PROBABILITY DISTRIBUTION OF SUM OF TWO CONTINUOUS VARIABLES AND CONVOLUTION

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Journal for Research| Volume 02| Issue 01 | March 2016 ISSN: 2395-7549

Probability Distribution of Sum of Two Continuous Variables and Convolution Rashmi R. Keshvani Professor Department of Mathematics Sarvajanik College of Engineering & Technology, Surat (Guj.) India

Yamini M. Parmar Assistant Professor Department of Mathematics Government Engineering College, Gandhi nagar (Guj), India

Abstract All physical subjects, involving random phenomena, something depending upon chance, naturally find their own way to theory of Statistics. Hence there arise relations between the results derived for hose random phenomena in different physical subjects and the concepts of Statistics. Convolution theorem has a variety of applications in field of Fourier transforms and many other situations, but it bears beautiful applications in field of statistics also .Here in this paper authors want to discuss some notions of Electrical Engineering in terms of convolution of some probability distributions. Keywords: A Probability Distribution, An Uniform Probability Distribution, Central Limit Theorem, Convolution, Mean And Variance of a Probability Distribution, Triangular Function, Unit Rectangle Function _______________________________________________________________________________________________________

I. INTRODUCTION Occurrence of resistance of a resistor, with its tolerance can be expressed as a probability distribution. In those circumstances, what would be the resultant distribution describing occurrence of resistances of resistors having different resistance and different tolerances, when combined in series? How means and variances are interrelated with that resultant distribution, is main focus of this paper. In other words, to obtain probability distribution of sum of two random variables is main objective of this paper.

II. SOME BASIC CONCEPTS OF STATISTICS A Probability Distribution, Its Mean and Variance: A real valued function đ?‘?(đ?‘Ľ) is said to be a probability distribution of a random variable đ?‘Ľ, if (1) đ?‘?(đ?‘Ľ) ≼ 0, ∀ đ?‘Ľ ∈ đ?‘… and ∞ ∑đ?‘Žđ?‘™đ?‘™ đ?‘Ľ đ?‘?(đ?‘Ľ) = 1 where đ?‘Ľ is a discrete random variable. or (2) đ?‘?(đ?‘Ľ) ≼ 0, ∀ đ?‘Ľ ∈ đ?‘… and âˆŤâˆ’âˆž đ?‘?(đ?‘Ľ)đ?‘‘đ?‘Ľ = 1 where đ?‘Ľ is a continuous random variable. The mean of probability distribution đ?‘?(đ?‘Ľ), denoted by Îź, is defined as ∞ (1) đ?œ‡ = ∑đ?‘Žđ?‘™đ?‘™ đ?‘Ľ đ?‘Ľ đ?‘?(đ?‘Ľ), if đ?‘Ľ is discrete, or (2) đ?œ‡ = âˆŤâˆ’âˆž đ?‘Ľ đ?‘?(đ?‘Ľ)đ?‘‘đ?‘Ľ, if đ?‘Ľ is continuous. (1) The variance of probability distribution đ?‘?(đ?‘Ľ), denoted by đ?œŽ 2 , is defined as ∞ (1) đ?œŽ 2 = ∑đ?‘Žđ?‘™đ?‘™ đ?‘Ľ ( đ?‘Ľ − đ?œ‡ )2 đ?‘?(đ?‘Ľ), if đ?‘Ľ is discrete, or (2) đ?œŽ 2 = âˆŤâˆ’âˆž( đ?‘Ľ − đ?œ‡ )2 đ?‘?(đ?‘Ľ)đ?‘‘đ?‘Ľ, if đ?‘Ľ is continuous. [1] (2) The Uniform Probability Distribution: The uniform probability distribution, with parameters đ?›ź and đ?›˝, has probability distribution đ?‘?(đ?‘Ľ) defined as 1 đ?‘–đ?‘“ đ?›ź < đ?‘Ľ < đ?›˝ đ?‘?(đ?‘Ľ) = { đ?›˝âˆ’đ?›ź 0 đ?‘’đ?‘™đ?‘ đ?‘’đ?‘¤â„Žđ?‘’đ?‘&#x;đ?‘’

(3)

III. THE CONCEPT OF CONVOLUTION The convolution of two functions đ?‘“(đ?‘Ľ) and đ?‘”(đ?‘Ľ), [2], denoted by đ?‘“(đ?‘Ľ) ∗ đ?‘”(đ?‘Ľ), or (đ?‘“ ∗ đ?‘”)(đ?‘Ľ) is defined as ∞ đ?‘“(đ?‘Ľ) ∗ đ?‘”(đ?‘Ľ) = âˆŤâˆ’âˆž đ?‘“(đ?‘˘)đ?‘”(đ?‘Ľ − đ?‘˘)đ?‘‘đ?‘˘ (4) One can easily check that the operation of convolution is commutative, associative and also distributive over addition. Convolution has additive property also. That is ∞ đ?‘“ ∗ (đ?‘”1 + đ?‘”2 )(đ?‘Ľ) = đ?‘“(đ?‘Ľ) ∗ (đ?‘”1 + đ?‘”2 )(đ?‘Ľ) = đ?‘“(đ?‘Ľ) ∗ (đ?‘”1 (đ?‘Ľ) + đ?‘”2 (đ?‘Ľ)) = âˆŤâˆ’âˆž đ?‘“(đ?‘˘) (đ?‘”1 (đ?‘Ľ − đ?‘˘) + đ?‘”2 ( đ?‘Ľ − đ?‘˘))đ?‘‘đ?‘˘ = −∞ −∞ âˆŤâˆž đ?‘“(đ?‘˘)đ?‘”1 (đ?‘Ľ − đ?‘˘)đ?‘‘đ?‘˘ + âˆŤâˆž đ?‘“(đ?‘˘)đ?‘”2 (đ?‘Ľ − đ?‘˘)đ?‘‘đ?‘˘ = ( đ?‘“ ∗ đ?‘”1 )(đ?‘Ľ) + ( đ?‘“ ∗ đ?‘”2 )(đ?‘Ľ) That is đ?‘“ ∗ (đ?‘”1 + đ?‘”2 ) = ( đ?‘“ ∗ đ?‘”1 ) + ( đ?‘“ ∗ đ?‘”2 )

