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?
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 $x = a - s$.
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 $\mathbf{E}X$.
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.
$E(X) - a = \int x f(x)dx - a\int f(x)dx = \int (x-a) f(x) dx =$
(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 $\int\limits_{-\infty}^{\infty}$.
