 x
وﻟﻣﺎ ﻛﺎن اﻟﺗوزﯾﻊ اﻟﺑﻌدى ل pﻟﻪ ﻣﻧوال وﺣﯾد ﻓﺈﻧﻪ ﺣﺻل ﻋﻠﻰ اﻋﻠﻰ ﻓﺗرة ﺛﻘﻪ. اﻟﻔﺘﺮات اﻟﺴﺎﺑﻘﺔ ﻳﺘﻢ اﻟﺤﺼﻮل ﻋﻠﻴﻬﺎ ﺑﺎﺳﺘﺨﺪام اﻟﺤﺎﺳﺐ اﻻﻟﻰ ﺑﺎﻟﻄﺮق اﻟﻌﺪدﻳﺔ . )ب( ﺗﻘدﯾر ﺑﯾﯾز ﻟﻣﻌﻠﻣﺔ اﻟﻘﯾﺎس واﻗﺻر ﻓﺗرة ﺛﻘﺔ اوﻻ :ﺗﻘﺪﻳﺮ ﺑﻴﻴﺰ ﻟﻠﻤﻌﻠﻤﺔ . ﲟﺎ ان اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي اﳌﺸﱰك ﻟﻠﻤﻌﻠﻤﺘﲔ , p ﻳﻌﻄﻰ ﻣﻦ اﻟﺼﻴﻐﺔ اﻟﺘﺎﻟﻴﺔ : p i
x k p 1 n 1 n 1 ,p x p e . n
٣٩٨
ﻫﻨﺎ ﺗﻌﺘﱪ pﻣﻘﻠﻘﻠﺔ و ﻟﺬﻟﻚ ﳒﺮي اﻟﺘﻜﺎﻣﻞ ﺑﺎﻟﻨﺴﺒﺔ ﳍﺎ ﻟﻨﺤﺼﻞ ﻋﻠﻰ 2 x و ﻫﻲ داﻟﺔ ﰲ اﳌﻌﻠﻤﺔ . ﺑﺎﻟﺘﻜﺎﻣﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟـ pﳓﺼﻞ ﻋﻠﻰ: p i
k n 1 p1 n 1 x 2 x 0 p e dp n
و اﻟﱵ ﳝﻜﻦ أن ﲢﻞ ﺑﺎﻟﻄﺮق اﻟﻌﺪدﻳﺔ أو ﺑﺎﺳﺘﺨﺪام اﳊﺎﺳﺐ. و ﻛﺬﻟﻚ ﳒﺪ أن اﻟﻘﻴﻤﺔ اﳌﺘﻮﻗﻌﺔ ﻟـ ﺗﻌﻄﻰ ﺑﺎﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ:
E x 2 x d 0
p i
k n p1 n 1 x p e dpd n 0 0 p i
k p1 n 1 n x p e ddp n 0 0 1 n k p1 n 1 p x pi n 1 dp n 0
n 1
k p1 p dp. n 1 0 x pi
واﻟﱴ ﰎ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﺑﺎﺳﺘﺨﺪام اﳊﺎﺳﺐ اﻻﱃ ﺑﺎﻟﻄﺮق اﻟﻌﺪدﻳﺔ وذﻟﻚ ﻣﻦ ﻗﺒﻞ Sinha and Gutman
).(1988
)ج( ﺗﻘدﯾر ﺑﯾﯾز ﻟداﻟﺔ اﻟﺻﻼﺣﯾﺔ وﻓﺗرة ﺛﻘﺔ اﳌﻘﺪر اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ) R(tﺣﻴﺚ :
dx
xp p 1
p x e
t
R(t) P(X t) 0 xp
, t 0. وﲟﺎ ان داﻟﺔ اﻟﺼﻼﺣﻴﺔ داﻟﺔ ﰱ p,ﻓﻴﻤﻜﻦ اﳊﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل ) R(tﻛﺎﻟﺘﺎﱃ : ٣٩٩
e
Let R(t ) R e
tp
,w p
tp tp ln R , w p ,w p ln R : وﻋﻠﻰ ذﻟﻚ
p w | J | w
p 1 R 0 R
tw . R(ln R) 2
w
t R(ln R) 2
: ﻳﻌﻄﻰ ﺑﺎﻟﺼﻴﻐﺔ اﻟﺘﺎﻟﻴﺔx ﲢﺖ ﺷﺮط , p وﲟﺎ ان اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻟﻜﻞ ﻣﻦ , p x p p 1
n 1
n 1
e
x ip
.
: ﻫﻰ w,R وﻋﻠﻰ ذﻟﻚ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل اﳌﺸﱰﻛﺔ ﻟﻜﻞ ﻣﻦ
tp w n 1 g w,R | x w ln R tp w ln R w
n 1
( n 1)
n 1
e
( x pi )(ln R ) t w
e
x ip tp ln R
tw R(ln R) 2
x ip w
w w n 1 t w (n 1) t w ln R t e R ( ln R) w
w R
n 1
(ln
1 n 1 wn ) t R R
٤٠٠
tw R(ln R) 2
x p i w t
w w n 1 1 (ln )n 1 t wn R R R
x w i t w
n
w
n 1 wn
w t
xi
( t ) 1 (ln ) n 1 R R
w
1
i1
n w i
1 n 1 y 1 x w n 1 wn w t (ln ) R , y i i ,0 R 1, w 0. R t i1
(١٢-٥) : وﻫﻰR ﳓﺼﻞ ﻋﻠﻰ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل لw ( ﺑﺎﻟﻨﺴﺒﺔ ل١٢-٥) ﺑﺘﻜﺎﻣﻞ n
g(R | x) k 3 0
w i
y 1 1 w w n 1t wn (ln ) n 1 R dw R i1
n w i
y 1 1 k 3 (ln )n 1 w w n 1t wn R dw, R 0 i 1
n
1 1 3
K 0
w i
y 1 1 (ln ) n 1 w w n 1t wn R dwdR R 0 i1
n
1 n 1 y 1 w n 1 wn w t (ln ) R dR dw R 0 0 1 Let u= ln u ln R e u R dR e u du R 0 ,1 0 w i
1
i 1
n
1 3
w
n 1 wn
K w t 0
n 1 u y 1 u e du dw u e 0 w i
i1
n
0
0
n 1 u y w n 1 wn w t u e du dw 0 (n) w w n 1t wn n dw yw i i 1 w i
i1
n 1 wn
(n) w t 0
n
x iw w . i 1 t n
w
٤٠١
ﺑﺈﺳﺘﺨﺪام داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﻟﺨﻄﺎ اﻣﻜﻦ اﻳﺠﺎد اﻟﻤﻘﺪر اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ واﻟﺘﻰ ﻳﺘﻢ اﻟﺤﺼﻮل ﻋﻠﻴﻬﺎ
ﺑﺎﺳﺘﺨﺪام اﻟﺤﺎﺳﺐ اﻻﻟﻰ ﺑﺎﻟﻄﺮق اﻟﻌﺪدﻳﺔ وذﻟﻚ ﻣﻦ ﻗﺒﻞ ).Sinha and Gutman (1988اﯾﺿﺎ ﻓﺗرﻩ اﺣﺗﻣﺎل ﺑﯾﯾز اﻟﻣﺗﻣﺎﺛﻠﻪ ذات اﻟﺟﺎﻧﺑﯾن ﻟداﻟﺔ اﻟﺻﻼﺣﯾﺔ . )د( ﺗﻘدﯾر ﺑﯾﯾز ﻟداﻟﺔ اﻟﺻﻼﺣﯾﺔ ﺑﺎﺳﺗﺧدام داﻟﺔ اﻟﺗﻧﺑؤ إذا ﻛﺎﻧﺖ
و )( | x
)f (y |
ﺗﻤﺜﻞ داﻟﺔ ﻛﺘﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻠﻤﺘﻐﻴﺮ Yﺗﺤﺖ ﺷﺮط
و )(
ﻫﻮ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ل ﻓﺈن داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻠﻤﺘﻐﻴﺮ Yﺗﺤﺖ ﺷﺮط
x
اﻟﺤﺼﻮل ﻋﻠﻴﻬﺎ ﻛﺎﻟﺘﺎﻟﻰ :
h(y | x) f (y | ) ( | x)d.
واﻟﺘﻰ ﺗﺴﻤﻰ داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻟﺘﻨﺒﺆﻳﺔ ل . Yﺗﺤﺖ ﻓﺮض داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﻟﺨﻄﺎ ﻓﺈن : R * (t) E |x [R(t ) | x] P(Y t | x)( | x)d
f (y | ) ( | x)d t
f (y | )( | x)d dy t
f (y | x)dy. t
واﻟﺗﻰ ﺗﻣﺛل داﻟﺔ اﻟﺻﻼﺣﯾﺔ اﻟﺧﺎﺻﺔ ﺑﺎﻟﻘﯾﻣﺔ اﻟﻣﺳﺗﻘﺑﻠﯾﺔ .yوﺑﻣﺎ ان :
ddp
x ip
yp
e
n 1
n 1
p 1
p
p p1 y e
0
h(y | x) 0
x y n 2 p 1 n p 1 py e d dp 0 )(n 1 p 1p n y p1 dp p p n 1 x y i p
p i
٤٠٢
0
0
اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ
ﻫﻮ
)h(y | x
ل
ﻳﻤﻜﻦ
1
h(y | x) k 3 (y) p 1 p n
x
0
k
0
0
1 3
( ) p 1 y p 1p n
() p 1 p n
y p 1
0
0
x
p 1p n 1
p i
dp,
n 1
dydp
p(n 1) p x y p i
x
0
n
p
n 1
dydp
n
0
0
n 1
x ip yp
(n 1) 1
p 1p n 1
yp 1
yp p i
p i
0
dp
n
dp
1 p 1 n 1 p (x ip ) n dp n0 p 1p n 1 1 k3 dp n(x ip ) n 0
n p n (y) p 1 h(y | x)
0
0
(n 1) n 1
x ip yp (n 1) p 1p n 1 dp p n x i
dp .
: وﻋﻠﻰ ذﻟك ﯾﻣﻛن اﻟﺣﺻول ﻋﻠﻰ اﻟﺗﻘدﯾر اﻟﺑﯾﯾزى ﻟداﻟﺔ اﻟﺻﻼﺣﯾﺔ ﻛﺎﻟﺗﺎﻟﻰ
٤٠٣
E(R(t) | x) h(y | x)dy t
np n y p 1
t
0
0
p 1
dpdy p p n 1 x y i (n 1) p1p n 1 dp p n x i p 1
np n y p 1 0 t x p y p n 1 dpdy i p 1 n 1 (n 1) 0 p x p n dp i
py p 1 np dy dp p p n 1 t p x i y p 1 n 1 (n 1) 0 p x p n dp i
n
0
0
0
0
0
0
p1
py p1 np dy dp p p n 1 t p x i y p 1 n 1 (n 1) 0 p x p n dp i p1
n
np n 1
p1
n dydp n x ip t p (n 1) p1p n 1 n dp x ip
p n 1
p1
n dydp tp . p 1 n 1 (n 1) p n dp x ip
x
p i
٤٠٤
) (٢-٢-٥ﺗﻘدﯾرات ﺑﯾﯾز ﻟﻣﻌﺎﻟم ﺗوزﯾﻊ واﯾﺑل ذو اﻟﺑﺗر اﻟﻣزدوج ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺔ اﻟﻛﺎﻣﻠﺔ
إذا ﻛﺎﻧﺖ Xﳍﺎ ﺗﻮزﻳﻊ اواﻳﺒﻞ ﺑﺒﻤﻌﻠﻤﺘﲔ ,bو ﻗﺪ ﰎ ﻋﻠﻴﻬﺎ ﺑﱰ ﻣﺰدوج ﺑﺬﻟﻚ ﺗﻜﻮن داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﳍﺎ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱄ :
( x b t1b ) b-1
, ) > 0 ,b>0, 0 <t1 t 2 ; t1 x t 2 .
ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ داﻟﺔ اﻟﺼﻼ ﺣﻴﺔ ﻛﺎﻟﺘﺎﱃ :
٤٠٥
bx e
) (t b2 t1b
(1 e
= ) f(x| t1 <X< t 2
e
R(t) =
t1b
t2
( t b2 t1b )
t1b
xb
dx
t
1 e
bx b-1e
b
2
t t 2 e e t 22 t1b 1 (e e )
e
2
b
1
t t 2 1 tb t2 2 1 (e e ) t1b e e e b
2
1
b
2
t t1 t 2 t 2 e -e e e b t2 t 2 e e b , t1 t t 2 . 2 t1 t 2 e -e
ﯾﻣﻛﻧﺗﺗت
: ﻫﻮX ﻟﻠﻤﺘﻐﲑm اﻟﻌﺰم اﻟﻼﻣﺮﻛﺰي ذو اﻟﺮﺗﺒﺔ e ` m
m
(x) E(X ) =
t1b
b
t2
t2
b (bm)1 x x e dx
(1 e
e
x1b
e
)
b
1
(t b2 t1b )
x 2b
t2
t2
x m b b-1 x ( x )e dx
٤٠٦
: ﻟﻴﻜﻦ 1 xb b z x (z) b dz x b1dx.
t b2 t1b
m
b (
m
z b e z dz m tb m tb 1, 2 ) ( 1, 1 ), b b
: ﻫﻰ داﻟﺔ ﺟﺎﻣﺎ اﻟﻐﲑ ﻛﺎﻣﻠﺔ واﳌﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﱃ (d,c) ﺣﻴﺚ (d, c)
c
0
z d 1e z dz, d 1.
: اﻟﻮﺳﻂ اﳊﺴﺎﰉ ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻴﻪ ﻛﺎﻟﺘﺎﱃ`m (x) ﰱm 1 ﺑﻮﺿﻊ 1 b
1 t b2 1 t1b E() ( 1, ) ( 1, ), b b
: ﻣﻦ اﻟﺘﻮزﻳﻊ اﻵﺳﻲ ذو اﻟﺒﱰ اﳌﺰدوج ﺗﻜﻮنn داﻟﺔ اﻹﻣﻜﺎن اﻷﻋﻈﻢ ﻟﻌﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﺣﺠﻤﻬﺎ n
L(x1 , x 2 ,..., x n ; ,b, t1 , t 2 ) L f (x i ) i 1
n
( n
b =
xib nt1b ) i 1
b1 e n ( x i )
i 1
1 e
tb tb ( 2 1 )
n
.
n
: وﻋﻠﻰ ذﻟﻚ x i ﺑﻔﺮض i 1
٤٠٧
n
(
xib nt1b ) i 1
n
b L b1e
n tb tb ( 2 1 ) 1e
.
: ﻫﻮ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ اﻟﻌﻜﺴﻲ ﻋﻠﻰ اﻟﺸﻜﻞ ﺑﻔﺮض أن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻟﻠﻤﻌﻠﻤﺔ
(,b) b e
h
; h, 0, > 0, 1,b B.
. ﺳﻮف ﻧﺴﺘﻔﻴﺪ ﻣﻦ داﻟﺔ ﺟﺎﻣﺎ اﻟﻨﺎﻗﺼﺔ اﻟﺘﺎﻟﻴﺔ ﰱ اﳊﺼﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻟﻠﻤﻌﻠﻤﺔ : ﻳﻜﻮن, b اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي اﳌﺸﱰك ﻟﻠﻤﻌﻠﻤﺘﲔ L(x; , b)(, b)
(, b | x)
b
b
.
L(x; , b)(,b) ddb 0
b
0
L(x; ,b)(, b) ddb 0
0
n
b
(
= b
n
b1
0
-( +n)
h
x ib nt1b )
i 1
e e
tb t b ( 2 1 ) 1e
0
n
b
=
b
n
b 1
0
(
n j1 j( . e j j=0
h -( +n-1)-1
e e
0
t b2 t1b
)
ddb
٤٠٨
xib nt1b ) i 1
n
ddb
-( +n-1)-1
0
b
n j1 n b1 = b j 0 j=0 n
) xib (n j)t1b jt 2b
(h
i 1
-
ddb.
.e
ﺑﻮﺿﻊ : b
n j1 n b1 I1 = (n+ -1) b (A j (b))n 1 db, j 0 j=0
ﺣﻴﺚ : 1
n
) xib (n j)t1b jt b2 i 1
A j (b) (h
ﻓﺈن اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي اﳌﺸﱰك ﻟﻠﻤﻌﻠﻤﺘﲔ , bﻳﺎﺧﺬ اﻟﺸﻜﻞ اﻟﺘﺎﱃ :
)(, b | x)= I -1 b n b1-( +n n
,
+n>1.
t b t b ( 2 1 ) 1e
n
) x ib (n j)t1b jt b2
i 1
(h -
.e
وﲟﺎ ان اﻟﺘﻜﺎﻣﻞ I1ﻻ ﳝﻜﻦ ﺗﻘﺪﻳﺮة ﲢﻠﻴﻠﻴﺎ وﻋﻠﻰ ذﻟﻚ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻻ ﳝﻜﻦ وﺿﻌﺔ ﰱ ﺻﻴﻐﺔ ﻣﻐﻠﻘﺔ .
اﳌﻨﻮال ) ( , bﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى
اﳌﻨﻮال ) ( , bﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﳝﻜﻦ إﳚﺎدﻩ ﺑﺎﺷﺘﻘﺎق داﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻛﻤﺎ ﻳﻠﻲ : ٤٠٩
ln ( | u) -ln I1 + (n-)lnb (b 1)ln ( +n)ln n
) x ib nt1b .
i 1
(h ]
-
)
t b2 t1b
(
-nln[1 e
اﳌﻨﻮال ) ( , bﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﳝﻜﻦ إﳚﺎدﻩ ﺑﺎﺷﺘﻘﺎق داﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻛﻤﺎ ﻳﻠﻲ : n
) (h x ib nt1b
i 1
2
) ln (, b | u) ( +n + t b2 t1b ) n( 2 . ]
t b2 t1b )
(
[1 e
اﳌﻨﻮال ) ( , bﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﳝﻜﻦ إﳚﺎدﻩ اﳊﺼﻮل ﻋﻠﻴﻪ ﻣﻦ اﳊﻞ ﻟﻨﻈﺎم اﳌﻌﺎدﻟﺘﲔ اﻟﺘﺎﻟﻴﺘﲔ:
)ا(اﻟﺘﺤﻠﻴﻞ اﻟﺒﻴﻴﺰى ﳌﻌﻠﻤﺔ اﻟﻘﻴﺎس
اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى اﳍﺎﻣﺸﻰ ﳌﻌﻠﻤﺔ اﻟﻘﻴﺎس ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻴﻪ ﻛﺎﻟﺘﺎﱃ :
٤١٠
B
1 ( | x)= (,b | x)db 0
B
b n b1 -( +n) I1
= 0
n
(h
x ib nt1b )
i 1
-
t b t b ( 2 1 ) 1e
e B -( +n) n j1 = I1 j=0 j 0
n
db
b n b1
n
(h
e
xib (n j) t1b jt b2 ) i 1
-
db
B
-( +n)
1 n j1 n b1 ( A j (b) / ) I b e , 0. j j=0 0 : ﻫﻮ ﲢﺖ ﻓﺮض داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺎ ﻓﺈن اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى ﻟﻠﻤﻌﻠﻤﺔ
1 1
B
n j1 1 E(|x) = I1 j 0 j=0
b
n
b 1
1 ( A 1 j (b)) -( +n-1)
e
ddb , 0
0
B
n j1 1 = I1 j 0 j=0 ﻻﳚﺎد I 2, I1
b n b1(+n-2)(A j 1 (b))-( +n-2) db
B
n j1 I 2 ( +n-2) j 0 j=0
b n b1 (A j (b))( +n-2) db.
: ﻧﺘﺒﻊ اﻻﺗﻰ اﻟﺘﺒﺎﻳﻦ ﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى اﳍﺎﻣﺸﻰ ﻟﻠﻤﻌﻠﻤﺔ
٤١١
B
n j1 1 E( |x) = I1 j 0 j=0 2
b
n
b 2
1 ( A j 1 (b)) -( +n-2)
e
ddb , 0
0
B
n j1 1 = I1 j 0 j=0 I 3, I1
b n b1( +n-3)(A 1 j (b))-( +n-3) db
B
n j1 I3 ( +n-3) j 0 j=0
b n b1 (A j (b))( +n-3) db. 2
Var() E(2 | x) E( | x) I3 I 22 2, I1 I1 B
n j1 Is1 ( +n-s-1) j 0 j=0 s 0,1, 2,...
b n b1 (A j (b)) ( +n-s-1) db,
. ﻋﺪدﻳﺎIs1 ,s 0,1, 2. ﻻﳚﺎد اﻟﻮﺳﻂ اﳊﺴﺎﰉ واﻟﺘﺒﺎﻳﻦ ﻻ ﺑﺪ ﻣﻦ ﺣﺴﺎب اﻟﺘﻜﺎﻣﻼت ﻓﱰة ﺗﻘﺪﻳﺮ ﺑﻴﻴﺰﻳﺔ ﻟﻠﻤﻌﻠﻤﺔ
: ﲝﻞ اﳌﻌﺎدﻟﺘﲔ اﳌﺘﻤﺎﺛﻠﺔ ﻟﻠﻤﻌﻠﻤﺔ100(1- )% ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﻓﱰة ﺗﻘﺪﻳﺮ ﺑﻴﻴﺰﻳﺔ t1
0
1 ( u ) d 2
,
t2
1
( u ) d
. 2
: أي أن، t 2 واﳊﺪ اﻷﻋﻠﻰt1 ﻟﻠﺤﺪ اﻷدﱏ P(t1 t 2 ) 1 .
: وﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﻛﺎﻟﺘﺎﱃ اﳌﺘﻤﺎﺛﻠﺔ ﻟﻠﻤﻌﻠﻤﺔ100(1- )% ( ﻫﻲ ﻓﱰة اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰﻳﺔt1 , t 2 ) اﻟﻔﱰة
٤١٢
t1
1 ( u ) d I1
1
j=0
0
t1
B
n j 1 j 0
b
n
b 1
-( +n) e
( A j 1 ( b ) / )
B
n j 1 I1 1 j 2 0 j=0
I 2
j=0
n j 1 j 0
b )) db t1
b n b 1 (A j (b )) ( +n-1) ( +n -1,A j (
B
b n b 1 , I
d db
0
I1 , (( +n-1)
: اﻳﻀﺎ I 2
j=0
B
n j1 j 0
b n b 1 (A j (b )) ( +n-1) ( +n-1,A j (
b ))db, t2
ﯾﻣﻛن
(a , z ) y a 1 e y dy, a 0, (a , z ) (a ) (a , z ) z
(a , z ) (a ) (a , z ).
وذﻟك ﺑﺈﺗﺑﺎع اﻟﺷروط اﻟﺗﺎﻟﯾﺔ اﻟﺣﺻول ﻋﻠﻰ أﻗﺻر ﻓﺗرة ﺛﻘﻪ )اﻟﻣرﻏوﺑﻪ او اﻟﻣﻌﺗﻣدة( ﻟﻠﻣﻌﻠﻣﺔ
:
1 t1 x 1 t 2 x () ا P(t1 t 2 ) 1 .
(ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺠﺰء )ا 1 t1 x 1 t 2 x
: ﻧﺘﺒﻊ اﻟﺘﺎﱃ
٤١٣
()ب
-( +n)
t1
-( +n)
t2
B
n j1 I j 0 j=0 1 1
b n b1e
B
n j1 I j 0 j=0 1 1
b
( +n)
n j1 j 0 j=0
j=0
n j1 B j
0
db
1 b 1 ( A j (b) / t 2 )
e
B
t 2 t1
n
( A j 1 (b) / t1 )
b n b1e
db
( A j 1 (b) / t 2 )
db .
b n b1e
( A j 1 (b) / t1 )
db
(ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺠﺰء )ب P(t1 t 2 ) 1 .
: ﻧﺘﺒﻊ اﻟﺘﺎﱃ t2
n j1 B
j=0
j
n b1 -(+n) (Aj1(b)/t1) ddb(1p)I1 b e 0 t1 -(+n) (A1(b)/t ) n b1 -(+n) (Aj1(b)/t1) j 1 b e d e d 0 t1 t2
n j1 B
(1p)I1 j=0
j
n j1 B
(1p)I1 j=0
j
n b1 1 1 b (n1),Aj (b)/t1 (n1),Aj (b)/t2db. 0
. ( ﺗﺳﻣﻰ اﻋﻠﻰ ﻓﺗرة ﺛﻘﺔt1 , t 2 ) ﻟﻬﺎ ﺗوزﯾﻊ ﺑﻌدى ﻟﻪ ﻣﻧوال وﺣﯾد ﻓﺈن اﻟﻔﺗرة إذا ﻛﺎﻧت
b )ب(اﻟﺘﺤﻠﻴﻞ اﻟﺒﻴﻴﺰى ﳌﻌﻠﻤﺔ اﻟﺸﻜﻞ : ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻴﻪ ﻛﺎﻟﺘﺎﱃb اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى اﳍﺎﻣﺸﻰ ﳌﻌﻠﻤﺔ اﻟﺸﻜﻞ
٤١٤
2 ( | x)= (, b | x)d 0
b n b1 = -( +n) I1 0 t b t b ( 2 1 ) n 1e (h x ib ( nt1b ) i 1
n
e
n n j1 b j I1 j=0
d
-( +n)
0
n h x ib (n j) t1b jt b2 d - i 1
e
n j1 = j j=0
n b n b1 (n 1) h x ib (n j)t1b jt b2 I1 i 1
(n 1)
n b1 n j1 b (A j (b))n+ -1 . j I j=0
: ﻫﻮb ﲢﺖ ﻓﺮض داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺎ ﻓﺈن اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى ﻟﻠﻤﻌﻠﻤﺔ B
n j1 -1 E(b | x) I b n b1 (A j (b)) n+ -1db j 0 j=0 B
J = 1 , J1 b n b1 (A j (b)) n+ -1db. I 0
: ﻧﺘﺒﻊ اﻻﺗﻰ ﻻﳚﺎد اﻟﺘﺒﺎﻳﻦ ﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى اﳍﺎﻣﺸﻰ ﻟﻠﻤﻌﻠﻤﺔ ٤١٥
B
n j1 -1 E(b | x) I b n 2 b1 (A j (b)) n+ -1db j 0 j=0 2
B J2 = ,J 2 b n 2 b1 (A j (b)) n+ -1db, I 0 2
Var(b) E(b 2 | x) E(b | x) 2
J J = 2 1 . I I
. ﻋﺪدﻳﺎI, J1 , J 2 ﻻﳚﺎد اﻟﻮﺳﻂ اﳊﺴﺎﰉ واﻟﺘﺒﺎﻳﻦ ﻻ ﺑﺪ ﻣﻦ ﺣﺴﺎب اﻟﺘﻜﺎﻣﻼت b ﻓﱰة ﺗﻘﺪﻳﺮ ﺑﻴﻴﺰﻳﺔ ﻟﻠﻤﻌﻠﻤﺔ
: ﲝﻞ اﳌﻌﺎدﻟﺘﲔb اﳌﺘﻤﺎﺛﻠﺔ ﻟﻠﻤﻌﻠﻤﺔ100(1- )% ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﻓﱰة ﺗﻘﺪﻳﺮ ﺑﻴﻴﺰﻳﺔ t1
0
2 (b u ) db 2
,
2
(b u ) d b
t2
. 2
: أي أن، t 2 واﳊﺪ اﻷﻋﻠﻰt1 ﻟﻠﺤﺪ اﻷدﱏ P(t1 b t 2 ) 1 .
: وﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﻛﺎﻟﺘﺎﱃ اﳌﺘﻤﺎﺛﻠﺔ ﻟﻠﻤﻌﻠﻤﺔ100(1- )% ( ﻫﻲ ﻓﱰة اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰﻳﺔt1 , t 2 ) اﻟﻔﱰة t1
0
t1
-1 n
2
(b u) db I b 0
n j1 (A j (b))n+ -1db j j=0 b 1
t
I 1 n b1 n j1 b (A j (b)) n+ -1db, j 2 0 j=0
: واﯾﺿﺎ B
I n j1 n b1 b (A j (b)) n+ -1db. j 2 t2 j=0
٤١٦
وﻓﻴﻤﺎ ﻳﻠﻰ اﳊﺎﻻت اﳋﺎﺻﺔ ﻣﻦ اﻟﻨﺘﺎﺋﺞ اﻟﺴﺎﺑﻘﺔ -اﻟﻨﺘﻴﺠﺔ ﻟﺘﻮزﻳﻊ واﻳﺒﻞ اﳌﺒﺘﻮر ﻣﻦ اﻟﻴﺴﺎر ﳓﺼﻞ ﻋﻠﻴﻬﺎ ﺑﻮﺿﻊ
t2
اﻟﻨﺘﻴﺠﺔ ﻟﺘﻮزﻳﻊ واﻳﺒﻞ ﰱ ﺣﺎﻟﺔ ﻋﺪم ﺣﺪوث ﺑﱰ ﳓﺼﻞ ﻋﻠﻴﻬﺎ ﺑﻮﺿﻊ-اﻟﻨﺘﻴﺠﺔ ﻟﻠﺘﻮزﻳﻊ اﻻﺳﻲ ﲟﻌﻠﻤﺔ واﺣﺪة ﳓﺼﻞ ﻋﻠﻴﻬﺎ ﺑﻮﺿﻊ
t1 0, t 2
b 1
) (٣-٥ﺗوزﯾﻊ ﻛوﺷﻰ) (١-٣-٥ﺗﻘدﯾرات اﻻﻣﻛﺎن اﻻﻛﺑر ﻟﻣﻌﺎﻟم ﺗوزﯾﻊ ﻛوﺷﻰ ﺑﻔــﺮض أن nﻣــﻦ اﻟﻮﺣــﺪات وﺿــﻌﺖ ﻟﻼﺧﺘﺒــﺎر وان اﻻﺧﺘﺒــﺎر ﻳﻨﺘﻬــﻲ ﺑﻌــﺪ ﻓﺸــﻞ ﻛــﻞ اﻟﻮﺣــﺪات .ﺑﻔــﺮض أن X1 , X 2 , , X nﻋﻴﻨــﺔ ﻋﺸﻮاﺋﻴﺔ ﲤﺜﻞ أزﻣﻨﺔ اﻟﻔﺸﻞ وان أزﻣﻨﺔ اﻟﻔﺸﻞ ﺗﺘﺒﻊ ﺗﻮزﻳﻊ ﻛﻮﺷﻰ ﺣﻴﺚ : 1
1
2
2
1 x f x; 1 2
1 2 x 2
2 2 1 . x
ﺣﻴﺚ ﻫﻮ وﺳﻴﻂ ا ﺘﻤﻊ )ﻣﻌﻠﻤﺔ اﳌﻮﻗﻊ( و ﻣﻌﻠﻤﺔ اﳌﻘﻴﺎس. ﺑﺎﻋﺘﺒﺎر أن اﻟﻌﻴﻨﺔ x 1 , x 2 ,...x nﰎ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﻓﺈﻧﻨﺎ ﻧﺮﻏﺐ ﰲ ﺗﻘﺪﻳﺮ ﻣﻌﺎﱂ ﺗﻮزﻳﻊ ﻛﻮﺷﻰ 2 ,
.
داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : 1
n
2 L n n 2 x i i 1
1
n
2 ln L n ln n ln ln 2 x i i 1 2
2 x ln L n 2 i 2 i 1 x i
٤١٧
2
2 n 1 x i 2 x i x i 2 ln L 2 2 2 i 1 2 x i 2
4 x i
n
i 1
2
2 x i 2 2 x i .2
2
n 2
i 1
2 x i 2
2
2 ln L n 2 i 1 2 x 2 i n x i 4 2 2 i 1 2 x i ln L n n 2 2 i 1 x i 2 2 2 2 ln L n n 2 x i 2.2 2 2 2 i 1 2 x i 2 n n n 1 4 2 2 2 2 2 2 2 i 1 x i 1 x i i n ln L x 2 2 i 2 i 1 x i
then n
x x i
2
2 0
i n ln L n 2 2 0 2 i 1 x i i 1
ﺣﻞ اﳌﻌﺎدﻟﺘﲔ اﻧﻴﺎ ﺑﺎﺳﺘﺨﺪام ﺑﺮﻧﺎﻣﺞ ﻋﻠﻰˆ وˆ أي ﳓﺼﻞ ﻋﻠﻰ و وﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﺗﻘﺪﻳﺮ ﻟﻜﻞ ﻣﻦ . Mathematica اﳊﺎﺳﺐ اﻻﱃ ﻣﺜﻞ ﺑﺮﻧﺎﻣﺞ : ﻣﻦ اﻟﻌﻼﻗﺔ اﻟﺘﺎﻟﻴﺔk,h ﻛﻤﺎ ﳝﻜﻦ ﺑﺎﺳﺘﺨﺪام اﺣﺪ اﻟﻄﺮق اﻟﻌﺪدﻳﺔ ﻛﻄﺮﻳﻘﺔ ﻧﻴﻮﺗﻦ راﻓﺴﻮن وذﻟﻚ ﺑﺎﳊﺼﻮل ﻋﻠﻰ 2 ln L 2 0 2 ln L 00
2 ln L ln L 0 0 h 0 2 ln L k ln L 02 0 ٤١٨
ln L 0 . ln L 0
1
2 ln L 0 0 2 ln L 02
2 ln L 2 0 h 0 k 2 ln L 0 00
ﺣﻴﺚ 0و 0ﻗﻴﻤﺘﲔ ﻣﺒﺪﺋﻴﺘﲔ .ﻧﺴﺘﻤﺮ ﰱ ﻋﻤﻠﻴﺔ اﻟﺘﻜﺮارات iteration techniqueﺣﱴ اﳊﺼﻮل ﻋﻠﻰ ˆ و ˆ ﻋﻨﺪﻣﺎ ﻳﻜﻮن اﻟﻔﺮق ﺑﲔ 0و 0ﺻﻐﲑ ﺟﺪاً . ) (٢-٣-٥ﺗﻘدﯾر اﻻﻣﻛﺎن اﻻﻛﺑر ﻟداﻟﺔ اﻟﺻﻼﺣﯾﺔ ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻟﺘﻮزﻳﻊ ﻛﻮﺷﻰ ﻛﺎﻟﺘﺎﱃ : 1 2 dx x 1
1 t
R t
x z z x
let
dx dz 1 1 1 1 dz tan z t t 1 z2 1 t tan 1 2 1 1 t tan 1 2
ﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﻣﻘﺪر اﻻﻣﻜﺎن اﻻﻛﱪ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ R t ﺑﺎﻟﺘﻌﻮﻳﺾ ﻋﻦ ﺗﻘﺪرات اﻻﻣﻜﺎن اﻷﻛﱪ ˆ ˆ , ﰱ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻛﺎﻟﺘﺎﱃ : 1 t ˆ . Rˆ t tan 1 2 ˆ
وﺑﺬﻟﻚ ﳓﺼﻞ ﻋﻠﻰ ﺗﻘﺪﻳﺮ اﻻﻣﻜﺎن اﻻﻛﱪ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ .
٤١٩
) (٣-٣-٥ﺗﻘدر ﺑﯾﯾز ﻟداﻟﺔ اﻟﺻﻼﺣﯾﺔ ﳝﻜﻦ اﳚﺎد ﺗﻘﺪر ﺑﻴﻴﺰ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ R t ﲢﺖ ﻓﺮض اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ل , ﻛﺎﻟﺘﺎﱄ: 1 g , .
و ﺑﺎﻟﺘﺎﱄ ﻓﺈن اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ل , ﻫﻮ : 1
n
2 n 1 2 x i
g1 , x
i 1
.
1 2
x i dd
n
2
n 1
i 1
0
ﻋﻨﺪ اﺳﺘﺨﺪام داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺄ ﻓﺈن اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰي ﻟـ R t ﻫﻮ: 1 n 2 1 1 1 t n 1 2 tan x i dd 0 2 i 1 E R(t) x 1 n 2 n 1 2 x i dd
i 1
0
)(١٣-٥ وﳝﻜﻦ وﺿﻊ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﻋﻠﻰ اﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ: d
u e
) L Q (
. d
) L Q (
e
E u x
)(١٤-٥ ﺣﻴﺚ :
و u داﻟﺔ ﰱ ) (1 , 2 ,..., mو L ln Lﻫﻮ ﻟﻮﻏﺎرﻳﺘﻢ داﻟﺔ ﻹﻣﻜﺎن و Q ln g و g ﻫﻮ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ل . أي أن ) (١٤-٥ﻫﻮ اﻟﺘﻮﻗﻊ ﻟﻠﺪاﻟﺔ u ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻟـ . وﻛﺤﺎﻟﺔ ﺧﺎﺻﺔ ﲢﺖ ﻓﺮض ﺗﻮزﻳﻊ ﻛﻮﺷﻰ ﻓﺎن ) u R(tو ﺗﺴﺘﺒﺪل ﺑﺎﳌﻌﺎﱂ
,
و g
ﺗﺴﺘﺒﺪل
ب
g , x و L
و ) (1 , 2 ) (, 2
ﺗﺴﺘﺒﺪل
ب
ln L , | x
وﳊﺴﺎب اﻟﺘﻘﺪﻳﺮ اﻟﺒﻴﻴﺰى ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ﺑﺎﺳﺘﺨﺪام ﺗﻘﺮﻳﺐ ﻟﻨﺪﱃ ﻓﺈﻧﻨﺎ ﳓﺘﺎج اﱃ ﺣﺴﺎب اﻟﻘﻴﻢ اﻟﺘﺎﻟﻴﺔ : ٤٢٠
2u u11 2 2u1u 2 2u u 22 2 2u1u 2 u11 2u u 21 u12 u 22 u12
: ﻣﻦ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻰ g ,
1
1 ln . Q Q 1 Q1 0 , Q2 . 1 1 t u R(t) tan 1 2 u 1 u1 . 2 2 t
Q , Q ln g , ln
:و ذﻟﻚ ﻷن 1 2 t 1 1 2 . 2 2 t
u1
du 1 . d
1 . 2 2 t
t 1 2. 2 t 1 1 2 t . 2 . 2 t 2
u2
du 1 . d
1 t . 2 2 t
: ﺑﻀﺮب اﻟﺒﺴﻂ و اﳌﻘﺎم ﰲ ٤٢١
1 t u2 . 2 . 2 t t u1
:( ﻳﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃ١٤-٥) اى ان اﻟﺘﻘﺪﻳﺮ اﻟﺘﻘﺮﻳﱮ ﻟﺪاﻟﺔ ا ﻟﺼﻼﺣﻴﺔ ﰱ 1 u B u u ij 2u i Q j ij Lijk u ijk 2 1 u [ u11 2u1Q1 11 u12 2u1Q 2 12 u 21 2u 2Q1 21 2 1 u 22 2u 2Q 2 22 ] [L111u11111 L111u 2 1112 L112 u11121 2 L112u 2 1122 L121u112 11 L121u 2 12 12 L122 u11221 L122 u 2 1222 L 211u12111 L 211u 2 2112 L 212 u12121 L 212 u 2 2122 L 221u122 11 L 221u 2 22 12 L 222 u122 21 L 222 u 2 22 22 ]. 1 2 2 E R t x u [{ u1111 u12 12 u12 u1 12 u 22 u 2 22 } 2 1 1 L111 112 u 2 1112 L112 3u11112 u 2 1122 2122 2 2 1 1 2 L122 u1 1122 2 122 3u 2 12 22 L 222 u 122 12 u 2 22 2 2 ˆ , ˆ , 2
.
ﰱ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﲟﻘﺪرات اﻻﻣﻜﺎن اﻻﻛﱪ ﺣﻴﺚ ﺗﺴﺘﺒﺪل اﳌﻌﺎﱂ : ﰲ اﳌﺼﻔﻮﻓﺔE L12 E L 21 0 1
L11 L12 L L 12 22
: واﻟﱴ ﺗﺴﺎوى ٤٢٢
1
2 lnL 2 lnL 2 11 12 2 lnL 2 lnL 12 22 2 وﻋﻠﻰ ذﻟﻚ : 1 2 1 1 u B u [u1111 u 22 u 2 22 ] L111112 L 222 u 2 222 2 2 2 2 ˆˆ , , ﰱ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﲟﻘﺪرات اﻻﻣﻜﺎن اﻻﻛﱪ . ﺣﻴﺚ ﺗﺴﺘﺒﺪل اﳌﻌﺎﱂ
ﺣﻴﺚ: 1 L11 1 22 L 22 11
L11 2i 42i2
ﺣﻴﺚ: 1
2 i 2 x i L12 4 x i i2 L 21 n L 22 2 L11 L111 4 x i 42i i2
2n 4 3 4 2i i2 3
L 222
)٤-٥اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻰ اﻟﻠوﻏﺎرﺗﻣﻰ ھﺬا اﻟﺒﻨﺪ ﻣﺎﺧﻮذ ﻣﻦ ﻗﺒﻞ ) Al-mobiad(2010ﻣﻦ ﺧﻼل رﺳﺎﻟﺔ اﻟﻤﺎﺟﺴﺘﯿﺮ اﻟﺨﺎﺻﺔ ﺑﺎﻟﻄﺎﻟﺒﺔ اﯾﻤﺎن اﻟﻤﺒﯿﺾ وذﻟﻚ ﺗﺤﺖ اﺷﺮاف اﻟﺪﻛﺘﻮرة ﺛﺮوت ﻣﺤﻤﺪ ﻋﺒﺪ اﻟﻤﻨﻌﻢ وﺳﻮف اﻗﺪﻣﮫ ﻛﻤﺎ ھﻮ وذﻟﻚ ﺣﺘﻰ ﻧﻤﻜﻦ ٤٢٣
اﻟﻘﺎرئ ﻣﻦ اﻟﺘﺪرب ﻋﻠﻰ ﻓﻚ اﻟﻤﻌﺎدﻻت ﻣﻦ ﺧﻼل ﻣﺎ اﻛﺘﺴﺒﮫ ﻣﻦ اﻻﺑﺤﺎث اﻟﺴﺎﺑﻘﺔ اﻟﺘﻰ ﺗﻨﺎوﻟﻨﺎھﺎ ﻓﻰ ھﺬا اﻟﻔﺼﻞ وﻟﻦ ﯾﺤﺘﺎج اﻟﺒﺎﺣﺚ اﻟﻰ ﺟﮭﺪ ﻛﺒﯿﺮ ﻓﻰ ﻓﻚ اﻟﻤﻌﺎدﻻت ﻻن اﻟﻄﺎﻟﺒﺔ ﺷﺮﺣﺖ اﻟﺨﻄﻮات ﺑﺎﻟﺘﻔﺼﯿﻞ.
)(١-٤-٥ﺗﻘدﯾرات اﻻﻣﻛﺎن اﻻﻛﺑر ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺎت اﻟﻣﺗﺗﺎﺑﻌﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ﺗﻢ ﺗﺨﺼﯿﺺ ھﺬا اﻟﺠﺰء ﻹﯾﺠﺎد ﻣﻘﺪرات اﻹﻣﻜﺎن اﻷﻛﺒﺮ ﻟﻤﻌﻠﻤﺘﻲ اﻟﺘﻮزﯾﻊ اﻟﻄﺒﯿﻌﻲ اﻟﻠﻮﻏﺎرﯾﺘﻤﻲ وﻛﺬﻟﻚ داﻟﺘﻲ اﻟﺼﻼﺣﯿﺔ و ﻣﻌﺪل اﻟﻔﺸﻞ ﻟﻨﻔﺲ اﻟﺘﻮزﯾﻊ وذﻟﻚ ﺑﺎﺳﺘﺨﺪام ﻋﯿﻨﺎت ذات ﻣﺮاﻗﺒﺔ ﻣﺘﺘﺎﺑﻌﺔ . ﺑﻔﺮض أﻧﻪ ﺗﻢ وﺿﻊ nﻣﻦ اﻟﻮﺣﺪات اﻟﻤﺴﺘﻘﻠﺔ ﺗﺤﺖ اﻟﺘﺠﺮﺑﺔ ،وﺗﻢ اﻟﺤﺼﻮل ﻋﻠﻰ أزﻣﻨﺔ اﻟﺤﻴﺎة ﻓﻘﻂ ﻟﻠﻌﻴﻨﺔ ذات اﻟﺤﺠﻢ (m n) mﺣﻴﺚ ﻳﺮﻣﺰ ﻟﻤﻔﺮدات اﻟﻌﻴﻨﺔ ﺑﺎﻟﺮﻣﺰ ) ، x (x1 , x 2 ,..., x mوﻫﻲ اﻟﻌﻴﻨﺔ ذات اﻟﻤﺮاﻗﺒﺔ اﻟﻤﺘﺘﺎﺑﻌﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﻧﻲ ذات اﻟﺤﺠﻢ mو اﻟﻤﺨﺘﺎرة ﻋﺸﻮاﺋﻴﺎً ﻣﻦ ﻋﻴﻨﺔ اﻟﻤﺸﺎﻫﺪات ذات اﻟﺤﺠﻢ ، nو اﻟﻤﺮﺗﺒﻄﺔ ﺑﻨﻈﺎم اﻟﻤﺮاﻗﺒﺔ ) (R1 ,R 2 ,..., R mو اﻟﺘﻲ ﺳﻮف ﻳﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ ) . x (x1 , x 2 ,..., x mداﻟﺔ اﻹﻣﻜﺎن ﻟﻠﻌﻴﻨﺔ xﺗﻌﻄﻰ ﻛﺎﻟﺘﺎﻟﻲ :
)(١٥-٥
,
Ri
m
f(x i , 2 , ) 1 F(x i , 2 , )
2
L(x | , ) c
i 1
ﺣﯿﺚ : c n(n 1 R1 )(n 2 R1 R 2 ) ... (n m+1 R1 ... R m1 ),
و ) F(.) , f(.ھﻤﺎ داﻟﺘﻲ اﻟﻜﺜﺎﻓﺔ و اﻟﺘﻮزﯾﻊ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ . اﻟﺘﻮزﯾﻊ اﻟﻠﻮﻏﺎرﯾﺘﻤﻲ اﻟﻄﺒﯿﻌﻲ ﻟﮫ داﻟﺘﻲ اﻟﻜﺜﺎﻓﺔ و اﻟﺘﻮزﯾﻊ اﻵﺗﯿﺘﯿﻦ : )(١٦-٥
و , x > 0 , > 0, - << .
1 (ln x ) 2 2 2
1
e
x 2
2
f (x; , )
2
)(١٧-٥
F(x, , ) F( ) (t) dt , -
ﺣﯿﺚ : ln x ) (tﺗﻤﺜﻞ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻠﺘﻮزﯾﻊ اﻟﻄﺒﯿﻌﻲ اﻟﻘﯿﺎﺳﻲ و ٤٢٤
.
