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I came across this statement in a book. While I know, how to prove $E(X) = a$ is using $f(x+a)=f(x-a)$. I cannot seem to prove it using $f(x-a)=f(-(x-a))$. I keep going on in a loop, no matter what I do. Any help will be very appreciated.

Doubts
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    Can you please be more clear about what you want to prove? – Ale Oct 20 '20 at 15:08
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    Can you share your source? Currently, $f(x-a)=f(-(x-a))$ doesn't mean anything other than $f(y)=f(-y)$ to me. – gunes Oct 20 '20 at 15:21
  • I need to show E(X)= a – Doubts Oct 20 '20 at 15:46
  • $f(x+a)=f(x-a) \not\equiv f(x-a)=f(-(x-a)) $ – Sextus Empiricus Oct 20 '20 at 15:54
  • @Doubts did you think about writing the expectation as an integral, and imagining how this relates to a graph of $f(x-a)$ and the surface under the graph? – Sextus Empiricus Oct 20 '20 at 15:56
  • Would you get that $E[X] = 0$ if $f(x) = f(-x)$? – Sextus Empiricus Oct 20 '20 at 15:58
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    The expected value may not exist if $f(x+a)=f(x-a)$, and neither under $f(y)=f(-y)$. Example: Cauchy distribution. So in general the statement is not true, regardless on how you write it down. Obviously it will be true *assuming that $E[X]$ exists*. – Christian Hennig Oct 20 '20 at 16:31
  • In light of the comment by @Lewian, what you *can* prove is that $E(X)=2a-E(X).$ From that it follows that when $E(X)$ is finite, $E(X)=a;$ otherwise $E(X)$ must be undefined. – whuber Oct 20 '20 at 17:34
  • @whuber How does your $E(X)=2a-E(X)$ make sense when $E(X)$ is undefined? – Dave Oct 21 '20 at 03:35
  • @Dave It equates two undefined quantities. That makes sense because (in this case) it is a deduction predicated on a false assumption; namely, that $E(X)$ *is* defined. – whuber Oct 21 '20 at 12:30

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