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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

$$\int_{-\infty}^af(x)dx$$

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

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$a$ 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 $d\epsilon \mapsto - d \epsilon$).

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