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58

Probability Distribution of Sum of Two Continuous Variables and Convolution (J4R/ Volume 02 / Issue 01 / 009)

IV. PROBABILITY DISTRIBUTION OF A SUM Suppose two probability distributions đ?‘?1 (đ?‘Ľ1 ) and đ?‘?2 (đ?‘Ľ2 ) are given and it is required to determine probability distribution đ?‘?(đ?‘Ľ) of random variable đ?‘Ľ = đ?‘Ľ1 + đ?‘Ľ2 , sum of these two random variables. That is đ?‘?(đ?‘Ľ) shows probability that sum of đ?‘Ľ1 and đ?‘Ľ2 remains đ?‘Ľ. That is đ?‘?(đ?‘Ľ) = đ?‘?( đ?‘Ľ = đ?‘Ľ1 + đ?‘Ľ2 ). As for example, suppose a large number of 100-ohm resisters , quoted as subject to a 10 percent tolerance are drawn from an infinite supply containing equal numbers of resisters in any 1-ohm interval between 90 and 110 and none outside this range. Suppose that the stock of resisters has been drawn from a supply in which the frequency of occurrence of resisters between đ?‘…1 and đ?‘…1 đ?‘‘đ?‘…1 is đ?‘ƒ1 (đ?‘…1 )đ?‘‘đ?‘…1 . [2] So đ?‘ƒ1 (đ?‘…1 ) has to be probability distribution, more precisely, uniform probability distribution over interval 90 to 110, ensuring that, 1 1 = đ?‘–đ?‘“ 90 ≤ đ?‘…1 ≤ 110 ∞ âˆŤâˆ’âˆž đ?‘ƒ1 (đ?‘…1 ) đ?‘‘đ?‘…1 = 1, that is đ?‘ƒ1 (đ?‘…1 ) = {(110−90) 20 0 đ?‘’đ?‘™đ?‘ đ?‘’đ?‘¤â„Žđ?‘’đ?‘&#x;đ?‘’ 1 đ?‘… −100 Expressing đ?‘ƒ1 (đ?‘…1 ) in terms of rectangle function, đ?‘ƒ1 (đ?‘…1 ) = âˆ? ( 1 ), where the rectangle function, denoted by âˆ?(đ?‘Ľ), 20 20 is defined as |đ?‘Ľ| < 1â „2 |đ?‘Ľ| > 1â „2

1 âˆ?(đ?‘Ľ) = { 0 ∞

1

The mean of this distribution is đ?œ‡1 = âˆŤâˆ’âˆž đ?‘…1 đ?‘ƒ1 (đ?‘…1 )đ?‘‘đ?‘…1 = ∞

âˆŤâˆ’âˆž(đ?‘…1 − 100)2 đ?‘ƒ(đ?‘…1 )đ?‘‘đ?‘…1 =

1

110

1

âˆŤ (đ?‘…1 − 100)2 đ?‘‘đ?‘…1 = 20 [ 20 90

20

110

1

đ?‘…2

110

âˆŤ90 đ?‘…1 đ?‘‘đ?‘…1 = 20 [ 21 ]

(đ?‘…1 −100)3 110

]

3

90

90

=

= 100, and the variance đ?œŽ12 =

100 3

Similarly for a stock of 50-ohm resisters, with 5-ohm tolerance, another probability distributionđ?‘ƒ2 (đ?‘…2 ), defined, as đ?‘ƒ2 (đ?‘…2 ) = 1 1 = đ?‘–đ?‘“ 45 ≤ đ?‘…2 ≤ 55 {(55−45) 10 will describe the frequency of occurrence of resisters. 0 đ?‘’đ?‘™đ?‘ đ?‘’đ?‘¤â„Žđ?‘’đ?‘&#x;đ?‘’ 1 đ?‘… −50 Expressing đ?‘ƒ2 (đ?‘…2 ) also, in terms of rectangle function, đ?‘ƒ2 (đ?‘…2 ) = âˆ? ( 2 ) and the mean of đ?‘ƒ2 (đ?‘…2 ), will be đ?œ‡2 = 50 10

25

10

and variance đ?œŽ22 = 3 If an electronic circuit, combining two of these stock resistors in series[2], and suppose the resulting distribution describing the probability of occurrence of resistances, is denoted by đ?‘ƒ(đ?‘…), then here problem is to know, what mathematical relation, đ?‘ƒ(đ?‘…) bears with đ?‘ƒ1 (đ?‘…1 ) and đ?‘ƒ2 (đ?‘…2 ) when đ?‘… = đ?‘…1 + đ?‘…2 . It is clear that đ?‘… = đ?‘…1 + đ?‘…2 will vary from 90 + 45 = 135 to 110 + 55 = 165, with mean equal to 150. It is also clear that the probability that đ?‘… assumes a value less than or equal to135, that is đ?‘ƒ(đ?‘… ≤ 135) = 0 and the probability that đ?‘… assumes a value greater than or equal to 165 , that is đ?‘ƒ(đ?‘… ≼ 165) = 0. Probability that đ?‘… assumes a particular value between 135 and 165, say, đ?‘… = 140, that is đ?‘…1 + đ?‘…2 = 140, denoted by đ?‘ƒ(140) will be that area where this sum assumes value 140.That is đ?‘…2 should remain 140 − đ?‘…1 . Required area will be 1 110 1 đ?‘… =50 1 đ?‘… =50 1 1 đ?‘ƒ(140) = âˆŤ90 đ?‘ƒ2 (140 − đ?‘…1 )đ?‘‘đ?‘…1 = âˆŤđ?‘… 2=30 đ?‘ƒ2 (đ?‘…2 )đ?‘‘ đ?‘…2 = âˆŤđ?‘… 2=45 đ?‘‘ đ?‘…2 = , 20

