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.

1. Journal for Research| Volume 02| Issue 01 | March 2016
ISSN: 2395-7549
All rights reserved by www.journalforresearch.org 58
Probability Distribution of Sum of Two
Continuous Variables and Convolution
Rashmi R. Keshvani Yamini M. Parmar
Professor Assistant Professor
Department of Mathematics Department of Mathematics
Sarvajanik College of Engineering & Technology, Surat (Guj.)
India
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 𝑝(𝑥) is said to be a probability distribution of a random variable 𝑥, if (1) 𝑝(𝑥) ≥ 0, ∀ 𝑥 ∈ 𝑅 and
∑ 𝑝(𝑥) = 1𝑎𝑙𝑙 𝑥 where 𝑥 is a discrete random variable. or (2) 𝑝(𝑥) ≥ 0, ∀ 𝑥 ∈ 𝑅 and ∫ 𝑝(𝑥)𝑑𝑥 = 1
∞
−∞
where 𝑥 is a
continuous random variable. The mean of probability distribution 𝑝(𝑥), denoted by μ, is defined as
(1) 𝜇 = ∑ 𝑥 𝑝(𝑥)𝑎𝑙𝑙 𝑥 , if 𝑥 is discrete, or (2) 𝜇 = ∫ 𝑥 𝑝(𝑥)𝑑𝑥,
∞
−∞
if 𝑥 is continuous. (1)
The variance of probability distribution 𝑝(𝑥), denoted by 𝜎2
, is defined as
(1) 𝜎2
= ∑ ( 𝑥 − 𝜇 )2
𝑝(𝑥)𝑎𝑙𝑙 𝑥 , if 𝑥 is discrete, or (2) 𝜎2
= ∫ ( 𝑥 − 𝜇 )2∞
−∞
𝑝(𝑥)𝑑𝑥, if 𝑥 is continuous. [1] (2)
The Uniform Probability Distribution:
The uniform probability distribution, with parameters 𝛼 and 𝛽, has probability distribution 𝑝(𝑥) defined as
𝑝(𝑥) = {
1
𝛽−𝛼
𝑖𝑓 𝛼 < 𝑥 < 𝛽
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
(3)
III.THE CONCEPT OF CONVOLUTION
The convolution of two functions 𝑓(𝑥) and 𝑔(𝑥), [2], denoted by 𝑓(𝑥) ∗ 𝑔(𝑥), or (𝑓 ∗ 𝑔)(𝑥) is defined as
𝑓(𝑥) ∗ 𝑔(𝑥) = ∫ 𝑓(𝑢)𝑔(𝑥 − 𝑢)𝑑𝑢
∞
−∞
(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
𝑓 ∗ (𝑔1 + 𝑔2)(𝑥) = 𝑓(𝑥) ∗ (𝑔1 + 𝑔2)(𝑥) = 𝑓(𝑥) ∗ (𝑔1(𝑥) + 𝑔2(𝑥)) = ∫ 𝑓(𝑢) (𝑔1(𝑥 − 𝑢) + 𝑔2( 𝑥 − 𝑢))𝑑𝑢
∞
−∞
=
∫ 𝑓(𝑢)𝑔1(𝑥 − 𝑢)𝑑𝑢
−∞
∞
+ ∫ 𝑓(𝑢)𝑔2(𝑥 − 𝑢)𝑑𝑢
−∞
∞
= ( 𝑓 ∗ 𝑔1)(𝑥) + ( 𝑓 ∗ 𝑔2)(𝑥)
That is 𝑓 ∗ (𝑔1 + 𝑔2) = ( 𝑓 ∗ 𝑔1) + ( 𝑓 ∗ 𝑔2)

2. Probability Distribution of Sum of Two Continuous Variables and Convolution
(J4R/ Volume 02 / Issue 01 / 009)
All rights reserved by www.journalforresearch.org 59
IV.PROBABILITY DISTRIBUTION OF A SUM
Suppose two probability distributions 𝑝1(𝑥1) and 𝑝2(𝑥2) are given and it is required to determine probability distribution
𝑝(𝑥) of random variable 𝑥 = 𝑥1 + 𝑥2, sum of these two random variables. That is 𝑝(𝑥) shows probability that sum of 𝑥1and 𝑥2
remains 𝑥. That is 𝑝(𝑥) = 𝑝( 𝑥 = 𝑥1 + 𝑥2). 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 𝑅1 and 𝑅1 𝑑𝑅1 is 𝑃1(𝑅1)𝑑𝑅1. [2]
