Convolution of Independent Random Variables¶

Theorem¶

For independent continuous random variables $X$ and $Y$ , the density of $X+Y$ is the convolution of their densities; independence is what lets the joint density factor into the two marginal densities:

\begin{aligned} f_{X+Y}(t)=\int f_X(x)\,f_Y(t-x)\,\mathrm{d}x. \end{aligned}

Interpretation¶

A sum can reach a value in many ways. For $X+Y$ to be near $t$ , $X$ takes some value $x$ and $Y$ must take the matching value $t-x$ $x$ ; convolution adds up all these splits, each weighted by $f_X(x)f_Y(t-x)$ $x)$ .

Exponential densities for X and Y, the density t times e to the minus t of their sum, and a shaded rectangle over x from 0 to 2 whose area gives the density of the sum at t = 2.

For independent $X,Y\sim\operatorname{Exp}(1)$ $Exp (1)$ , the product $\mathrm{e}^{-x}\mathrm{e}^{-(2-x)}=\mathrm{e}^{-2}$ is constant for $0<x<2$ $2$ , so the green region is a rectangle of height $\mathrm{e}^{-2}\approx0.135$ and width $2$ . Its area $f_{X+Y}(2)=2\mathrm{e}^{-2}\approx0.27$ is the height of the red point. More generally, $f_{X+Y}(t)=t\mathrm{e}^{-t}$ for $t>0$ $0$ .

Independence is what turns the joint density $f_{X,Y}(x,t-x)$ into the product $f_X(x)f_Y(t-x)$ $x)$ . Without it the formula fails: for $X\sim\operatorname{Unif}(0,1)$ and $Y=X$ $X$ , the sum $2X$ is uniform on $(0,2)$ with density $1/2$ , whereas the convolution of two uniform densities gives the triangular density. The integral only runs over the $x$ for which both factors are positive, which for exponentials is $0<x<t$ $t$ .

The discrete version is the sum used for two dice in Probability Theory (sum of independent random variables). Moment generating functions give a second route: by the MGF of an independent sum, $M_{X+Y}=M_XM_Y$ $M_{X} M_{Y}$ , which avoids the integral when the product is a recognizable MGF. The density $t\mathrm{e}^{-t}$ found here is the Gamma distribution with shape $2$ and rate $1$ .

Adding values is not adding densities: $f_X+f_Y$ has total integral $2$ , and the product $f_X(t)f_Y(t)$ is not a density of the sum either.

Key Results¶

For independent discrete variables, $\displaystyle p_{X+Y}(t)=\sum_xp_X(x)p_Y(t-x)$ $x)$ .
For independent variables with densities, the integral is over $x$ for which both density factors are nonzero.

Examples¶

For independent $\operatorname{Exp}(1)$ variables, $f_{X+Y}(t)=\int_0^t\mathrm{e}^{-x}\mathrm{e}^{-(t-x)}\,\mathrm{d}x=t\mathrm{e}^{-t}$ for $t>0$ $0$ .
For independent fair Bernoulli variables, the sum has masses $1/4,1/2,1/4$ at $0,1,2$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 3

Aliases: convolution formula, convolution integral Symbols: integral, sum

Last updated: 20 October 2026 03:02