Convolution of probability distributions

Example derivation

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.

Convolution of Bernoulli distributions

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

To show this let $${\displaystyle X_i\sim \mathrm{B}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}(p),0<p<1,1\le i\le 2}$$ and define $${\displaystyle Y=\sum _{i=1}^2{X}_i}$$ Also, let Z denote a generic binomial random variable: $${\displaystyle Z\sim \mathrm{B}\mathrm{i}\mathrm{n}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{a}\mathrm{l}(2,p)}$$

Using probability mass functions

As ${\displaystyle X_1\text{ and }{X}_2}$ are independent, $${\displaystyle \begin{array}{cc}Pr[Y=n]& =Pr\left[\sum _{i=1}^2{X}_i=n\right]\\ & =\sum _{m\in \mathbb{Z}}Pr[X_1=m]\times Pr[X_2=n-m]\\ & =\sum _{m\in \mathbb{Z}}\left[(\genfrac{}{}{0ex}{}{1}{m})p^m{\left(1-p\right)}^{1-m}\right]\left[(\genfrac{}{}{0ex}{}{1}{n-m})p^{n-m}{\left(1-p\right)}^{1-n+m}\right]\\ & =p^n{\left(1-p\right)}^{2-n}\sum _{m\in \mathbb{Z}}(\genfrac{}{}{0ex}{}{1}{m})(\genfrac{}{}{0ex}{}{1}{n-m})\\ & =p^n{\left(1-p\right)}^{2-n}\left[(\genfrac{}{}{0ex}{}{1}{0})(\genfrac{}{}{0ex}{}{1}{n})+(\genfrac{}{}{0ex}{}{1}{1})(\genfrac{}{}{0ex}{}{1}{n-1})\right]\\ & =(\genfrac{}{}{0ex}{}{2}{n})p^n{\left(1-p\right)}^{2-n}=Pr[Z=n]\end{array}}$$

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

Using characteristic functions

The characteristic function of each ${\displaystyle X_k}$ and of ${\displaystyle Z}$ is $${\displaystyle {\phi }_{X_k}(t)=1-p+p{e}^{it}{\phi }_Z(t)={\left(1-p+p{e}^{it}\right)}^2}$$ where t is within some neighborhood of zero.

$${\displaystyle \begin{array}{cc}{\phi }_Y(t)& =\mathrm{E}\left(e^{it\sum _{k=1}^2{X}_k}\right)=\mathrm{E}\left(\prod _{k=1}^2{e}^{it{X}_k}\right)\\ & =\prod _{k=1}^2\mathrm{E}\left(e^{it{X}_k}\right)=\prod _{k=1}^2\left(1-p+p{e}^{it}\right)\\ & ={\left(1-p+p{e}^{it}\right)}^2={\phi }_Z(t)\end{array}}$$

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.