So 𝑃1(𝑅1) has to be probability distribution, more precisely, uniform probability distribution over interval 90 to 110, ensuring
that,
∫ 𝑃1(𝑅1)
∞
−∞
𝑑𝑅1 = 1, that is 𝑃1(𝑅1) = {
1
(110−90)
=
1
20
𝑖𝑓 90 ≤ 𝑅1 ≤ 110
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
Expressing 𝑃1(𝑅1) in terms of rectangle function, 𝑃1(𝑅1) =
1
20
∏ (
𝑅1−100
20
), where the rectangle function, denoted by ∏(𝑥),
is defined as
∏(𝑥) = {
1 |𝑥| < 1
2⁄
0 |𝑥| > 1
2⁄
The mean of this distribution is 𝜇1 = ∫ 𝑅1 𝑃1(𝑅1)𝑑𝑅1 =
1
20
∫ 𝑅1 𝑑𝑅1 =
1
20
[
𝑅1
2
2
]
90
110
= 100
110
90
∞
−∞
, and the variance 𝜎1
2
=
∫ (𝑅1 − 100)2
𝑃(𝑅1)𝑑𝑅1 =
1
20
∫ (𝑅1 − 100)2
𝑑𝑅1 =
1
20
[
(𝑅1−100)3
3
]
90
110
=
100
3
110
90
∞
−∞
Similarly for a stock of 50-ohm resisters, with 5-ohm tolerance, another probability distribution𝑃2(𝑅2), defined, as 𝑃2(𝑅2) =
{
1
(55−45)
=
1
10
𝑖𝑓 45 ≤ 𝑅2 ≤ 55
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
will describe the frequency of occurrence of resisters.
Expressing 𝑃2(𝑅2) also, in terms of rectangle function, 𝑃2(𝑅2) =
1
10
∏ (
𝑅2−50
10
) and the mean of 𝑃2(𝑅2), will be 𝜇2 = 50
and variance 𝜎2
2
=
25
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 𝑃(𝑅), then here problem is to know, what mathematical relation, 𝑃(𝑅)
bears with 𝑃1(𝑅1) and 𝑃2(𝑅2) when 𝑅 = 𝑅1 + 𝑅2. It is clear that 𝑅 = 𝑅1 + 𝑅2 will vary from 90 + 45 = 135 to 110 +
55 = 165, with mean equal to 150. It is also clear that the probability that 𝑅 assumes a value less than or equal to135, that is
𝑃(𝑅 ≤ 135) = 0 and the probability that 𝑅 assumes a value greater than or equal to 165 , that is 𝑃(𝑅 ≥ 165) = 0.
Probability that 𝑅 assumes a particular value between 135 and 165, say, 𝑅 = 140, that is 𝑅1 + 𝑅2 = 140, denoted by 𝑃(140)
will be that area where this sum assumes value 140.That is 𝑅2 should remain 140 − 𝑅1. Required area will be
𝑃(140) =
1
20
∫ 𝑃2(140 − 𝑅1)𝑑𝑅1
110
90
=
1
20
∫ 𝑃2(𝑅2)𝑑
𝑅2=50
𝑅2=30
𝑅2 =
1
20
∫
1
10
𝑑
𝑅2=50
𝑅2=45
𝑅2 =
1
40
,
As 𝑅1 is nonzero if 90 ≤ 𝑅1 ≤ 110 and 𝑅2 is nonzero, if 45 ≤ 𝑅2 ≤ 55.
One can understand that, this is nothing but convolution of 𝑃1(𝑅1) and𝑃2(𝑅2), that is
𝑃(140) = (𝑃1 ∗ 𝑃2)(140) = ∫ 𝑃1(𝑅1)𝑃2(140 − 𝑅1)𝑑𝑅1
∞
−∞
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 𝑅1 have greater resistance 𝑎1 with
tolerance 𝑏1 , and resistor 𝑅2 have smaller resistance 𝑎2 with tolerance 𝑏2. As 𝑎1 > 𝑎2 and, as 𝑎1 > 𝑏1, 𝑎2 > 𝑏2, there is
no loss of generality any, if we assume 𝑎1 > 𝑎2 > 𝑏1 > 𝑏2 > 0.