داﻟﺔ اﻟﺼﻼﺣﯿﺔ ﻟﻠﺘﻮزﯾﻊ اﻟﻠﻮﻏﺎرﯾﺘﻤﻲ اﻟﻄﺒﯿﻌﻲ ھﻲ : ] ln[t )
)(١٨-٥
( R(t)
ﺣﯿﺚ ) (.داﻟﺔ اﻟﺘﻮزﯾﻊ ﻟﻤﺘﻐﯿ ﺮ ﻋﺸ ﻮاﺋﻲ ﯾﺘﺒ ﻊ اﻟﺘﻮزﯾ ﻊ اﻟﻄﺒﯿﻌ ﻲ اﻟﻘﯿﺎﺳ ﻲ .أو ﯾﻤﻜ ﻦ ﻛﺘﺎﺑﺘﮭ ﺎ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﻟﻲ : ln[t] )
( R(t) 1
وداﻟﺔ ﻣﻌﺪل اﻟﻔﺸﻞ ھﻲ : 2
1 ln t 2
e ln t (2 t 1 )
)(١٩-٥
H(t)
و ﺑﺎﻟﺘﻌﻮﯾﺾ ﻣﻦ اﻟﻤﻌﺎدﻟﺘﯿﻦ ) (١٦-٥و ) (١٧-٥ﻓﻲ اﻟﻤﻌﺎدﻟﺔ ) (١٥-٥ﻓﺈن داﻟﺔ اﻹﻣﻜﺎن ﻓﻲ ﺣﺎﻟﺔ اﻟﺘﻮزﯾﻊ اﻟﻠﻮﻏﺎرﯾﺘﻤﻲ اﻟﻄﺒﯿﻌﻲ ﺗﺄﺧﺬ اﻟﺼﯿﻐﺔ اﻟﺘﺎﻟﯿﺔ : 2
m
Ri
1 F(x ) i
m
1 ln[ x i ] ( ) 2 i 1
e
i 1
1 m
m x i i1
ﻟﻮﻏﺎرﯾﺘﻢ داﻟﺔ اﻹﻣﻜﺎن ھﻮ : ln L 2 m )(٢٠-٥ 1 m ln[x i ] ( m ln[] ln[x i ] ) R i ln 1 F(x i ) 2 i1 i 1 i 1 m
ﺣﯿﺚ : ٤٢٥
L(x; , )
) ( iداﻟﺔ اﻟﺘﻮزﯾﻊ اﻟﻄﺒﯿﻌﻲ اﻟﻘﯿﺎﺳﻲ . i
Fi F(x i ) F( i ) (t) dt , -
ln x ,
, i
t2 2
-
1
(t) ( 2 ) e
و ﻟﻠﻮﺻﻮل ﻟﺪاﻟﺔ اﻹﻣﻜﺎن ) L(x; 2 , إﻟﻲ ﻧﮭﺎﯾﺘﮭﺎ ﻋﻨﺪ ﻧﻘﻄﺔ داﺧﻠﯿﺔ ﻓﻲ ﻓﻀﺎء اﻟﻤﻌﺎﻟﻢ ) (2 , ﯾﺘﻢ إﯾﺠﺎد اﻟﺘﻔﺎﺿﻞ اﻟﺠﺰﺋﻲ ﻟﻠﻮﻏﺎرﯾﺘﻢ داﻟﺔ اﻹﻣﻜﺎن ﺑﺎﻟﻨﺴﺒﺔ ﻟﻜﻞ ﻣﻌﻠﻤﺔ ﻣﻊ اﻟﻤﺴﺎواة ﺑﺎﻟﺼﻔﺮ ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﻣﻌﺎدﻟﺘﻲ اﻹﻣﻜﺎن اﻟﺘﺎﻟﯿﺘﯿﻦ : ) ln L(x; 2 , ) ln L(x; 2 , 0 , 0 2
و ذﻟﻚ ﺑﺎﻟﺘﻔﺎﺿﻞ اﻟﺠﺰﺋﻲ ﻟﻠﻤﻌﺎدﻟﺔ ) (٢٠-٥ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻤﺔ ﺣﯿﺚ ﯾﺘﻢ اﻟﺤﺼﻮل ﻋﻠﻰ : m m ln L 1 R i Fi 2 (2) (ln[x] ) 2 i 1 i 1 1 Fi
m m 1 1 Ri 2 (2) (ln[x] ) ( i ) 2 i 1 i 1 1 Fi
) ( i ﺑﻮﺿﻊ 1 Fi
zi ﻓﺈن : m ln L 1 1 m 2 (2) (ln[x] ) R i z i 2 i1 i 1
m 1 m [ ](ln[x ) ] R izi 2 i1 i 1
)(٢١-٥
٤٢٦
. ﺗﻤﺜﻞ داﻟﺔ اﻟﻔﺸﻞzi
( i ) ﺣﯿﺚ 1 Fi
: ﯾﺘﻢ اﻟﺤﺼﻮل ﻋﻠﻰ2 ( ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻤﺔ٢٠-٥) وﺑﺎﻟﺘﻔﺎﺿﻞ اﻟﺠﺰﺋﻲ ﻟﻠﻤﻌﺎدﻟﺔ m ln L m 1 m R i Fi 2 3 (ln[x] ) ( 2 ) 2 2 i1 1 F i 1 i
m 1 m ln[x] 2 m R i d i ( ) ( i ) i1 d i 1 1 Fi
m 1 m ln[x] 2 m Ri ( ) ( i ) i i1 i 1 1 Fi
m 1 m ln[x] 2 m R i ( ) zi i i1 i 1 m 1 m 2 2 2 3 [ (ln[x] ) m R i i z i ] i1 i 1
m 1 m 2 2 [ (ln x ) ( R i i z i m)] . 3 i1 i 1
(٢٢-٥)
: ( ﺑﺎﻟﺼﻔﺮ ﯾﺘﻢ اﻟﺤﺼﻮل ﻋﻠﻰ٢٢-٥) ( و٢١-٥) و ﺑﻤﺴﺎواة m nL 1 m Rz]=0 2 [ (ln[x] ) i i i1 i 1
(٢٣-٥)
2 m ln L 1 m 2 3 [ (ln[x] ) ( R i i z i m)] = 0 i1 2 i 1
(٢٤-٥) 2 وﯾﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﻰ ﺗﻘﺪﯾﺮات اﻹﻣﻜﺎن اﻷﻛﺒﺮ ( و٢٣-٥) ﻣﻦ اﻟﻤﻌﺎدﻟﺘﯿﻦ , ٤٢٧
Newton
( ﺑﺎﺳ ﺘﺨﺪام أﺣ ﺪ اﻟﻄ ﺮق اﻟﻌﺪدﯾ ﺔ ﻛﻄﺮﯾﻘ ﺔ )أﺳ ﻠﻮب ( ﻧﯿ ﻮﺗﻦ راﻓﺴ ﻮن٢٤-٥) : ﻣﻦ اﻟﻌﻼﻗﺔ اﻟﺘﺎﻟﯿﺔk,h وذﻟﻚ ﺑﺎﻟﺤﺼﻮل ﻋﻠﻰraphson method
2 ln L 2 0 2 ln L 0 0
2 ln L 0 0
ln L h 0 2 k ln L ln L 0 2 0
:ﺣﯿﺚ 2 ln L 2 0 h 0 k 2 ln L 0 0 0
1
2 ln L ln L 0 0 0 . 2 ln L ln L 0 2 0
ﺣﺘ ﻰiteration technique ﻧﺴ ﺘﻤﺮ ﻓ ﻲ ﻋﻤﻠﯿ ﺔ اﻟﺘﻜ ﺮارت. ﻗﯿﻤﺘ ﯿﻦ ﻣﺒ ﺪﺋﯿﺘﯿﻦ0 و0 أو ﺣﻞ اﻟﻤﻌﺎدﻟﺘﯿﻦ. ً ﺻﻐﯿﺮ ﺟﺪاk وh ﻋﻨﺪﻣﺎ ﯾﻜﻮنˆ وˆ اﻟﺤﺼﻮل ﻋﻠﻰ ( آﻧﯿ ﺎ ً ﺑﺎﺳ ﺘﺨﺪام ﺑﺮﻧ ﺎﻣﺞ ﻋﻠ ﻰ اﻟﺤﺎﺳ ﺐ اﻵﻟ ﻲ ﻣﺜ ﻞ ﺑﺮﻧ ﺎﻣﺞ٢٤-٥) ( و٢٣-٥) 2 ln L 2 ln L 2 ln L : ﻧﺘﺒﻊ اﻵﺗﻲ ﻹﯾﺠﺎد. 5 اﻻﺻﺪارMathematica , , 2 2 m 2 ln L 1 z [ m R i i ], 2 2 i 1
( i ) ( i ) ( ) ( i ) i i (1 i ( i )) 1 i ( i ) z i , 2 1 i (i ) ( i ) i ( i ) i ( i ) ( i ) , , ٤٢٨
z i
i ( i ) ( i ) ( i ) 2 1 i (i )
(1 i ( i ))
2
1 (1 i ( i )) i ( i ) ( i ) 2 1 i ( i )
2 ( i ) 1 (1 i ( i )) i ( i ) 2 1 i ( i )2 1 i (i )
zi 1 i zi i zi zi 2
1 1 A i zi z i i
: ﺣﯿﺚ A i zi zi i
m 2 ln L 1 Ai [ m R ] i 2 2 i 1
m 1 2 [m R i A i ], i 1
m 2 ln L m R iAi [1 ]. 2 2 m i 1
m m 2 ln L 1 z 2 3 [2 (ln[x] ) R i ( i zi i i )] 2 i 1 i 1
m m 1 z 2 [ 2 (ln[x] ) R i ( i z i i i )] 3 i 1 i 1
٤٢٩
m m 1 z A 2 3 [2 (ln[x] ) R i ( i (1) i i )] i 1 i 1
m m 1 z A 2 [ 2 (ln[x] ) R i ( i (1) i i )] 3 i 1 i 1 m m 1 3 [2 (ln[x] ) R i (z i i A i )] i 1 i 1
m m 1 [ 2 (ln[x] ) R i Bi ] . 3 i 1 i 1
: ﺣﯿﺚ Bi (zi i Ai ). m 2 ln L 3 m 2 2 [ (ln[x] ) ( R i i z i m)] [ 2 ]2 4 i1 i 1 m m 1 z 2 [2( R i i z i m) R i ( 2i z i i i2 )], 3 i 1 i 1
i ( i ) ( i ) i i , ( i ) i 2 , 2i . 2 2
z i 2
z i 2
( i ) ( ) ( i ) i i 2 1 i (i )
(1 i ( i ))
( i ) i 2 ( i ) i ( i ) 2 1 i ( i )
(1 i ( i ))
z i 1 ( i ) i 2 i ( i )2 2 1 i ( i ) 1 i ( i )2 ٤٣٠
z i 1 2 i z i i zi 2 2 z i i zi z i 2 z i i Ai . 2 m 2 ln L 3 m 2 2 [ (ln[x] ) ( R i i z i m)] [ 2 ]2 4 i1 i 1 m m 1 2 [2( R i i z i m) R i ([ i ]z i i [ i A i ])] 3 i 1 i 1
m 3 m 2 2 [ (ln[x] ) ( R i i z i m)] 4 i1 i 1
m m 1 i z i Ai i 2 2 [2( R i i z i m) R i ( )] 3 i 1 i 1
3 m 3 m 3 2 (ln[x] ) R z 2 m 4 2 i i i i1 i1 2 m 2m 1 m R z R i i (z i A i i )] i i i 2 2 2 i1 i 1
3 m 1 m m 2 (ln[x] ) Rz 2 4 2 i i i i1 i1 1 m 2 R i i Bi i1
3 m 1 m m 2 4 (ln[x] ) 2 (R i i zi R i i Bi ) 2 i1 i1
3 m 1 m m 2 (ln[x] ) R (z i Bi ) 2 4 2 i i i1 i1 ٤٣١
m 3 m 1 2 )) 4 (ln[x] ) 2 ( m R i i (z i Bi i1 i 1
m 3 m 1 2 ](ln[x ) ( m ) R i Ci 4 i1 2 i 1
ﺣﯿﺚ : Ci i (zi Bi ). m 1 m ln[x] 2 1 3 ( ) ( r r C ) i i . 2 i1 2 2 i 1
2 ln L 2
2
ˆ ˆ ) H(tﻟﻜ ﻞ ﻣ ﻦ داﻟ ﺔ اﻟﺼ ﻼﺣﯿﺔ ) R(tوداﻟ ﺔ ﻣﻌ ﺪل اﻟﻔﺸ ﻞ ) R(tو ﻣﻘﺪرات اﻹﻣﻜﺎن اﻷﻛﺒ ﺮ
) H(tﻟﻠﺘﻮزﯾ ﻊ اﻟﻠﻮﻏ ﺎرﯾﺘﻤﻲ اﻟﻄﺒﯿﻌ ﻲ ﺑﺎﺳ ﺘﺨﺪام ﻋﯿﻨ ﺔ ذات ﻣﺮاﻗﺒ ﺔ ﻣﺘﺘﺎﺑﻌ ﺔ ﺗﻌﻄ ﻰ ﻣ ﻦ اﻟﻤﻌ ﺎدﻟﺘﯿﻦ ) (١٨-٥و ) (١٩-٥ﻋﻠ ﻰ اﻟﺘ ﻮاﻟﻲ ﺑﻌ ﺪ اﻟﺘﻌ ﻮﯾﺾ ﻋ ﻦ ﻛ ﻞ ﻣ ﻦ 2و ﺑﻘﯿﻤ ﺔ ﻣﻘﺪرات اﻹﻣﻜﺎن اﻷﻛﺒﺮ ˆ 2و ˆ . أي أن : )(٢٥-٥
ˆ ( ˆ ln[t] ), )R(t ˆ
2 1 ln[t ] 2
e . ]ln[t ( 1 2 t )
)(٢٦-٥ ﻓﻲ ﺣﺎﻟﺔ اﻟﻨﻈﺎم اﻟﺨﺎﻟﻲ ﻣﻦ اﻟﻤﺮاﻗﺒﺔ أو اﻟﻌﯿﻨﺔ اﻟﻜﺎﻣﻠﺔ ) (complete sampleﺣﯿﺚ
٤٣٢
ˆ )H(t
ﻓﺈن داﻟﺔ اﻹﻣﻜﺎن اﻷﻛﺒﺮ وﻣﻌﺎدﻻت اﻹﻣﻜﺎن ﺗﺼﺒﺢ ﻋﻠﻰR1 R 2 ... R m 0 , n=m :اﻟﺸﻜﻞ اﻟﺘﺎﻟﻲ n
L(x; 2 , )
-
n! n 2 2
e
1 ln[x i ] 2 i 1
2
(٢٧-٥)
2
n ln L ln[x i ] ˆ 0 ˆ i1
n ln L 2 2 0 ˆ ln[x ] n i 2 i1
: ﻛﺎﻟﺘﺎﻟﻲH(t) وR(t) و و2 و ﺑﺎﻟﺘﺎﻟﻲ ﻧﺤﺼﻞ ﻋﻠﻰ ﻣﻘﺪرات اﻹﻣﻜﺎن اﻷﻛﺒﺮ ﻟـ : ( ﻓﺈن٢٥-٥) ﻣﻦ n
ln[x i ] nˆ 0 i 1
ˆ
1 n ln[x i ]. n i1
: ( ﻓﺈن٢٦-٥) ﻣﻦ n
2 2 ln[x i ] ˆ n i 1
2
1 n 2 ln[x i ] ˆ n i1
٤٣٣
2
1 n 1 n ln ln[x i ] ln[x i ] . n i1 n i1 2
وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن: ˆ ( ˆ ln[t] ), )R(t ˆ 2 1 ln[t ] 2
e . ]ln[t ( 1 2 t )
ˆ )H(t
)(٢-٤-٥ﺗﻘدﯾرات ﺑﯾﯾز ﺗﺣت ﻓرض ﺗوزﯾﻊ ﻗﺑﻠﻰ ﻣﻌﻠم ﻟﻣﻌﻠﻣﺔ اﻟﻘﯾﺎس 2ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺎت اﻟﻛﺎﻣﻠﺔ )ا(ﺗﻘدﯾرات ﺑﯾﯾز ﻟﻣﻌﻠﻣﺔ اﻟﻘﯾﺎس 2
ﺳﻮف ﻳﻔﱰض أن ﻣﻌﻠﻤﺔ اﻟﻘﻴﺎس 2ﻗﻴﻤﺔ ﳌﺘﻐﲑ ﻋﺸﻮاﺋﻲ وأن اﳌﻌﻠﻤﺔ ﻣﻌﻠﻮﻣـﺔ و ﺛﺎﺑﺘـﺔ .ﻗﻴﻤـﺔ ﻣﻔﺮوﺿـﺔ ﻣﻦ ﺑﻴﺎﻧﺎت اﺧﺘﺒﺎر ﺣﻴﺎة ﺳﺎﺑﻘﺔ أو ﺑﻴﺎﻧﺎت ﻋﻠﻰ ﻧﻔﺲ اﻟﻮﺣﺪات ﰲ ﺧـﻂ اﻹﻧﺘـﺎج ﻣﻮﺿـﻊ اﻟﺪراﺳـﺔ .ﺑﺎﺳـﺘﺨﺪام ﺗﻮزﻳـﻊ ﻣﻌﻜﻮس ﺟﺎﻣﺎ ﺑﺎﳌﻌﻠﻤﺘﲔ ) ( , ﻛﺘﻮزﻳـﻊ ﻗﺒﻠـﻲ ﻣﺮاﻓـﻖ ﻟﻠﻤﻌﻠﻤـﺔ 2 InverseGamma( , ) 2ﻳﻜـﻮن اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱄ : )(٢٨-٥
2 )( 2 ) 1 e , 2 0,( , 0
1
g( ; , ) () 2
ﻫـﺬا اﻟﺘﻮزﻳـﻊ اﺳـﺘﺨﺪم ﰲ دراﺳـﺔ ﺗﻮزﻳـﻊ واﻳﺒـﻞ ﻣـﻦ ﻗﺒـﻞ )Padgett & Johnson , Tsokos (1972 ). (1983 و ﺑﺎﺳﺘﺨﺪام داﻟﺔ اﻹﻣﻜﺎن اﳌﻌﺮﻓﺔ ﰲ ﺣﺎﻟﺔ اﻟﻌﻴﻨﺎت اﻟﻜﺎﻣﻠﺔ ) (٢٧-٥وﻣﻦ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ) (٢٨-٥ﳝﻜﻦ اﳊﺼـﻮل ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪي ﻛﻤﺎ ﻳﻠﻲ : ٤٣٤
g(2 ) L(x;, 2 ) ( | x) f (x) 2
n
1
()
2 ( 2 ) 1 e
1
() 0
2 ( 2 ) 1 e
(ln xi ) 2 1
-1
n
i 1 2 -n 2 x i e 2 i=1 n (ln x i )2 -1 1 n n i 1 2 2 x i -n e 2 d2 i=1
n
n
(ln x i ) 2 1
1 1 2 -n 2 i 1 2 ( 2) e e n (ln x i )2 1 i 1 1 1 2 -n 2 2 ( d 2 0 2 ) e e
n ( 1 ) 2 2
( )
n 2 (ln x i )2 1 i 1 2 2
e
n 2 (ln x i )2 1 i 1 2 2
n ( 1 ) 2 2
( )
e
d2
0
: ﺑﻮﺿﻊ n
2 (ln x i )2 i 1
y
22 n
2
2 (ln x i ) 2 i 1
2y n
d2
2 (ln x i )2 i 1
dy
2y 2
n ( 1 ) 2 2
( )
( 2 | x)
n 2 (ln x i )2 1 i 1 2 2
e
n
( 0
n
2 (ln x i ) 2 i 1
2y
)
n ( 1 ) 2
e y
٤٣٥
2 (ln x i ) 2 i 1
2y 2
dy
n 2 (ln x i )2 1 i 1 2 2
n ) ( 1 2
e ydy
n ) ( 2
y 0
e
n ) ( 1 2 2
) (
1 n 2 (ln x ) i 2 i1
n 2 (ln x i )2 1 i 1 2 2
n ) ( 2
n ) ( 2
e
n ) ( 1 2 2
) (
1 n 2 (ln x ) i 2 i 1
ﺑﻮﺿﻊ : 1 n n (ln x i ) 2 , * 2 i 1 2 * 2
) 2 0, * , * 0. (٢٩-٥
* 2
* 1
e
*
(2 ) 1 e ) * ( *
2
*
* *
*
( | x) 2
( 2 | x) *(* )
ﻣ ــﻦ اﳌﻌﺎدﻟ ــﺔ ) (٢٩-٥ﻳﻼﺣ ــﻆ أن اﻟﺘﻮزﻳ ــﻊ اﻟﺒﻌ ــﺪي ﻟﻠﻤﻌﻠﻤ ــﺔ 2ﻫ ــﻮ أﻳﻀ ــﺎً ﺗﻮزﻳ ــﻊ ﻣﻌﻜ ــﻮس ﺟﺎﻣ ــﺎ ﺑـ ــﺎﳌﻌﻠﻤﺘﲔ 1 n n * و (ln x i ) 2 2 i1 2
، * و ﳑــﺎ ﳚــﺪر اﻹﺷــﺎرة إﻟﻴــﻪ إﻧﻨــﺎ ﺳــﻮف ﻧﻮﺟــﺪ اﻟﺘﻘــﺪﻳﺮات
اﻟﺒﻴﺰﻳــﺔ ﰲ ﺣﺎﻟــﺔ اﻟﻌﻴﻨــﺔ اﻟﻜﺎﻣﻠــﺔ ،وﻛﻨــﺎ ﻧــﻮد اﺳــﺘﺨﺪام ﻋﻴﻨــﺔ ﻣﺘﺘﺎﺑﻌــﺔ ﻣــﻦ اﻟﻨــﻮع اﻟﺜــﺎﱐ ،و ﻟﻜــﻦ ﻋﻨــﺪ ﺣﺴــﺎب اﻟﺘﻮزﻳــﻊ اﻟﺒﻌﺪي وﺟﺪﻧﺎ أﻧﻪ ﻻ ﻳﻨﺘﻤﻲ ﻟﻨﻔﺲ اﻟﻌﺎﺋﻠﺔ اﻟﱵ ﻳﻨﺘﻤﻲ ﳍﺎ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ اﻧﻪ ﻏﲑ ﻣﺮاﻓﻖ وﻟﺬﻟﻚ اﻗﺘﺼﺮت اﻟﺪراﺳﺔ ﻋﻠـﻰ اﻟﻌﻴﻨـﺎت اﻟﻜﺎﻣﻠـﺔ .ﻧﻔـﺲ اﳌﺸـﻜﻠﺔ ﺻـﺎدﻓﺖ اﻟﺒﺎﺣـﺚ ) Calabria and PuLcini (1992ﻋﻨـﺪ دراﺳـﺘﻪ ﻟﺘﻮزﻳﻊ ﻣﻌﻜـﻮس واﻳﺒـﻞ ﻓﻠـﻢ ﻳﺴـﺘﻄﻊ إﳚـﺎد ﺗﻮزﻳـﻊ ﻗﺒﻠـﻲ ﻣﺮاﻓـﻖ ﻟﻠﻌﻴﻨـﺎت اﳌﺘﺘﺎﺑﻌـﺔ واﻛﺘﻔـﻰ ﺑﺪراﺳـﺔ ﺗﻮزﻳـﻊ ﻗﺒﻠـﻲ ﻻ ﻣﻌﻠﻤـﻲ ﻟﻠﻌﻴﻨـﺎت اﳌﺘﺘﺎﺑﻌـﺔ .ﺑﻴﻨﻤــﺎ اﺳـﺘﻄﺎع ) AL- Ohali (2006و AL- Hussaini and Jaheen ) (1994إﳚﺎد اﻟﺘﻘﺪﻳﺮات اﻟﺒﻴﺰﻳﺔ ﰲ ﺣﺎﻟﺔ اﻟﻌﻴﻨﺎت اﳌﺘﺘﺎﺑﻌﺔ . ﰲ اﳉﺰء اﻟﺘﺎﱄ ﺳﻮف ﳓﺼﻞ ﻋﻠﻰ ﻣﻘﺪر ﺑﻴﻴﺰ ﳌﻌﻠﻤﺔ اﳌﻘﻴﺎس 2ﲢﺖ ﻓﺮض دوال ﺧﺴـﺎرة ﳐﺘﻠﻔـﺔ .ﻫـﺬا وﳝﻜـﻦ اﺳ ــﺘﺨﺪام اﻻﺳ ــﻠﻮب اﳌﺴ ــﺘﺨﺪم ﰱ اﳚ ــﺎد ﻣﻘ ــﺪرات ﺑﻴﻴ ــﺰ ﻟﺪاﻟ ــﺔ اﻟﺼ ــﻼﺣﻴﺔ وداﻟ ــﺔ اﻟﻔﺸ ــﻞ ﻛﻤ ــﺎ ﺟ ــﺎء ﰱ رﺳ ــﺎﻟﺔ اﳌﺎﺟﺴﺘﲑ. ٤٣٦
)ا(ﻣﻘﺪر ﺑﻴﻴﺰ اﳌﻀﺒﻮط ﲢﺖ ﻓﺮض داﻟﺔ ﺧﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺎ
d 2
* 2
* 1
e
*
2
E[ ] * * 2 ( ) 0 2
d 2 .
* 2
*
e
2 sq
*
*
( ) 2
*
0
)(٣٠-٥ *
*
* 2 و d 2 dy ﺑﻮﺿﻊ y 2و ﺑﺎﻟﺘﺎﱄ ﻓﺈن y y 2
و ﺑﺎﻟﺘﻌﻮﻳﺾ ﰲ ) (٣٠-٥ﻓﺈن : * dy y2
* y
e
* y
*
*
(* ) 0
2
E[ ]
* * 2 y e y dy * ( ) 0 * ) (* 1 * ) ( * * ) (* 1 * )( 1) ( 1 * * E[ ] , )( 1 2
و ﺑﺎﻟﺘﺎﱄ ﻓﺈن ﻣﻘﺪر ﺑﻴﻴﺰ ﳌﻌﻠﻤﺔ اﳌﻘﻴﺎس 2ﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة ﻣﺮﺑﻊ اﳋﻄﺄ ﻫﻮ : * * . )( 1 2 sq
)ب( ﻣﻘﺪر ﺑﻴﻴﺰ ﳌﻌﻠﻤﺔ اﳌﻘﻴﺎس 2ﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة اﻻﻧﱰوﺑﻴﺎ اﳌﻌﻤﻤﺔ: 1 q
٤٣٧
2 q
2 int E
* 1
2 int E 2
*
*
q
*
*
*
*
2 * e 2 ( 2 ) q * * d 2 ( ) 0
2 (q 1) e ( )
2
( ) 0
d 2 .
: ﺑﻮﺿﻊ * y 2
: ﻓﺈن
* * 2 d 2 dy y y 2
*
*
* (q * 1) y * ( ) e dy , (* ) 0 y y2
*
q
*
*
( )
*
q 1 y (y) e dy , 0
q
*
( )
(q * ) ,
: ﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة اﻷﻧﱰوﺑﻴﺎ اﳌﻌﻤﻤﺔ ﻫﻮ2 ﺑﺎﻟﺘﺎﱄ ﻓﺈن ﻣﻘﺪر ﺑﻴﻴﺰ ﳌﻌﻠﻤﺔ اﳌﻘﻴﺎس
1
* q q 2 * int (q ) . (* )
: (LINEX) ﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة اﳋﻄﻴﺔ اﻷﺳﻴﺔ2 )ج( ﻣﻘﺪر ﺑﻴﻴﺰ ﳌﻌﻠﻤﺔ اﳌﻘﻴﺎس 2 1 2 inx ln E[e c ] . c
*
* 1
2 inx
E[e
c 2
] e c 0
2
2 * e 2 d 2 * * ( ) * *
e ( )
* c2 2
*
0
٤٣٨
2
*
( ) ( 1) d 2 ,
(٣١-٥)
ﺑﺎﻻﺳﺘﻔﺎدة ﻣﻦ داﻟﺔ ﺑﺴﻞ اﻟﱵ ﻋﻠﻰ اﻟﺸﻜﻞ : K (r-1) (2 ab) .
r 1 2
a dx 2 b
b ax x
e
r
x 0
ﻋﻨﺪﻣﺎ * r * 1, a c, b ﻓﺈن ) (٣١-٥ﺗﺼﺒﺢ ﻋﻠﻰ اﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ: .
وﺑﺎﻟﺘﺎﱄ ﻓﺈن :
*
*
c 2 2 ) * K (* ) (2 c * * ( ) *
* * * 1 c 2 * ln K (* ) (2 c ) . 2 c (* ) *
2 inx
دراﺳﺔ اﶈﺎﻛﺎة Simulation Study
ﻟﻠﻮﻗﻮف ﻋﻠﻰ ﻣﺪى ﻛﻔﺎءة وﺳﻠﻮك ﻃﺮق اﻟﺘﻘﺪﻳﺮ اﻟﱵ ﰎ اﻟﻮﺻﻮل ﻋﻠﻴﻬـﺎ ﰲ ﻫـﺬا اﳉـﺰء ﻻﺑـﺪ ﻣـﻦ ﻋﻤـﻞ ﻣﻘﺎرﻧـﺎت ﻋﺪدﻳــﺔ ﺑﺎﺳــﺘﺨﺪام دراﺳــﺔ اﶈﺎﻛــﺎة ﺣﻴــﺚ ﻳــﺘﻢ ﺗﻮﻟﻴــﺪ ﻋﻴﻨــﺔ ﻋﺸـﻮاﺋﻴﺔ ﻣــﻦ اﳊﺠــﻢ nﺗﺘﺒــﻊ اﻟﺘﻮزﻳــﻊ اﻟﻠﻮﻏــﺎرﲤﻲ اﻟﻄﺒﻴﻌــﻲ ﺑﺎﺗﺒﺎع اﳋﻄﻮات اﻵﺗﻴﺔ : ) (١ﻷي ﻗﻴﻤﺘ ــﲔ ﻟﻠﻤﻌﻠﻤﺘ ــﲔ ) (,و ) ( > 0, > 0ﻳ ــﺘﻢ ﺗﻮﻟﻴ ــﺪ 2ﻣ ــﻦ ﺗﻮزﻳ ــﻊ ﺟﺎﻣ ــﺎ اﻟﻌﻜﺴ ــﻲ اﻟـ ـﻮارد ﰲ اﳌﻌﺎدﻟﺔ ).(٢٨-٥ ) (٢ﺑﺎﺳﺘﺨﺪام ﻗﻴﻤﺔ 2اﻟﻨﺎﲡـﺔ ﻣـﻦ ) (١و ﻣﻌﻠﻮﻣﻴـﺔ اﳌﻌﻠﻤـﺔ ﻳـﺘﻢ ﺗﻮﻟﻴـﺪ ﳎﻤﻮﻋـﺎت ﻣـﻦ اﻟﺒﻴﺎﻧـﺎت ذات أﺣﺠـﺎم ﳐﺘﻠﻔﺔ ﻣﻦ اﻟﺘﻮزﻳﻊ اﻟﻠﻮﻏﺎرﲤﻲ اﻟﻄﺒﻴﻌﻲ . (10)50 وذﻟﻚ ﺑﺎﺳﺘﺨﺪام ﻧﻈﺎم ﺗﻮﻟﻴﺪ اﻟﻌﻴﻨﺎت ذات اﳌﺮاﻗﺒﺔ اﳌﺘﺘﺎﺑﻌﺔ وﺑﺎﺳﺘﺨﺪام ﺑﺮﻧﺎﻣﺞ ﻳﺘﻢ إﻋﺪادﻩ ﳍﺬا اﻟﻐﺮض . ﰲ ﻫﺬا اﳌﺜﺎل وﺑﺎﻋﺘﺒﺎر ) 2.25ﻣﻌﻠﻮﻣﺔ( ﺳﻴﺘﻢ ﺗﻮﻟﻴﺪ ﻗﻴﻢ ﳐﺘﻠﻔﺔ ﻟﻠﻤﻌﻠﻤﺔ 2اﻋﺘﻤﺎد ﻋﻠـﻰ ﻗـﻴﻢ ﳐﺘﻠﻔـﺔ ﳌﻌـﺎﱂ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ) (,ﻟﻠﻤﻌﻠﻤﺔ 2ﻛﺎﻵﰐ : ) (١ﺑﺎﺧﺘﻴـ ــﺎر اﻟﻘـ ــﻴﻢ ) (, )=(9,1),,(10,1),(11,1),(9,2),(10,2),(11,2ﰎ ﺗﻮﻟﻴـ ــﺪ ﺳـ ــﺘﺔ ﻗـ ــﻴﻢ ﻟﻠﻤﻌﻠﻤ ـ ـ ـ ـ ـ ـ ـ ــﺔ 2ﺑﺎﺳ ـ ـ ـ ـ ـ ـ ـ ــﺘﺨﺪام ﺗﻮزﻳ ـ ـ ـ ـ ـ ـ ـ ــﻊ ﺟﺎﻣـ ـ ـ ـ ـ ـ ـ ـ ــﺎ اﻟﻌﻜﺴ ـ ـ ـ ـ ـ ـ ـ ــﻲ ﻓﻜﺎﻧ ـ ـ ـ ـ ـ ـ ـ ــﺖ ﻋﻠ ـ ـ ـ ـ ـ ـ ـ ــﻰ اﻟﱰﺗﻴ ـ ـ ـ ـ ـ ـ ـ ــﺐ )(٢
2 0.0930109,0.15663,0.12688,0.172181,0.185025,0.179918. (2 , ) (0.0930109, 2.25),(0.15663,2.25),(0.12688,2.25), ﰎ ﺗﻮﻟﻴ ــﺪ ﻟﻘــﻴﻢ (0.172181,2.25),(0.185025,2.25),(0.179918,2.25).
