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I have normally distributed $X\sim\mathcal{N}(0, 1)$, and I want to compute \begin{equation*} \mathbb{E}[1/(1+e^X)] = \int_{-\infty}^\infty \frac{e^{-x^2/2}/\sqrt{2\pi}}{1+e^x} dx \end{equation*}

I found numerically (and confirmed with Mathematica) that $\mathbb{E}[1/(1+e^X)] = 1/2$; this result continues to hold for arbitrary variances but breaks down once I select non-zero mean for $X$.

How can I prove this result? The integration trick to use is not jumping out to me.

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  • $\begingroup$ Substitute x as -t. The integration result must be same. Let the both be C, and calculate 2C by adding those two integrals. They should give the answer. $\endgroup$
    – Joshua Woo
    Jul 17, 2020 at 4:43

2 Answers 2

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\begin{equation} \frac{2}{1+e^X} = 1 + \frac{1-e^X}{1+e^X} = 1 - \tanh(X/2) \end{equation} Hyperbolic tangent is odd, hence the expected is $0$ if mean of $X$ is $0$.

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$$N=\left( \int_{-\infty}^{\infty}\sqrt{2\pi}~ e^{-x^2/2} dx\right)^{-1}=(2\pi)^{-1}$$ So $$E[1/(1+e^x)]=N\sqrt{2\pi} \int_{-\infty}^{\infty} \frac{e^{-x^2/2}}{1+e^x}dx =I$$ Use $$\int_{-a}^{a} f(x) dx= \int_{0}^{a} [f(x)+f(-x)] dx,$$ to get $$I=N\sqrt{2\pi} \int_{0}^{\infty} e^{-x^2/2}~ dx=N\sqrt{2\pi} \sqrt{\frac{\pi}{2}}= \frac{1}{2}$$

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