In terms of rectangle functions 𝑅1 =
1
2𝑏1
∏ (
𝑅1−𝑎1
2𝑏1
) and 𝑅2 =
1
2𝑏2
∏ (
𝑅2−𝑎2
2𝑏2
)
It is quite obvious that 𝜇1 = ∫ 𝑅1 𝑃1(𝑅1)𝑑𝑅1
∞
−∞
=
1
2𝑏1
∫ 𝑅1 𝑑𝑅1
𝑎1+𝑏1
𝑎1−𝑏1
=
4𝑎1 𝑏1
4𝑏1
= 𝑎1
𝜇2 = ∫ 𝑅2 𝑃2(𝑅2)𝑑𝑅2
∞
−∞
=
1
2𝑏2
∫ 𝑅2 𝑑𝑅2
𝑎2+𝑏2
𝑎2−𝑏2
=
4𝑎2 𝑏2
4𝑏2
= 𝑎2
𝜎1
2
= ∫ (𝑅1 − 𝑎1)2
𝑃1(𝑅)𝑑𝑅1
∞
−∞
=
1
2𝑏1
∫ (𝑅1 − 𝑎1)2
𝑑𝑅1
𝑎1+𝑏1
𝑎1−𝑏1
=
1
2𝑏1
[
(𝑅1 − 𝑎1 )3
3
] 𝑎1−𝑏1
𝑎1+𝑏1
=
𝑏1
2
3
𝜎2
2
= ∫ ( 𝑅2 − 𝑎2)2
𝑃2(𝑅)𝑑𝑅
∞
−∞
=
1
2𝑏2
∫ ( 𝑅2 − 𝑎2)2
𝑑𝑅
𝑎2+𝑏2
𝑎2−𝑏2
=
1
2𝑏2
[
(𝑅2 − 𝑎2 )3
3
] 𝑎2−𝑏2
𝑎2+𝑏2
=
𝑏2
2
3
To determine 𝑃(𝑅), following cases are to be considered separately,
If 𝑅 ≤ 𝑎1 + 𝑎2 − 𝑏1 − 𝑏2 or 𝑅 ≥ 𝑎1 + 𝑎2 + 𝑏1 + 𝑏2, then 𝑃(𝑅) = 0
If (𝑎1 + 𝑎2) − (𝑏1 + 𝑏2) < 𝑅 < (𝑎1 + 𝑎2) − 𝑏1 + 𝑏2
𝑃(𝑅) = ∫ 𝑃1(𝑅1)𝑃2(𝑅 − 𝑅1)𝑑𝑅1 =
1
2𝑏1
∫ 𝑃2(𝑅 − 𝑅1)𝑑𝑅1 =
1
2𝑏1
∫ 𝑃2(𝑅2)𝑑𝑅2
𝑅−(𝑎1−𝑏1)
𝑅−(𝑎1 + 𝑏1 )
𝑎1+ 𝑏1
𝑎1−𝑏1
∞
−∞
=
1
4𝑏1 𝑏2
∫ 𝑑𝑅2
𝑅−(𝑎1−𝑏1)
𝑎2−𝑏2

3. Probability Distribution of Sum of Two Continuous Variables and Convolution
(J4R/ Volume 02 / Issue 01 / 009)
All rights reserved by www.journalforresearch.org 60
=
1
4𝑏1 𝑏2
(𝑅 − (𝑎1 − 𝑏1) − (𝑎2 − 𝑏2)) as (𝑏1 > 𝑏2 𝑎𝑛𝑑 𝑅 < 𝑎1 + 𝑎2 − 𝑏1 + 𝑏2) ⇒ {
𝑅 − (𝑎1 − 𝑏1) < 𝑎2 + 𝑏2
𝑅 − (𝑎1 + 𝑏1) < 𝑎2 − 𝑏2
If (𝑎1 + 𝑎2) − 𝑏1 + 𝑏2 ≤ 𝑅 ≤ ( 𝑎1 + 𝑎2) + 𝑏1 − 𝑏2