ﳎﻤﻮﻋﺎت ﻣﻦ اﻟﺒﻴﺎﻧﺎت ذات اﳊﺠﻢ ). n=10(50 ٤٣٩
)(٣
ﺑﺎﺳﺘﺨﺪام اﻟﻨﺘﺎﺋﺞ ﰲ ﰎ ﺣﺴﺎب ﻣﻘﺪرات اﻹﻣﻜﺎن اﻷﻛﱪ ﻟﻠﻤﻌﻠﻤﺔ MLو . ML
)(٤ )(٥
ﻣﻘﺪرات ﺑﻴﻴﺰ اﳌﺨﺘﻠﻔﺔ ﰎ ﺣﺴﺎ ﺎ ﳌﻌﻠﻤﺔ اﻟﺸﻜﻞ . 2 اﳋﻄ ـﻮات اﻟﺴــﺎﺑﻘﺔ ﻣــﻦ ) (١إﱃ ) (٥ﰎ ﺗﻜﺮارﻫــﺎ N 1000ﻣــﺮﻩ و ﰎ ﺣﺴــﺎب ﻣﺘﻮﺳــﻂ اﻟﺘﻘــﺪﻳﺮ ) (AVو ﻣﺘﻮﺳﻂ ﻣﺮﺑﻊ اﳋﻄﺄ ) (MSEﻟﻜﻞ ﺣﺎﻟﺔ ﻣﻦ ﺣﺎﻻت ﺣﺠﻢ اﻟﻌﻴﻨﺔ ﺣﻴﺚ : 2
i N
N
MSE i 1
i AV i 1 N N
,
ﺣﻴﺚ ﲤﺜﻞ اﳌﻌﻠﻤﺔ ،ﲤﺜﻞ اﳌﻘﺪر . ) (٦ﰎ اﻋﺘﺒــﺎر ﻗــﻴﻢ اﳌﻌــﺎﱂ اﳌﺴــﺘﺨﺪﻣﺔ ﰲ ﺗﻮﻟﻴــﺪ اﻟﺒﻴﺎﻧــﺎت و اﻟــﺪوال اﳌﻌﺘﻤــﺪة ﻋﻠﻴﻬــﺎ ﻫــﻲ اﻟﻘــﻴﻢ اﻟﺼــﺤﻴﺤﺔ true valuesﻟﻠﻤﻌﺎﱂ و اﻟﻨﺘﺎﺋﺞ اﻟﻌﺪدﻳﺔ ﻟﻠﺤﺴﺎﺑﺎت ﰎ ﻋﺮﺿﻬﺎ ﰱ اﳉﺪاول اﻟﺘﺎﻟﻴﺔ: ﻣﻦ اﳊﺠﻢ n=10 ˆ li 0.108647 )(0.00010415 0.211341
ˆ int 0.0794765 )(0.000309492 0.154931
)(0.00134569 0.122054 )(0.000379084 0.202669
)(0.000691752 0.0910037 )(0.00461683 0.15134
(0.00143392 0.105328 )(0.000189408 0.184616
)(0.00159009 0.0798214 )(0.0025859 0.140105
(0.00112254
)(0.00192993
ˆ sq 0.108897 )(0.000489507 0.212283 )(0.00234816 0.122351 )(0.00173543 0.203471 )(0.00116285 0.105532 )(0.00088222 0.185234 )(0.000631114
٤٤٠
ˆ ML 0.0784758 )(0.00170063 0.0143794
)(0.00541244 0.135181 )(0.00459072 0.16043 )(0.00668462 0.110152 )(0.00326042 0.147243 )(0.00601216
) (, )(9, 1 )(9,2 )(10,1 )(10,2 )(11,1 )(11,2
n=20
(, )
ˆ ML
ˆ int
ˆ li
(9, 1)
0.0862957
ˆ sq 0.104778
(9,2)
(0.000854259) 0.160646
(0.000396112) 0.202594
(0.000269803) 0.159565
(10,1)
(0.00271308) 0.144278
(0.00174491) 0.13079
(0.000667797) 0.104149
(0.0012335) 0.130544
(10,2)
(0.00255872) 0.172934
(0.00135819) 0.198747
(0.00319207) 0.158264
(0.000434896) 0.19819
(11,1)
(0.00340466) 0.117434
(0.00111611) 0.110297
(0.00130474) 0.104149
(0.00137617) 0.130544
(0.00161124) 0.167083
(0.000724654) 0.185733
(0.00319207) 0.149388
(0.000434896) 0.185272
(0.00312005)
(0.000795962)
(0.00142512)
(0.00113556)
ˆ int
ˆ li
(11,2)
0.0825242
0.104613 (0.0000968908) 0.201984
n=30
(, )
ˆ ML
(9, 1)
0.0889398
(9,2)
(0.000601277) 0.164182
(10,1)
(0.00190742) 0.149646
(10,2)
(0.00181175) 0.174009
(11,1)
(0.00241778) 0.121656
(11,2)
(0.00111271) 0.168834 (0.00220439)
ˆ sq 0.102529 (0.000343757) 0.19585 (0.00135996) 0.136961 (0.00108784) 0.194094 (0.00100279) 0.11429 (0.000599336) 0.183085 (0.000779926) ٤٤١
0.0845821
0.102407
(0.000243329) 0.161568
(0.0000929052) 0.195408
(0.000656907) 0.1138
(0.00115551) 0.13675
(0.00231833) 0.161272
(0.000475868) 0.193678
(0.00119996) 0.0955973
(0.00131546) 0.11415
(0.00136927) 0.15314
(0.000222249) 0.182729
(0.0012557)
(0.00110557)
n=40
(, ) (9, 1)
0.0881042
(9,2)
(0.000405069) 0.164698
(10,1)
(0.00151922) 0.151816
(10,2)
(0.00131461) 0.178481
(11,1)
(0.00167176) 0.123204
ˆ sq 0.0994079 (0.000240066) 0.190597 (0.00109967) 0.140625 (0.000877889) 0.193611 (0.000860005) 0.116389
(11,2)
(0.000721703) 0.173769
(0.000464539) 0.183971
(0.00105241) 0.158121
(0.000229043) 0.183674
(0.0016212)
(0.000729982)
(0.00100223)
(0.001115)
ˆ int
ˆ li
ˆ ML
ˆ int
0.0846061
ˆ li
0.0993148
(0.000214896) 0.162217
(0.000087045) 0.190255
(0.000650191) 0.120293
(0.00109549) 0.140444
(0.00177532) 0.165618
(0.000499071) 0.193271
(0.000952209) 0.100035
(0.00130555) 0.11627
n=50
(, )
ˆ ML
(9, 1)
0.0900798
(9,2)
(0.000326625) 0.166934
ˆ sq 0.0992093 (0.000223454) 0.188277
(10,1)
(0.00111628) 0.152065
0.0863554
0.0991311
(0.000184489) 0.163883
(0.0000866091) 0.187997
(0.000892079) 0.142358
(0.000548439) 0.12438
(0.00106643) 0.142201
(10,2)
(0.000986908) 0.179083
(0.000732338) 0.191807
(0.00144362) 0.167585
(0.000508991) 0.191524
(11,1)
(0.00132919) 0.126169
(0.000755173) 0.119631
(0.000845707) 0.1004896
(0.00127996) 0.119523
(11,2)
(0.000684085) 0.17436
(0.000430318) 0.182974
(0.000833835) 0.160437
(0.000242328) 0.182724
(0.0013101)
(0.000670562)
(0.000887872)
(0.00110207)
٤٤٢
اﻟﺘﻌﻠﻴﻖ ﻋﻠﻰ اﻟﻨﺘﺎﺋﺞ : ﺑﺼﻮرة ﻋﺎﻣﻪ ﻓﺈن ﻣﻘﺪرات ﺑﻴﻴﺰ أﻓﻀﻞ ﻣﻦ ﻣﻘﺪرات اﻹﻣﻜﺎن وذﻟﻚ ﻋﻨﺪﻣﺎ n 30وﻫﺬا ﻣﺎ ﻛﺎن ﻣﺘﻮﻗﻊ .
ﺑﺰﻳﺎدة ﺣﺠﻢ اﻟﻌﻴﻨﺔ ﻳﻘﻞ ﻣﺘﻮﺳﻂ ﻣﺮﺑﻊ اﳋﻄﺄ وﻫﺬا ﻣﺎ ﻛﺎن ﻣﺘﻮﻗﻊ . ﻋﻨﺪ اﳌﻘﺎرﻧﺔ ﺑﲔ دوال اﳋﺴـﺎرة ﰲ اﻟﻐﺎﻟـﺐ ﺗﻜـﻮن داﻟـﺔ اﳋﺴـﺎرة اﻻﻧﱰوﺑﻴـﺎ اﳌﻌﻤﻤـﺔ ﳍـﺎ أﻗـﻞ ﻣﺘﻮﺳـﻂ ﻣﺮﺑـﻊ ﺧﻄﺄ ﺑﻴﻨﻤﺎ داﻟﱵ اﳋﺴﺎرة اﳌﺮﺑﻌﺔ واﻵﺳﻴﺔ اﳋﻄﻴﺔ ﻣﺘﻘﺎرﺑﺘﲔ ﺟﺪاً ﰲ ﺳﻠﻮﻛﻬﻤﺎ . وﻧﻼﺣﻆ أﻧﻪ ﻋﻨﺪ ﺛﺒﺎت إذا ﻛﺎﻧﺖ 1ﻓﺈن اﻟﺘﻘﺪﻳﺮ أﻓﻀﻞ ﻣﻨﻪ ﻋﻨﺪﻣﺎ 2و ﻧﺴﺘﺨﻠﺺ ﻣﻨﻪ اﻧﻪ ﻛﻠﻤﺎ ﻗﻠﺖ ﻗﻴﻤﺔ ﻓﺈن اﻟﺘﻘﺪﻳﺮ ﻳﺼﺒﺢ أﻓﻀﻞ ) ﻋﻼﻗﺔ ﻋﻜﺴﻴﺔ ( . ﳒﺪ أﻧﻪ ﻋﻨﺪ ﺛﺒﺎت و ﺗﻐﲑ ﻓﺈن أﻓﻀﻞ ﺗﻘﺪﻳﺮ ﻳﻜﻮن ﻋﻨﺪﻣﺎ 9ﰒ 11و أﺧﲑاً 10و ﺑﺎﻟﺘﺎﱄ ﻓﺈﻧﻪ ﻋﻨﺪ ﺗﺘﻐﲑ ﻗﻴﻤﺔ ﻓﺈن دوال اﳋﺴﺎرة ﻟﻴﺲ ﳍﺎ ﺳﻠﻮك ﻣﻌﲔ .
وﳌﺰﻳﺪ ﻣﻦ اﳌﻌﻠﻮﻣﺎت ﳝﻜﻦ اﻟﺮﺟﻮع اﱃ رﺳﺎﻟﺔ اﳌﺎﺟﺴﺘﲑ اﳋﺎﺻﺔ ﺑﺎﻟﺒﺎﺣﺜﺔ. )ب(ﺗﻘدﯾر ﺑﯾﯾز ﻟداﻟﺔ اﻟﺻﻼﺣﯾﺔ ﺑﺎﺳﺗﺧدام ﺗﻘرﯾﯾب ﺗﯾرﻧﻰ وﻛﺎدﯾن ﯾﻤﻜ ﻦ اﺳ ﺘﺨﺪام طﺮﯾﻘ ﺔ ﺗﯿﺮﻧ ﻲ وﻛ ﺎدﯾﻦ واﳌﻘﺪﻣـﺔ ﻣـﻦ ) Tiernsy and Kadane (1986وذﻟـﻚ ﻻﳚﺎد ﻣﻘﺪر ﺑﻴﻴﺰ ﻷي داﻟﺔ ﰲ ) 2اى داﻟﺔ ﰱ ﻣﻌﻠﻤﺔ واﺣﺪة( ﻋﻠﻰ اﻟﺼﻴﻐﺔ اﻟﺘﺎﻟﻴﺔ : d 2 .
* 2
* 1
e
u( 2 ) * E[u( 2 )] ( * ) * 2 0
)(٣٢-٥
ﺣﻴﺚ ان اﻟﺘﻜﺎﻣﻞ ) (٣٢-٥ﻻ ﳝﻜﻦ ﺣﻠﻪ ﺑﺎﻟﻄﺮق اﻟﺘﺤﻠﻴﻠﻴﺔ .ﻓﻰ اﻟﺑﻧد اﻟذى ﯾﻠﯾﻪ ﺳوف ﻧﻬﺗم ﺑﻬذا اﻟﺗﻘرﯾب ﻓﻰ وﺟود ﻣﻌﻠﻣﺗﯾن.
ﺗﻘرﯾب ﺗﯾرﻧﻰ وﻛﺎدﯾن
The Tierney-Kadane Approximate
ﻧظرﯾـﺔ:
٤٤٣
ﻋﻧدﻣﺎ ﺗﻛون nﻛﺑﯾرة ﺑدرﺟﺔ ﻛﺎﻓﯾﺔ ٕواذا ﻛﺎن اﻟﺗوزﯾﻊ اﻟﺑﻌدي ﻟﻠداﻟﺔ )) u(ﯾﻌطﻰ
ﺑﯾﺎﻧﺎت( ﺗﺗرﻛز ﻋﻠﻲ ﻧﺻف ﺧط اﻷﻋداد اﻟﻣوﺟب )أو اﻟﺳﺎﻟب( ﺣﯿﺚ ) L(ھﻮ ﻟﻮﻏﺎرﯾﺘﻢ داﻟﺔ اﻹﻣﻜﺎن ) L(x | و ) (ھﻮ ﻟﻮﻏﺎرﯾﺘﻢ اﻟﺘﻮزﯾﻊ اﻟﻘﺒﻠﻰ ) g(ﻓﺈن L g ﺗﺗرﻛز
ﺣول ﻗﯾﻣﺔ ﻋظﻣﻰ وﺣﯾدة ﻓﺈن اﻟﺗﻛﺎﻣل اﻟﻣﻌرف ﻓﻲ ) (٣٢-٥ﯾﻣﻛن اﻟﺗﻌﺑﯾر ﻋﻧﻪ ﺑﺎﻟﺻورة اﻵﺗﯾﺔ:
d
) n *(
e
E[u() | x]
. d
) n (
e
)(٣٣-٥
ﺣﯾث :
1 1 L , * () ln u L n n
()
وﻟﻤﺎ ﻛﺎن ﻓﻰ ﻛﺜﻴﺮ ﻣﻦ اﻟﺤﺎﻻت ﻣﻦ اﻟﺼﻌﺐ اﻟﺤﺼﻮل ﻋﻠﻰ ﺷﻜﻞ ﻣﺤﺪد ﻟﻨﺘﻴﺠﺔ ﻫﺬﻩ اﻟﺘﻜﺎﻣﻼت وﻟذﻟك
اﻟﻣﻌﺎدﻟﺔ ) (٣٣-٥ﯾﻣﻛن ﻛﺗﺎﺑﺗﻬﺎ ﺑﺎﺳﺗﺧدام ﺷﻛل ﺗﻘرﯾﺑﻰ ﺑﺻﯾﻐﺔ ﺗﯾرﻧﻰ وﻛﺎدﯾن ) ﻓﻰ ﺣﺎﻟﺔ
وﺟود ﻣﻌﻠﻣﺔ واﺣدة ﻧﻬﺗم ﺑﻬﺎ( ﻋﻠﻰ اﻟﺻورة اﻟﺗﺎﻟﯾﺔ :
* n e . *
*
uˆ BT
)(٣٤-٥
ﺣﯾث :
* ﻫﻲ اﻟﻘﯾﻣﺔ اﻟﻌظﻣﻰ ﻟـ * و ﻫو ﻣﻧوال اﻟﺗوزﯾﻊ اﻟﺑﻌدي واﻟﻘﯾﻣﺔ اﻟﻌظﻣﻰ ﻟـ
و:
2 2* *2 2 , 2 2
*
٤٤٤
اﻟﺘﻮزﯾﻊ اﻟﺒﻌﺪي ﻓﻰ ) (٢٩-٥ﳛﻘﻖ اﻟﺸﺮط ﻟﺘﻄﺒﻴﻖ ﻫﺬﻩ اﻟﻄﺮﻳﻘﺔ وﻫﻮ اﻧﻪ آﺣ ﺎدي اﻟﻤﻨ ﻮال وﻣﻌ ﺮف ﻋﻠ ﻰ اﻟﺠﺰء اﻟﻤﻮﺟﺐ ﻟﺨﻂ اﻷﻋ ﺪاد .ﻟﺘﻄﺒﻴـﻖ ﺻـﻴﻐﺔ ﺗﻘﺮﻳـﺐ ﺗـﲑﱐ و ﻛـﺎدﻳﻦ ﻹﳚـﺎد ﻣﻘـﺪرات ﺑﻴﻴـﺰ ﻷي داﻟـﺔ ﰲ 2 ﻧﺘﺒﻊ اﻵﰐ : ﻟﻮﻏﺎرﻳﺘﻢ داﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﻘﺒﻠﻲ ل 2ﻫﻮ :
2 ( 2 ) 1 e ,
1
( ) = ln g( ) () 2
2
و ﻟﻮﻏﺎرﻳﺘﻢ داﻟﺔ اﻹﻣﻜﺎن ﻫﻮ :
2
] ,
n
1 ln[x i ] ln 2 i 1
-
e
!n n 2 2
[ln L(x | 2 , ) ln
2
وﳊﺴﺎب ﺻﻴﻐﺔ ﺗﯿﺮﻧﻲ وﻛﺎدﯾﻦ اﻟﻤﻌﺮﻓﺔ ﻓﻲ ) (٣٤-٥ﻧﺘﺒﻊ اﳋﻄﻮات اﻟﺘﺎﻟﻴﺔ : * ;n ( 1)ln 2 , 2
*
)(٣٥-٥ *
* *, 2
n* ln u (* 1)ln 2
)(٣٦-٥ وﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﻘﯿﻤﺔ اﻟﺘﻰ ﺗﻌﻈﻢ اﻟﺪاﻟﺔ ﻓﻰ ) (٣٥-٥ﻧﺘﺒﻊ اﻟﺘﺎﻟﻰ : 2
* (* 1) 4, 2 2 * *( 1) 0 4 0. 2 2
* * )( 1 2
ﺣﯿﺚ
2
ھﻰ ﻗﯿﻤﺔ اﻟﺘﻰ ﺗﻌﻈﻢ اﻟﺪاﻟﺔ ﻓﻰ ).(٣٥-٥ 2
وﻟﻠﺤﺼﻮل ﻋﻠﻰ ﻧﺤﺴﺐ اوﻻ : 2
2
2
ﺣﯿﺚ : ٤٤٥
2 2
2
(* 1) * 2 3, 2 واﻟﺘﻌﻮﯾﺾ ﻋﻦ : ﻓﻰ اﻟﻤﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﻧﺤﺼﻞ ﻋﻠﻰ 2
2
2
2 2 2
(v* 1) * 2 * * 2 ( * ) ( * )3 (v 1) (v 1) : ﻛﻤﺎ ﯾﻠﻰ
*
2
وذﻟﻚ ﻟﻠﺤﺼﻮل ﻋﻠﻰ 2
واﻟﺘﻰ ﻧﺴﺎوﯾﮭﺎ ب
*
( 1) 2 , * * ( * )2 ( * )3 ( 1) (v 1)
(v* 1)3 * (v* 1)3 2 , *2 *2 2
(* 1)3 2* (* 1)3 , *2 2
(1 2 * )(* 1)3 , *2 2
*2 , (1 2 * )(* 1)2 2
1 * 2
(1 2 ) (* 1) . *
2
: ( ﻧﺘﺒﻊ اﻟﺘﺎﻟﻰ٣٦-٥) ﻓﻰ اﻟﺘﻰ ﺗﻌﻈﻢ اﻟﺪاﻟﺔ وﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﻘﯿﻤﺔ *
* u (* 1) * 4, 2 u 2
(٣٧-٥) : ( ﺑﺎﻟﺼﻔﺮ ﻧﺤﺼﻞ ﻋﻠﻰ٣٧-٥) ﺑﻤﺴﺎواة * u (* 1) * 0 4 0 2 u 2 ٤٤٦
اﻟﻤﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ اﻣﻜﻦ ﺣﻠﮭﺎ ﻋﺪدﯾﺎ ﺑﺎﺳﺘﺨﺪام اﻟﺤﺎﺳﺐ اﻻﻟﻰ وذﻟﻚ ﺑﺎﺳﺘﺨﺪام ﺑﺮﻧﺎﻣﺞ Mathematicaﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﻘﯿﻤﺔ
2
اﻟﺘﻰ ﺗﻌﻈﻢ اﻟﺪاﻟﺔ . وﻟﻠﺤﺼﻮل ﻋﻠﻰ ﻧﺤﺴﺐ اوﻻ : *
*
2
* 2
2
ﺣﯿﺚ :
)u u u2 (* 1 * 2 4, u2 4
2
* 2
2
)(٣٨-٥ ﺣﯿﺚ u,uاﻟﻤﺸﺘﻘﺔ اﻻوﻟﻰ واﻟﺜﺎﻧﯿﺔ ﻟﻠﺪاﻟﺔ ) u(2ﺑﺎﻟﻨﺴﺒﺔ ل . 2وھﻨﺎ ﺳﻮف ﻧﻘﺘﺼﺮ ﻋﻠﻰ ﺣﺎﻟﺔ ﺧﺎﺻﺔ وذﻟﻚ ﺑﻮﺿﻊ )) u R(tوﻟﻜﻦ ﯾﻤﻜﻦ ﺗﻄﺒﯿﻘﮭﺎ ﻻى داﻟﺔ ﻓﻰ 2ﻣﺜﻼ ﻓﻰ اﯾﺠ ﺎد ﻣﻘﺪر ﻟﺪاﻟﺔ اﻟﻔﺸﻞ( وﻋﻠﻰ ذﻟﻚ ﻓﺈن اﻟﻤﺸﺘﻘﺎت اﻟﺠﺰﺋﯿﺔ ﻟـ uﺗﻜﻮن :
,
)(ln[t] 3 2
2
واﻟﺘﻌﻮﯾﺾ ﻋﻦ
2
2
) 2 (ln[ t ] 3 e 2
(ln[t] ) .
(ln[ t] )2 2
2
e
u
2 2
) 1 (ln2 u 7 e 8 2
) (ln[t ] 2 2 (ln[t] ) e 2
2
ﻓﻰ ) (٣٨-٥ﺑﺎﻟﻘﯿﻤﺔ ﻧﺤﺼﻞ ﻋﻠﻰ : 2
* 2 2 2
* 2
) ( وذﻟﻚ ﻟﻠﺤﺼﻮل ﻋﻠﻰ واﻟﺘﻰ ﻧﺴﺎوﯾﮭﺎ ب وذﻟﻚ ﺑﺎﺳﺘﺨﺪام ﺑﺮﻧﺎﻣﺞ . Mathematica
*
2 *
.ھﺬا و ﻗﺪ ﺗﻢ اﻟﺤﺼﻮل ﻋﻠﻰ ﻋﺪدﯾﺎ ﺑﺎﺳﺘﺨﺪام اﻟﺤﺎﺳﺐ اﻻﻟﻰ
٤٤٧
وﻓﯿﻤﺎ ﯾﻠﻰ ﺧﻄﻮات اﳊﺼﻮل ﻋﻠﻰ ﻣﻘﺪر ﺑﻴﻴﺰ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﻴﺔ ﲢﺖ ﻓﺮض داﻟـﺔ اﳋﺴـﺎرة ﻣﺮﺑـﻊ اﳋﻄـﺄ ﺑﺎﺳـﺘﺨﺪام : (٣٤-٥) ﺻﯿﻐﺔ ﺗﯿﺮﻧﻲ وﻛﺎدﯾﻦ * n ( ) ( ) R(t) * u BT e E[R(t)] * * 2 0 ( ) *
2
2
* 1
* 2
e d2 .