20

20

2

2

10

40

As đ?‘…1 is nonzero if 90 ≤ đ?‘…1 ≤ 110 and đ?‘…2 is nonzero, if 45 ≤ đ?‘…2 ≤ 55. One can understand that, this is nothing but convolution of đ?‘ƒ1 (đ?‘…1 ) andđ?‘ƒ2 (đ?‘…2 ), that is ∞

đ?‘ƒ(140) = (đ?‘ƒ1 ∗ đ?‘ƒ2 )(140) = âˆŤ đ?‘ƒ1 (đ?‘…1 )đ?‘ƒ2 (140 − đ?‘…1 )đ?‘‘đ?‘…1 −∞

This is an instance of the basic convolution relation between the probability distribution describing the sum of two quantities and the probability distributions of the given quantities. Now taking in general, let resistor đ?‘…1 have greater resistance đ?‘Ž1 with tolerance đ?‘?1 , and resistor đ?‘…2 have smaller resistance đ?‘Ž2 with tolerance đ?‘?2 . As đ?‘Ž1 > đ?‘Ž2 and, as đ?‘Ž1 > đ?‘?1 , đ?‘Ž2 > đ?‘?2 , there is no loss of generality any, if we assume đ?‘Ž1 > đ?‘Ž2 > đ?‘?1 > đ?‘?2 > 0. In terms of rectangle functions đ?‘…1 = It is quite obvious that đ?œ‡1 =

1 2đ?‘?1

�(

đ?‘…1 −đ?‘Ž1 2đ?‘?1

∞ âˆŤâˆ’âˆž đ?‘…1 đ?‘ƒ1 (đ?‘…1 )đ?‘‘đ?‘…1 ∞

=

) and đ?‘…2 =

1 2đ?‘?2

�(

đ?‘Ž1 +đ?‘?1 đ?‘… đ?‘‘đ?‘…1 âˆŤ 2đ?‘?1 đ?‘Ž1 −đ?‘?1 1 1

=

đ?‘Ž2 +đ?‘?2

đ?‘…2 −đ?‘Ž2 2đ?‘?2 4đ?‘Ž1 đ?‘?1 4đ?‘?1

) = đ?‘Ž1

1 4đ?‘Ž2 đ?‘?2 âˆŤ đ?‘… đ?‘‘đ?‘… = = đ?‘Ž2 2đ?‘?2 đ?‘Ž2−đ?‘?2 2 2 4đ?‘?2 ∞ 1 đ?‘Ž1+đ?‘?1 1 (đ?‘…1 − đ?‘Ž1 )3 đ?‘Ž1+đ?‘?1 đ?‘?12 (đ?‘…1 − đ?‘Ž1 )2 đ?‘‘đ?‘…1 = đ?œŽ12 = âˆŤ (đ?‘…1 − đ?‘Ž1 )2 đ?‘ƒ1 (đ?‘…)đ?‘‘đ?‘…1 = âˆŤ [ ]đ?‘Ž1−đ?‘?1 = 2đ?‘?1 đ?‘Ž1−đ?‘?1 2đ?‘?1 3 3 −∞ ∞ đ?‘Ž2 +đ?‘?2 2 3 (đ?‘… ) 1 1 − đ?‘Ž đ?‘? 2 2 2 đ?‘Ž +đ?‘? ( đ?‘…2 − đ?‘Ž2 )2 đ?‘‘đ?‘… = đ?œŽ22 = âˆŤ ( đ?‘…2 − đ?‘Ž2 )2 đ?‘ƒ2 (đ?‘…)đ?‘‘đ?‘… = âˆŤ [ ]đ?‘Ž22 −đ?‘?22 = 2đ?‘?2 đ?‘Ž2−đ?‘?2 2đ?‘?2 3 3 −∞ đ?œ‡2 = âˆŤ đ?‘…2 đ?‘ƒ2 (đ?‘…2 )đ?‘‘đ?‘…2 = −∞

To determine đ?‘ƒ(đ?‘…), following cases are to be considered separately, If đ?‘… ≤ đ?‘Ž1 + đ?‘Ž2 − đ?‘?1 − đ?‘?2 or đ?‘… ≼ đ?‘Ž1 + đ?‘Ž2 + đ?‘?1 + đ?‘?2 , then đ?‘ƒ(đ?‘…) = 0 If (đ?‘Ž1 + đ?‘Ž2 ) − (đ?‘?1 + đ?‘?2 ) < đ?‘… < (đ?‘Ž1 + đ?‘Ž2 ) − đ?‘?1 + đ?‘?2 ∞ đ?‘Ž1 + đ?‘?1 đ?‘…−(đ?‘Ž1 −đ?‘?1 ) 1 1 đ?‘ƒ(đ?‘…) = âˆŤâˆ’âˆž đ?‘ƒ1 (đ?‘…1 )đ?‘ƒ2 (đ?‘… − đ?‘…1 )đ?‘‘đ?‘…1 = âˆŤđ?‘Ž −đ?‘? đ?‘ƒ2 (đ?‘… − đ?‘…1 )đ?‘‘đ?‘…1 = âˆŤđ?‘…−(đ?‘Ž + đ?‘? ) đ?‘ƒ2 (đ?‘…2 )đ?‘‘đ?‘…2 = 2đ?‘?1