𝑃(𝑅) = ∫ 𝑃1(𝑅1)𝑃2(𝑅 − 𝑅1)𝑑𝑅1 =
1
2𝑏1
∫ 𝑃2(𝑅 − 𝑅1)𝑑𝑅1 =
1
2𝑏1
∫ 𝑃2(𝑅2)𝑑𝑅2
𝑅−(𝑎1−𝑏1)
𝑅−(𝑎1 + 𝑏1 )
𝑎1+ 𝑏1
𝑎1−𝑏1
∞
−∞
=
1
4𝑏1 𝑏2
∫ 𝑑𝑅2
𝑎2+𝑏2
𝑎2−𝑏2
=
2𝑏2
4𝑏1 𝑏2
=
1
2𝑏1
If 𝑎1 + 𝑎2 + 𝑏1 − 𝑏2 ≤ 𝑅 ≤ 𝑎1 + 𝑎2 + 𝑏1 + 𝑏2,
𝑃(𝑅) = ∫ 𝑃1(𝑅1)𝑃2(𝑅 − 𝑅1)𝑑𝑅1 =
1
2𝑏1
∫ 𝑃2(𝑅 − 𝑅1)𝑑𝑅1 =
1
2𝑏1
∫ 𝑃2(𝑅2)𝑑𝑅2
𝑅−(𝑎1−𝑏1)
𝑅−(𝑎1 + 𝑏1 )
𝑎1+ 𝑏1
𝑎1−𝑏1
∞
−∞
=
1
4𝑏1 𝑏2
∫ 𝑑𝑅2
𝑎2+𝑏2
𝑅−(𝑎1 + 𝑏1)
=
1
4𝑏1 𝑏2
(𝑎2 + 𝑏2 + 𝑎1 + 𝑏1 − 𝑅 ),
as (𝑎1 + 𝑎2 + 𝑏1 − 𝑏2 ≤ 𝑅 𝑎𝑛𝑑 𝑏1 > 𝑏2) ⇒ {
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
𝑃(𝑅) =
{
0 𝑖𝑓 𝑅 ≤ 𝑎1+ 𝑎2−𝑏1−𝑏2
1
4𝑏1 𝑏2
(𝑅−(𝑎1+𝑎2−𝑏1− 𝑏2)) 𝑖𝑓 𝑎1+ 𝑎2−𝑏1−𝑏2 ≤ 𝑅 ≤ 𝑎1+ 𝑎2−𝑏1+ 𝑏2
1
2𝑏1
𝑖𝑓 𝑎1 + 𝑎2 − 𝑏1 + 𝑏2 ≤ 𝑅 ≤ 𝑎1 + 𝑎2 + 𝑏1 − 𝑏2
1
4𝑏1 𝑏2
(𝑎1+ 𝑎2+ 𝑏1+ 𝑏2−𝑅 ) 𝑖𝑓 𝑎1+ 𝑎2+𝑏1− 𝑏2 ≤ 𝑅 ≤ 𝑎1 + 𝑎2+ 𝑏1 + 𝑏2
0 𝑖𝑓 𝑅 ≥ 𝑎1+ 𝑎2+𝑏1+ 𝑏2
(5)
Now ∫ 𝑃(𝑅)𝑑𝑅
∞
−∞
=
1
4𝑏1 𝑏2
{ [
(𝑅 − (𝑎1 − 𝑏1 + 𝑎2 − 𝑏2))
2
2
]
(𝑎1+ 𝑎2−𝑏1−𝑏2)
(𝑎1+ 𝑎2−𝑏1+ 𝑏2)
+ 2𝑏2[ 𝑅 ](𝑎1+ 𝑎2−𝑏1+ 𝑏2)
(𝑎1+ 𝑎2+𝑏1− 𝑏2)
+ [
−(𝑎1 + 𝑎2 + 𝑏1 + 𝑏2 − 𝑅 )2
2
]
(𝑎1+ 𝑎2+ 𝑏1− 𝑏2)
(𝑎1+ 𝑎2+ 𝑏1+ 𝑏2)
}
=
1
4𝑏1 𝑏2
{ 2 𝑏2
2
+ 2𝑏2(2𝑏1 − 2𝑏2) + 2𝑏2
2 } =
1
4𝑏1 𝑏2
{ 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)
𝜎2
= ∫ (𝑅 − 𝜇)2
𝑃(𝑅)𝑑𝑅
∞
−∞
= ∫ 𝑅2
𝑃(𝑅)𝑑𝑅
∞
−∞
− 𝜇2
= {∫ 𝑅2(𝑃1 ∗ 𝑃2)(𝑅)𝑑𝑅
∞
−∞
