# Solved – Prove that the mean value of a convolution is the sum of the mean values of its individual parts

Prove that the mean value of a density function convolution, if the mean values exist (they may not for fat-tailed distributions) is the sum of the mean values of the density functions used to make that convolution.

This is a simple one, but, I would like to see what the compact formal notation looks like for it.

Contents

Like many demonstrations involving convolutions, it comes down to applying Fubini's Theorem.

### Let's establish notation and assumptions.

Let \$f\$ and \$g\$ be integrable real-valued functions defined on \$mathbb{R}^n\$ having unit integrals (with respect to Lebesgue measure): that is,

\$\$1=int_{mathbb{R}^n} f(x) dx = int_{mathbb{R}^n} g(x) dx.\$\$

(For convenience, let's drop the "\$mathbb{R}^n\$" subscript, because all integrals will be evaluated over this entire space.)

The convolution \$fstar g\$ is the function defined by

\$\$(fstar g)(x) = int f(x-y) g(y) dy.\$\$

(This is guaranteed to exist when \$f\$ and \$g\$ are both bounded or whenever \$f\$ and \$g\$ are both probability density functions.)

The mean of any integrable function is

\$\$E[f] = int x f(x) dx.\$\$

It might be infinite or undefined.

### Solution

The question asks to compute \$E[fstar g]\$ (in the special case where \$f\$ and \$g\$ are nonnegative–but this assumption doesn't matter). Apply the definitions of \$E\$ and \$star\$ to obtain a double integral; switch the order of integration according to Fubini's Theorem (which requires assuming \$E[fstar g]\$ is finite), then substitute \$x-yto u\$ and exploit linearity of integration (which is a basic property established immediately whenever any theory of integration is developed). The result will appear because both \$f\$ and \$g\$ have unit integrals.

For those who want to see the details, here they are:

\$\$eqalign{ E[fstar g] &= int x (fstar g)(x) dx &text{Definition of }E\ &= int x left(int f(x-y) g(y) dyright) dx &text{Definition of convolution}\ &= int g(y) left(int x f(x-y) dxright) dy &text{Fubini}\ &= int g(y) left(int (x-y)f(x-y) + yf(x-y) dxright) dy&text{Expand }x=(x-y)+y \ &= int g(y) left(int (x-y)f(x-y) dx + yint f(x-y) dxright) dy &text{Linearity of integration}\ &= int g(y) left(int u f(u) du + y int f(u) duright) dy &text{Substitution } x-yto u\ &= int g(y) (E[f] + y(1)) dy &text{Assumptions about }f\ &= E[f]int g(y) dy + int y g(y) dy &text{Linearity of integration}\ &= E[f](1) + E[g] &text{Assumptions about }g\ &= E[f] + E[g]. }\$\$

These calculations are legitimate provided all three expectations \$E[fstar g], E[f], E[g]\$ are defined and finite. Fubini's Theorem requires only the finiteness of \$E[fstar g],\$ but the steps at the end (involving linearity) also need the finiteness of the other two expectations.

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