1

1

2đ?‘?1

1

1

đ?‘…−(đ?‘Ž1 −đ?‘?1 ) 1 đ?‘‘đ?‘…2 âˆŤ 4đ?‘?1 đ?‘?2 đ?‘Ž2 −đ?‘?2

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59

Probability Distribution of Sum of Two Continuous Variables and Convolution (J4R/ Volume 02 / Issue 01 / 009)

𝑅 − (𝑎1 − 𝑏1 ) < 𝑎2 + 𝑏2 (𝑅 − (𝑎1 − 𝑏1 ) − (𝑎2 − 𝑏2 )) as (𝑏1 > 𝑏2 𝑎𝑛𝑑 𝑅 < 𝑎1 + 𝑎2 − 𝑏1 + 𝑏2 ) ⇒ { 𝑅 − (𝑎1 + 𝑏1 ) < 𝑎2 − 𝑏2 If (𝑎1 + 𝑎2 ) − 𝑏1 + 𝑏2 ≤ 𝑅 ≤ ( 𝑎1 + 𝑎2 ) + 𝑏1 − 𝑏2 ∞ 𝑎1 + 𝑏1 𝑅−(𝑎1 −𝑏1 ) 1 1 𝑃(𝑅) = ∫−∞ 𝑃1 (𝑅1 )𝑃2 (𝑅 − 𝑅1 )𝑑𝑅1 = ∫𝑎 −𝑏 𝑃2 (𝑅 − 𝑅1 )𝑑𝑅1 = ∫𝑅−(𝑎 + 𝑏 ) 𝑃2 (𝑅2 )𝑑𝑅2 =

1

4𝑏1 𝑏2

=

1 4𝑏1 𝑏2

𝑎2 +𝑏2 𝑑𝑅2 2 −𝑏2

∫𝑎

2𝑏2

=

=

4𝑏1 𝑏2

2𝑏1 1

1

2𝑏1

1

𝑎2 +𝑏2 1 𝑑𝑅2 ∫ 4𝑏1 𝑏2 𝑅−(𝑎1 + 𝑏1 )

=

1

1

1 4𝑏1 𝑏2

1

2𝑏1

If 𝑎1 + 𝑎2 + 𝑏1 − 𝑏2 ≤ 𝑅 ≤ 𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 , ∞ 𝑎1 + 𝑏1 1 𝑃(𝑅) = ∫−∞ 𝑃1 (𝑅1 )𝑃2 (𝑅 − 𝑅1 )𝑑𝑅1 = 𝑃2 (𝑅 − 𝑅1 )𝑑𝑅1 = ∫ 2𝑏 𝑎 −𝑏 =

1

1

𝑅−(𝑎1 −𝑏1 ) 1 𝑃 (𝑅2 )𝑑𝑅2 ∫ 2𝑏1 𝑅−(𝑎1 + 𝑏1 ) 2

(𝑎2 + 𝑏2 + 𝑎1 + 𝑏1 − 𝑅 ),

𝑅 − (𝑎1 − 𝑏1 ) > 𝑎2 + 𝑏2 𝑅 − (𝑎1 + 𝑏1 ) > 𝑎2 − 𝑏2 Thus the resulting distribution 𝑃(𝑅), describing the probability of resistances in the electronic circuit, combining two of these stock resistors in series, 𝑅 = 𝑅1 + 𝑅2 will be obtained as as (𝑎1 + 𝑎2 + 𝑏1 − 𝑏2 ≤ 𝑅 𝑎𝑛𝑑 𝑏1 > 𝑏2 ) ⇒ {

0

𝑖𝑓 𝑅 ≤ 𝑎1 + 𝑎2 −𝑏1 −𝑏2

1 4𝑏1 𝑏2

𝑃(𝑅) =

1

4𝑏1 𝑏2

(𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 −𝑅 )

{ 0 ∞ ∫−∞ 𝑃(𝑅)𝑑𝑅

=

4𝑏1 𝑏2

{2

𝑏22

(5)

𝑖𝑓 𝑎1 + 𝑎2 +𝑏1 − 𝑏2 ≤ 𝑅 ≤ 𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 𝑖𝑓 𝑅 ≥ 𝑎1 + 𝑎2 +𝑏1 + 𝑏2

2 (𝑎1 + 𝑎2 −𝑏1 + 𝑏2 )

(𝑅 − (𝑎1 − 𝑏1 + 𝑎2 − 𝑏2 )) 1 = { [ ] 4𝑏1 𝑏2 2 1

𝑖𝑓 𝑎1 + 𝑎2 −𝑏1 −𝑏2 ≤ 𝑅 ≤ 𝑎1 + 𝑎2 −𝑏1 + 𝑏2

𝑖𝑓 𝑎1 + 𝑎2 − 𝑏1 + 𝑏2 ≤ 𝑅 ≤ 𝑎1 + 𝑎2 + 𝑏1 − 𝑏2

2𝑏1 1

Now

(𝑅−(𝑎1 +𝑎2 −𝑏1 − 𝑏2 ))

(𝑎 + 𝑎 + 𝑏 + 𝑏 )

(𝑎 + 𝑎 +𝑏 − 𝑏 )

+ 2𝑏2 [ 𝑅 ](𝑎11+ 𝑎22−𝑏11 + 𝑏22) + [ (𝑎1 + 𝑎2 −𝑏1 −𝑏2 )

+ 2𝑏2 (2𝑏1 − 2𝑏2 ) +

2𝑏22

1

1

}=

−(𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 − 𝑅 )2 1 2 1 2 ] } 2 (𝑎 + 𝑎 + 𝑏 − 𝑏 )

4𝑏1 𝑏2

2

1

2

{ 4𝑏1 𝑏2 } = 1, Ensuring that 𝑃(𝑅) is a probability distribution.

