$E(X)= a$ for some constant $a$ Let $X$ be a random variable and let $a$ be a constant such that
$f_X(a-y) = f_X(a+y)$ for all $y$. Prove  then that $E(X) = a$. 
the hint Iam given is to show that $E(X) - a = \int_{-\infty}^{\infty+} (x-a)f_X(x)dx = 0$ but i still don't know how to show it, can someone help me?
 A: Hints: Do you know how to express $E(X)$ as an integral?
And what do you know about $\int_{-\infty}^\infty f_X(x)\ dx$?
Once you have $E(X) - a = \int_{-\infty}^\infty (x - a) f_X(x)\ dx$, use the changes of variable $x = a + t$ and $x = a - s$.
A: Assuming I understand what you need (and I don't understand how to use the identity for $f_X(x+a)$ if you are given the equality, then
$$
0=\int_{D}x f dx-\int_{D}a f dx = \int_{D}x f dx -a \cdot 1
$$ 
hence $\mathbf{E}X=a$. On RHS we have the definition of $\mathbf{E}X$.
A: First, recall that
$$
1=\int f_X(y)\,\mathrm{d}y\tag{1}
$$
and
$$
\mathrm{E}(X)=\int yf_X(y)\,\mathrm{d}y\tag{2}
$$
A change of variables yields
$$
\mathrm{E}(X)=\int (a+y)f_X(a+y)\,\mathrm{d}y\tag{3}
$$
Another change of variables gives
$$
\mathrm{E}(X)=\int (a-y)f_X(a-y)\,\mathrm{d}y\tag{4}
$$
Apply the given identity to $(4)$ to get
$$
\mathrm{E}(X)=\int (a-y)f_X(a+y)\,\mathrm{d}y\tag{5}
$$
Average $(3)$ and $(5)$ and you're almost done.
Note that the given identity simply states that the distribution of $X$ is symmetric about $a$. The computation above simply shows what seems obvious from a diagram.
A: $E(X) - a = \int x f(x)dx - a\int f(x)dx = \int (x-a) f(x) dx =$ 
(since $\int f(x)dx = 1$ for $f$  probability density function.
Making $t = x-a$ we have $dt=dx$ and:)
$= \int t f(a+t) dt$ =
(Making now $g(t)=tf(a+t)$ we have $g(-t)=-tf(a-t)=-tf(a+t)=-g(t) \ \forall t \in \mathbb{R}$, so since g is an odd function $\int g(t) dt = 0$ .)
$= \int g(t)dt =0$.
Therefore $E(X) - a = 0 \iff E(X)=a$.
I used $\int$ to denote the integral over the real line, so all used integrals are $\int\limits_{-\infty}^{\infty}$.