} − 𝜇2
= { ∫ 𝑅2
(∫ 𝑃1(𝑅1)𝑃2(𝑅 −
∞
−∞
𝑅1)𝑑𝑅1) 𝑑𝑅 }
∞
−∞
− 𝜇2
= { ∫ 𝑃1(𝑅1) (∫ 𝑅2
𝑃2(𝑅 −
∞
−∞
𝑅1)𝑑𝑅)
∞
−∞
𝑑𝑅1 } − 𝜇2
= ∫ 𝑃1(𝑅1) (∫ ((𝑅 − 𝑅1) + 𝑅1) 2
𝑃2(𝑅 −
∞
−∞
𝑅1)𝑑𝑅)
∞
−∞
𝑑𝑅1 − 𝜇2
= ∫ 𝑃1(𝑅1) (∫ {(𝑅 − 𝑅1) 2
+ 2 𝑅1(𝑅 − 𝑅1) + 𝑅1
2
} 𝑃2(𝑅 −
∞
−∞
𝑅1)𝑑𝑅 )
∞
−∞
𝑑𝑅1 − 𝜇2
= ∫ 𝑃1(𝑅1) ( {∫ (𝑅 − 𝑅1)2
𝑃2(𝑅 −
∞
−∞
𝑅1)𝑑(𝑅 − 𝑅1)} + 2𝑅1 𝜇2 + 𝑅1
2
)
∞
−∞
𝑑𝑅1 − 𝜇2

4. Probability Distribution of Sum of Two Continuous Variables and Convolution
(J4R/ Volume 02 / Issue 01 / 009)
All rights reserved by www.journalforresearch.org 61
= ∫ 𝑃1(𝑅1)(𝜎2
2
+ 𝜇2
2
+ 2𝑅1 𝜇2 + 𝑅1
2)
∞
−∞
𝑑𝑅1 − 𝜇2
= (𝜎2
2
+ 𝜇2
2) + 2𝜇1 𝜇2 + (𝜎1
2
+ 𝜇1
2) − (𝜇1 + 𝜇2)2
= 𝜎1
2
+ 𝜎2
2
(7)
VI.VERIFICATION OF THE RESULTS
For combined resistor in series,
The mean 𝜇 = ∫ 𝑅 𝑃(𝑅)𝑑𝑅 =
1
4𝑏1 𝑏2
∫ (𝑅2
− 𝑅(𝑎1 + 𝑎2 − 𝑏1 − 𝑏2) )
(𝑎1+ 𝑎2−𝑏1+ 𝑏2)
(𝑎1+ 𝑎2−𝑏1−𝑏2)
𝑑𝑅
∞
−∞
=
1
4𝑏1 𝑏2
∫ (𝑅2
− 𝑅(𝑎1 + 𝑎2 − 𝑏1 − 𝑏2) )
(𝑎1+ 𝑎2−𝑏1+ 𝑏2)
(𝑎1+ 𝑎2−𝑏1−𝑏2)
𝑑𝑅 +
1
2𝑏1
∫ 𝑅𝑑𝑅
(𝑎1+ 𝑎2+𝑏1− 𝑏2)
(𝑎1+ 𝑎2−𝑏1+ 𝑏2)
+
1
4𝑏1 𝑏2
∫ { 𝑅(𝑎1 + 𝑎2 +
(𝑎1+ 𝑎2+ 𝑏1+ 𝑏2)
(𝑎1+ 𝑎2+ 𝑏1− 𝑏2)
𝑏1 + 𝑏2) − 𝑅2
} 𝑑𝑅
=
1
4𝑏1 𝑏2
{ [
𝑅3
3
−
𝑅2
2
(𝑎1 + 𝑎2 − 𝑏1 − 𝑏2)]
(𝑎1+ 𝑎2−𝑏1−𝑏2)
(𝑎1+ 𝑎2−𝑏1+ 𝑏2)
+ 2𝑏2 [
𝑅2
2
]
(𝑎1+ 𝑎2−𝑏1+ 𝑏2)
(𝑎1+ 𝑎2+𝑏1− 𝑏2)
}
+
1
4𝑏1 𝑏2
{ [