This result shows that composite resistance can be readily found using convolution. The composite resistance is distributed trapezoid ally as described in the expression (5).

V. CONSEQUENCES OF THE CONVOLUTION RELATION Let 𝜇, 𝜇1 , 𝜇2 be mean of probability distributions 𝑃, 𝑃1 , 𝑃2 respectively. As 𝑃 is convolution of 𝑃1 and 𝑃2 , ∞

∞

∞

∞

𝜇 = ∫ 𝑅𝑃(𝑅)𝑑𝑅 = ∫ 𝑅(𝑃1 ∗ 𝑃2 )(𝑅) 𝑑𝑅 = ∫ 𝑅 (∫ 𝑃1 (𝑅1 )𝑃2 (𝑅 − 𝑅1 ) 𝑑𝑅1 ) 𝑑𝑅 −∞

∞

−∞

∞

∞

−∞

−∞ ∞

= ∫ 𝑃1 (𝑅1 )( ∫ 𝑅𝑃2 (𝑅 − 𝑅1 )𝑑𝑅 ) 𝑑𝑅1 = ∫ 𝑃1 (𝑅1 )(∫ (𝑅−𝑅1 + 𝑅1 )𝑃2 (𝑅 − 𝑅1 )𝑑𝑅 ) 𝑑𝑅1 −∞

∞

−∞

−∞

∞

−∞

∞

= ∫ 𝑃1 (𝑅1 ) ( ∫ (𝑅−𝑅1 )𝑃2 (𝑅 − 𝑅1 )𝑑𝑅 + 𝑅1 ∫ 𝑃2 (𝑅 − 𝑅1 )𝑑𝑅 ) 𝑑𝑅1 −∞

∞

−∞

∞

−∞ ∞

= ∫ 𝑃1 (𝑅1 ) ( ∫ (𝑅−𝑅1 )𝑃2 (𝑅 − 𝑅1 )𝑑(𝑅 − 𝑅1 ) + 𝑅1 ∫ 𝑃2 (𝑅 − 𝑅1 )𝑑(𝑅 − 𝑅1 )] 𝑑𝑅1 −∞

−∞

−∞

(as in the innermost integration 𝑅1 is constant, 𝑑(𝑅 − 𝑅1 ) = 𝑑𝑅 = 𝑑𝑅2 ) ∞ ∞ ∞ = ∫−∞ 𝑃1 (𝑅1 ) {𝜇2 + 𝑅1 }𝑑𝑅1 = 𝜇2 ∫−∞ 𝑃1 (𝑅1 ) 𝑑𝑅1 + ∫−∞ 𝑅1 𝑃1 (𝑅1 ) 𝑑𝑅1 = 𝜇2 + 𝜇1 ∞

∞

(6)

∞

𝜎 2 = ∫ (𝑅 − 𝜇)2 𝑃(𝑅)𝑑𝑅 = ∫ 𝑅2 𝑃(𝑅)𝑑𝑅 − 𝜇 2 = {∫ 𝑅2 (𝑃1 ∗ 𝑃2 )(𝑅)𝑑𝑅 } − 𝜇 2 −∞

−∞ ∞

∞

−∞

= { ∫ 𝑅2 (∫ 𝑃1 (𝑅1 )𝑃2 (𝑅 − 𝑅1 )𝑑𝑅1 ) 𝑑𝑅 } − 𝜇 2 −∞ ∞

= { ∫

∞

= ∫ −∞ ∞

= ∫ −∞

∞

𝑃1 (𝑅1 ) (∫

−∞

𝑅2 𝑃2 (𝑅 − 𝑅1 )𝑑𝑅) 𝑑𝑅1 } − 𝜇 2

−∞

((𝑅 − 𝑅1 ) + 𝑅1 ) 2 𝑃2 (𝑅 − 𝑅1 )𝑑𝑅) 𝑑𝑅1 − 𝜇 2

−∞

∞

𝑃1 (𝑅1 ) (∫

∞

𝑃1 (𝑅1 ) (∫

−∞

∞

= ∫

−∞

{(𝑅 − 𝑅1 ) 2 + 2 𝑅1 (𝑅 − 𝑅1 ) + 𝑅12 } 𝑃2 (𝑅 − 𝑅1 )𝑑𝑅 ) 𝑑𝑅1 − 𝜇 2

−∞ ∞

𝑃1 (𝑅1 ) ( {∫

(𝑅 − 𝑅1 )2 𝑃2 (𝑅 − 𝑅1 )𝑑(𝑅 − 𝑅1 )} + 2𝑅1 𝜇2 + 𝑅12 ) 𝑑𝑅1 − 𝜇 2

−∞

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60

Probability Distribution of Sum of Two Continuous Variables and Convolution (J4R/ Volume 02 / Issue 01 / 009) ∞

𝑃1 (𝑅1 )(𝜎22 + 𝜇22 + 2𝑅1 𝜇2 + 𝑅12 ) 𝑑𝑅1 − 𝜇 2

= ∫ −∞

= (𝜎22 + 𝜇22 ) + 2𝜇1 𝜇2 + (𝜎12 + 𝜇12 ) − (𝜇1 + 𝜇2 )2 = 𝜎12 + 𝜎22

(7)

VI. VERIFICATION OF THE RESULTS For combined resistor in series, ∞ The mean 𝜇 = ∫−∞ 𝑅 𝑃(𝑅)𝑑𝑅 = (𝑎1 + 𝑎2 −𝑏1 + 𝑏2 ) (𝑅2 = ∫ 4𝑏1 𝑏2 (𝑎1 + 𝑎2 −𝑏1 −𝑏2 ) 𝑏1 + 𝑏2 ) − 𝑅2 } 𝑑𝑅 1