* n * ( 2 ) ( 2 ) e * * [ ln[u]( * 1)ln[ 2 ] 2 ( * 1)ln[ 2 ] 2 ] 2 2 2 2
*
1 * 2
e
(1 2 ) (* 1) *
** 1 * 2
* * [ln[u ] ( * 1)ln[ 2 ] 2 ( * 1)ln[ 2 ] 2 ]
e
(1 2 ) (* 1)
** 1 * 2
* * [ln[u ] ( * 1)ln[ 2 ] 2 ( * 1)ln[ * ] ( 1)
e
* ] * ( * 1)
(1 2 ) (v* 1) * *
1 * 2
* * [ln[u] ( * 1)ln[ 2 ] 2 ( * 1)ln[ * ]( * 1)] ( 1)
e
(1 2 ) (* 1)
٤٤٨
* * * [ln[ 2 ] 2 ( * 1) ln ] ]1 ( 1)
* *
e
1 * 2
)(1 2 ) (* 1
) (٥-٥ﺗوزﯾﻊ ﺑﯾﯾر اﻻﺛﻧﻰ ﻋﺷر )(١-٥-٥ﺗﻘدﯾرات اﻻﻣﻛﺎن اﻻﻛﺑر ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺎت اﻟﻣراﻗﺑﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ٤ﻗﺪم ﻫﺬا اﻟﺒﺤﺚ ﻣﻦ ﻗﺒﻞ ) L- Hussaini & Jaheen, (1994وﻗﺪ ﰎ ﺗﻘﺪﳝﻪ ﺑﺸﻜﻞ ﻣﻔﺼﻞ ﻛﺎﻟﺘﺎﱃ : ﺗﻢ ﺗﺨﺼﯿﺺ ھﺬا اﻟﺒﻨﺪ ﻹﯾﺠﺎد ﻣﻘﺪرات اﻹﻣﻜﺎن اﻷﻛﺒﺮ و ﻣﻘﺪرات ﺑﯿﯿﺰ اﻟﺘﻘﺮﯾﺒﯿﮫ ﻟﻤﻌﻠﻤﺘﻲ ﺗﻮزﯾﻊ ﺑﯿﯿﺮ اﻻﺛﻨﻰ ﻋﺸﺮ و ﻛﺬﻟﻚ داﻟﺘﻲ اﻟﺼﻼﺣﯿﺔ و ﻣﻌﺪل اﻟﻔﺸﻞ ﻟﻨﻔﺲ اﻟﺘﻮزﯾﻊ ،و ذﻟﻚ ﺑﺎﺳﺘﺨﺪام اﻟﻤﻌﺎﯾﻨﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﻧﻰ .
إذا ﻛﺎن ﻟﺪﻳﻨﺎ اﺧﺘﺒﺎر اﻟﺤﻴﺎة ﻣﻦ ﻋﻴﻨﺔ ﻣﺮاﻗﺒﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﻧﻲ ،ﻓﺈن داﻟﺔ اﻹﻣﻜﺎن ﻷول rﻣﻦ وﺣﺪات اﻟﻔﺸﻞ اﻟﻤﺮﺗﺒﺔ ﺗﻈﻬﺮ ﻣﻦ ﻋﻴﻨﺔ ﻣﻜﻮﻧﺔ ﻣﻦ nﻣﻦ اﻟﻮﺣﺪات ﺗﻜﻮن :ﺑﺎﻋﺘﺒﺎر أن y1 y 2 y rﺗﻢ اﻟﺤﺼﻮل ﻋﻠﻴﻬﺎ واﻟﻤﻄﻠﻮب اﻳﺠﺎد ﻣﻘﺪرات اﻻﻣﻜﺎن اﻻﻛﺒﺮ ﻟﻤﻌﻠﻤﺘﻲ ﺗﻮزﯾﻊ ﺑﯿﯿﺮ اﻻﺛﻨﻰ ﻋﺸﺮ و ﻛﺬﻟﻚ داﻟﺘﻲ اﻟﺼﻼﺣﯿﺔ و ﻣﻌﺪل اﻟﻔﺸﻞ ﻟﻨﻔﺲ اﻟﺘﻮزﯾﻊ . ،داﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﺗﻲ : r !n f (y i )[1 F(y r )]n r . (n r)! i 1
L(y1 , y 2 ,..., y n | c, k) L
)(٣٩-٥ و ) F(.) , f(.ھﻤﺎ داﻟﺘﻲ اﻟﻜﺜﺎﻓﺔ و اﻟﺘﻮزﯾﻊ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ . ﺗﻮزﯾﻊ ﺑﯿﯿﺮ اﻟﺜﺎﻧﻰ ﻋﺸﺮ ﻟﮫ داﻟﺘﻲ اﻟﻜﺜﺎﻓﺔ و اﻟﺘﻮزﯾﻊ اﻵﺗﯿﺘﯿﻦ : ;f (x;c, k) ckx c1 (1 x c ) (k 1) , x 0, c 0, k 0 F(x;c, k) 1 (1 x c ) k , x 0.
داﻟﺔ اﻟﺼﻼﺣﯿﺔ ﻟﺘﻮزﯾﻊ ﺑﯿﯿﺮ اﻟﺜﺎﻧﻰ ﻋﺸﺮ ھﻲ : ٤٤٩
R(t) (1 t c ) k , t 0.
: وداﻟﺔ ﻣﻌﺪل اﻟﻔﺸﻞ ھﻲ H(t)
kct c 1 , t 0. (1 t c )
: ﻓﺈن داﻟﺔ اﻹﻣﻜﺎن ﻓﻲ ﺣﺎﻟﺔ اﻟﻤﻌﺎﯾﻨﺔ ﻣﻦ اﻟﻨﻮع اﻟﺜﺎﻧﻰ ﺗﺄﺧﺬ اﻟﺼﯿﻐﺔ اﻟﺘﺎﻟﯿﺔ L(y | c, k) L
r n! f (y i )[1 F(yr )]n r . (n r)! i 1
: داﻟﺔ اﻹﻣﻜﺎن ﻓﻲ ﺣﺎﻟﺔ ﺗﻮزﯾﻊ ﺑﯿﯿﺮ اﻟﺜﺎﻧﻰ ﻋﺸﺮ ﺗﺄﺧﺬ اﻟﺼﯿﻐﺔ اﻟﺘﺎﻟﯿﺔ L
n r n! r r c r k r yci 1 (1 yic ) (k 1) (1 y cr ) k (n r)! i 1 i1
nr r n! yic1 (1 ycr ) k c r k r c (k 1) (n r)! i 1 (1 yi ) nr r n! yci 1 (1 y cr ) k c r k r c c k (n r)! i 1 (1 yi )(1 yi ) nr r y c1 r n! c r k r i c (1 yic ) k (1 ycr ) k (n r)! i 1 (1 yi ) i 1
٤٥٠
r
r
r ln(1 yic ) k ln(1 yci ) !n yic1 k (n r) r r i1 e i1 c k e c !)(n r (1 y ) i 1 i r k ln(1 yic ) k (n r) ln(1 yrc ) !n r r c k v(c; y) e i1 !)(n r !n c r k r v(c; y)e kT , !)(n r r r y c1 v(c; y) i c , T T(c | y) (n r) ln(1 ycr ) ln(1 yic ). i 1 i 1 (1 y i )
ﻟﻮﻏﺎرﯾﺘﻢ داﻟﺔ اﻹﻣﻜﺎن ھﻮ : ln L r ln k r ln c ln v(c; y) kT.
)(٤٠-٥
ﺑﺎﻓﺘﺮاض ان cﻣﻌﻠﻮﻣﺔ ﻓﺈن ﻣﻘﺪر اﻻﻣﻜﺎن اﻻﻛﺒﺮ ل kﯾﻜﻮن ﻛﺎﻟﺘﺎﻟﻰ : r kˆ ML . T
إذا ﻛﺎن ﻛﻼ اﻟﻤﻌﻠﻤﺘﯿﻦ c,kﻣﺠﮭﻮﻟﺘﯿﻦ ﻓﺈن ﺗﻘﺪﯾﺮ اﻻﻣﻜﺎن اﻻﻛﺒﺮ ﻟﻜﻞ ﻣﻦ kو cﯾﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﯿﮫ و ذﻟﻚ ﺑﺎﻟﺘﻔﺎﺿﻞ اﻟﺠﺰﺋﻲ ﻟﻠﻤﻌﺎدﻟﺔ ) (٤٠-٥ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻤﺔ cو ﻛﺬﻟﻚ ﺑﺎﻟﺘﻔﺎﺿﻞ اﻟﺠﺰﺋﻲ ﻟﻠﻤﻌﺎدﻟﺔ ) (٤٠-٥ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻤﺔ kﻣﻊ اﻟﻤﺴﺎواة ﺑﺎﻟﺼﻔﺮ ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﻣﻌﺎدﻟﺘﻲ اﻹﻣﻜﺎن اﻟﺘﺎﻟﯿﺘﯿﻦ : ln L ln L 0, 0 c k
)(٤١-٥
٤٥١
وﺑﺎﺳ ﺘﺨﺪام أﺣ ﺪ اﻟﻄ ﺮق اﻟﻌﺪدﯾ ﺔ ﻛﻄﺮﯾﻘ ﺔ )أﺳ ﻠﻮب ( ﻧﯿ ﻮﺗﻦ راﻓﺴ ﻮن newton raphson methodوﯾﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﻰ ﺗﻘ ﺪﯾﺮات اﻹﻣﻜ ﺎن اﻷﻛﺒ ﺮ ﻣ ﻦ اﻟﻤﻌ ﺎدﻟﺘﯿﻦ اﻟﺴ ﺎﺑﻘﺘﯿﻦ )-٥ .(٤١ ﺗﻘﺪﯾﺮات اﻹﻣﻜﺎن اﻷﻛﺒﺮ
ﻟﻜﻞ ﻣﻦ داﻟﺔ اﻟﺼﻼﺣﯿﺔ ) R(tوداﻟﺔ ﻣﻌﺪل اﻟﻔﺸﻞ ) H(tﻟﺘﻮزﯾﻊ ﺑﯿﯿﺮ اﻟﺜﺎﻧﻰ
ﻋﺸﺮ ﺗﻌﻄﻰ اﻟﺘﻌﻮﯾﺾ ﻋﻦ ﻛﻞ ﻣﻦ kو cﺑﻘﯿﻤﺔ ﻣﻘﺪرات ﺑﺎﻹﻣﻜﺎن اﻷﻛﺒﺮ ˆ cˆ ,kﻓﻰ ﻛﻞ ﻣﻦ اﻟﺪاﻟﺘﯿﻦ .أي أن ﻣﻘﺪر اﻻﻣﻜﺎن اﻻﻛﺒﺮ ﻟﺪاﻟﺔ اﻟﺼﻼﺣﯿﺔ ھﻮ :
ˆ (1 t cˆ ) kˆ , t 0. )R(t
وﻣﻘﺪر اﻻﻣﻜﺎن اﻻﻛﺒﺮ ﻟﺪاﻟﺔ ﻣﻌﺪل اﻟﻔﺸﻞ ھﻮ : ˆ ˆ cˆ 1 kct ˆ H(t) , t 0. ) ˆ(1 t c
)(٢-٥-٥ﺗﻘدﯾرات ﺑﯾﯾز ﻓﻰ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺎت اﻟﻣراﻗﺑﺔ ﻣن اﻟﻧوع اﻟﺛﺎﻧﻰ ﻓﻰ ھﺬا اﻟﺠﺰء ﺳﻮف ﯾﻔﺘﺮض ﺗﻮزﯾﻊ ﻗﺒﻠﻰ ﻣﺸﺘﺮك ﻟﻤﻌﻠﻤﺘﻲ اﻟﻤﻮﻗﻊ و اﻟﻘﯿﺎس ﯾﻌﻄﻰ ﺑﺎﻟﺼﻮرة اﻟﺘﺎﻟﯿﺔ : g(k, c) g1 (k | c)g 2 (c), kc
c 1 k e , ( 1), 0, 1 ( 1) c
g1 (k | c)
1 1 g 2 (c) c e , 0, 0, ()
اى ان :
٤٥٢
kc
g(k, c)
c
c 1 1 1 k e c e ( 1) 1 () k 1 c( )
c k e ( 1) () 1
Rc k k e
k 1 c( )
, 1
R ( 1) () 1 , ( 1), 0, 0, 0
(٤٢-٥) : ﻓﻰ ﻣﻌﺎﻟﻢ ﺗﻮزﯾﻊ ﺑﯿﯿﺮ اﻟﺜﺎﻧﻰ ﻋﺸﺮ ﳝﻜﻦ ﻛﺘﺎﺑﺘﻪ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﺘﺎﱃu(c,k) اﻟﻤﻘﺪر اﻟﺒﯿﯿﺰى ﻻى داﻟﺔ
E(u(c,k))
0
u(c,k)q(c,k | y)dcdk,
0
(٤٣-٥)
: وﯾﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﯿﮫ ﺑﺎﺗﺒﺎع اﻟﺨﻄﻮات اﻟﺘﺎﻟﯿﺔc,k اﻟﺘﻮزﯾﻊ اﻟﺒﻌﺪى لq(c, k | y) ﺣﯿﺚ
q(c, k | y) L(y | c, k)g(c,k) r
r
c k v(c; y)e
kT
c
cr k r v(c; y)e c
r
k
r
v(c; y)e
k e
k 1 c( )
k 1 kT c( ) c c k (T )
,c 0, k 0 (٤٤-٥) : ﻟﻮﻏﺎرﻳﺘﻢ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﺗﻌﻄﻰ ﻛﺎﻟﺘﺎﱃ
٤٥٣
)Q(c, k | y) ln q(c,k | y r
( r)ln c +( r)ln k (c 1) ln yi i 1
r
c c ln(1 yic ) k(T ) . i1 )(٤٥-٥ وﻟﻤﺎ ﻛﺎﻧﺖ اﻟﺼﯿﻐﺔ ﻓﻰ ) (٤٣-٥ﺻﻌﺐ اﳊﺼﻮل ﻋﻠﻴﻬﺎ ﻓﺴﻮف ﻧﺴﺘﺨﺪم ﺗﻘﺮﻳﺒﱭ ﳊﺴﺎ ﻤﺎ .
)أ( ﺗﻘرﯾب ﻟﻧدﻟﻰ ﺑﺎﺳﺘﺨﺪام ﺗﻘﺮﻳﺐ ﻟﻨﺪﱃ ﺳﻮف ﻳﺴﺘﺨﺪم اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى اﳌﺸﱰك ل ) (٤٤-٥) (c, kﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺗﻘﺮﻳﺐ ﻟﻠﻤﻘﺪر اﻟﺒﻴﻴﺰى ﰱ ) (٤٥-٥ﻟﻜﻞ ﻣﻦ اﳌﻌﺎﻟﻠﻤﺘﲔ c,kو داﻟﺔ اﻟﺼﻼﺣﻴﺔ وداﻟﺔ اﻟﻔﺸﻞ ﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة اﳌﺮﺑﻌﺔ .ﺑﺼﻮرة ﻋﺎﻣﺔ ﻟﻠﺪاﻟﺔ ) u u( و (1 , 2 ,..., k ) وﲢﺖ ﻓﺮض داﻟﺔ اﳋﺴﺎرة اﳌﺮﺑﻌﺔ ﻓﺈن ﻣﻘﺪر ﺑﻴﻴﺰ ﻻى داﻟﺔ ﰱ ﰱ ) (٤٣-٥ﻳﺎﺧﺬ اﻟﺸﻜﻞ اﻟﺘﺎﱃ :
d
) L( ) (
u()e
E u()
. ) L( ) ( e d
)(٤٦-٥ ﺣﯿ ﺚ ) L( ھ ﻮ ﻟﻮﻏ ﺎرﯾﺘﻢ داﻟ ﺔ اﻹﻣﻜ ﺎن ) L(y | و ) ( ھ ﻮ ﻟﻮﻏ ﺎرﯾﺘﻢ اﻟﺘﻮزﯾ ﻊ اﻟﻘﺒﻠ ﻰ ) . g( اﻟﻤﻌﺎدﻟﺔ ) (٤٦-٥ﻳﻤﻜﻦ وﺿﻌﻬﺎ ﻋﻠﻰ اﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ :
d
) Q(
u()e
E u( )
. d
) Q(
e
)(٤٧-٥ ﺣﻴﺚ : ٤٥٤
Q( ) L( ) ( ) ln L(y | ) g( ) .
(٤٨-٥) : ﺗﻜﻮن ﻋﻠﻰ اﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ (1 , 2 ) ( ﻓﺈن ﺻﻴﻐﺔ ﺗﻘﺮﻳﺐ ﻻﻧﺪﱃ ﰱ ﺣﺎﻟﺔ ﻣﻌﻠﻤﺘﲔ٤٧-٥) ﻟﻠﺼﻴﻐﺔ ﰱ
uˆ B u( )
1 A Q30 B12 Q21C12 Q12C21 Q03B21 , 2 (٤٩-٥) : ﺣﻴﺚ
Q A u ijij ,Q , , 0,1, 2,3, 3,i, j 1, 2, 1 2 i 1 j1 2
2
u 2u for i, j 1,2, u i , u ij and for i j. i i j Bij (u i ii u j ij )ii ,Cij 3u i ii ij u j (ii jj 2ij2 ), :( ﰱ ﻣﻌﻜﻮس اﳌﺼﻔﻮﻓﺔij) اﻟﻌﻨﺼﺮij و
Q* (Q*ij ), i, j 1,2, 2Q Q . i j * ij
(1 , 2 ) (c, k) وﰱ ﺣﺎﻟﺘﻨﺎ ﻓﺈن. ( اﳌﻨﻮال ﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى 1 , 2 ) ( ﺗﻘﺪر ﻋﻨﺪ٤٩-٥) اﻟﺼﻴﻐﺔ ( ﳓﺼﻞc ,k ) ﻣﻨﻮال اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى واﻟﺬى ﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ.(٤٥-٥) اﳌﻌﻄﺎﻩ ﰱQ Q(c, k | y) و D
D
: ﻋﻠﻴﻪ ﲝﻞ اﳌﻌﺎدﻟﺘﲔ 0
Q r c (T ), k k
(٥٠-٥) ٤٥٥
r Q r r 1 1 dT 0 ln yi - yic a i - -k( ) , c k dc i 1 i 1
)(٥١-٥ ﺣﻴﺚ : ln yi r ,i 1,,...,r, k . c c 1 yi T
ai
ﺑﺎﻟﺘﻌﻮﻳﺾ ﻋﻦ ﻗﻴﻤﺔ kاﻟﱴ ﳓﺼﻞ ﻋﻠﻴﻬﺎ ﻣﻦ ) (٥٠-٥ﰱ ) (٥١-٥ﳓﺼﻞ ﻋﻠﻰ اﳌﻌﺎدﻟﺔ اﻟﻐﲑ ﺧﻄﻴﺔ اﻟﺘﺎﻟﻴﺔ ﰱ cاﻟﺘﺎﻟﻴﺔ : h(c) 0. ﺣﻴﺚ : r r r 1 (r+) 1 dT h(c) ln yi - yic a i -( ), c (c+ )T dc i1 i1 وﻗﺪ اﺛﺒﺖ اﻟﺒﺎﺣﺜﺎن اﻧﻪ ﻻى ﻓﺌﺔ ﻣﻦ ﻗﻴﻢ yiﻓﺈن ) h(cﻣﺘﻨﺎﻗﺼﺔ ﺑﺎﻃﺮاد ﰱ ) (0, وﺗﻘﻄﻊ اﶈﻮر اﻻﻓﻘﻰ ﻣﺮة واﺣﺪة وﻟﺘﻜﻦ * cﲝﻴﺚ ان . h(c* ) 0وﻋﻠﻰ ذﻟﻚ ﻣﻦ ) (٥٠-٥ﻓﺈن ) (c ,kﻫﻮ اﳌﻨﻮال اﻟﻮﺣﻴﺪ
D
D
ﻟﻠﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﰱ )(٤٤-٥
اﻻن ﻟﺘﻄﺒﻴﻖ ﺻﻴﻐﺔ ﻻﻧﺪﱃ ﰱ ) (٤٩-٥اوﻻ ﳓﺴﺐ اﻟﻘﻴﻢ ij ,i, j 1,2واﻟﱴ ﲤﺜﻞ ﻋﻨﺎﺻﺮ ﻣﻌﻜﻮس اﳌﺼﻔﻮﻓﺔ Q* (Q*ij ),i, j 1, 2,ﺣﻴﺚ :
2Q Q , i j * ij
2Q r r c 2 d 2T Q 2 - y i a i - k 2 , c c2 dc i1 * 11
d 2T r c 2 = yi a i (n r)y rc a r2 , 2 dc i1 ٤٥٦
: ﺑﻮﺿﻊ
d 2T r c 2 M=k 2 yi a i , dc i 1 r c 2 r r
M= (n r)ky a (k 1)N, N yci a i2 , i 1
r -(k+1)N-k(n r)ycr a 2r , 2 c r (r ) * * Q11 M,Q , 12 c2 k2 r r * 1 * Q 22 2 ,Q12 yci a i (n r)ycr a r B, k i 1 * Q11
r 1 dT 1 c B= = (n r)y r a r yic a i , dc i 1
: وﻋﻠﻰ ذﻟﻚ
(r ) c2 * Q B
1
B 11 (r ) 12 k 2
12 . 22 : اﶈﺪد ﻟﻠﻤﺼﻔﻮﻓﺔ اﻟﺴﺎﺑﻘﺔ ﻫﻮ
c 2 k 2 det(Q ) 2 2 2 , B k c (r )(c 2 M ) *
٤٥٧
(r )
11
1 2 (r )(Mc ) c2 (r )c2 (r )c 2 , B2 k 2 c 2 (r )(Mc 2 ) D B2 k 2
D= c2 k 2 det Q* (r )(r Mc 2 ) (Bck) 2 .
B
12
(r )(M B2
( ) c2
k2
Bk 2 c 2 Bk 2c 2 , B2 k 2c 2 (r )(Mc2 ) D
( ) 2 Bk 2c 2 c 22 ( ) D (r )(M 2 c B2 2 k M
٤٥٨
Mk 2 c2 k 2 ( ) 22 B2 k 2 c 2 ( )(Mc 2 ) k 2 (Mc 2 ) B2 k 2 c2 ( )(Mc 2 ) k 2 (Mc2 ) . D
: وﺑﺎﻻﺿﺎﻓﺔ اﱃ ذﻟﻚ
Q12 0,Q 21 (N (n r)y rc a r2 1 M (k 1)N (N ) M N, k k 2(r ) 2(r ) M Q03 ,Q , 30 k3 c2 c
ﲝﻴﺚuˆ B ( وذﻟﻚ ﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﺘﻘﺪﻳﺮ٤٩-٥) ﺑﺎﻟﺘﻌﻮﻳﺾ ﻋﻦ اﻟﻘﻴﻢ اﻟﺴﺎﺑﻘﺔ اﳌﺬﻛﻮرة اﻋﻼﻩ ﰱ ﺻﻴﻐﺔ ﻻﻧﺪﱃ . (c D ,k D ) ( ﺗﻘﺪر ﻋﻨﺪ ﻣﻨﻮال اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى٤٩-٥) ان ﻛﻞ اﻟﺪوال ﰱ : (ﳓﺼﻞ ﻋﻠﻰ٤٩-٥) ﰱu u(c, k) ﻟﻠﺪاﻟﺔ W 1 uˆ B E u(c,k | y) u cW1u1 kW2 u 2 , 2D 2D 2 (٥٢-٥)
٤٥٩
W c 2 (r )u11 Bk 2 (u12 u 21 ) k 2 (r Mc 2 )u 22 , W1 (r ) 2 2(r ) sc3 3(M N)Bkc3 (r ) 2Bkc(r )(r Mc2 ), W2 2(r )(r Mc 2 )2 Bck(r ) [2(r ) sc3 ] 2B2 k 2c 4 (M N) c 2 (r )(M N)(r Mc2 ).
