I'm trying to prove that if $f(a+\epsilon) = f(a-\epsilon)$, i.e. $f(x)$ is symmetric about $a$, then $a$ is the median of a continuous random variable with pdf $f(x)$. Using the fact that $a$ being the median means that $$\int_{-\infty}^a f(x)dx = \int_{a}^{\infty}f(x)dx = 1/2$$ I thought I could do something like

$$1 = \int_{-\infty}^{\infty}f(x)dx = \int_{-\infty}^a f(x)dx + \int_{a}^{\infty}f(x)dx,$$

but couldn't see why I could claim that the two integrals had to then be equal/ both then must be $1/2$.

I saw a similar version of what I'm trying to prove with $a=0$ here, but couldn't figure out how to apply a transform to my integrals so that it worked out the same way. I tried using $x=a-y$, but still just couldn't get the negative signs to work out correctly.

Any advice is apricated, and thank you in advance.


1 Answer 1


You start by knowing that $\int_{-\infty}^{a} f(x) \,dx + \int_{a}^{\infty} f(x) \,dx =1 $ where $a$ is the point of symmetry. Then use substitution $y=x-a$ and you will get

$\int_{-\infty}^{0} f(a+y) \,dy + \int_{0}^{\infty} f(a+y) \,dy =1 $

Now another substitution only for the first integral $y=-z$

$\int_{-\infty}^{0} f(a+y) \,dy = \int_{\infty}^{0}- f(a-z) \,dz = \int_{0}^{\infty} f(a-z) \,dz = \int_{0}^{\infty} f(a+z) \,dz$

So you can rewrite the equation using the first integral as above

$\int_{0}^{\infty} f(a+y) \,dy + \int_{0}^{\infty} f(a+y) \,dy =1 $

$2*\int_{0}^{\infty} f(a+y) \,dy =1 $

$\int_{0}^{\infty} f(a+y) \,dy =1/2 $

Now undo the substitution

$\int_{a}^{\infty} f(x) \,dx =1/2 $

and rewrite the equation

$\int_{-\infty}^{a} f(x) \,dx + 1/2 =1 $

$\int_{-\infty}^{a} f(x) \,dx = 1/2 $

that means that the two integrals are equal and both must be 1/2

$\int_{-\infty}^{a} f(x) \,dx = \int_{a}^{\infty} f(x) \,dx = 1/2 $


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