=

(𝑎1 + 𝑎2 −𝑏1 + 𝑏2 ) 1 (𝑅2 ∫ 4𝑏1 𝑏2 (𝑎1 + 𝑎2 −𝑏1 −𝑏2 )

− 𝑅(𝑎1 + 𝑎2 − 𝑏1 − 𝑏2 ) ) 𝑑𝑅

− 𝑅(𝑎1 + 𝑎2 − 𝑏1 − 𝑏2 ) ) 𝑑𝑅 +

(𝑎1 + 𝑎2 +𝑏1 − 𝑏2 ) 1 𝑅𝑑𝑅 ∫ 2𝑏1 (𝑎1 + 𝑎2 −𝑏1 + 𝑏2 )

(𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 ) 1 ∫ 4𝑏1 𝑏2 (𝑎1 + 𝑎2 + 𝑏1 − 𝑏2 )

+

(𝑎 + 𝑎 −𝑏 + 𝑏 )

{ 𝑅(𝑎1 + 𝑎2 +

(𝑎 + 𝑎 +𝑏 − 𝑏 )

1 2 1 2 1 𝑅3 𝑅2 𝑅2 1 2 1 2 (𝑎1 + 𝑎2 − 𝑏1 − 𝑏2 )] { [ − + 2𝑏2 [ ] } 4𝑏1 𝑏2 3 2 2 (𝑎 + 𝑎 −𝑏 + 𝑏 ) (𝑎 + 𝑎 −𝑏 −𝑏 ) 1

2

1

2

1

2

(𝑎 + 𝑎 + 𝑏 + 𝑏 )

1

2

1 2 1 2 1 𝑅2 𝑅3 + { [ (𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 ) − ] } 4𝑏1 𝑏2 2 3 (𝑎 + 𝑎 + 𝑏 − 𝑏 ) 1

2

1

2

1 (𝑎1 + 𝑎2 − 𝑏1 + 𝑏2 )2 (𝑏1 + 5𝑏2 − 𝑎1 − 𝑎2 ) (𝑎1 + 𝑎2 − 𝑏1 − 𝑏2 )3 = { + + 4𝑏2 (𝑎1 + 𝑎2 )(𝑏1 − 𝑏2 )} 4𝑏1 𝑏2 6 6 3 2 (𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 ) 1 (𝑎1 + 𝑎2 + 𝑏1 − 𝑏2 ) (𝑎1 + 𝑎2 + 𝑏1 + 5𝑏2 ) + { − } 4𝑏1 𝑏2 6 6 2(𝑎1 + 𝑎2 ) { (𝑎1 + 𝑎2 )2 + 3(𝑏1 + 𝑏2 )2 ) + 12𝑏2 (𝑏1 − 𝑏2 ) − 2(𝑏1 − 𝑏2 )(𝑏1 + 5𝑏2 ) − (𝑎1 + 𝑎2 )2 + (𝑏1 − 𝑏2 )2 } = 24𝑏1 𝑏2 2(𝑎1 +𝑎2 ) {3(𝑏1 + 𝑏2 )2 + 12𝑏2 (𝑏1 − 𝑏2 ) − 2(𝑏1 − 𝑏2 )(𝑏1 + 5𝑏2 ) − (𝑏1 − 𝑏2 )2 } = (𝑎1 + 𝑎2 ) = 𝜇1 + 𝜇2 . The variance = 24𝑏1 𝑏2

∞

𝑎1 +𝑎2 −𝑏1 +𝑏2

=∫

𝜎 2 = ∫−∞(𝑅 − (𝑎1 + 𝑎2 ))2 𝑃(𝑅)𝑑𝑅

𝑎1 +𝑎2 −𝑏1 −𝑏2

={

=

𝑎1 +𝑎2 +𝑏1 −𝑏2 1 1 (𝑅 − (𝑎1 + 𝑎2 − 𝑏1 − 𝑏2 ))𝑑𝑅 + ∫ (𝑅 − (𝑎1 + 𝑎2 ))2 𝑑𝑅 4𝑏1 𝑏2 2𝑏1 𝑎1 +𝑎2 −𝑏1 +𝑏2 𝑎1 +𝑎2 + 𝑏1 +𝑏2 1 + ∫ (𝑅 − (𝑎1 + 𝑎2 ))2 ((𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 ) − 𝑅)𝑑𝑅 4𝑏1 𝑏2 𝑎1 +𝑎2 + 𝑏1 −𝑏2

(𝑅 − (𝑎1 + 𝑎2 ))2

𝑎1 +𝑎2 −𝑏1 +𝑏2 1 (( ∫ 4𝑏1 𝑏2 𝑎1 +𝑎2 −𝑏1 −𝑏2

3

2

(𝑅 − (𝑎1 + 𝑎2 )) + (𝑅 − (𝑎1 + 𝑎2 )) (𝑏1 + 𝑏2 )) 𝑑𝑅 +

𝑎1 +𝑎2 + 𝑏1 +𝑏2 1 ∫ 4𝑏1 𝑏2 𝑎1 +𝑎2 + 𝑏1 −𝑏2 4 𝑎1 +𝑎2 −𝑏1 +𝑏2 (𝑅−(𝑎1 +𝑎2 )) 1 {[ ] + (𝑏1 + 4𝑏1 𝑏2 4 𝑎1 +𝑎2 −𝑏1 −𝑏2 3

