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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.

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  • $\begingroup$ Convolution of densities arises when you add independent r.v.s; but the result you ask for proof of doesn't require independence -- it's quite general that E(X+Y) = E(X)+E(Y) $\endgroup$
    – Glen_b
    Apr 22, 2018 at 8:42
  • $\begingroup$ @Glen_b I know, I just want to see it. $\endgroup$
    – Carl
    Apr 22, 2018 at 9:02
  • $\begingroup$ There are two calculations involved: one is to demonstrate that integration is additive (which is clear for Riemann integration and involves technical subtleties for other integrals, such as Lebesgue integrals) and the other is to demonstrate that the density of a sum of RVs is the convolution of their densities. Your question implicitly assumes not only that a demonstration would be uncomplicated, but also that there exists some unique "compact formal notation." There is not: consult a textbook that is appropriate for your interests and background and use its notation. $\endgroup$
    – whuber
    Apr 22, 2018 at 14:29
  • $\begingroup$ @whuber Is the answer below correct? $\endgroup$
    – Carl
    Apr 22, 2018 at 14:34
  • $\begingroup$ It's a nice answer. Note that it assumes much of the essential matter: namely, that the PDF of a sum of RVs is given by the convolution of their PDFs. Although one has to start somewhere in any demonstration, the way your question was asked suggested to me that you were looking primarily for a demonstration of this particular fact. In comparison, linearity of expectation is obvious and almost trivial to demonstrate, and so would appear to be by far the lesser part of your question. $\endgroup$
    – whuber
    Apr 22, 2018 at 14:48

2 Answers 2

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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.

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  • $\begingroup$ Like your's better. Has the iff's. $\endgroup$
    – Carl
    Apr 22, 2018 at 16:05
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    $\begingroup$ If two of the means exist, the third exists. $\endgroup$
    – Carl
    Apr 23, 2018 at 2:55
  • $\begingroup$ Moreover, if $E[f\star g]$ is finite, then $𝐸[𝑓]$ and $𝐸[𝑔]$ are defined and finite. $\endgroup$
    – Carl
    Apr 5, 2020 at 1:51
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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 $Y$:

$$E[X+Y] = E[X] + E[Y]$$

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