Finding the mean from a probability density function

Question

This one is worded a little weirdly, since the question I found seems to maintain that there exists an unknown mean for the PDF and a "mystery number" represented by $\mu$. Just something to be aware of when reading the following question:

Let f be a continuous density having a finite mean and $\mu$ be any number.

Suppose that $f(x) = f(-x)$, i.e., $f$ is symmetric about 0.

Convince yourself that $f(x-\mu)$ is a valid density.

What is its associated mean?

The following is the furthest progress I've made on this problem:

1. $E[X] = \int_{-\infty}^{\infty}(x-\mu)f(x-\mu)dx$
2. $= \int_{-\infty}^{\infty}xf(x-\mu)dx - \int_{-\infty}^{\infty}\mu f(x-\mu)dx$
3. $= \int_{-\infty}^{\infty}xf(x-\mu)dx - \mu$; which is accomplished by treating $\mu$ as a constant that gets removed outside of the integral, leaving us with the integration of the pdf from negative to positive infinity, which should equal 1.

I am unsure how to proceed from here with the remaining first term. Trying to solve it as an "integration by parts" gives me the following:

$xf'(x-\mu)-1-\mu$

Assuming this is correct, how should I proceed? The question also gives me the option of choosing, "It can't be ascertained", as a solution, but I suspect that's not true in this case.

Answer 1

First, a change of variable $t=x-\mu$ enables you to finish your computations.

Second, your definition of the mean is wrong: if $X$ has density $x\mapsto f(x-\mu)$, then $$ \mathbb{E}(X)= \int x f(x-\mu) \mathrm{d}x = \int (t+\mu) f(t) \mathrm{d}t = \int t f(t) \mathrm{d}t + \int \mu f(t) \mathrm{d}t = 0 + \mu = \mu. $$

Edit: thanks Henry for the remark (I thought $\mu$ was the mean). Now fixed!

Answer 2

The condition $f_{X}(x) = f_{X}(-x)$ implies that $X \stackrel{d}{=} -X$ i.e.$P(X\leq x)= P(-X\leq x)$ for all $x$. In particular $EX=-EX$ i.e. $EX=0$.

Note that $Y=X+\mu$ has the density $f_{Y}(y)=f_{X}(y-\mu)$ by the change of variables formula (the jacobian determinant is $1$)

It follows that $EY = EX+\mu = \mu$.