Given PDF find Mean of the distribution

Question

pdf: $ f(x) = \frac{e^{-x}}{(1+e^{-x})^2} $

Find the mean of the distribution using: $ \int_{-\infty}^{\infty} xf(x) \ \mathrm{d}x $

Is it possible?

I have run the integral in a calculator only to find out that it diverges.

I got: $ \int x\frac{e^{-x}}{(1+e^{-x})^2} \ \mathrm{d}x = \frac{x}{1+e^{-x}}-\ln \left|e^{-x}+1\right|+C $

which should basically diverge. Am I missing something here?

@Dr.SonnhardGraubner Thank You! turns out as I switched the calculator I too got the same result (=0). Interestingly $\int xf(x)dx$ is symmetric about y- axis, so the mean must be zero. — DubsVeer23, May 15 '19 at 10:58
Density is symmetric about zero, so of course mean (which exists) equals zero. — StubbornAtom, May 15 '19 at 10:59
Also asked here: https://math.stackexchange.com/questions/2849517/how-to-calculate-mean-and-variance-for-logistic-distribution. — StubbornAtom, May 15 '19 at 11:22

score 1 · Accepted Answer · answered May 15 '19 at 11:19

First of all, notice that $|e^{-x}+1| = e^{-x}+1.$ Moreover, observe that for any $a > 0$:

$$\int_{-a}^a x\frac{e^{-x}}{(1+e^{-x})^2} \ \mathrm{d}x = -\left[\log(e^{-x} + 1) + \frac{x}{e^{x} + 1}\right]_{x=-a}^{x=a} = \\ -\left[\log(e^{-a} + 1) + \frac{a}{e^{a} + 1} - \log(e^{a} + 1) - \frac{-a}{e^{-a} + 1}\right] = \\ -\left[\log\left(\frac{e^{-a} + 1}{e^{a} + 1}\right) + \frac{a}{e^{a} + 1} + \frac{a}{e^{-a} + 1}\right] = \\ -\left[\log\left(\frac{e^{-a}(e^{a} + 1)}{e^{a} + 1}\right) + \frac{a}{e^{a} + 1} + \frac{ae^{a}}{e^{a}(e^{-a} + 1)}\right] = \\ -\left[\log e^{-a} + \frac{a}{e^{a} + 1} + \frac{ae^{a}}{e^{a} + 1}\right] = \\ -\left[-a + \frac{a(e^a + 1)}{e^{a} + 1} \right] = -[-a+a] = 0.\\ $$

Given PDF find Mean of the distribution

1 Answers1