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How could I compute the expected value E(x) given that its cdf is,

$F(x) = \dfrac{\exp(x)}{1+\exp(x)}$.

I know the method to derive E(x), as in this post. I still find it very hard to derive the result. Thanks.

19aksh
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Carl
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1 Answers1

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I prefer to refer to this question.

We have $$F(x)=\frac{e^x}{e^x+1}\implies f(x) = \frac{dF(x)}{dx}=\frac{e^x}{\left(e^x+1\right)^2}$$ $$E[X] = \int_{1}^{\infty} x f(x)\,dx$$ $$I = \int x f(x)\,dx=\frac{x \,e^x }{e^x+1}-\log \left(e^x+1\right)$$ Just use the bounds.

Claude Leibovici
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  • Thank you so much. I thought it was a numerical value. Why it is not $$E[X] = \int_{-\infty}^{\infty} x f(x),dx$$ – Carl Mar 04 '21 at 05:26
  • Could I use a software to compute it? – Carl Mar 04 '21 at 15:05
  • Have a look at https://www.wolframalpha.com/input/?i=Integrate%5Bx+Exp%5Bx%5D%2F%281%2BExp%5Bx%5D%29%5E2%2Cx%5D and https://www.wolframalpha.com/input/?i=Integrate%5Bx+Exp%5Bx%5D%2F%281%2BExp%5Bx%5D%29%5E2%2C%7Bx%2C0%2CInfinity%7D%5D – Claude Leibovici Mar 04 '21 at 15:25
  • Thank you so much. – Carl Mar 04 '21 at 16:22