𝑅2
2
(𝑎1 + 𝑎2 + 𝑏1 + 𝑏2) −
𝑅3
3
]
(𝑎1+ 𝑎2+ 𝑏1− 𝑏2)
(𝑎1+ 𝑎2+ 𝑏1+ 𝑏2)
}
=
1
4𝑏1 𝑏2
{
(𝑎1 + 𝑎2 − 𝑏1 + 𝑏2)2(𝑏1 + 5𝑏2 − 𝑎1 − 𝑎2)
6
+
(𝑎1 + 𝑎2 − 𝑏1 − 𝑏2)3
6
+ 4𝑏2(𝑎1 + 𝑎2)(𝑏1 − 𝑏2)}
+
1
4𝑏1 𝑏2
{
(𝑎1 + 𝑎2 + 𝑏1 + 𝑏2)3
6
−
(𝑎1 + 𝑎2 + 𝑏1 − 𝑏2)2(𝑎1 + 𝑎2 + 𝑏1 + 5𝑏2)
6
}
=
2(𝑎1 + 𝑎2)
24𝑏1 𝑏2
{ (𝑎1 + 𝑎2)2
+ 3(𝑏1 + 𝑏2)2) + 12𝑏2(𝑏1 − 𝑏2) − 2(𝑏1 − 𝑏2)(𝑏1 + 5𝑏2) − (𝑎1 + 𝑎2)2
+ (𝑏1 − 𝑏2)2
}
=
2(𝑎1+𝑎2)
24𝑏1 𝑏2
{3(𝑏1 + 𝑏2)2
+ 12𝑏2(𝑏1 − 𝑏2) − 2(𝑏1 − 𝑏2)(𝑏1 + 5𝑏2) − (𝑏1 − 𝑏2)2} = (𝑎1 + 𝑎2 ) = 𝜇1 + 𝜇2. The variance
𝜎2
= ∫ (𝑅 − (𝑎1 + 𝑎2))2∞
−∞
𝑃(𝑅)𝑑𝑅
= ∫ (𝑅 − (𝑎1 + 𝑎2))2
1
4𝑏1 𝑏2
(𝑅 − (𝑎1 + 𝑎2 − 𝑏1 − 𝑏2))𝑑𝑅
𝑎1+𝑎2−𝑏1+𝑏2
𝑎1+𝑎2−𝑏1−𝑏2
+ ∫ (𝑅 − (𝑎1 + 𝑎2))2
1
2𝑏1
𝑑𝑅
𝑎1+𝑎2+𝑏1−𝑏2
𝑎1+𝑎2−𝑏1+𝑏2
+ ∫ (𝑅 − (𝑎1 + 𝑎2))2
1
4𝑏1 𝑏2
((𝑎1 + 𝑎2+ 𝑏1 + 𝑏2) − 𝑅)𝑑𝑅
𝑎1+𝑎2+ 𝑏1+𝑏2
𝑎1+𝑎2+ 𝑏1−𝑏2
= {
1
4𝑏1 𝑏2
∫ (( (𝑅 − (𝑎1 + 𝑎2))
3
+ (𝑅 − (𝑎1 + 𝑎2))
2
(𝑏1 + 𝑏2)) 𝑑𝑅
𝑎1+𝑎2−𝑏1+𝑏2
𝑎1+𝑎2−𝑏1−𝑏2
+
1
2𝑏1
∫ (𝑅 − (𝑎1 + 𝑎2))2
𝑑𝑅
𝑎1+𝑎2+𝑏1−𝑏2
𝑎1+𝑎2−𝑏1+𝑏2
+
1
4𝑏1 𝑏2
∫ ( ((𝑎1 + 𝑎2) − 𝑅)3
+ (𝑅 − (𝑎1 + 𝑎2))2( 𝑏1 + 𝑏2))𝑑𝑅
𝑎1+𝑎2+ 𝑏1+𝑏2
𝑎1+𝑎2+ 𝑏1−𝑏2
}
=
1
4𝑏1 𝑏2
{[
(𝑅−(𝑎1+𝑎2))
4
4
]
𝑎1+𝑎2−𝑏1−𝑏2
𝑎1+𝑎2−𝑏1+𝑏2
+ (𝑏1 + 𝑏2) [
(𝑅−(𝑎1+𝑎2))
3
3
]
𝑎1+𝑎2−𝑏1−𝑏2
𝑎1+𝑎2−𝑏1+𝑏2
} +
1
2𝑏1
[
(𝑅−(𝑎1+𝑎2))
3
3
]
𝑎1+𝑎2−𝑏1+ 𝑏2
𝑎1+𝑎2+𝑏1−𝑏2
