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Ok, suppose we have a random variable, X, on $(\Omega,\mathcal{B},P)$ and $r>0$.

I am trying to prove the following 4 things:

1- If $E(|X|^r)<\infty$ then $E(|X|^s)<\infty \;\;\;\forall s\in(0,r]$

2- If $E(e^{t|X|})<\infty$ for some $t>0$, then $E(|X|^s)<\infty \;\;\;\forall s\in(0,\infty)$

side note for #2...is this just saying that if the moment-generating function is finite somewhere, then all of the moments are finite?

3- If $\sup_{x\geq 0}(x^{r+\delta}P(|X|>x))<\infty$ for some $\delta>0$ then $E(|X|^r)<\infty$

4- $E(|X|^r)<\infty\Rightarrow \lim_{x\to\infty}x^rP(|X|>x)=0$


Here is what I have for the first 2:

1: For $0<s<r, |x|^s\leq\max(|x|^r,1)<|x|^r+1$

$E(|X|^s)\leq E(|X|^r)+1<\infty$

2: This follows from the above and the relation:

$|x|^n\leq n!e^{t|x|}t^{-n}$ for $n\geq 1$


Could anyone please offer some detail into how to complete 3 and 4? For some reason limits and sup/inf often elude me. Also, if I have made any mistakes in the first 2, please note those as well. This is all being learned on my own, so the more detail and critique, the better for me!

Thanks.

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1 Answer 1

up vote 2 down vote accepted

For 3., a consequence of Fubini's theorem is that $$E|X|^r=\int_0^{+\infty}rt^{r-1}P(|X|\geq t)dt.$$ (the equality has to be understood as follows: the LHS is finite if and only if the RHS is finite, and if this is so, they are equal).

Call $M$ the supremum; then $$E|X|^r\leq M\int_1^{+\infty}rt^{-\delta-1}dt+\int_0^1rt^{r-1}P(|X|\geq t)dt\leq M\int_1^{+\infty}rt^{-\delta-1}dt+r\int_0^1\frac 1{t^{1-r}}dt,$$ and the last integrals are convergent.

For 4., we have $$x^rP(|X|\geq x)=x^rP(|X|^r\geq x^r)\leq \int_{\Omega}|X|^r\chi_{\{|X|\geq x\}}dP,$$ and use monotone convergence theorem.

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"the last integral is convergent"? Are you sure about that? –  Byron Schmuland Aug 3 '12 at 16:59
    
@ByronSchmuland You are right, I missed a step. I hope I have corrected it now. Thanks! –  Davide Giraudo Aug 3 '12 at 17:09
    
Not to be picky, but I don't like this equation either: $x^rP(|X|\geq x)=x^rP(|X|^r\geq x)$ –  Byron Schmuland Aug 3 '12 at 17:14
    
I missed the power $r$, thanks again. –  Davide Giraudo Aug 3 '12 at 18:00

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