أي. i 1,2,...,m ( ﺣﻴﺚc D ,k D ) ( ﺗﺤﺴﺐ ﻋﻨﺪ ﻣﻨﻮال اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى٥٢-٥) ﻛﻞ اﻟﺪوال ﻓﻰ
. ﻋﻠﻰ اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪيu ( ﻫﻲ ﺻﻴﻐﺔ ﺗﻘﺮﻳﺒﻴﺔ ﻟﻠﺘﻮﻗﻊ ﻟﻠﺪاﻟﺔ٥٢-٥) أن
ﺣﺎﻻت ﺧﺎﺻﺔ
: ( ﻓﺈن٥٢-٥) ﰱu u(c, k) c اذا ﻛﺎﻧﺖ W c BL c 1 12 , 2D
k BL
. (c D ,k D ) واﳌﻘﺪرة ﻋﻨﺪ
: ( ﻓﺈن٥٢-٥) ﰱu u(c,k) k اذا ﻛﺎﻧﺖ W c 1 22 , 2D . (c D ,k D ) واﳌﻘﺪرة ﻋﻨﺪ
: ( ﻓﺈن٥٢-٥) ﰱu u(c,k) R(t) (1 t c ) k اذا ﻛﺎن
R BL
z1 kt c ln t R(t) 1 k k BL ln(1 t c ), c c BL c (1 t ) 2D . (c D ,k D ) واﳌﻘﺪرة ﻋﻨﺪ : ﺣﻴﺚ
٤٦٠
2
ln t z1 c [(r )kt (kt 1) c 1 t c
c
2
2 t c ln t 2 2 c 2Bk k ln(1 t ) 1 k (r c )M ln(1 t ) c ]. 1 t ckt c1 u u(c,k) H(t) ﰱ ) (٥٢-٥ﻓﺈن : اذا ﻛﺎن ) (1 t c c2 z 2 1 ln t 1 H BL H(t) 1 c c k k BL , BL c 2D c (1 t ) k ﻋﻨﺪ اﻟﻘﻴﻤﺔ ) . (c D ,k D ﺣﻴﺚ : c
2
2 2ln t ln t 1 ln t c z 2 (r ) 1 t 2kB c c c 1 t c . c(1 t ) (1 t )
)ب(ﺗﻘرﯾب ﺗﯾرﻧﻰ وﻛﺎدﯾن The Tierney-Kadane Approximate ﯾﻤﻜﻦ اﺳﺘﺨﺪام طﺮﯾﻘﺔ ﺗﯿﺮﻧﻲ وﻛﺎدﯾﻦ واﳌﻘﺪﻣﺔ ﻣﻦ )Tiernsy and Kadane (1986 وذﻟﻚ ﰱ وﺟﻮد اﻛﺜﺮ ﻣﻦ ﻣﻌﻠﻤﺔ. ﻧظرﯾـﺔ:
ﻋﻧدﻣﺎ ﺗﻛون nﻛﺑﯾرة ﺑدرﺟﺔ ﻛﺎﻓﯾﺔ ٕواذا ﻛﺎن اﻟﺗوزﯾﻊ اﻟﺑﻌدي ﻟﻠداﻟﺔ )u(
و ) ) (, ,..., mﯾﻌطﻰ ﺑﯾﺎﻧﺎت( ﺗﺗرﻛز ﻋﻠﻲ ﻧﺻف ﺧط اﻷﻋداد اﻟﻣوﺟب )أو اﻟﺳﺎﻟب( ﺣﯿﺚ ) L( ھﻮ ﻟﻮﻏﺎرﯾﺘﻢ داﻟﺔ اﻹﻣﻜﺎن و ) ( ھﻮ ﻟﻮﻏﺎرﯾﺘﻢ اﻟﺘﻮزﯾﻊ اﻟﻘﺒﻠﻰ ) g( ﻓﺈن L g ﺗﺗرﻛز ﺣول ﻗﯾﻣﺔ ﻋظﻣﻰ وﺣﯾدة ﻓﺈن اﻟﺗﻛﺎﻣل اﻟﻣﻌرف ﻓﻲ ) (٤٦-٥ﯾﻣﻛن
اﻟﺗﻌﺑﯾر ﻋﻧﻪ ﺑﺎﻟﺻورة اﻵﺗﯾﺔ:
* n
d
e
. d
n
e
)(٥٣-٥ ٤٦١
E[u( ) | x]
ﺣﯾث : 1 1 L g , * ln u L g n n 1 1 Q( ) , * ln u Q( ) ,Q( ) L g . n n
وﺑﺎﺗﺒﺎع ) Tiernsy and Kadane (1986اﻟﻣﻌﺎدﻟﺔ ) (٥٣-٥ﯾﻣﻛن ﻛﺗﺎﺑﺗﻬﺎ ﺑﺻورة
ﺗﻘرﯾﺑﯾﺔ ﻋﻠﻰ اﻟﺻورة اﻟﺗﺎﻟﯾﺔ :
1 2
*
det n . u BT e det *
*
)(٥٤-٥
ﺣﯾث :
* ﻫﻲ اﻟﻘﯾﻣﺔ اﻟﻌظﻣﻰ ﻟـ * * , ﻫﻲ ﻣﻧوال اﻟﺗوزﯾﻊ اﻟﺑﻌدي واﻟﻘﯾﻣﺔ اﻟﻌظﻣﻰ ﻟـ و
* , ﻫﻣﺎ ﻣﻌﻛوس ﺳﺎﻟب اﻟﻣﺻﻔوﻓﺎت ﻟﻠﺗﻔﺎﺿل اﻟﺛﺎﻧﻰ ﻟﻛل ﻣن * , ﻋﻧد * , ﻋﻠﻰ
اﻟﺗواﻟﻰ .و det * ,det ﳘﺎ اﶈﺪدان ﻟﻠﻤﺼﻔﻮﻓﺘﺎن . * , وﻋﻠﻰ ذﻟﻚ u BTﰱ ﻫﺬﻩ اﳊﺎﻟﺔ ﻫﻰ ﺗﻘﺮﻳﺐ ﺗﲑﱏ وﻛﺎدﻳﻦ ﻟﻠﺪاﻟﺔ . u ﻟﺘﻮزﻳﻊ ﺑﻴﲑ اﻟﺜﺎﱏ ﻋﺸﺮ ﻓﺈن * , ﻋﻠﻰ اﻟﺘﻮاﱃ ﳘﺎ :
)(٥٥-٥
c c r ln c (r )ln k ln v(c | y) k ( T) , B 1 * ln u(c,k) . n
وﻋﻠﻰ ذﻟﻚ ) (٥٤-٥ﳝﻜﻦ ﻛﺘﺎﺑﺘﻬﺎ ﻋﻠﻰ اﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ ؛ 1 2
*
det n c ,k c,k uˆ BT . e det *
*
*
)(٥٦-٥
اﻟﺘﻘﺪﻳﺮ اﻟﺴﺎﺑﻖ ) (٥٦-٥ﳛﺘﺎج اﱃ ﺣﺴﺎب اﻟﻘﻴﻢ اﻟﺘﺎﻟﻴﺔ : ﳊﺴﺎب det ﳓﺴﺐ اﻟﻘﻴﻢ اﻟﺘﺎﻟﻴﺔ ٤٦٢
1 r 1 M , B, 2 n n n c ck 1 r . 2 n k 2 k
1 (r ) M n c2 B n
c 2 n(r ) D Bc 2 k 2 n D
B n (r ) nk 2
1
, 2 k n(cM r ) D Bc 2 k 2 n
c 2 k 2 n 2 (ckn) 2 det 2 2 . D Bc k (r )(c2 M r )
(٥٧-٥)
(c D ,k D ) واﻟذى ﯾﻘدر ﻋﻧد اﻟﻣﻧوال ﻟﻠﺗوزﯾﻊ اﻟﺑﻌدى
ﻳﻜﻮن ﻟﻪc,k وﳚﺐ اﻟﺘﺬﻛﲑ ان اﺳﺘﺨﺪام ﻫﺬﻩ اﻟﻄﺮﻳﻘﺔ ﻳﺘﻄﻠﺐ ان اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻟﻠﻤﻌﻠﻤﺘﲔ ﰱ اﳉﺰء اﻟﺘﺎﱃ ﱂ ﻧﻔﻚ اﳌﻌﺎدﻻت ﺑﺎﻟﺘﻔﺼﻴﻞ. ﻣﻨﻮال وﺣﻴﺪ وﻗﺪ ﲢﻘﻖ ﻟﻠﺘﻮزﻳﻊ ﲢﺖ اﳌﻨﺎﻗﺸﺔ . وﻛﻦ ﻧﱰﻛﻬﺎ ﻟﻠﻘﺎرئ ﻟﻜﻰ ﻳﺘﻌﻠﻢ ﻓﻚ اﳌﻌﺎدﻻت ﺣﺎﻻت ﺧﺎﺻﺔ (٥٦-٥) ﰱu u(c, k) c اذا ﻛﺎن
c BT
* c
1 2
det n * c1* ,k 1* c D ,k D . e det ٤٦٣
: ﻓﺈن
: ﺣﯾث det *c
c2k 2 n 2 , pp
pp (r )(c 2 M r ) B2c 2 k 2 , . u u(c,k) c ( ﺣﻴﺚ٥٥-٥) * ( اﳌﻨﻮال لc1* ,k 1* ) واﻟﱴ ﺗﻘﺪر ﻋﻨﺪ اﻟﻘﻴﻤﺔ : ( ﻓﺈن٥٦-٥) ﰱu u(c,k) k اذا ﻛﺎن * k
1 2
* * * k det en c 2 ,k 2 c D ,k D . BT det
det *k
: ﺣﯾث
c2 k 2n 2 , pp
pp (r )(c 2 M R r ) (BR ck)2 ,
: ﺣﯾث
u u(c,k) k ( ﺣﻴﺚ٥٥-٥) * ( اﳌﻨﻮال لc1* ,k 1* ) واﻟﱴ ﺗﻘﺪر ﻋﻨﺪ اﻟﻘﻴﻤﺔ
: ( ﻓﺈن٥٦-٥) ﰱu u(c,k) R(t) (1 t c ) k اذا ﻛﺎن
٤٦٤
* R
1 2
det n * cˆ *3 ,kˆ *3 cˆ D ,kˆ D R BT (t) . e det
det *R
nck 2
r r M R c 2 BR ck
: ﺣﯾث ,
2 c 2 c ln t M R k 1 N k n r x r a r t , c 1 t r 1 ln t BR (n r) x rca r x ica i t c , c 1 t i 1
u u(c,k) R(t) ( ﺣﻴﺚ٥٥-٥) * ( اﳌﻨﻮال لc *3 ,k *3 ) واﻟﱴ ﺗﻘﺪر ﻋﻨﺪ اﻟﻘﻴﻤﺔ : ( ﻓﺈن٥٦-٥) ﰱu u(c, k) H(t) * H
, اذا ﻛﺎن
1 2
det n * cˆ *4 ,kˆ *4 cˆ D ,kˆ D H BT (t) . e det det *H
ckt c 1 1 t c
nck 2 r 1 r 1 M H c 2 Bck 2
: ﺣﻴﺚ ,
M H k 1 N k(n r)x cr a r2
. u u(c,k) H(t) ( ﺣﻴﺚ٥٥-٥) * ( اﳌﻨﻮال لc *4 ,k *4 ) واﻟﱴ ﺗﻘﺪر ﻋﻨﺪ اﻟﻘﻴﻤﺔ
( ﺗﻧﺑؤ اﻟﻌﯾﻧﺗﯾن ﻟﺑﯾﯾز ﺗﺣت ﻓرض ﺗوزﯾﻊ ﺑﺎرﯾﺗو٦-٥) ٤٦٥
) (١-٦-٥ﻣﻌﻠﻣﺔ اﻟﺷﻛل ﻏﯾر ﻣﻌﻠوﻣﺔ وﺣﺟم اﻟﻌﯾﻧﺔ ﺛﺎﺑت ﻫﺬا اﻟﺒﻨﺪ ﻣﺎﺧﻮذ ﻣﻦ ﻗﺒﻞ ) AL-Ohaly(2000ﻣﻦ ﺧﻼل رﺳﺎﻟﺔ اﳌﺎﺟﺴﺘﲑ ﻣﻨﺎل داوود اﺑﺮاﻫﻴﻢ اﻟﻌﻮﻫﻠﻰ ﲢﺖ اﺷﺮاف اﻟﺪﻛﺘﻮر اﲪﺪ اﺑﻮ ا ﺪ ﺳﻠﻴﻤﺎن اﻳﻀﺎ ) Soliman and AL-Ohaly (2000وﺳﻮف ﻧﻘﺪﻣﻪ ﻛﻤﺎ ﻫﻮ وذﻟﻚ ﺣﱴ ﳕﻜﻦ اﻟﻘﺎرئ ﻣﻦ اﻟﺘﺪرب ﻋﻠﻰ ﻓﻚ اﳌﻌﺎدﻻت ﻣﻦ ﺧﻼل ﻣﺎ اﻛﺘﺴﺒﻪ ﻣﻦ اﻻﲝﺎث اﻟﺴﺎﺑﻘﺔ اﻟﱴ ﺗﻨﺎوﻟﻨﺎﻫﺎ ﰱ ﻫﺬا اﻟﻔﺼﻞ . ﺑﻔـ ــﺮض أن nﻣـ ــﻦ اﻟﻮﺣ ـ ــﺪات ﻣـ ــﻦ اﻟﻌﻴﻨـ ــﺔ اﻻوﱃ وﺿ ـ ــﻌﺖ ﻟﻼﺧﺘﺒـ ــﺎر وان اﻻﺧﺘﺒ ـ ــﺎر ﻳﻨﺘﻬـ ــﻲ ﺑﻌـ ــﺪ ﻓﺸ ـ ــﻞ rﻣـ ــﻦ اﻟﻮﺣ ـ ــﺪات .اى أن X1 , X 2 , , X nﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﲤﺜﻞ أزﻣﻨﺔ اﻟﻔﺸﻞ وان أزﻣﻨﺔ اﻟﻔﺸﻞ ﺗﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﺎرﻳﺘﻮ ﺣﻴﺚ :
)f (x) x -( +1
;,>0, >0,x
)(٨٥-٥
,>0, >0,x
F(x) 1 x
)(٨٦-٥ ﺣﻴﺚ ﻣﻌﻠﻤﺔ اﳌﻘﻴﺎس و ﻣﻌﻠﻤﺔ اﻟﺸﻜﻞ. ﺑﺎﻋﺘﺒﺎر ﻣﻌﻠﻮﻣﻴﺔ ﻣﻌﻠﻤﺔ اﻟﻘﻴﺎس وﻋﺪم ﻣﻌﻠﻮﻣﻴﺔ ﻣﻌﻠﻤﺔ اﻟﺸﻜﻞ و ﺑﻔﺮض ان اﻟﻌﻴﻨﺔ اﻻوﱃ ذات اﳊﺠﻢ nﻣﻌﻠﻮم ﻋﻨﻬﺎ اﳌﺸﺎﻫﺪات . y y1 y 2 y rداﻟﺔ اﻹﻣﻜﺎن ﺗﻌﻄﻰ ﻛﺎﻵﰐ : r !n f (y i )[1 F(y r )]n r (n r)! i 1
L(, | y) L
r !n ) (n r r y [ yi (1) ] n . r !)(n r i 1
)(٨٧-٥ ﺑﺎﺳﺘﺨﺪام ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﲟﻌﻠﻤﺘﲔ ) (c,dﻛﺘﻮزﻳﻊ ﻗﺒﻠﻰ ﻣﺮاﻓﻖ ﻟﻠﻤﻌﻠﻤﺔ اى ان ) Gamma(c,dواﻟﺬى ﻳﺎﺧﺬ اﻟﺸﻜﻞ : ٤٦٦
c 1e dd c ( ) )(c
0 , c,d 0 .
)(٨٨-٥ ﺑﺎﺳﺘﺨﺪام ﻧﻈﺮﻳﺔ ﺑﻴﻴﺰ واﳌﻌﺎدﻻت ) (٨٧-٥و) (٨٨-٥ﻳﻜﻮن اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻟﻠﻤﻌﻠﻤﺔ ﻋﻠﻰ اﻟﺼﻮرة اﻟﺘﺎﻟﻴﺔ :
) ( 1
r y e yr i 1 ) ( 1 r n (n r) d yr e yr d i 1 (n r ) d r
n
(r c) 1
(r c)1
( | y)
0
) (d u)r c e (d u , )(r c xi x n ln r . xr
(r c) 1
r
u ln i 1
)(٨٩-٥ ﻳﻼﺣﻆ ان اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻟﻠﻤﻌﻠﻤﺔ ( | y) ﻫﻮ اﻳﻀﺎ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﺑﺎﳌﻌﻠﻤﺘﲔ ) (d u),(r cاى ان ). Gamma((r c),(d u ﺑﻔﺮض ان z z1 z 2 z rﻋﻴﻨﺔ ﻣﺴﺘﻘﺒﻠﻴﺔ وﻣﺴﺘﻘﻠﺔ ﻋﻦ اﻟﻌﻴﻨﺔ اﻻوﱃ وذات اﳊﺠﻢ اﻟﺜﺎﺑﺖ mواﳌﻄﻠﻮب اﳚﺎد ﻓﱰات ﺗﻨﺒﺆ ﺑﻴﻴﺰ ﻻى ﻣﺸﺎﻫﺪة z sﻣﻦ اﳌﺸﺎﻫﺪات zﺣﻴﺚ : 1 s m ﲢﺖ ﻓﺮض ان z sﻫﻰ اﳌﺸﺎﻫﺪة ذات اﻟﱰﺗﻴﺐ sﰱ اﻟﻌﻴﻨﺔ اﳌﺮﺗﺒﺔ zﻓﺈن داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻻﺣﺘﻤﺎﻟﻴﺔ ل z sﺗﻌﻄﻰ ﻣﻦ :
)f (z s | ,
)(s 1
F(zs | , )
)(m s
h * (z s | , ) D1 (s) 1 F(z s | , )
ﺣﻴﺚ : m D1 (s) s , s
وﰱ ﺣﺎﻟﺔ ﺗﻮزﻳﻊ ﺑﺎرﻳﺘﻮ ﺗﺎﺧﺬ داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻟﺼﻴﻐﺔ اﻟﺘﺎﻟﻴﺔ : s 1
)(٩٠-٥
1 zs
) ( m s 1
h (z s | , ) D1 (s) z s z s *
٤٦٧
اﱃh* (z s | , ) ﻋﺪد ﺻﺤﻴﺢ ﻣﻮﺟﺐ ( ﺗﺆول داﻟﺔ اﻟﻜﺜﺎﻓﺔs وﺑﺎﺳﺘﺨﺪام ﻣﻔﻜﻮك ذات اﳊﺪﻳﻦ )ﺣﻴﺚ : اﻟﺸﻜﻞ اﻟﺘﺎﱃ s 1
s 1 h (z s | , ) D1 (s) 1 j z s z s j 0 *
( m s j1)
j
, z s .
(٩١-٥) : ﺗﻜﻮن ﻋﻠﻰ اﻟﺼﻮرةzs ( ﻓﺈن داﻟﺔ ﻛﺜﺎﻓﺔ ﺑﻴﻴﺰ اﻟﺘﻨﺒﺆﻳﺔ ﻟﻠﻤﺸﺎﻫﺪة٩١-٥) (٨٩-٥)وﺑﺎﺳﺘﺨﺪام اﳌﻌﺎدﻟﺘﲔ p1 (z s | y) h * (z s | , ) ( )d
s 1 (d u) ( r c) s 1 j D1 (s) 1 j ( rc)e (d u) z s (r c) j0 0 zs
(m s j1)
s 1 (d u) ( r c) D1 (s) (r c) a j (d u) m j ln zs j 0 zs
d (r c 1)
,
(٩٢-٥) : ﺣﻴﺚ s 1 j a j 1 , m j (m s j 1). j : ﻳﺘﻄﻠﺐ ﺣﺴﺎب داﻟﺔ اﻟﺼﻼﺣﻴﺔ اﻻﺗﻴﺔzs ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺣﺪود ﺗﻨﺒﺆ ﺑﻴﻴﺰ ﻟﻠﻤﺸﺎﻫﺪة
R(t) P z s t | y p1 (z s | y)dzs . t
: ( وﺑﻌﺾ اﳌﻌﺎﳉﺎت اﻟﺮﻳﺎﺿﻴﺔ ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻰ٩١-٥) ﺑﺎﻟﺘﻌﻮﻳﺾ ﻣﻦ اﳌﻌﺎدﻟﺔ s 1
1 P z s t | y D1 (s)(d u) (r c) (r c) a j z j 0 t s aj t (d u) m j ln j 0 m j s 1
D1 (s)(d u)
(r c)
(d u) m j ln zs
(r c 1)
dz s
(r c)
.
(٩٣-٥) : ﻳﻌﻄﻴﺎ ﻣﻦ100% ﲟﺴﺘﻮى ﺛﻘﺔzs ﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔU(y) واﻻﻋﻠﻰL(y) ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ
٤٦٨
1 , 2 1 P z s U(y) | y . 2 P z s L(y) | y
)(٩٤-٥ اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﳝﻜﻦ ﺣﻠﻬﺎ ﻋﺪدﻳﺎ ﺑﺎﺳﺘﺨﺪام ) (٩٣-٥ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺣﺪود وﻓﱰات ﺗﻨﺒﺆ ﺑﻴﻴﺰ ﻟﻘﻴﻢ اﳌﺨﺘﻠﻔﺔ .
ﺣﺎﻻت ﺧﺎﺻﺔ )ا( ﻋﻨﺪﻣﺎ s 1 ﻋﺎدة ﻳﻜﻮن ﻣﻦ اﳌﻬﻢ اﻟﺘﻨﺒﺆ ﺑﺎﻗﻞ ﻣﺸﺎﻫﺪة ﻋﻦ اﳌﺸﺎﻫﺪات اﳌﺴﺘﻘﺒﻠﻴﺔ ذات اﳊﺠﻢ ، mوﳝﻜﻨﻨﺎ اﳊﺼﻮل ﻋﻠﻰ ذﻟﻚ ﺑﻮﺿﻊ s 1ﰱ اﳌﻌﺎدﻟﺔ ) (٩٣-٥ﳓﺼﻞ ﻋﻠﻰ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻟﻠﻤﺸﺎﻫﺪة ) z1اﻗﻞ ﻣﺸﺎﻫﺪة( وﻣﻦ اﳌﻌﺎدﺗﲔ ﰱ) (٩٤-٥ﳓﺼﻞ ﻋﻠﻰ ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ ) L(yواﻻﻋﻠﻰ )U(y ﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔ z1ﲟﺴﺘﻮى ﺛﻘﺔ 100%وﺳﻮف اﺗﺮﻛﻬﺎ ﻛﺘﻤﺮﻳﻦ . )ب( ﻋﻨﺪﻣﺎ s m ﻟﻮ اﻓﱰﺿﻨﺎ ان ﻋﻨﺪﻣﺎ (i 1,2,...m),ziﲤﺜﻞ اوﻗﺎت اﻟﻌﻤﻞ ﺑﺪون اﻋﻄﺎل ﻤﻮﻋﺔ ﺣﺠﻤﻬﺎ ﻋﻨﺪﻣﺎ m ﻣﻦ اﻟﻮﺣﺪات ﻓﺈن اﻟﺘﻨﺒﺆ ﺑﺎﳌﺸﺎﻫﺪة ziﻳﻌﲎ اﻟﺘﻨﺒﺆ ﺑﻌﻤﺮ اﻟﻮﺣﺪة اﻻﺧﲑة ﰱ ﻫﺬﻩ ا ﻤﻮﻋﺔ وﻫﺬا ﻣﻔﻴﺪ اﱃ ﺣﺪ ﻛﺒﲑ ﰱ اﳊﻴﺎة اﻟﻌﻤﻠﻴﺔ .وﺑﻮﺿﻊ s mﰱ اﳌﻌﺎدﻟﺔ ) (٩٣-٥ﳓﺼﻞ ﻋﻠﻰ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻟﻠﻤﺸﺎﻫﺪة اﻻﺧﲑة z mوﻣﻦ اﳌﻌﺎدﺗﲔ ﰱ) (٩٤-٥ﳓﺼﻞ ﻋﻠﻰ ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ ) L(yواﻻﻋﻠﻰ ) U(yﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔ z mﲟﺴﺘﻮى ﺛﻘﺔ 100%وﺳﻮف اﺗﺮﻛﻬﺎ ﻛﺘﻤﺮﻳﻦ . ) (٢-٦-٥ﻣﻌﻠﻣﺔ اﻟﺷﻛل ﻏﯾر ﻣﻌﻠوﻣﺔ وﺣﺟم اﻟﻌﯾﻧﺔﻋﺷواﺋﻰ
ﻣﻦ اﳌﻌﻠﻮم ﰱ ﻛﺜﲑ ﻣﻦ اﻟﺒﺤﻮث اﻟﺒﻴﻮﻟﻮﺟﻴﺔ واﻟﺰراﻋﻴﺔ وﲡﺎرب اﻟﺘﺤﻜﻢ ﰱ ﺟﻮدة اﻻﻧﺘﺎج ان ﻫﻨﺎك اﺣﺘﻤﺎل ﻛﺒﲑ ﻟﻔﻘﺪ اﺣﺪ او ﺑﻌﺾ اﻋﻀﺎء اﻟﻌﻴﻨﺔ ﲢﺖ اﻟﺪراﺳﺔ ﻻﺳﺒﺎب ﻟﻴﺴﺖ ﲢﺖ اﻟﺴﻴﻄﺮة ،ﻟﺬا ﳝﻜﻦ اﻻﺧﺬ ﰱ اﻻﻋﺘﺒﺎر ﻋﻨﺪ ﺣﺴﺎب ﻓﱰات اﻟﺘﻨﺒﺆ ﻟﻌﻴﻨﺔ ﻣﺴﺘﻘﺒﻠﻴﺔ ان ﻳﻜﻮن ﺣﺠﻢ ﻫﺬﻩ اﻟﻌﻴﻨﺔ ﻏﲑ ﺛﺎﺑﺖ اى ﻣﺘﻐﲑ ﻋﺸﻮاﺋﻰ ،وﻣﺎ ﺗﺎﺛﲑ ذﻟﻚ ﻋﻠﻰ ﻓﱰات اﻟﺘﻨﺒﺆ اﻟﻨﺎﲡﺔ ؟ ﻫﺬا ﻣﺎ ﺳﻨﺤﺎول اﻻﺟﺎﺑﺔ ﻋﻠﻴﻪ ﺧﻼل ﻫﺬا اﳉﺰء . ٤٦٩
ﺑﻔﺮض ان z z1 z2 zrﻋﻴﻨﺔ ﻣﺴﺘﻘﺒﻠﻴﺔ وﻣﺴﺘﻘﻠﺔ ﻋﻦ اﻟﻌﻴﻨﺔ اﻻوﱃ ﺣﺠﻤﻬﺎ m ،وﺑﻔﺮض ان mﻣﺘﻐﲑ ﻋﺸﻮاﺋﻰ ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﻮاﺳﻮن ﻏﲑ اﻟﺘﺎم ﺑﺎﳌﻌﻠﻤﺔ واﻟﺬى داﻟﺔ ﻛﺜﺎﻓﺘﻪ اﻻﺣﺘﻤﺎﻟﻴﺔ ﻋﻠﻰ اﻟﺸﻜﻞ :
ﻏﲑ ﺛﺎﺑﺖ اى ﻣﺘﻐﲑ ﻋﺸﻮاﺋﻰ
e m p (m) , m 1, 2,3,... ) m!(1 e *
)(٩٥-٥ ﺑﺎﻻﺳﺘﻌﺎﻧﺔ ﺑﺒﺤﺜﻰ ) Gupta and Gupta (1984),Consul (1984ﻓﺈن داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻟﺘﻨﺒﺆﻳﺔ ﻻى ﻣﺸﺎﻫﺪة ﻣﺴﺘﻘﺒﻠﻴﺔ zﺣﻴﺚ 1 s mﻋﻨﺪﻣﺎ ﺗﻜﻮن mﻣﺘﻐﲑ ﻋﺸﻮاﺋﻰ ﺗﻌﻄﻰ ﻣﻦ : s
1 )p* (m)p(z s | m P(m s) m s
g(z s | y)
)(٩٦-٥ ﺣﻴﺚ :
) p(z s | mداﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻻﺣﺘﻤﺎﻟﻴﺔ ﻟﻠﻤﺸﺎﻫﺪة zﻋﻨﺪﻣﺎ ﺗﻜﻮن mﺛﺎﺑﺘﺔ . ﻣﻦ اﳌﻌﺎدﻻت ) (٩١-٥و) (٩٥-٥و) (٩٦-٥ﻓﺈن داﻟﺔ ﻛﺜﺎﻓﺔ ﺑﻴﻴﺰ اﻟﺘﻨﺒﺆﻳﺔ ﻟﻠﻤﺸﺎﻫﺪة ﺣﺠﻢ اﻟﻌﻴﻨﺔ mﻣﺘﻐﲑ ﻋﺸﻮاﺋﻰ ﻟﻪ داﻟﺔ اﻟﻜﺘﻠﺔ اﻻﺣﺘﻤﺎﻟﻴﺔ ) (٩٥-٥ﻫﻰ : s
) (r c 1
ﺣﻴﺚ :
zs
(r c)H r c s 1 m g(z s | y) a jD1 (s) H m j ln( ) k 1z s m! zs m s j 0 w H (d u), k1 e . !w 0 w s 1
وﺗﻜﻮن داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﰱ ﻫﺬﻩ اﳊﺎﻟﺔ : P zs t | y g1 (z s | y)dz s ) (r c
.
m t D1 (s) H m j ln( ) m! m s j0 m j aj
)(٩٧-٥
٤٧٠
s 1
r c
H k1
ﰱ ﺣﺎﻟﺔ
ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ ) L(yواﻻﻋﻠﻰ ) U(yﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔ zsﲟﺴﺘﻮى ﺛﻘﺔ 100%ﳓﺼﻞ ﻋﻠﻴﻬﻤﺎ ﺑﺎﳊﻞ اﻟﻌﺪدى ﺑﺎﻟﻨﺴﺒﺔ اﱃ tﻟﻠﻤﻌﺎدﻟﺔ : ) (r c
m t D )(s H m j ln( ) 1 m! m s j 0 m j aj
s 1
H rc k1
ﻟﻘﻴﻢ: 1 1 , 2 2
ﻋﻠﻰ اﻟﱰﺗﻴﺐ . ﺣﺎﻟﺔ ﺧﺎﺻﺔ ﻋﻨﺪﻣﺎ s 1 ﻋﺎدة ﻳﻜﻮن ﻣﻦ اﳌﻬﻢ اﻟﺘﻨﺒﺆ ﺑﺎﻗﻞ ﻣﺸﺎﻫﺪة ﻋﻦ اﳌﺸﺎﻫﺪات اﳌﺴﺘﻘﺒﻠﻴﺔ ذات اﳊﺠﻢ ، mوﳝﻜﻨﻨﺎ اﳊﺼﻮل ﻋﻠﻰ ذﻟﻚ ﺑﻮﺿﻊ s 1ﰱ اﳌﻌﺎدﻟﺔ ) (٩٧-٥ﳓﺼﻞ ﻋﻠﻰ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ﻟﻠﻤﺸﺎﻫﺪة ) z1اﻗﻞ ﻣﺸﺎﻫﺪة( .واﳊﺼﻮل ﻋﻠﻰ ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ ) L(yواﻻﻋﻠﻰ ) U(yﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔ z1ﲟﺴﺘﻮى ﺛﻘﺔ 100%ﻣﱰوك ﻛﺘﻤﺮﻳﻦ او اﻟﺮﺟﻮع اﱃ اﻟﺮﺳﺎﻟﺔ.