+
1
4𝑏1 𝑏2
{( 𝑏1 +
𝑏2) [
(𝑅−(𝑎1+𝑎2))
3
3
]
𝑎1+𝑎2+ 𝑏1−𝑏2
𝑎1+𝑎2+𝑏1+𝑏2
− [
((𝑎1+𝑎2)−𝑅)
4
4
]
𝑎1+𝑎2+ 𝑏1−𝑏2
𝑎1+𝑎2+𝑏1+𝑏2
}
=
1
4𝑏1 𝑏2
{
(𝑏2 − 𝑏1)4
4
−
(𝑏1 + 𝑏2)4
4
+ (𝑏1 + 𝑏2) (
(𝑏2 − 𝑏1)3
3
+
(𝑏1 + 𝑏2)3
3
)} +
1
2𝑏1
{
2(𝑏1 − 𝑏2)3
3
}
+
1
4𝑏1 𝑏2
{( 𝑏1 + 𝑏2) {
(𝑏1 + 𝑏2)3
3
−
(𝑏1 − 𝑏2)3
3
} –
(𝑏1 + 𝑏2)4
4
+
(𝑏2 − 𝑏1)4
4
}
=
1
4𝑏1 𝑏2
{
(𝑏1 + 𝑏2)4
6
+
(𝑏2 − 𝑏1)3
6
(𝑏1 + 7𝑏2) −
8𝑏2(𝑏2 − 𝑏1)3
6
}
=
1
24𝑏1 𝑏2
{(𝑏1 + 𝑏2)4
− (𝑏2 − 𝑏1)4} =
(𝑏1
2
+ 𝑏2
2)
3
= 𝜎1
2
+ 𝜎2
2
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:
𝑃(𝑅) = {
0 𝑖𝑓 𝑅 ≤ 2(𝑎−𝑏)
1
4𝑏2(𝑅−2( 𝑎−𝑏)) 𝑖𝑓 2(𝑎−𝑏)≤ 𝑅 ≤ 2𝑎
1
4𝑏2(2(𝑎+𝑏)−𝑅 ) 𝑖𝑓 2𝑎 ≤ 𝑅 ≤ 2( 𝑎+𝑏)
0 𝑖𝑓 𝑅 ≥2( 𝑎+𝑏)
(8)
That is, 𝑃(𝑅) =
1
2𝑏
Λ (
𝑅−2𝑎
2𝑏
), where Λ(𝑥), known as triangular function, is defined as

5. Probability Distribution of Sum of Two Continuous Variables and Convolution
(J4R/ Volume 02 / Issue 01 / 009)
All rights reserved by www.journalforresearch.org 62
Λ(𝑥) = {
0 𝑖𝑓 |𝑥| > 1
1 − |𝑥| 𝑖𝑓 |𝑥| ≤ 1
In this case, 𝜇1 = 𝜇2, 𝜎1 = 𝜎2 and
∫ 𝑃(𝑅)𝑑𝑅 = ∫
1
4𝑏2 (𝑅 − 2( 𝑎 − 𝑏))𝑑𝑅 + ∫
1
4𝑏2
(2( 𝑎 + 𝑏) − 𝑅)𝑑𝑅
2(𝑎+𝑏)
2𝑎
2𝑎
2(𝑎−𝑏)
∞
−∞
=
1
4𝑏2 {[
(𝑅−2(𝑎−𝑏))
2
2
]
2(𝑎−𝑏)
2𝑎
+
[−
(2(𝑎+𝑏)−𝑅)2
2
]
2𝑎
2(𝑎+𝑏)
} =
1
4𝑏2
{ 2𝑏2
+ 2𝑏2} = 1.
