Find the mathematical expectation

Question

Find the expectation of

$$f(x) = a(1+x)^{-(1+a)}, \quad x>0.$$

The answer given is $\frac{1}{a-1}$.

I am not getting the answer. Please help.

Answer 1

Your problem is simply an integration problem. The average value can be written as $$<x>=\int_{0}^{\infty}xf(x)dx=a\int_{0}^{\infty}x(1+x)^{-(1+a)}dx$$ you can integrate by parts, calling $u=x$ and $dv=(1+x)^{-(1+a)}$, finding $$<x>=\int_{0}^{\infty}xf(x)dx=\left.-(1+a)^{-a}x\right|_{0}^{\infty}+\int_{0}^{\infty}(1+x)^{-a}dx=\frac{-1}{1-a}$$ considering $a>1$ (otherwise the function do not converge)

Edit: After looking at Did's comment, I have realized that there was wrong sign mistake in the answer: a positive variable can never have a negative expectation value

Answer 2

If $X$ is a random variable with density $f(x)$, then $$E(X) = \int_{-\infty}^\infty xf(x) dx = a\int_0^\infty x(1+x)^{-(1+a)}$$

First of all, this converges only if $a > 1$

Then the integral can be done fairly easily with the substitution $ u = x +1$, unless I'm mistaken