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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Sign up to join this communityGiven $X$ 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!
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!)