Thursday, 8 May 2014

probability theory - Finite expectation of a random variable

$X \geq 0$ be a random variable defined on $(\Omega,\mathcal{F},P)$. Show that $\mathbb{E}[X]<\infty \iff \Sigma_{n=1}^\infty P(X>n) < \infty $.



I got the reverse direction but I am struggling with the $"\implies"$ direction. So far, I have the following worked out:




$\mathbb{E}[X]<\infty$



$\implies \int_0^\infty (1-F(x)) dx < \infty$ (where $F$ is the distribution function of the random variable X)



$\implies \int_0^\infty (1-P(X\leq x)) dx < \infty$



$\implies \int_0^\infty P(X>x) dx < \infty$



Consider $\int_0^\infty P(X>x) dx$




$= \Sigma_{n=1}^\infty \int_{n-1}^n P(X>x) dx$



This is the point I am stuck at. Any help will be deeply appreciated!

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...