Convolution of probability distributions

There are several ways of deriving formulae for the convolution of probability distributions. Often the manipulation of integrals can be avoided by use of some type of generating function. Such methods can also be useful in deriving properties of the resulting distribution, such as moments, even if an explicit formula for the distribution itself cannot be derived.

One of the straightforward techniques is to use characteristic functions, which always exists and are unique to a given distribution.^{[citation needed]}

The convolution of two independent identically distributed Bernoulli random variables is a binomial random variable. That is, in a shorthand notation, $${\displaystyle \sum _{i=1}^{2}\mathrm {Bernoulli} (p)\sim \mathrm {Binomial} (2,p)}$$

To show this let $${\displaystyle X_{i}\sim \mathrm {Bernoulli} (p),\quad 0<p<1,\quad 1\leq i\leq 2}$$ and define $${\displaystyle Y=\sum _{i=1}^{2}X_{i}}$$ Also, let Z denote a generic binomial random variable: $${\displaystyle Z\sim \mathrm {Binomial} (2,p)\,\!}$$

As ${\displaystyle X_{1}{\text{ and }}X_{2}}$ are independent, $${\displaystyle {\begin{aligned}\Pr[Y=n]&=\Pr \left[\sum _{i=1}^{2}X_{i}=n\right]\\&=\sum _{m\in \mathbb {Z} }\Pr[X_{1}=m]\times \Pr[X_{2}=n{-}m]\\&=\sum _{m\in \mathbb {Z} }\left[{\binom {1}{m}}p^{m}\left(1-p\right)^{1-m}\right]\left[{\binom {1}{n-m}}p^{n-m}\left(1-p\right)^{1-n+m}\right]\\&=p^{n}\left(1-p\right)^{2-n}\sum _{m\in \mathbb {Z} }{\binom {1}{m}}{\binom {1}{n-m}}\\&=p^{n}\left(1-p\right)^{2-n}\left[{\binom {1}{0}}{\binom {1}{n}}+{\binom {1}{1}}{\binom {1}{n-1}}\right]\\&={\binom {2}{n}}p^{n}\left(1-p\right)^{2-n}=\Pr[Z=n]\end{aligned}}}$$

Here, we used the fact that ${\displaystyle {\tbinom {n}{k}}=0}$ for k>n in the last but three equality, and of Pascal's rule in the second last equality.

The characteristic function of each ${\displaystyle X_{k}}$ and of ${\displaystyle Z}$ is $${\displaystyle \varphi _{X_{k}}(t)=1-p+pe^{it}\qquad \varphi _{Z}(t)=\left(1-p+pe^{it}\right)^{2}}$$ where t is within some neighborhood of zero.

$${\displaystyle {\begin{aligned}\varphi _{Y}(t)&=\operatorname {E} \left(e^{it\sum _{k=1}^{2}X_{k}}\right)=\operatorname {E} \left(\prod _{k=1}^{2}e^{itX_{k}}\right)\\&=\prod _{k=1}^{2}\operatorname {E} \left(e^{itX_{k}}\right)=\prod _{k=1}^{2}\left(1-p+pe^{it}\right)\\&=\left(1-p+pe^{it}\right)^{2}=\varphi _{Z}(t)\end{aligned}}}$$

The expectation of the product is the product of the expectations since each ${\displaystyle X_{k}}$ is independent. Since ${\displaystyle Y}$ and ${\displaystyle Z}$ have the same characteristic function, they must have the same distribution.

• List of convolutions of probability distributions
• Sum of normally distributed random variables

• Hogg, Robert V.; McKean, Joseph W.; Craig, Allen T. (2004). Introduction to mathematical statistics (6th ed.). Upper Saddle River, New Jersey: Prentice Hall. p. 692. ISBN 978-0-13-008507-8. MR 0467974.