) (٣-٦-٥وﺟود ﻗﯾﻣﺔ ﻣﻧﻌزﻟﺔ
ﻣﻦ اﳉﺪﻳﺮ ﺑﺎﻟﺬﻛﺮ ان دراﺳﺔ اﳊﺎﻻت اﳋﺎﺻﺔ ﺑﻮﺟﻮد اﻛﺜﺮ ﻣﻦ ﻗﻴﻤﺔ ﻣﻨﻌﺰﻟﺔ ﰱ اﻟﻌﻴﻨﺔ اﳌﺴﺘﻘﺒﻠﻴﺔ ﻳﻨﺘﺞ ﻋﻨﻪ ﻋﻼﻗﺎت رﻳﺎﺿﻴﺔ ﻏﺎﻳﺔ ﻣﻦ اﻟﺘﻌﻘﻴﺪ ﳑﺎ ﳚﻌﻞ اﻣﻜﺎﻧﻴﺔ اﳊﺼﻮل ﻋﻠﻰ ﻧﺘﺎﺋﺞ ﻋﻤﻠﻴﺔ ﳍﺬﻩ اﳊﺎﻻت ﻏﺎﻳﺔ ﰱ اﻟﺼﻌﻮﺑﺔ ). Banett and Lewis(1994 ﺑﻔﺮض أن nﻣﻦ اﻟﻮﺣﺪات ﻣﻦ اﻟﻌﻴﻨﺔ اﻻوﱃ وﺿﻌﺖ ﻟﻼﺧﺘﺒﺎر وان اﻻﺧﺘﺒﺎر ﻳﻨﺘﻬﻲ ﺑﻌﺪ ﻓﺸﻞ rﻣﻦ اﻟﻮﺣﺪات .اى أن y y1 , y 2 ,...y rﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﲤﺜﻞ أزﻣﻨﺔ اﻟﻔﺸﻞ وان أزﻣﻨﺔ اﻟﻔﺸﻞ ﺗﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﺎرﻳﺘﻮ ﲟﻌﻠﻤﺘﲔ ﺑﺎﻋﺘﺒﺎر ﻣﻌﻠﻮﻣﻴﺔ ﻣﻌﻠﻤﺔ اﻟﻘﻴﺎس وﻋﺪم ﻣﻌﻠﻮﻣﻴﺔ ﻣﻌﻠﻤﺔ اﻟﺸﻜﻞ و ﺑﻔﺮض ان اﻟﻌﻴﻨﺔ اﻻوﱃ ذات اﳊﺠﻢ nﻣﻌﻠﻮم ﻋﻨﻬﺎ اﳌﺸﺎﻫﺪات
. y y1 y 2 y rﺑﻔﺮض ان z z1 z 2 z rﻋﻴﻨﺔ ﻣﺴﺘﻘﺒﻠﻴﺔ وﻣﺴﺘﻘﻠﺔ ﻋﻦ اﻟﻌﻴﻨﺔ اﻻوﱃ وذات
اﳊﺠﻢ اﻟﺜﺎﺑﺖ m
ﺗﺎﺑﻌﺔ ﻟﻨﻔﺲ اﻟﺘﻮزﻳﻊ اﻻﺣﺘﻤﺎﱃ وإذا اﻓﱰﺿﻨﺎ وﺟﻮد ﻗﻴﻤﺔ ﻣﻨﻌﺰﻟﺔ ﻣﻦ اﻟﻨﻮع oﰱ اﻟﻌﻴﻨﺔ. ٤٧١
: 1 s m ﺣﻴﺚz ﻣﻦ اﳌﺸﺎﻫﺪاتzs واﳌﻄﻠﻮب اﳚﺎد ﻓﱰات ﺗﻨﺒﺆ ﺑﻴﻴﺰ ﻻى ﻣﺸﺎﻫﺪة ﺗﻌﻄﻰ ﻣﻦ اﻟﺼﻴﻐﺔz s ﻓﺈن داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻻﺣﺘﻤﺎﻟﻴﺔ لz ﰱ اﻟﻌﻴﻨﺔ اﳌﺮﺗﺒﺔs ﻫﻰ اﳌﺸﺎﻫﺪة ذات اﻟﱰﺗﻴﺐz s ﲢﺖ ﻓﺮض ان : اﻟﺘﺎﻟﻴﺔ m s m1 h(ys | ) [(s 1)Fs2 1 F F* f (ys | ) s1 m s
Fs1 1 F
f * (ys | ) m s 1
(m s)Fs1 1 F
1 F* f (ys | )].
اﻟﻐﯾر اﻟﻣﻧﻌزﻟﺔ ﺑﯾﻧﻣﺎy ﻫﻣﺎ داﻟﺗﻰ اﻟﻛﺛﺎﻓﺔ واﻟﺗوزﯾﻊ ﻟﻛل ﻗﯾمf (ys | ),F(ys | ) ﺣﯾث
.
. اﻟﻣﻧﻌزﻟﺔ ﻋﻠﻰ اﻟﺗرﺗﯾبy ﻫﻣﺎ داﻟﺗﻰ اﻟﻛﺛﺎﻓﺔ واﻟﺗوزﯾﻊ ﻟﻛل ﻗﯾمf * (ys | ),F* (ys | ) : وﰱ ﺣﺎﻟﺔ ﺗﻮزﻳﻊ ﺑﺎرﻳﺘﻮ ﺗﺎﺧﺬ داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻟﺼﻴﻐﺔ اﻟﺘﺎﻟﻴﺔ (s 2) (m s) o 1 1 h(z s | , ) [ (s 1) 1 s 1 zs zs z z s s m 1
1 zs
(s 1)
zs
( m s)
o o o 1 zs
(s 1)
(m s 1) o (m s) 1 1 ], z s . zs zs zs zs اﱃh* (z s | , ) ﻋﺪد ﺻﺤﻴﺢ ﻣﻮﺟﺐ ( ﺗﺆول داﻟﺔ اﻟﻜﺜﺎﻓﺔs وﺑﺎﺳﺘﺨﺪام ﻣﻔﻜﻮك ذات اﳊﺪﻳﻦ )ﺣﻴﺚ
: اﻟﺸﻜﻞ اﻟﺘﺎﱃ
m 1 h(z s | , ) [ s 1 z s s 2
s2 (s 1) 1 j zs j0
m oj
j
s 1
s2 (s 1) 1 j j0
s 1 o m s 1 j zs j 0 j
s2
j
(moj 1)
].
: ﺣﻴﺚ ٤٧٢
m oj o m j .
وﺑﻌﺪ ﺑﻌﺾ اﻻﺧﺘﺼﺎرات اﳉﱪﻳﺔ ﺗﺎﺧﺬ اﻟﺪاﻟﺔ اﻟﺴﺎﺑﻘﺔ اﻟﺸﻜﻞ اﻟﺘﺎﱃ :
s 2 m z , m z , h(zs | , ) D2 (s) [ s 1 b j e j s e oj s j0 zs
],zs
m 1 z , s
s1
o m s a je
oj
j 0
)(٩٨-٥ ﺣﻴﺚ : s 2 j s 2 D 2 (s) , b j 1 , (z s , ) ln . j j zs
ﻋﻨﺪﻣﺎ ﺗﻜﻮن ﻣﻌﻠﻤﺔ اﻟﺸﻜﻞ ﻏﲑ ﻣﻌﻠﻮﻣﺔ وﻣﻌﻠﻤﺔ اﻟﻘﻴﺎس ﻣﻌﻠﻮﻣﺔ ﻓﻔﻰ ﻫﺬﻩ اﳊﺎﻟﺔ ﻳﺴﺘﺨﺪم ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﺑﺎﳌﻌﻠﻤﺘﲔ ) (c,dﻛﺘﻮزﻳﻊ ﻗﺒﻠﻰ ﻣﺮاﻓﻖ ﻟﻠﻤﻌﻠﻤﺔ واﻟﺬى ﻳﺎﺧﺬ اﻟﺸﻜﻞ اﻟﺘﺎﱃ : c 1e dd c () )(c
0 , c,d 0 .
ﺑﻌﺪ اﳊﺼﻮل ﻋﻠﻰ اﻟﻌﻴﻨﺔ yﻓﺈن اﻟﺘﻮزﻳﻊ اﻟﺒﻌﺪى ﻟﻠﻤﻌﻠﻤﺔ ﻳﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ : H
,
) ( r c
( r c) 1H e ( | y) ) (r c
xi x ln r , H d u. xr
r
u ln i 1
)(٩٩-٥
ﻣﻦ اﳌﻌﺎدﻟﺘﲔ ) (٩٨-٥و) (٩٧-٥ﻓﺈن داﻟﺔ ﻛﺜﺎﻓﺔ ﺑﻴﻴﺰ اﻟﺘﻨﺒﺆﻳﺔ ﻟﻠﻤﺸﺎﻫﺪة zsﺗﻜﻮن ﻋﻠﻰ اﻟﺼﻮرة :
٤٧٣
f1 ( z s | y )
h ( z s | , ) ( )d
D 2 (s )[ 0
s2 ( H ) (r c ) ( H m j ( z s , )) ( H m o j ( z s , )) [(s 1) b j e e (r c) j 0
s2
( o m s ) a je
( H ( m o j 1) ( z s , ))
]d ,
j 0
: وﺑﺎ ﺎء اﻟﺘﻜﺎﻣﻼت وﻋﺪد ﻣﻦ اﻻﺧﺘﺼﺎرات ﺗﺎﺧﺬ اﻟﺪاﻟﺔ اﻟﺴﺎﺑﻘﺔ اﻟﺸﻜﻞ اﻟﺘﺎﱃ f 1 ( z s | y ) D 2 (s )
s2 (rc) ( H ) ( r c ) (r c ) [(s 1) b j (H m j (z s , ) zs j 0
(H m oj (z s , )
( r c 1)
s 1
( o m s ) a j [( H (m oj 1) (z s , )]
( r c 1 )
, z s .
j 0
: ﻳﺘﻄﻠﺐ ﺣﺴﺎب داﻟﺔ اﻟﺼﻼﺣﻴﺔ اﻻﺗﻴﺔz s ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺣﺪود ﺑﻴﻴﺰ ﻟﻠﻤﺸﺎﻫﺪة
R(t) P z s t | y p1 (z s | y)dzs . t
: ( وﺑﻌﺾ اﳌﻌﺎﳉﺎت اﻟﺮﻳﺎﺿﻴﺔ ﻳﺘﻢ اﳊﺼﻮل ﻋﻠﻰ١٠٠-٥) ﺑﺎﻟﺘﻌﻮﻳﺾ ﻣﻦ اﳌﻌﺎدﻟﺔ
P z s t | y D 2 (s)(H)
(r c)
(r c)[ t
t
(r c 1) (s 1) s 2 b j{ (H m j (z s , ) }dz s z s j0
s 1
(r c 1) ( o m s) a j (H (m oj 1) (z s , ) dz s ] zs j 0 s2
D 2 (s)(H)
(r c)
{(s 1) b j ((t,m j ) (t, m oj ) j 0
s 1
( o m s) a j ( (t, m oj 1))], j 0
٤٧٤
)(١٠١-٥ ﺣﻴﺚ: 1 ) (r c H x(t, ) . x ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ ) L(yواﻻﻋﻠﻰ ) U(yﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔ zsﲟﺴﺘﻮى ﺛﻘﺔ 100%ﻳﻌﻄﻴﺎ ﻣﻦ : (t, x)
1 , 2 1 P z s U(y) | y . 2 P z s L(y) | y
اﳌﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﳝﻜﻦ ﺣﻠﻬﺎ ﻋﺪدﻳﺎ ﺑﺎﺳﺘﺨﺪام ) (١٠١-٥ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﺣﺪود وﻓﱰات ﺗﻨﺒﺆ ﺑﻴﻴﺰ ﻟﻘﻴﻢ
اﳌﺨﺘﻠﻔﺔ .
ﺣﺎﻻت ﺧﺎﺻﺔ )ا( ﻋﻨﺪﻣﺎ s 1 ﻋﺎدة ﻳﻜﻮن ﻣﻦ اﳌﻬﻢ اﻟﺘﻨﺒﺆ ﺑﺎﻗﻞ ﻣﺸﺎﻫﺪة ﰱ اﻟﻌﻴﻨﺔ اﳌﺴﺘﻘﺒﻠﻴﺔ اﻟﱴ ﲢﺘﻮى ﻋﻠﻰ اﺣﺪ اﻟﻘﻴﻢ اﳌﻨﻌﺰﻟﺔ ﻣﻦ اﻟﻨﻮع oﻧﻀﻊ s 1ﰱ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ) (١٠١-٥ﳓﺼﻞ ﻋﻠﻰ داﻟﺔ اﻟﺼﻼﺣﻴﺔ اﻟﻼ زﻣﺔ ﳊﺴﺎب ﺣﺪود ﺗﻨﺒﺆ ﺑﻴﻴﺰ ﻟﻠﻤﺸﺎﻫﺪة ) z1اﻗﻞ ﻣﺸﺎﻫﺪة( ﰱ اﻟﺼﻮرة: .
) (r c
P zs t | y H(rc) H (o m 1)(t, )
وﳝﻜﻦ اﳊﺼﻮل ﻋﻠﻰ ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ ) L(yواﻻﻋﻠﻰ ) U(yﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔ z1ﲟﺴﺘﻮى ﺛﻘﺔ
100%وﺳﻮف اﺗﺮﻛﻬﺎ ﻛﺘﻤﺮﻳﻦ .
)ب( ﻋﻨﺪﻣﺎ ﻟﻠﺘﻨﺒﺆ ﺑﺎﳌﺸﺎﻫﺪة اﻻﺧﲑة ﰱ اﻟﻌﻴﻨﺔ اﳌﺴﺘﻘﺒﻠﻴﺔ اﻟﱴ ﲢﺘﻮى ﻋﻠﻰ اﺣﺪ اﻟﻘﻴﻢ اﳌﻨﻌﺰﻟﺔ ﻣﻦ اﻟﻨﻮع oﻧﻀﻊ s mﰱ داﻟﺔ اﻟﺼﻼﺣﻴﺔ ) (١٠١-٥ﳓﺼﻞ ﻋﻠﻰ داﻟﺔ اﻟﺼﻼﺣﻴﺔ اﻟﻼ زﻣﺔ ﳊﺴﺎب ﺣﺪود ﺗﻨﺒﺆ ﺑﻴﻴﺰ ﻟﻠﻤﺸﺎﻫﺪة z mوﺑﺎﻟﺘﺎﱃ ﻋﻠﻰ ﺣﺪى ﺗﻨﺒﺆ ﺑﻴﻴﺰ اﻻدﱏ ) L(yواﻻﻋﻠﻰ ) U(yﻟﻠﻤﺸﺎﻫﺪة اﳌﺴﺘﻘﺒﻠﻴﺔ z m ﲟﺴﺘﻮى ﺛﻘﺔ 100%وﺳﻮف اﺗﺮﻛﻬﺎ ﻛﺘﻤﺮﻳﻦ . sm
٤٧٥
٤٧٦
اﻟﻤﺮاﺟﻊ
اﻟﻤﺮاﺟﻊ اﻟﻌﺮﺑﯿﺔ :
o
-١أﺣﻣد ﻋودة ، (١٩٩١) ،ﻣﻘدﻣﺔ ﻓﻲ اﻟﻧظرﯾﺔ اﻹﺣﺻﺎﺋﯾﺔ – ﺟﺎﻣﻌﺔ اﻟﻣﻠك ﺳﻌود -ﻋﻣﺎدة ﺷؤون اﻟﻣﻛﺗﺑﺎت .
o o
-٢أﻣﯾر ﺣﻧﺎ ﻫرﻣز ، (١٩٩٠) ،اﻹﺣﺻﺎء اﻟرﯾﺎﺿﻲ – ﻣدﯾرﯾﺔ دار اﻟﻛﺗب ﻟﻠطﺑﺎﻋﺔ واﻟﻧﺷر –
o o
-٣ﺛروت ﻣﺣﻣد ﻋﺑد اﻟﻣﻧﻌم ، (٢٠٠٠) ،ﻧظرﯾﺔ اﻻﺣﺗﻣﺎﻻت – اﻟطﺑﻌﺔ اﻟﺛﺎﻧﯾﺔ -ﻣﻛﺗﺑﺔ اﻟﻣﺗﻧﺑﻰ
اﻟﺟﻣﻬورﯾﺔ اﻟﻌراﻗﯾﺔ – اﻟﻣوﺻل .
– اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ .
o o
-٤ﺛروت ﻣﺣﻣد ﻋﺑد اﻟﻣﻧﻌم ، (٢٠٠٨) ،ﻣدﺧل ﺣدﯾث ﻟﻺﺣﺻﺎء واﻻﺣﺗﻣﺎﻻت – اﻟطﺑﻌﺔ اﻟﺛﺎﻟﺛﺔ – ﻣﻛﺗﺑﺔ اﻟﻌﺑﯾﻛﺎن – اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ .
o o
-٥ﺛروت ﻣﺣﻣد ﻋﺑد اﻟﻣﻧﻌم ، (٢٠٠٩) ،اﻟﻣدﺧل اﻟﺣدﯾث ﻟﻺﺣﺻﺎء واﻻﺣﺗﻣﺎﻻت ﻣﻊ اﻟﺣﻠول ل ٧٠٨ﻣﺳﺎﻟﺔ -اﻟطﺑﻌﺔ اﻻوﻟﻰ– ﻣﻛﺗﺑﺔ اﻟﻣﺗﻧﺑﻰ -اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ .
o o
-٦ﺛروت ﻣﺣﻣد ﻋﺑد اﻟﻣﻧﻌم ، (٢٠١٠) ،ﻧظرﯾﺔ اﻻﺣﺗﻣﺎﻻت ﻣﻊ اﻟﺣﻠول ﻟﺣواﻟﻰ ١٠٣٥
o o
-٦ﺟﻼل اﻟﺻﯾﺎد ، (١٩٨٨) ،ﻧظرﯾﺔ اﻻﺣﺗﻣﺎﻻت – اﻟطﺑﻌﺔ اﻟﺛﺎﻧﯾﺔ – دار اﻟﺷروق – ﺟدة –
ﻣﺳﺎﻟﺔ– اﻟطﺑﻌﺔ اﻻوﻟﻰ -ﻣﻛﺗﺑﺔ اﻟﻣﺗﻧﺑﻲ – اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ .
اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ .
o o
-٧ﺟﻼل اﻟﺻﯾﺎد ، (١٩٩٣) ،اﻻﺳﺗدﻻل اﻻﺣﺻﺎﺋﻰ – اﻟطﺑﻌﺔ اﻻوﻟﻰ -دار اﻟﻣرﯾﺦ ﻟﻠﻧﺷر – اﻟرﯾﺎض– اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ . ٤٧٧
o
-٨ﺳﻠﯾم ذﯾﺎب اﻟﺳﻌدي ، (١٩٨٣) ،طرق اﻹﺣﺻﺎء – اﻟﺟﻣﻬورﯾﺔ اﻟﻌراﻗﯾﺔ – و ازرة اﻟﺗﻌﻠﯾم اﻟﻌﺎﻟﻲ واﻟﺑﺣث اﻟﻌﻠﻣﻲ .
o o
-٩ﻋﻠﻲ ﻋﺑد اﻟﺳﻼم اﻟﻌﻣﺎوي وﻋﻠﻲ ﺣﺳﯾن اﻟﻌﺟﯾﻠﻲ ، (١٩٩٨) ،أﺳﺎﺳﯾﺎت اﻻﺣﺻﺎء اﻟرﯾﺎﺿﻲ – إدارة اﻟﻣطﺑوﻋﺎت واﻟﻧﺷر – ﺟﺎﻣﻌﺔ اﻟﻔﺎﺗﺢ .
-١٠ oﻋﺒﺪ اﻟﺤﻔﯿﻆ ﻣﺤﻤﺪ ﻓﻮزي ﻣﺼﻄﻔﻰ ‘)٢٠٠٠أ ( ،اﻻﺳﺘﺪﻻل اﻻﺣﺼﺎﺋﻲ ) ( ١ﻧﻈﺮﯾﺔ اﻟﺘﻘﺪﯾﺮ ،ﻣﺠﻤﻮﻋﺔ اﻟﻨﯿﻞ اﻟﻌﺮﺑﯿﺔ -اﻟﻘﺎھﺮة – ﻣﺪﯾﻨﺔ ﻧﺼﺮ . o -١١ oﻋﺒﺪ اﻟﺤﻔﯿﻆ ﻣﺤﻤﺪ ﻓﻮزي ﻣﺼﻄﻔﻰ ٢٠٠٠) ،ب ( ،اﻻﺳﺘﺪﻻل اﻻﺣﺼﺎﺋﻲ ) ( ٢ﻧﻈﺮﯾﺔ اﻟﺘﻘﺪﯾﺮ ،ﻣﺠﻤﻮﻋﺔ اﻟﻨﯿﻞ اﻟﻌﺮﺑﯿﺔ -اﻟﻘﺎھﺮة – ﻣﺪﯾﻨﺔ ﻧﺼﺮ .
o o
-١٢ﻣﺣﻣد إﺑراﻫﯾم ﻋﻘﯾل و ﻋﺑد اﻟرﺣﻣن ﻣﺣﻣد أﺑو ﻋﻣﻪ ، (٢٠٠٠)،ﻧظرﯾﺔ اﻻﺣﺗﻣﺎﻻت و ﺗطﺑﯾﻘﺎﺗﻬﺎ – اﻟﻧﺷر اﻟﻌﻠﻣﻲ و اﻟﻣطﺎﺑﻊ – ﺟﺎﻣﻌﺔ اﻟﻣﻠك ﺳﻌود – اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ.
o اﻟﻤﺮاﺟﻊ اﻷﺟﻨﺒﯿﺔ :
1-Ashour , S.K and Salem, S.A (1990) An Introduction to Mathematical Statistics, I.S.S.R ,Cairo University. 2-Abdel Moneim, T.M. (1998), Bayesian estimation of the reliability function of a two-parameter Cauchy distribution , The Egyptian Statistical Journal, 42(1),1-9. 3- Aitchison, J.& Brown J.A.C (1957), The Lognormal Distribution, Cambridge:Cambridge University Press. 4-AL- Braheem ,F . M . (1990) Regression Models in Testing, Msc. Girls College , Dammam.
