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Given $X$ has density function $$f(x)= \frac{e^{x}}{(1+e^{x})^2},\quad -\infty<x<\infty$$ How to show $\mathbb{E}(X)$ is finite.

Thanks for your help. I am able to answer it now after getting help on stackexchange!

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    $\begingroup$ Can you show us any calculations, or attempts at solving this problem, that you have tried? It is easy for us to give you hints if we know what you have already tried. $\endgroup$ – user25658 Sep 18 '13 at 17:15
  • $\begingroup$ I sure hope there is a typo in your hint and that's not the way it was actually written on your assignment! (You need to find an upper bound for $\mathbb E|X|$, not $\mathbb E X$.) $\endgroup$ – cardinal Sep 18 '13 at 17:30
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    $\begingroup$ Hint: Use symmetry. Bound just(!) the density part. Recognize a very simple and familiar integrand. Declare success. $\endgroup$ – cardinal Sep 18 '13 at 17:31
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Please, fill in the details marked with a $\star$. First of all, remember that to prove that $\mathrm{E}[X]$ is finite it is enough $\star$ to check that $\mathrm{E}[|X|]$ is finite. Symmetry $\star$ shows that $$ \mathrm{E}[|X|] = \int_{-\infty}^\infty \frac{|x|\, e^x}{(1+e^x)^2} \, dx = 2 \int_0^\infty \frac{x\, e^x}{(1+e^x)^2} \, dx \, . $$ For $x>0$, we have $\star$ $$ \frac{x\,e^x}{(1+e^x)^2} < \frac{x\,e^x}{e^{2x}} = x\,e^{-x} \, . $$ Therefore, if we let $Y$ be a r.v. with $\mathrm{Exp}(1)$ distribution, it follows $\star$ that $\mathrm{E}[|X|]<2\, \mathrm{E}[Y] = 2$.

(Now, taking @cardinal's advice, speak out loudly and proudly: Success!)

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    $\begingroup$ This is, almost verbatim, what I had in mind. (+1) $\endgroup$ – cardinal Sep 18 '13 at 18:51
  • $\begingroup$ That means that E[X] is less that 2 and hence finite. $\endgroup$ – user30438 Sep 18 '13 at 18:56

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