Compute variance, using explicit PDF

Question

I'm trying to get $\text{Var}(x)$ of $f(x) = 2(1+x)^{-3},\ x>0$.

Please tell me if my working is correct and/or whether there is a better method I can use to get this more easily.

$$ \text{Var}(x) = E[X^2] - [E[X]]^2 $$ and $$E[X] = \int_0^{\infty}2\frac{x}{(x+1)^3}dx$$

Let $u = x+1$

$du = dx$

When $x=0, u = 0 + 1 = 1$ \begin{align} $E[X] &= 2\int_1^\infty \frac{u-1}{u^3}du\\ &= 2[\int_1^\infty\frac{1}{u^2}du-\int_1^{\infty}\frac{1}{u^3}du]\\ &= 2[[-\frac{1}{u}]_1^\infty-[-\frac{1}{2u^2}]_1^\infty]\\ &=2[[0-(-\frac{1}{1})]-[0-(-\frac{1}{2(1)^2})]]\\ &=2[1-\frac{1}{2}]\\ &=1\\ \end{align} $$E[X^2] = \int_0^{\infty}2\frac{x^2}{(x+1)^3}dx$$

Now I'm stuck

I tried to say. $u = x+1$. $x = u-1$. $x^2 = u^2-2u+1$ and substitute but then I get $\int_1^\infty \frac{1}{u} = [\ln|u|]_0^\infty$ for one of the terms?

Answer 1

By definition, $$E(X^2)=\int_0^\infty x^2f(x)\mathrm dx.$$ If, as is the case here, when $x\to+\infty$, $$f(x)\sim\frac{c}{x^a},$$ for some positive $c$, with $a\leqslant3$, then the integral defining $E(X^2)$ diverges hence the variance of $X$ does not exist.

Answer 2

Since the integral defining $\text{E}\left[X^2\right]$ diverges, one sometimes expresses that by saying the second moment does not exist. Consequently, its variance also doesn't exist.

Answer 3

Note $x^2 = (x+1)^2 - 2 x - 1$. Writing it like that should make the integral much easier. Though I'm sure you'll find what Did said.