𝜇 = ∫ 𝑅𝑃(𝑅)𝑑𝑅
∞
−∞
= { ∫
1
4𝑏2 𝑅(𝑅 − 2( 𝑎 − 𝑏))𝑑𝑅
2𝑎
2(𝑎−𝑏)
+ ∫
1
4𝑏2 𝑅(2(𝑎 + 𝑏) − 𝑅 )𝑑𝑅
2(𝑎+𝑏)
2𝑎
} =
1
4𝑏2 { [
𝑅3
3
]
2(𝑎−𝑏)
2𝑎
− 2(𝑎 − 𝑏) [
𝑅2
2
]
2(𝑎−𝑏)
2𝑎
+
2(𝑎 + 𝑏) [
𝑅2
2
]
2𝑎
2(𝑎+𝑏)
− [
𝑅3
3
]
2𝑎
2(𝑎+𝑏)
} =
1
4𝑏2 {
16𝑎3
3
−
8(𝑎−𝑏)3
3
+ 4(𝑎 − 𝑏)3
+ 4(𝑎 + 𝑏)3
−
8(𝑎+𝑏)3
3
− 8𝑎3
}
=
1
4𝑏2
{
−8𝑎3
3
+
4(𝑎 − 𝑏)3
3
+
4(𝑎 + 𝑏)3
3
} =
1
3𝑏2
{−2𝑎3 + 2𝑎3 + 6𝑎𝑏2} = 2𝑎 = 2𝜇1
𝜎2
= ∫ ( 𝑅 − 2𝑎)2
𝑃(𝑅)𝑑𝑅
∞
−∞
= ∫ ( 𝑅 − 2𝑎)2
1
4𝑏2
(𝑅 − 2( 𝑎 − 𝑏))𝑑𝑅 + ∫
1
4𝑏2
( 𝑅 − 2𝑎)2(2(𝑎 + 𝑏) − 𝑅 )𝑑𝑅
2(𝑎+𝑏)
2𝑎
2𝑎
2(𝑎−𝑏)
=
1
4𝑏2
{ ∫ (( 𝑅 − 2𝑎)3 + 2𝑏( 𝑅 − 2𝑎)2)𝑑𝑅 + ∫ ((2𝑎 − 𝑅)3 + 2𝑏(2𝑎 − 𝑅)2)𝑑𝑅
2(𝑎+𝑏)
2𝑎
2𝑎
2(𝑎−𝑏)
}
=
1
4𝑏2
{ [
(𝑅 − 2𝑎)4
4
]
2(𝑎−𝑏)
2𝑎
+ 2𝑏 [
(𝑅 − 2𝑎)3
3
]
2(𝑎−𝑏)
2𝑎
− [
(2𝑎 − 𝑅)4
4
]
2𝑎
2(𝑎+𝑏)
− 2𝑏 [
(2𝑎 − 𝑅)3
3
]
2𝑎
2(𝑎+𝑏)
}
=
1
4𝑏2
{4𝑏4
+ 16
𝑏4
3
− 4𝑏4
+ 16
𝑏4
3
} =
8𝑏2
3
= 2 𝜎1
2
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 one”,
where the normal distribution is defined as
𝑝( 𝑋 ≤ 𝑥) =
1
𝜎√2𝜋
∫ 𝑒−
1
2
(
𝑥−𝜇
𝜎
)2
𝑥
−∞
𝑑𝑥 ,
Where 𝑋 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
If 𝑋̅ is the mean of a sample of size 𝑛 taken from a population having the mean 𝜇 and the finite variance 𝜎2
, then 𝑧 =
𝑋̅−𝜇
𝜎
√𝑛
⁄
is
a random variable whose distribution function approaches that of the standard normal distribution as 𝑛 → ∞. [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] Richard A. Johnson, Miller & Freund’s “Probability and Statistics for Engineers”, Sixth edition
[2] Ronald N. Bracewell, “The Fourier transform and its applications”, International edition 2000, McGrow-Hill Education.
[3] 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