Let $f(x)$ be a pdf and let $a$ be a number such that, for all $\epsilon>0$, $f(a+\epsilon=f(a-\epsilon)$. Such a pdf is said to be symmetric about the point a. Show that If $X \sim f(x)$ is symmetric then the median of X is $a$.

The solutions show that:

$$\int_a^{\infty}f(x)dx = \int_0^{\infty}f(a+\epsilon)d\epsilon$$

by changing the variable with $\epsilon = x-a$. I do not understand why the $a$ disappears in the integral sign.

However I need to get


I do not understand why the change of variable $x = a-\epsilon$ produces the integral sign.


1 Answer 1


$a$ disappears from the integral sign because of the change of variables that you performed. If $\epsilon = x-a\ \ (d \epsilon = dx)$, then, because of the equality $f(a - \epsilon) = f(a + \epsilon)$, we have:

$$ \int_{x = a}^{x = \infty} f(x) dx =\\ \int_{\epsilon=0}^{\epsilon = \infty} f(a+\epsilon) d\epsilon =\\ \int_{\epsilon=0}^{\epsilon = \infty} f(a-\epsilon) d\epsilon =\\ - \int_{\epsilon=0}^{\epsilon = -\infty} f(a+\epsilon) d\epsilon =\\ \int_{\epsilon=-\infty}^{\epsilon = 0} f(a+\epsilon) d\epsilon =\\ \int_{x = -\infty}^{x = a} f(x) dx $$ where I also changed variables one more time in the third equality $ \epsilon \mapsto - \epsilon$ (whence $d\epsilon \mapsto - d \epsilon$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.