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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

alish
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  • the original equation holds for all $x$ or for a specific $x$? – gt6989b Oct 03 '18 at 16:57
  • sorry could you tell me more which equation you mentioned? if you talk about $f(x-\mu) = f(x+\mu)$ the answer is yes . it is true for all x . – alish Oct 03 '18 at 17:02
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    the question said that $F(\mu) = \frac{1}{2}$ no $F(x)$ and $\mu$ is a constant . – alish Oct 03 '18 at 17:21
  • For example, consider the normal distribution form. Since, for every constant value of the axis of curvature symmetry, the value of this graph is equal, then it applies to the condition of the question. Now the area under the curve is equal to one. Intuitively, it can be understood that the area under the curve is equal to the axis of symmetry with the area under the curve after the axis of symmetry. Because these two areas are equal and the total of these areas is equal to one, then the area of each is halved. – alish Oct 03 '18 at 17:36
  • When you change variable, don't forget the limits of integration. – Fabio Somenzi Oct 03 '18 at 17:41
  • On a more general level, if the distribution is symmetric, then $F(\mu) = 1/2$, but the converse is not true. Hence it's not properly an equivalence that you are proving, but an implication. – Fabio Somenzi Oct 03 '18 at 17:43
  • Yes, that's right, but this does not make a difference. because $-\infty < t < \infty$ and if i put $t = x-\mu$ it change to $-\infty < x-\mu < \infty \rightarrow -\infty + \mu < x < \infty + \mu$ and $\mu$ is constant because of that we can say $-\infty < x < \infty $ – alish Oct 03 '18 at 17:46
  • could tell me more about your last comment Fabio Somenzi . i can't understand it is not true generally . – alish Oct 03 '18 at 17:52
  • Intuitively, from the condition of the question, one can find that the function has a symmetric axis. The axis of symmetry is equal to the line $x = \mu$ – alish Oct 03 '18 at 17:54
  • Your comment about the integration limits is incorrect. when $x=\mu$, $x-\mu=0$ and that's the key. – Fabio Somenzi Oct 03 '18 at 17:58
  • Concerning the second comment, I'm only observing that a distribution with $F(\mu)=1/2$ needs not be symmetric. – Fabio Somenzi Oct 03 '18 at 18:00
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    I'm not seeing how the statement can be correct. Take $\mu=0$: for every probability function, we have $f(x-\mu)=f(x+\mu)$ for all $x$ (this is a vacuous statement). It's certainly not true that all real distributions functions satisfy $F(0)=1/2$. – Clement C. Oct 03 '18 at 18:02
  • i think because $f$ is probability density function it couldn't be every $f$ because we know that $\int_{-\infty}^{\infty}f(t).dt = 1$ – alish Oct 03 '18 at 18:06
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    @alish Read my comment. EVERY pdf satisfies $f(x)=f(x)$ for all $x$. There is no condition there! – Clement C. Oct 03 '18 at 18:07
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    No, @ClementC. is right. I overlooked the simple mistake that you have $f(x-\mu)=f(x+\mu$ instead of $f(\mu - x)=f(\mu+x)$. You still need to fix the integration bounds. – Fabio Somenzi Oct 03 '18 at 18:09
  • Cross posted on https://stats.stackexchange.com/q/369982/119261. – StubbornAtom Oct 03 '18 at 18:38
  • In fact, if $f(x-\mu)=f(x+\mu)$ for all $x,$ and $\mu\neq0,$ then $f$ is periodic and cannot be a PDF. – David K Oct 03 '18 at 20:35

1 Answers1

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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 $1/2$.

D. Ungaretti
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