# Equivalence of the probability distribution of a symmetric function = 1/2

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

• Did you mean to write $f(\mu - x) = f(\mu + x)?$ That would be the condition for the distribution being symmetric about $\mu$ – Bridgeburners Oct 3 '18 at 18:07
• 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. – whuber Oct 3 '18 at 18:29
• Cross posted on math.stackexchange.com/q/2940974/321264. – StubbornAtom Oct 3 '18 at 18:37

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.

• 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 – alish Oct 3 '18 at 18:17
• @alish You made a few mistakes doing the substitutions that I prescribed – Bridgeburners Oct 3 '18 at 18:20
• $\int_{-\infty}^{2\mu} f(x-\mu).dx +\int_{0}^{\infty} f(x+\mu).dx = 1$ Do i write correctly? @Bridgeburners – alish Oct 3 '18 at 18:21
• @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. – Bridgeburners Oct 3 '18 at 18:25
• i see . but i can't understand one matter we know that $f(x-\mu) = f(x+\mu)$ but we have $\int_{0}^{\infty} -f(\mu-x).dx + \int_{0}^{\infty}f(\mu + x).dx = 1$ – alish Oct 3 '18 at 18:31