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.
 A: 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 $f\star g$ is the function defined by
$$(f\star g)(x) = \int f(x-y) g(y) dy.$$
(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 $E[f\star g]$ (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 $E[f\star g]$ is finite), then substitute $x-y\to u$ 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:
$$\eqalign{
E[f\star g] &= \int x (f\star g)(x) dx &\text{Definition of }E\\
&= \int x \left(\int f(x-y) g(y) dy\right) dx &\text{Definition of convolution}\\
&= \int g(y) \left(\int x f(x-y) dx\right) dy &\text{Fubini}\\
&= \int g(y) \left(\int (x-y)f(x-y) + yf(x-y) dx\right) dy&\text{Expand }x=(x-y)+y \\
&= \int g(y) \left(\int (x-y)f(x-y) dx + y\int f(x-y) dx\right) dy &\text{Linearity of integration}\\
&= \int g(y) \left(\int u f(u) du + y \int f(u) du\right) dy &\text{Substitution } x-y\to u\\
&= \int g(y) (E[f] + y(1)) dy &\text{Assumptions about }f\\
&= E[f]\int g(y) dy + \int y g(y) dy &\text{Linearity of integration}\\
&= E[f](1) + E[g] &\text{Assumptions about }g\\
&= E[f] + E[g].
}$$ 
These calculations are legitimate provided all three expectations $E[f\star g], E[f], E[g]$ are defined and finite.  Fubini's Theorem requires only the finiteness of $E[f\star g],$ but the steps at the end (involving linearity) also need the finiteness of the other two expectations.
A: Convolving distributions corresponds to adding independent random variables. Given PDFs $f_X$ and $f_Y$, let $f_Z = f_X * f_Y$ denote their convolution. $f_Z$ is the PDF of a random variable $Z = X + Y$, where $X \sim f_X$, $Y \sim f_Y$, and $X$ and $Y$ are independent. By linearity of expectation, $E[Z] = E[X+Y] = E[X] + E[Y]$.
Proof for linearity of expectation:
Let $f_{X,Y}$ be the joint distribution of $X$ and $Y$, with marginal distributions $f_X$ and $f_Y$. $X$ and $Y$ need not be independent.
The expected value of $X+Y$ is:
$$E[X+Y] =
\int_{-\infty}^\infty \int_{-\infty}^\infty
(x+y) \ f_{X,Y}(x, y) \ dx \ dy$$
Re-arranging terms gives:
$$E[X+Y] =
\int_{-\infty}^\infty x
\underbrace{\int_{-\infty}^\infty f_{X,Y}(x, y) \ dy}_{f_X} \ dx
+ \int_{-\infty}^\infty y
\underbrace{\int_{-\infty}^\infty f_{X,Y}(x, y) \ dx}_{f_Y} dy$$
As indicated under the terms above, integrating over the joint distribution gives the marginal distributions for $X$ and $Y$, so:
$$E[X+Y] =
\int_{-\infty}^\infty x \ f_X(x) \ dx
+ \int_{-\infty}^\infty y \ f_Y(y) \ dy$$
This corresponds to the sum of the expected values of $X$ and $Y$:
$$E[X+Y] = E[X] + E[Y]$$