٤٧٨
5-AL- Hussaini, E. K. & Jaheen, Z. F. (1994), Approximate bayes estimators applied to the Burr model , Communications in Statistics-Theory and Methods, 23 (1), 99 - 121. 6-AL- Ohali , M . E . (2000) , Some Bayesian Predictions Based on Pareto Distribution , Msc. Girls College , Dammam. 7-AL- mobaudh , E . A . (2010) , Estimates of Lognormal Distribution Parameters Based on Bayesian Approach, Msc. Girls College , Dammam . 8- Balasooriya , U. & Balakrishnan, N. (2000) ,Reliability sampling plans for lognormal distribution , based on progressively-censored samples , IEEE Transactions on Reliability ,49(2),199-203 . 9-Basu , A . B . & Ebrahimi , N . (1991) , Bayesian approach to life testing and reliability estimation using asymmetric, Journal of Statistical Planning and Inference, 29, 21-31. 10-Barnett, V . B . &Lewis, T . (1994) ,Outliers in Statistical Data 3 ed.,Jon Wiley ,New York. 11-Bekker, A., Roux, J.J.J & Mostert, P.J. (2000), A generalization of the compound Ra10yleigh distribution: using bayesian methods on cancer survival times , Communications of Statistics – Theory and Methods, 29(7), 1419-1433. 12-Bhattacharya, S . K . (1967) , Bayesian approach to life testing and reliability estimation , Journal of American Statistical Association , 48-62 . 13-Box , G . E . P . , & Tiao , G . C . (1973) , Bayesian Inference in Statistical Analysis , Reading , MA : Addison – Wesley .
٤٧٩
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٤٨٤
٤٨٥
اﻟﻤﻼﺣﻖ ﻣﻠﺤﻖ ) ( ١ﺟﺪول ﺣﺴﺎب
r ) p (x; x 0
ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ﺑﻮاﺳﻮن .
ﻣﻠﺤﻖ ) ( ٢ﺟﺪول اﻟﻤﺴﺎﺣﺎت ﺗﺤﺖ اﻟﻤﻨﺤﻨﻰ اﻟﻄﺒﯿﻌﻲ اﻟﻘﯿﺎﺳﻲ ). P(0 Z z ﻣﻠﺤﻖ ) ( ٣ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) f (1, 2ﻟﺘﻮزﯾﻊ Fﻋﻨﺪ ) . ( 0.05 ﻣﻠﺤﻖ ) (٤ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) f (1, 2ﻟﺘﻮزﯾﻊ Fﻋﻨﺪ ) . ( 0.01 ﻣﻠﺤﻖ ) ( ٥ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ 2ﻟﺘﻮزﯾﻊ . 2 ﻣﻠﺤﻖ ) ( ٦ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ t ﻟﺘﻮزﯾﻊ . t
٤٨٦
ﻣﻠﺤﻖ ): (١ ﺟﺪول ﺣﺴﺎب
r ) p (x; x 0
ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ﺑﻮاﺳﻮن
1.0 .368 .736 .920 .981
.9 .407 .772 .937 .987
.8 .449 .809 .953 .991
.7 .497 .844 .966 .994
.6 .549 .878 .977 .997
.5 .607 .910 .986 .998
.4 .670 .938 .992 .999
.996 .999 1.00
.998 1.00
.999 1.00
.999 1.00
1.00
1.00
1.00
اﻟﻤﺼﺪر ﻋﻦ [Devore (1995)] :
٤٨٧
.3 .741 .963 .996 1.00
.2 .819 .982 .999 1.00
.1 .905 .995 1.00
0 1 2 3 4 5 6
x
ﺗﺎﺑﻊ ﻣﻠﺤﻖ ) : (١ﺟﺪول ﺣﺴﺎب
r ) p (x; x 0
ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ﺑﻮاﺳﻮن
20.0
15.0
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
.000
.000
.000
.000
.000
.001
.002
.007
.018
.050
.135
0
.000
.000
.000
.001
.003
.007
.017
.040
.092
.199
.406
1
.000
.000
.003
.006
.014
.030
.062
.125
.238
.423
.677
2
.000
.000
.010
.021
.042
.082
.151
.265
.433
.647
.857
3
.000
.001
.029
.055
.100
.173
.285
.440
.629
.815
.947
4
.000
.003
.067
.116
.191
.301
.446
.616
.785
.916
.983
5
.000
.008
.130
.207
.313
.450
.606
.762
.889
.966
.995
6
.001
.018
.220
.324
.453
.599
.744
.867
.949
.988
.999
7
.002
.037
.333
.456
.593
.729
.847
.932
.979
.996
1.00
8
.005
.070
.458
.587
.717
.830
.916
.968
.992
.999
9
.011
.118
.583
.706
.816
.901
.957
.986
.997
1.00
10
.021
.185
.697
.803
.888
.947
.980
.995
.999
11
.039
.268
.792
.876
.936
.973
.991
.998
1.00
12
.066
.363
.864
.926
.966
.987
.996
.999
13
.105
.466
.917
.959
.983
.994
.999
1.00
14
.157
.568
.951
.978
.992
.998
.999
15
.221
.664
.973
.989
.996
.999
1.00
16
.297
.749
.986
.995
.998
1.00
.381
.819
.993
.998
.999
18
.470
.875
.997
.999
1.00
19
17
20 21 22 23 24 25 26
1.00 .998 .917 .559 .999 .947 .644 1.00 .967 .721 .981 .787 .989 .843 .994 .888 .997 .922 .948
.998
27
.966
.999
28
.978
1.00
29
.987
30
.992
31
.995
32
.997
33
.999
34
.999
35
1.00
36
٤٨٨
X
ﻣﻠﺤﻖ )(٢ ﺟدول اﻟﻣﺳﺎﺣﺎت ﺗﺣت اﻟﻣﻧﺣﻧﻰ اﻟطﺑﯾﻌﻲ اﻟﻘﯾﺎﺳﻲ )P(0<Z<z .09 .0359 .0753 .1141 .1517 .1879 .2224 .2549 .2852 .3133 .3389 .3621 .3830 .4015 .4177 .4319 .4441 .4545 .4633 .4706 .4767 .4817 .4857 .4890 .4916 .4936 .4952 .4964 .4974 .4981 .4986 .4990
.08 .0319 .0714 .1103 .1480 .1844 .2190 .2517 .2823 .3106 .3365 .3599 .3810 .3997 .4162 .4306 .4429 .4535 .4625 .4699 .4761 .4812 .4854 .4887 .4913 .4934 .4951 .4963 .4973 .4980 .4986 .4990
.07 .0279 .0675 .1064 .1443 .1808 .2157 .2486 .2794 .3078 .3340 .3577 .3790 .3980 .4147 .4292 .4418 .4525 .4616 .4693 .4756 .4808 .4850 .4884 .4911 .4932 .4949 .4962 .4972 .4979 .4985 .4989
.06 .0239 .0636 .1026 .1406 .1772 .2123 .2454 .2764 .3051 .3315 .3554 .3770 .3962 .4131 .4279 .4406 .4515 .4608 .4686 .4750 .4803 .4846 .4881 .4909 .4931 .4948 .4961 .4971 .4979 .4985 .4989
.05 .0199 .0596 .0987 .1368 .1736 .2088 .2422 .2734 .3023 .3289 .3531 .3749 .3944 .4115 .4265 .4394 .4505 .4599 .4678 .4744 .4798 .4842 .4878 .4906 .4929 .4946 .4960 .4970 .4978 .4984 .4989
.04 .0160 .0557 .0948 .1331 .1700 .2054 .2389 .2704 .2995 .3264 .3508 .3729 .3925 .4099 .4251 .4382 .4495 .4591 .4671 .4738 .4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 .4988
اﻟﻤﺼﺪر :ﻋﻦ ])[Daniel (1978
٤٨٩
.03 .0120 .0517 .0910 .1293 .1664 .2019 .2357 .2673 .2967 .3238 .3485 .3708 .3907 .4082 .4236 .4370 .4484 .4582 .4664 .4732 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .4988
.02 .0080 .0478 .0871 .1255 .1628 .1985 .2324 .2642 .2939 .3212 .3461 .3686 .3888 .4066 .4222 .4357 .4474 .4573 .4656 .4726 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .4987
.01 .0040 .0438 .0832 .1217 .1591 .1950 .2291 .2611 .2910 .3186 .3438 .3665 .3869 .4049 .4207 .4345 .4463 .4564 .4649 .4719 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .4987
.00 .0000 .0398 .079 .1179 .1554 .1915 .2257 .2580 .2881 .3159 .3413 .3643 .3849 .4032 .4192 .4332 .4452 .4554 .4641 .4713 .4772 .4821 .4861 .4893 .4918 .4938 .4953 .4965 .4974 .4981 .4987
Z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0
ﻣﻠﺤﻖ )(٣ ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) f (1, 2ﻟﺘﻮزﯾﻊ Fﻋﻨﺪ
120
60
40
30
24
20
15
12
10
9
8
)( 0.05 7
236. 238. 240. 241. 243. 245. 248. 249. 250. 251. 252. 253.3 254. 8 19.3 9 19.3 5 19.4 9 19.4 9 19.4 9 19.4 0 19.4 1 19.4 1 19.4 1 19.4 2 19.49 19.5 3 19.3 5 8.85 7 8.81 8 8.79 0 8.74 1 8.70 3 8.66 5 8.64 5 8.62 6 8.59 7 8.57 8 0 8.89 8.55 8.53
6
5
4
3
2
1
161. 199. 215. 224. 230. 234. 4 19.0 5 19.1 7 19.2 6 19.3 2 19.3 0 18.5 1 9..55 0 9.28 6 9.12 5 9.01 0 8.94 3 10.1 3 7.71 6.94 6.59 6.39 6.26 6.16
1
2 1 2 3 4
6.09 6.04 6.00 5.96 5.91 5.86 5.80 5.77 5.75 5.72 5.69 5.66 5.63 6. 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.68 4.62 4.56 4.53 4.50 4.46 4.43 4.40 4.36
5
5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 4.06 4.00 3.94 3.87 3.84 3.81 3.77 3.74 3.70 3.67
6
5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 3.64 3.57 3.51 3.44 3.41 3.38 3.34 3.30 3.27 3.23
7
5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 3.35 3.28 3.22 3.15 3.12 3.08 3.04 3.01 2.97 2.93
8
5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 3.07 3.01 2.94 2.90 2.86 2.83 2.79 2.75 2.71
9
4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98 2.91 2.85 2.77 2.74 2.70 2.66 2.62 2.58 2.54
10
4.84 3.98 3.59 3.36 3.20 3.09 3.01 2.95 2.90 2.85 2.79 2.72 2.65 2.61 2.57 2.53 2.49 2.45 2.40
11
4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 2.75 2.69 2.62 2.54 2.51 2.47 2.43 2.38 2.34 2.30
12
4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 2.67 2.60 2.53 2.46 2.42 2.38 2.34 2.30 2.25 2.21
13
4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 2.60 2.53 2.46 2.39 2.35 2.31 2.27 2.22 2.18 2.13
14
4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 2.48 2.40 2.33 2.29 2.25 2.20 2.16 2.11 2.07
15
4.49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 2.49 2.42 2.35 2.28 2.24 2.19 2.15 2.11 2.06 2.07
16
4.45 3.59 3.20 2.96 2.81 2.70 2.61 2.55 2.49 2.45 2.38 2.31 2.23 2.19 2.15 2.10 2.06 2.01 1.96
17
4.41 3.55 3.16 2.93 2.77 2.66 2.58 2.51 2.46 2.41 2.34 2.27 2.19 2.15 2.11 2.06 2.02 1.97 1.92
18
4.38 3.52 3.13 2.90 2.74 2.63 2.54 2.48 2.42 2.38 2.31 2.23 2.16 2.11 2.07 2.03 1.98 1.93 1.88
19
4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 2.28 2.20 2.12 2.08 2.04 1.99 1.95 1.90 1.84
20
4.32 3.47 3.07 2.84 2.68 2.57 2.49 2.42 2.37 2.32 2.25 2.18 2.10 2.05 2.01 1.96 1.92 1.87 1.81
21
4.30 3.44 3.05 2.82 2.66 2.55 2.46 2.40 2.34 2.30 2.23 2.15 2.07 2.03 1.98 1.94 1.89 1.84 1.78
22
4.28 3.42 3.03 2.80 2.64 2.53 2.44 2.37 2.32 2.27 2.20 2.13 2.05 2.01 1.96 1.91 1.86 1.81 1.76
23
4.26 3.40 3.01 2.78 2.62 2.51 2.42 2.36 2.30 2.25 2.18 2.11 2.03 1.98 1.94 1.89 1.84 1.79 1.73
24
2.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 2.24 2.16 2.09 2.01 1.96 1.92 1.87 1.82 1.77 1.71
25
4.23 3.37 2.98 2.74 2.59 2.47 2.39 2.32 2.27 2.22 2.15 2.07 1.99 1.95 1.90 1.58 1.80 1.75 1.69
26
4.21 3.35 2.96 2.73 2.57 2.46 2.37 2.31 2.25 2.20 2.13 2.06 1.97 1.93 1.88 1.84 1.79 1.73 1.67
27
4.20 3.34 2.95 2.71 2.56 2.45 2.36 2.29 2.24 2.19 2.12 2.04 1.96 1.91 1.87 1.82 1.77 1.71 1.65
28
4.18 3.33 2.93 2.70 2.55 2.43 2.35 2.28 2.22 2.18 2.10 2.03 1.94 1.90 1.85 1.81 1.75 1.70 1.64
29
4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 2.21 2.16 2.09 2.01 1.93 1.89 1.84 1.79 1.74 1.68 1.62
30
4.08 3.23 2.48 2.61 2.45 2.34 2.25 2.18 2.12 2.08 2.00 1.92 1.84 1.79 1.74 1.69 1.64 1.58 1.51
40
4.00 3.15 2.76 2.53 2.37 2.25 2.17 2.10 2.04 1.99 1.92 1.84 1.75 1.70 1.65 1.59 1.53 1.47 1.39
60
3.92 3.07 2.68 2.45 2.29 2.17 2.09 2.02 1.96 1.91 1.83 1.75 1.66 1.61 1.55 1.50 1.43 1.35 1.25
120
3.84 3.84 2.60 2.37 2.21 2.10 2.01 1.94 1.88 1.83 1.75 1.67 1.57 1.52 1.46 1.39 1.32 1.22 1.00
اﻟﻤﺼﺪر :ﻋﻦ ])[Devore (1995
٤٩٠
ﻣﻠﺤﻖ )(٤ ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) f (1, 2ﻟﺘﻮزﯾﻊ
120
60
40
30
24
20
15
12
10
Fﻋﻨﺪ ) ( 0.01 9
8
7
6
5
4
3
2
1
4052 5000 5403 5625 5764 5859 5928 5981 6022 6056 6106 6157 6209 6235 6261 6287 6313 6339 6366
1
2 1
98.50 99.00 99.17 99.25 99.30 99.33 99.36 99.37 99.39 99.40 99.42 99.43 99.45 99.46 99.47 99.47 99.48 99.49 99.50
2
34.12 30.82 29.46 28.71 28.24 27.91 27.67 27.49 27.35 27.23 27.05 26.87 26.69 26.60 26.50 26.41 26.32 26.22 26.13
3
21.20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 14.66 14.55 14.37 14.20 14.02 13.93 13.84 13.75 13.65 13.56 13.46
4
9.02
9.11
9.20
9.29
9.38
9.47
9.55
9.72
16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 10.16 10.05 9.89
5
6.88
6.97
7.06
7.14
7.23
7.31
7.40
7.56
7.72
7.87
7.98
8.10
8.26
8.47
8.75
9.15
13.57 10.92 9.78
6
5.65
5.74
5.82
5.91
5.99
6.07
6.16
6.31
6.47
6.62
6.72
6.84
6.99
7.19
7.46
7.85
8.45
12.25 9.55
7
4.86
4.95
5.03
5.12
5.20
5.28
5.36
5.52
5.67
5.81
5.91
6.03
6.18
6.37
6.63
7.01
7.59
11.26 8.65
8
4.31
4.40
4.48
4.57
4.65
4.73
4.81
4.96
5.11
5.26
5.35
5.47
5.61
5.80
6.06
6.42
6.99
10.56 8.02
9
3.91
4.00
4.08
4.17
4.25
4.33
4.41
4.56
4.71
4.85
4.94
5.06
5.20
5.39
5.64
5.99
6.55
10.04 7.56
10
3.60
3.69
3.78
3.86
3.94
4.02
4.10
4.25
4.40
4.54
4.63
4.74
4.89
5.07
5.32
5.67
6.22
7.21
9.65
11
3.45 3.369 .07 3.25 3.17
3.54
3.62
3.70
3.78
3.86
4.01
4.16
4.30
4.39
4.50
4.64
4.82
5.06
5.41
5.95
6.93
9.33
12
3.34
3.43
3.51
3.59
3.66
3.82
3.96
4.10
4.19
4.30
4.41
4.62
4.86
5.21
5.74
6.70
9.07
13
3.00
3.09
3.18
3.27
3.35
3.43
3.51
3.66
3.80
3.94
4.03
4.14
4.28
4.46
4.69
5.04
5.56
6.51
8.86
14
2.87
2.96
3.05
3.13
3.21
3.29
3.37
3.52
3.67
3.80
3.89
4.00
4.14
4.32
4.56
4.89
5.42
6.36
8.68
15
2.75
2.84
2.93
3.02
3.10
3.18
3.26
3.41
3.55
3.69
3.78
3.89
4.03
4.20
4.44
4.77
5.29
6.23
8.53
16
2.65
2.75
2.83
2.92
3.00
3.08
3.16
3.31
3.46
3.59
3.68
3.79
3.93
4.10
4.34
4.67
5.18
6.11
8.40
17
2.57
2.66
2.75
2.84
2.92
3.00
3.08
3.23
3.37
3.51
3.60
3.71
3.84
4.01
4.25
4.58
5.09
6.01
8.29
18
2.49
2.58
2.67
2.76
2.84
2.92
3.00
3.15
3.30
3.43
3.52
3.63
3.77
3.94
4.17
4.50
5.01
5.93
8.18
19
2.42
2.52
2.61
2.69
2.78
2.86
2.94
3.09
3.23
3.37
3.46
3.56
3.70
3.87
4.10
4.43
4.94
5.85
8.10
20
2.36
2.46
2.55
2.64
2.72
2.80
2.88
3.03
3.17
3.31
3.40
3.51
3.64
3.81
4.04
4.37
4.87
5.78
8.02
21
2.31
2.40
2.50
2.58
2.67
2.75
2.83
2.98
3.12
3.26
3.35
3.45
3.59
3.76
3.99
4.31
4.82
5.72
7.95
22
2.26
2.35
2.45
2.54
2.62
2.70
2.78
2.93
3.07
3.21
3.30
3.41
3.54
3.71
3.94
4.26
4.76
5.66
7.88
23
2.21
2.31
2.40
2.49
2.58
2.66
2.74
3.03 2..89
3.17
3.26
3.36
3.50
3.67
3.90
4.22
4.72
5.61
7.82
24
2.17
2.27
2.36
2.45
2.54
2.62
2.70
2.85
2.99
3.13
3.22
3.32
3.46
3.63
3.85
4.18
4.68
5.57
7.77
25
2.13
2.23
2.33
2.42
2.50
2.58
2.66
2.81
2.96
3.09
3.18
3.29
3.42
3.59
3.82
4.14
4.64
5.53
7.72
26
2.10
2.20
2.29
2.38
2.47
2.55
2.63
2.78
2.93
3.06
3.15
3.26
3.39
3.56
3.78
4.11
4.60
5.49
7.86
27
2.06
2.17
2.26
2.35
2.44
2.52
2.60
2.75
2.90
3.03
3.12
3.23
3.36
3.53
3.75
4.07
4.57
5.45
7.64
28
2.03
2.14
2.23
2.33
2.41
2.49
2.57
2.73
2.87
3.00
3.09
3.20
3.33
3.50
3.73
4.04
4.54
5.42
7.60
29
2.01
2.11
2.21
2.30
2.39
2.47
2.55
2.70
2.84
2.98
3.07
3.17
3.30
3.47
3.70
4.02
4.51
5.39
7.56
30
1.80
1.92
2.02
2.11
2.20
2.29
2.37
2.52
2.66
2.80
3.89
2.99
3.12
3.29
3.51
3.83
4.31
5.18
7.31
40
1.60
1.73
1.84
1.94
2.03
2.12
2.20
2.35
2.50
2.63
3.72
2.82
2.95
3.12
3.34
3.65
4.13
4.98
7.08
60
1.38
1.53
1.66
1.76
1.86
1.95
2.03
2.19
2.34
2.47
3.56
2.66
2.79
2.96
3.17
3.48
3.95
4.79
6.85
120
1.00
1.32
1.47
1.59
1.70
1.79
1.88
2.04
2.18
2.32
3.41
2.51
2.64
2.80
3.02
3.32
3.78
4.61
6.63
٤٩١
ﻣﻠﺤﻖ )(٥ ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ 2ﻟﺘﻮزﯾﻊ 2 .005
.01
.025
.05
.10
.90
.95
.975
.99
.995
7.882 10.59 12.83 14.86 16.74 18.54 20.27 21.95 23.58 25.18 26.75 28.30 29.81 31.31 32.79 34.26 35.71 37.15 38.58 39.99 41.39 42.79 44.17 45.55 46.92 48.29 49.64 50.99 52.33 53.67 55.00 56.32 57.64 58.96 60.27 61.58 62.88 64.18 65.47 66.76
6.637 9.210 11.34 13.27 15.08 16.81 18.47 20.09 21.66 23.20 24.72 26.21 27.68 29.14 30.57 32.00 33.40 34.80 36.19 37.56 38.93 40.28 41.63 42.98 44.31 45.64 46.96 48.27 49.58 50.89 52.19 53.48 54.77 56.06 57.34 58.61 59.89 61.16 62.42 63.69
5.025 7.378 9.348 11.14 12.83 14.44 16.01 17.53 19.02 20.48 21.92 23.33 24.73 26.11 27.48 28.84 30.19 31.52 32.85 34.17 35.47 36.78 38.07 39.36 40.64 41.92 43.19 44.46 45.77 46.97 48.23 49.48 50.72 51.96 53.20 54.43 55.66 56.89 58.11 59.34
3.843 5.992 7.815 9.488 11.07 12.59 14.06 15.50 16.91 18.30 19.67 21.02 22.36 23.68 24.99 26.29 27.58 28.86 30.14 31.41 32.67 33.92 35.17 36.41 37.65 38.88 40.11 41.33 42.55 43.77 44.98 46.19 47.40 48.60 49.80 50.99 52.19 53.38 54.57 55.75
2.706 4.605 6.251 7.779 9.236 10.64 12.01 13.36 14.68 15.98 17.27 18.54 19.81 21.06 22.30 23.54 24.76 25.98 27.20 28.41 29.61 30.81 32.00 33.19 34.38 35.56 36.74 37.91 39.08 40.25 41.42 42.58 43.74 44.90 46.05 47.21 48.36 49.51 50.66 51.80
0.016 0.211 0.584 1.064 1.610 2.204 2.833 3.490 4.168 4.865 5.578 6.304 7.041 7.790 8.547 9.312 10.08 10.86 11.65 12.44 13.24 14.04 14.84 15.65 16.47 17.29 18.11 18.93 19.76 20.59 21.43 22.27 23.11 23.95 24.79 25.64 26.49 27.34 28.19 29.05
0.004 0.103 0.352 0.711 1.145 1.635 2.167 2.733 3.325 3.940 4.575 5.226 5.892 6.571 7.261 7.962 8.682 9.390 10.11 10.85 11.59 12.33 13.09 13.84 14.61 15.37 16.15 16.92 17.70 18.49 19.28 20.07 20.86 21.66 22.46 23.26 24.07 24.88 25.69 26.50
0.001 0.051 0.216 0.484 0.831 1.237 1.690 2.180 2.700 3.247 3.816 4.404 5.009 5.629 6.262 6.908 7.564 8.231 8.906 9.591 10.28 10.98 11.68 12.40 13.12 13.84 14.57 15.30 16.14 16.79 17.53 18.29 19.04 19.80 20.56 21.33 22.10 22.87 23.65 24.43
0.000 0.020 0.115 0.297 0.554 0.872 1.239 1.646 2.088 2.558 3.053 3.571 4.107 4.660 5.229 5.812 6.407 7.015 7.632 8.260 8.897 9.542 10.19 10.85 11.52 12.19 12.87 13.56 14.25 14.95 15.65 16.36 17.07 17.78 18.50 19.23 19.96 20.69 21.42 22.16
0.000 0.010 0.072 0.207 0.412 0.676 0.989 1.344 1.735 2.156 2.603 3.074 3.565 4.075 4.600 5.142 5.697 6.265 6.843 7.434 8.033 8.643 9.260 9.886 10.51 11.16 11.80 12.46 13.12 13.78 14.45 15.13 15.81 16.50 17.19 17.88 18.58 19.28 19.99 20.70
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
اﻟﻤﺼﺪر :ﻋﻦ ])[Devore(1995 ٤٩٢
ﻣﻠﺤﻖ )(٦ ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ t ﻟﺘﻮزﯾﻊ
t
.0005 636.62 31.598 12.924 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 4.015 3.965 3.922 3.883 3.850 3.819 3.792 3.767 3.745 3.725 3.707 3.690 3.674 3.659 3.646 3.551 3.460 3.373 3.291
اﻟﻤﺼﺪر :
.001 318.31 22.326 10.213 7.173 5.893 5.208 4.785 4.501 4.297 4.144 4.025 3.930 3.852 3.787 3.733 3.686 3.646 3.610 3.579 3.552 3.527 3.505 3.485 3.467 3.450 3.435 3.421 3.408 3.396 3.385 3.307 3.232 3.160 3.090
.005 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.704 2.660 2.617 2.576
.01 31.821 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.423 2.390 2.358 2.326
ﻋﻦ ])[Devore (1995
٤٩٣
.025 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 2.093 2.086 2.080 2.074 2.069 2.064 2.060 2.056 2.052 2.048 2.045 2.042 2.021 2.000 1.980 1.960
.05 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 1.721 1.717 1.714 1.711 1.708 1.706 1.703 1.701 1.699 1.697 1.684 1.671 1.658 1.645
.10 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 1.328 1.325 1.323 1.321 1.319 1.318 1.316 1.315 1.314 1.313 1.311 1.310 1.303 1.296 1.289 1.282
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 60 120
٤٩٤