𝑎1 +𝑎2 +𝑏1 −𝑏2 (𝑅 1 +𝑎2 −𝑏1 +𝑏2

∫𝑎

− (𝑎1 + 𝑎2 ))2 𝑑𝑅 +

( ((𝑎1 + 𝑎2 ) − 𝑅)3 + (𝑅 − (𝑎1 + 𝑎2 ))2 ( 𝑏1 + 𝑏2 ))𝑑𝑅 } 𝑏2 ) [

(𝑅−(𝑎1 +𝑎2 )) 3

3 𝑎1 +𝑎2 −𝑏1 +𝑏2

]

}+

𝑎1 +𝑎2 −𝑏1 −𝑏2

3 𝑎1 +𝑎2 +𝑏1 +𝑏2

(𝑅−(𝑎1 +𝑎2 ))

𝑏2 ) [

1 2𝑏1

]

−[

4

2𝑏1

[

(𝑅−(𝑎1 +𝑎2 )) 3

]

3 𝑎1 +𝑎2 +𝑏1 −𝑏2

]

+

𝑎1 +𝑎2 −𝑏1 + 𝑏2

4 𝑎1 +𝑎2 +𝑏1 +𝑏2

((𝑎1 +𝑎2 )−𝑅)

𝑎1 +𝑎2 + 𝑏1 −𝑏2

1

1 4𝑏1 𝑏2

{( 𝑏1 +

}

𝑎1 +𝑎2 + 𝑏1 −𝑏2 (𝑏1 + 𝑏2 )3

(𝑏2 − 𝑏1 )4 (𝑏1 + 𝑏2 )4 (𝑏2 − 𝑏1 )3 1 1 2(𝑏1 − 𝑏2 )3 { − + (𝑏1 + 𝑏2 ) ( + )} + { } 4𝑏1 𝑏2 4 4 3 3 2𝑏1 3 (𝑏1 + 𝑏2 )3 (𝑏1 − 𝑏2 )3 1 (𝑏1 + 𝑏2 )4 (𝑏2 − 𝑏1 )4 + {( 𝑏1 + 𝑏2 ) { − }– + } 4𝑏1 𝑏2 3 3 4 4 (𝑏2 − 𝑏1 )3 1 (𝑏1 + 𝑏2 )4 8𝑏2 (𝑏2 − 𝑏1 )3 (𝑏1 + 7𝑏2 ) − = { + } 4𝑏1 𝑏2 6 6 6 (𝑏12 + 𝑏22 ) 1 = {(𝑏1 + 𝑏2 )4 − (𝑏2 − 𝑏1 )4 } = = 𝜎12 + 𝜎22 24𝑏1 𝑏2 3 Thus the results (6) and (7) hold true. What happens if both the resistors are having same resistance as well as same tolerance? That is if 𝑎1 = 𝑎2 = 𝑎 and 𝑏1 = 𝑏2 = 𝑏 then 𝑃(𝑅) will be defined as follows: =

0

𝑃(𝑅) = { That is, 𝑃(𝑅) =

1 2𝑏

Λ(

1 (𝑅−2( 𝑎−𝑏)) 4𝑏2 1 (2(𝑎+𝑏)−𝑅 ) 4𝑏2

0 𝑅−2𝑎 2𝑏

𝑖𝑓 𝑅 ≤ 2(𝑎−𝑏)

𝑖𝑓 2(𝑎−𝑏)≤ 𝑅 ≤ 2𝑎 𝑖𝑓

(8)

2𝑎 ≤ 𝑅 ≤ 2( 𝑎+𝑏) 𝑖𝑓 𝑅 ≥2( 𝑎+𝑏)

), where Λ(𝑥), known as triangular function, is defined as

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61

Probability Distribution of Sum of Two Continuous Variables and Convolution (J4R/ Volume 02 / Issue 01 / 009)

Λ(�) = {

[−

2đ?‘Ž

2(đ?‘Ž+đ?‘?) 1

1

= âˆŤ2(đ?‘Žâˆ’đ?‘?) 2 (đ?‘… − 2( đ?‘Ž − đ?‘?))đ?‘‘đ?‘… + âˆŤ2đ?‘Ž 4đ?‘?

(2(đ?‘Ž+đ?‘?)−đ?‘…)2 2

�� |�| > 1 �� |�| ≤ 1

đ?œ‡1 = đ?œ‡2 , đ?œŽ1 = đ?œŽ2 and

In this case, ∞ âˆŤâˆ’âˆž đ?‘ƒ(đ?‘…)đ?‘‘đ?‘…

0 1 − |đ?‘Ľ|

2(đ?‘Ž+đ?‘?)

]

2đ?‘Ž

}=

∞

1 4đ?‘? 2

4đ?‘? 2

2đ?‘Ž

đ?‘…2 2(đ?‘Ž+đ?‘?) 2 2đ?‘Ž

2(đ?‘Ž+đ?‘?) 1

1

đ?‘…3 2(đ?‘Ž+đ?‘?)

−[ ]

3 2đ?‘Ž

=

=

1 4đ?‘? 2

2 2đ?‘Ž

{[

(đ?‘…−2(đ?‘Žâˆ’đ?‘?)) 2

]

+ 2(đ?‘Žâˆ’đ?‘?)

{ 2đ?‘? 2 + 2đ?‘? 2 } = 1.

đ?œ‡ = âˆŤâˆ’âˆž đ?‘…đ?‘ƒ(đ?‘…)đ?‘‘đ?‘… = { âˆŤ2(đ?‘Žâˆ’đ?‘?) 2 đ?‘…(đ?‘… − 2( đ?‘Ž − đ?‘?))đ?‘‘đ?‘… + âˆŤ2đ?‘Ž 4đ?‘? 2(đ?‘Ž + đ?‘?) [ ]

(2( đ?‘Ž + đ?‘?) − đ?‘…)đ?‘‘đ?‘…

}=

1 4đ?‘?2

{

16đ?‘Ž3 3

−

8(đ?‘Žâˆ’đ?‘?)3 3

4đ?‘?2

đ?‘…(2(đ?‘Ž + đ?‘?) − đ?‘… )đ?‘‘đ?‘… } =

+ 4(đ?‘Ž − đ?‘?)3 + 4(đ?‘Ž + đ?‘?)3 −

8(đ?‘Ž+đ?‘?)3 3

1 4đ?‘?2

đ?‘…3 2đ?‘Ž

{[ ]

3 2(đ?‘Žâˆ’đ?‘?)

