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Sums of independent random variables

by , PhD

This lecture discusses how to derive the distribution of the sum of two independent random variables.

We explain:

Given two variables having a discrete distribution, how do I compute the distribution of their sum?

Table of Contents

Cumulative distribution function of a sum

The next proposition characterizes the cumulative distribution function (cdf) of the sum.

Proposition Let X and Y be two independent random variables. Denote their cdfs by [eq1] and [eq2]. Let[eq3]and denote the cdf of Z by [eq4]. Then,[eq5]or[eq6]

Proof

The first formula is derived as follows:[eq7]The second formula is symmetric to the first.

Example Let X be a uniform random variable with support [eq8] and probability density function[eq9]Let Y be another uniform random variable, independent of X, with support [eq10] and probability density function[eq11]The cdf of X is[eq12]The cdf of $Z=X+Y$ is[eq13]There are four cases to consider:

  1. If [eq14], then[eq15]

  2. If [eq16], then[eq17]

  3. If [eq18], then[eq19]

  4. If $z>2$, then[eq20]

By combining these four possible cases, we obtain[eq21]

Probability mass function of a sum

When the two summands are discrete random variables, the probability mass function (pmf) of their sum can be derived as follows.

Proposition Let X and Y be two independent discrete random variables. Denote their respective pmfs by [eq22] and [eq23], and their supports by R_X and $R_{Y}$. Let[eq3]and denote the pmf of Z by [eq25]. Then,[eq26]or[eq27]

Proof

The first formula is derived as follows:[eq28]The second formula is symmetric to the first.

The two summations above are called convolutions (of two pmfs).

Example Let X be a discrete random variable with support [eq29] and pmf[eq30]Let Y be another discrete random variable, independent of X, with support [eq31] and pmf[eq32]Define[eq3]Its support is [eq34]The pmf of Z, evaluated at $z=0$ is[eq35]Evaluated at $z=1$, it is[eq36]Evaluated at $z=2$, it is[eq37]Therefore, the pmf of Z is[eq38]

Probability density function of a sum

When the two summands are continuous variables, the probability density function (pdf) of their sum can be derived as follows.

Proposition Let X and Y be two independent continuous random variables and denote their respective pdfs by [eq39] and [eq40]. Let[eq3]and denote the pdf of Z by [eq42]. Then,[eq43]or[eq44]

Proof

The distribution function of a sum of independent variables is[eq5]Differentiating both sides and using the fact that the density function is the derivative of the distribution function, we obtain[eq46]The second formula is symmetric to the first.

The two integrals above are called convolutions (of two pdfs).

Example Let X be an exponential random variable with support [eq47] and pdf[eq48]Let Y be another exponential random variable, independent of X, with support [eq49] and pdf[eq50]Define [eq3]The support of Z is[eq52]When $zin R_{Z}$, the pdf of Z is[eq53]Therefore, the pdf of Z is[eq54]

Sum of n independent random variables

We have discussed above how to work out the distribution of the sum of two independent random variables.

How do we derive the distribution of the sum of more than two mutually independent random variables?

Suppose that X_1, X_2, ..., X_n are n mutually independent random variables and let Z be their sum:[eq55]

The distribution of Z can be derived recursively, using the results for sums of two random variables given above:

  1. first, define[eq56]and compute the distribution of $Y_{2}$;

  2. then, define[eq57]and compute the distribution of $Y_{3}$;

  3. and so on, until the distribution of Z can be computed from[eq58]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let X be a uniform random variable with support [eq59] and pdf[eq9]

Let Y be an exponential random variable, independent of X, with support [eq61] and pdf[eq50]

Derive the pdf of the sum [eq3]

Solution

The support of Z is[eq64]When $zin R_{Z}$, the pdf of Z is[eq65]Therefore, the pdf of Z is[eq66]

Exercise 2

Let X be a discrete random variable with support [eq67] and pmf[eq68]

Let Y be another discrete random variable, independent of X, with support [eq69] and pmf[eq70]

Derive the pmf of the sum[eq3]

Solution

The support of Z is [eq72]The pmf of Z, evaluated at $z=1$ is[eq73]Evaluated at $z=2$, it is[eq74]Evaluated at $z=3$, it is[eq75]Evaluated at $z=4$, it is[eq76]Therefore, the pmf of Z is[eq77]

How to cite

Please cite as:

Taboga, Marco (2021). "Sums of independent random variables", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-probability/sums-of-independent-random-variables.

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