# X integrable

• Jun 6th 2010, 06:24 AM
Veve
X integrable
I have a problem and I can not figure it out,

Let c be a fixed constant, $\displaystyle c>0$. Then $\displaystyle \mathbb{E}(|X|)<\infty$ if and only if $\displaystyle \sum_{n=1}^\infty\mathbb{P}(|X|\ge cn)<\infty$. In particular, if the last series converges for one value of c, it converges for all values of c.(Thinking)

I appreciate any help.
• Jun 6th 2010, 10:52 AM
Moo
Hello,

$\displaystyle \sum_{n=1}^\infty \mathbb{P}(|X|\geq cn)=\sum_{n=1}^\infty \mathbb{E}\left[\bold{1}_{\{|X|\geq cn\}}\right]=\mathbb{E}\left[\sum_{n=1}^\infty \bold{1}_{\{|X|/c\geq n\}}\right]$

But $\displaystyle \sum_{n=1}^\infty \bold{1}_{\{|X|/c\geq n\}}=\left\lfloor \frac{|X|}{c}\right\rfloor$ (counting the number of integers that are less or equal to |X|/c)

and since $\displaystyle \mathbb{E}\left[\left\lfloor\frac{|X|}{c}\right\rfloor\right]\leq \mathbb{E}\left[\frac{|X|}{c}\right]<\mathbb{E}\left[\left\lfloor\frac{|X|}{c}\right\rfloor\right]+1$, you can conclude.

If you think it's not correct, just tell me, because I just wrote this, I don't know if it's a correct proof...
• Jun 6th 2010, 02:42 PM
matheagle