The Convolution Formula

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Let $X$ and $Y$ be discrete random variables and let $S = X+Y$. We know that a good way to find the distribution of $S$ is to partition the event ${ S = s}$ according to values of $X$. That is,

If $X$ and $Y$ are independent, this becomes the discrete convolution formula:

This formula has a straightforward continuous analog. Let $X$ and $Y$ be continuous random variables with joint density $f$, and let $S = X+Y$. Then the density of $S$ is given by

which becomes the convolution formula when $X$ and $Y$ are independent:

Sum of Two IID Exponential Random Variables

Let $X$ and $Y$ be i.i.d. exponential $(\lambda)$ random variables and let $S = X+Y$. For the sum to be $s > 0$, neither $X$ nor $Y$ can exceed $s$. The convolution formula says that the density of $S$ is given by

That’s the gamma $(2, \lambda)$ density, consistent with the claim made in the previous chapter about sums of independent gamma random variables.

Sometimes, the density of a sum can be found without the convolution formula.

Sum of Two IID Uniform $(0, 1)$ Random Variables

Let $S = U_1 + U_2$ where the $U_i$’s are i.i.d. uniform on $(0, 1)$. The gold stripes in the graph below show the events ${ S \in ds }$ for various values of $S$.

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The joint density surface is flat. So the shape of the density of $S$ depends only on the lengths of the stripes, which increase linearly between $s = 0$ and $s = 1$ and then decrease linearly between $s = 1$ and $s = 2$. So the joint density of $S$ is triangular. The height of the triangle is 1 since the area of the triangle has to be 1.

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At the other end of the difficulty scale, the integral in the convolution formula can sometimes be quite intractable. In the rest of the chapter we will develop a different way of deriving distributions of sums.