Convolutions

Let $X,Y$ be independent RVs. Suppose that we know the distribution of the random variables. What is the distribution of $X+Y$?

Discrete

Suppose $X,Y:\Omega \to \mathbb{Z}$ and are independent. We have:

$$\begin{array}{rl}{f}_{X+Y}(n)& =\mathbb{P}(X+Y=n)\\ & =\sum _{k\in \mathbb{Z}}P(X=k,Y=n-k)\\ & =\sum _{k\in \mathbb{Z}}P(X=k)\cdot P(Y=n-k)\\ & =\sum _{k\in \mathbb{Z}}{P}_X(k)\cdot P_Y(n-k)\\ & =P_X\ast P_Y(n)\end{array}$$

Indeed,

$$P_{X+Y}=P_X\ast P_Y$$

Continuous

Given $X,Y$ independent RVs and densities $f_X,f_y$ are given, then:

$$\begin{array}{rl}{F}_{X+Y}(t)& =\mathbb{P}(X+Y\le t)\\ & ={\int }_{x+y\le t}{f}_{X,Y}(x,y)dxdy\\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{t-x}{f}_X(x)f_Y(y)dydx\\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^t{f}_X(x)f_Y(s-x)dsdx\\ & ={\int }_{-\infty }^t{\int }_{-\infty }^{\infty }{f}_X(x)f_Y(s-x)dsdx\end{array}$$

Then,

$$f_{X+Y}(s)={\int }_{-\infty }^{\infty }{f}_X(x)f_Y(s-x)dx$$

or, $f_X\ast f_Y(s)$

Example 1

$X,Y\sim \text{Unif}[0,1]$ and independent. What is $f_{X+Y}$? Let $s=X+Y$.

We have:

$$f_X(x)=\{\begin{array}{ll}1& 0\le x\le 1\\ 0& \text{otherwise}\end{array}{f}_Y(s-x)=\{\begin{array}{ll}1& 0\le s-x\le 1\\ 0& \text{otherwise}\end{array}$$

Indeed, we have the case when $0\le x\le 1$ and the case where

$$0\le s-x\le 1⟺s-1\le x\le 1$$

We calculate the convolutions:

$$\begin{array}{rl}{f}_{X+Y}(s)& ={\int }_0^s{f}_X(x)f_Y(s-x)dx\\ & ={\int }_0^s\frac{1}{1-0}\cdot \frac{1}{1-0}dx\\ & ={\left[x\right]}_0^s\\ & =s\end{array}\begin{array}{rl}{f}_{X+Y}(s)& ={\int }_{s-1}^{1}1\cdot 1dx\\ & ={\left[x\right]}_{s-1}^1\\ & =1-(s-1)\\ & =2-s\end{array}$$

such that

$$f_{X+Y}(s)=\{\begin{array}{ll}s& 0\le s\le 1\\ 2-s& 1\le s\le 2\\ 0& \text{otherwise}\end{array}$$

Example 2

Convolutions prove Sum of 2 Independent RVs.