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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

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    $\begingroup$ Did you mean to write $f(\mu - x) = f(\mu + x)?$ That would be the condition for the distribution being symmetric about $\mu$ $\endgroup$ – Bridgeburners Oct 3 '18 at 18:07
  • $\begingroup$ Because this is an immediate consequence of the symmetry, see stats.stackexchange.com/questions/28992/… for ideas. For instance, my answer there arrives at this conclusion in a single line. @Bridgeburners is correct. Your formulation implies $f$ is periodic with period $2\mu.$ There does not exist any such distribution. $\endgroup$ – whuber Oct 3 '18 at 18:29
  • $\begingroup$ Cross posted on math.stackexchange.com/q/2940974/321264. $\endgroup$ – StubbornAtom Oct 3 '18 at 18:37
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Hint:

Start with the expression, $$ \int_{-\infty}^\mu f(t) dt + \int_\mu^\infty f(t) dt = 1. $$

On the first integral, make the substitution: $$ t = \mu - x. $$

On the second integral, make the substitution: $$ t = \mu + x. $$

Once you do that, you should be able to see how you can use the symmetry condition to make the proof.

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  • $\begingroup$ i do this $\int_{-\infty}^{0} f(x+\mu).dx +\int_{2\mu}^{\infty} f(x-\mu).dx = 1$ what i should do next @Bridgeburners $\endgroup$ – alish Oct 3 '18 at 18:17
  • $\begingroup$ @alish You made a few mistakes doing the substitutions that I prescribed $\endgroup$ – Bridgeburners Oct 3 '18 at 18:20
  • $\begingroup$ $\int_{-\infty}^{2\mu} f(x-\mu).dx +\int_{0}^{\infty} f(x+\mu).dx = 1$ Do i write correctly? @Bridgeburners $\endgroup$ – alish Oct 3 '18 at 18:21
  • $\begingroup$ @alish The second one is correct, the first one still has some mistakes in both the argument and the limits. Keep in mind, the sign of the argument matters. $\endgroup$ – Bridgeburners Oct 3 '18 at 18:25
  • $\begingroup$ i see . but i can't understand one matter we know that $f(x-\mu) = f(x+\mu)$ but we have $\int_{0}^{\infty} -f(\mu-x).dx + \int_{0}^{\infty}f(\mu + x).dx = 1$ $\endgroup$ – alish Oct 3 '18 at 18:31

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