đ?‘…2 2đ?‘Ž

− 2(đ?‘Ž − đ?‘?) [ ]

2 2(đ?‘Žâˆ’đ?‘?)

+

− 8đ?‘Ž3 }

1 −8đ?‘Ž 3 4(đ?‘Ž − đ?‘?)3 4(đ?‘Ž + đ?‘?)3 1 { + + } = 2 {−2đ?‘Ž3 + 2đ?‘Ž3 + 6đ?‘Žđ?‘? 2 } = 2đ?‘Ž = 2đ?œ‡1 4đ?‘? 2 3 3 3 3đ?‘? ∞

đ?œŽ 2 = âˆŤ ( đ?‘… − 2đ?‘Ž)2 đ?‘ƒ(đ?‘…)đ?‘‘đ?‘… −∞

2đ?‘Ž

2(đ?‘Ž+đ?‘?) 1 1 2 (2(đ?‘Ž ( đ?‘… − 2đ?‘Ž) ( =âˆŤ (đ?‘… − 2( đ?‘Ž − đ?‘?))đ?‘‘đ?‘… + âˆŤ + đ?‘?) − đ?‘… )đ?‘‘đ?‘… 2 2 đ?‘… − 2đ?‘Ž) 4đ?‘? 4đ?‘? 2(đ?‘Žâˆ’đ?‘?) 2đ?‘Ž 2đ?‘Ž 2(đ?‘Ž+đ?‘?) 1 ((2đ?‘Ž − đ?‘…)3 + 2đ?‘?(2đ?‘Ž − đ?‘…)2 )đ?‘‘đ?‘… } = {âˆŤ (( đ?‘… − 2đ?‘Ž)3 + 2đ?‘?( đ?‘… − 2đ?‘Ž)2 )đ?‘‘đ?‘… + âˆŤ 2 4đ?‘? 2(đ?‘Žâˆ’đ?‘?) 2đ?‘Ž 2

2đ?‘Ž

=

(đ?‘… − 2đ?‘Ž)4 1 { [ ] 2 4đ?‘? 4 2(đ?‘Žâˆ’đ?‘?) =

2đ?‘Ž

+ 2đ?‘? [

2(đ?‘Ž+đ?‘?)

(đ?‘… − 2đ?‘Ž)3 (2đ?‘Ž − đ?‘…)4 ] − [ ] 3 4 2(đ?‘Žâˆ’đ?‘?) 2đ?‘Ž

2(đ?‘Ž+đ?‘?)

− 2đ?‘? [

1 đ?‘?4 đ?‘?4 8đ?‘? 2 {4đ?‘? 4 + 16 − 4đ?‘? 4 + 16 } = = 2 đ?œŽ12 2 4đ?‘? 3 3 3

(2đ?‘Ž − đ?‘…)3 ] 3 2đ?‘Ž

}

By (5) and (8), it can be observed that the original distributions had been more rounded by convolution. If more elements were connected in series then the tendency toward smoothness would be more advanced. If a large number of functions are convolved together, the resultant may be very smooth and as the number increases indefinitely, the resultant may approach Gaussian form.[2] In Statistics Gaussian distribution is referred to as “normal distribution with zero mean and standard deviation oneâ€?, where the normal distribution is defined as đ?‘Ľ 1 đ?‘Ľâˆ’đ?œ‡ 2 1 đ?‘?( đ?‘‹ ≤ đ?‘Ľ) = âˆŤ đ?‘’ −2( đ?œŽ ) đ?‘‘đ?‘Ľ , đ?œŽâˆš2đ?œ‹ −∞ Where đ?‘‹ is a continuous random variable. The rigorous statement of the above stated tendency of protracted convolution is nothing but the theorem known as the central-limit theorem.[1].

VII. CENTRAL LIMIT THEOREM đ?‘‹Ě…−đ?œ‡

If đ?‘‹Ě… is the mean of a sample of size đ?‘› taken from a population having the mean đ?œ‡ and the finite variance đ?œŽ 2 , then đ?‘§ = đ?œŽ

â „ √đ?‘›

is

a random variable whose distribution function approaches that of the standard normal distribution as đ?‘› → ∞. [1] From the central limit theorem, it follows that if several, n functions are convolved together or a function is self-convolved n times, the result approaches Gaussian distribution.[3] Therefore it follows that if several random quantities are added, then the frequency distribution of the sum will approach a Gaussian distribution.

VIII. CONCLUSION   

Distribution of sum of two random variables is convolution of distributions of those random variables. Composite resistance of two resistors, combined in series, is obtained by convolution of distributions of those resistors. Convolution has additive property of means of the components. The mean value of the composite resistance of two resistors, combined in series, is the sum of the means of those two resistors. Variances are also additive under convolution. Variance of composite resistance of two resistors, combined in series, is addition of variances of those two resistors.

REFERENCES [1] [2] [3]

Richard A. Johnson, Miller & Freund’s “Probability and Statistics for Engineers�, Sixth edition Ronald N. Bracewell, “The Fourier transform and its applications�, International edition 2000, McGrow-Hill Education. Rashmi R. Keshvani, Yamini M Parmar, “An approach to various Probability Distributions through Convolution.�, international journal of Physical, Chemical & Mathematical Sciences. Vol-3, pp-1-8, July-Dec- 2014

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