Let $\mu \in \mathbb{R}$ and suppose the probability density function $f$ of the random variable $X$ satisfies $$f(x-\mu) = f(x+\mu) \quad \forall x \in \mathbb R.$$ Show that $F(\mu) = \frac{1}{2}$, where $F$ denotes the probability distribution function of $X$, $$F(x) = \int_{-\infty}^x f(t)\ dt$$
My Approach
$$F(+\infty) = 1 \implies \int_{-\infty}^{+\infty} f(t)\ dt = \int_{-\infty}^{\mu} f(t)\ dt + \int_{\mu}^{+\infty} f(t)\ dt = 1 $$
i change my variable in $f$ this means that $x-\mu = t \to dx = dt$
$$\int_{-\infty}^{2\mu} f(x-\mu)\ dx + \int_{2\mu}^{+\infty} f(x-\mu)\ dx = 1 $$
and we know that $f(x-\mu) = f(x+\mu)$
$$\int_{-\infty}^{2\mu} f(x+\mu)\ dt + \int_{2\mu}^{+\infty} f(x-\mu)\ dx = 1 $$
but i don't know what i should do . I think if I can prove in a way that two integrals are equal, the question is solved, but I have no idea to prove their equality. Please help me.
we can understand from $f(x+\mu) = f(x-\mu)$ our function $f$ is a symmetric function.
Thanks a lot
I believe the expression $f(x + \mu) = f(x - \mu)$ for every $x$ should be actually $f(\mu + x) = f(\mu - x)$ for every $x$. Indeed, if you consider the previous condition and that $\mu = 0$, your condition tells us nothing at all, so your claim is false.
Assuming the condition is the one I stated above, notice that the substitution $t = \mu - x$ implies $$ \int_{-\infty}^{\mu} f(t)dt = \int_{+\infty}^{0} -f(\mu - x) \, dx = \int_{0}^{\infty} f(\mu - x)\, dx, $$ and the substitution $t = \mu + x$ implies $$ \int_{\mu}^{\infty} f(t)dt = \int_{0}^{\infty} f(\mu + x) \, dx. $$ Since by hypothesis $f(\mu - x) = f(\mu + x)$ for every $x$, we have that the expressions above are equal. But we know they sum to one, so each of them is $1/2$.