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So the problem asks: If $X$ is a continuous random variable with a density that is symmetric about some point, $ξ$ , show that $E(X) = ξ$ , provided that $E(X)$ exists.

The solution given is: enter image description here

But I am confused that why is $X-ξ$ is symmetric about $0$? Isn't $X-ξ$ the distance between $X-ξ$?

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    what is your definition of $X$ being symmetric about some point $a$ ? – 2017-02-07
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    Yes, but the proof looks nicer if the distribution is symmetrical about $0$ as for $Y$ instead of symmetrical about $\xi$ as for $X.$ – 2017-02-07
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    @BruceET Thanks I got it!! – 2017-02-07

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But I am confused that why is $X-ξ$ is symmetric about $0$?

That $X$ is symmetrically distributed about $\xi$ means exactly that $f_X(\xi+s)=f_X(\xi-s)$ for all $s\in\Bbb R^+$.

$Y=X-\xi$ is a linear transformation; a shift of axis.   As such $f_Y(y)=f_X(y+\xi$).

Then clearly $f_Y(y)=f_Y(-y)$, so therefore $Y$ is symmetric about $0$.

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    I got it! So when $Y=X-\xi$, $\xi = \xi-\xi = 0$! – 2017-02-07