Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm faced with the following problem: I have to lower bound the expected value of the n-th root of an arbitrary distributed real random variable using its expected value. So I'm looking for something that has a similar form as the Jensen inequalty but goes the other way around.

I can assume the variable satisfies 0< X< 2 so I thought I could lower bound the root by a line but that approximation is to strong.

Does any one know a way of lower bounding the expected value of a root?

share|improve this question
    
If you rule out the estimate $\sqrt{x}\geq \frac{x}{\sqrt{2}}$, then please define when an approximation is not too strong. –  Rasmus Oct 7 '11 at 7:59
    
To strong was in referring to my application. I tried to bound an error probability and I ended up with a value >1. So I was hoping for any other approximation that is not strictly worse. btw something for $0<X<1$ would also help me, but I guess that does not really make a difference. –  Andreas Mueller Oct 7 '11 at 10:23
    
I seems that, in order to get a helpful answer, you should describe what you actually want to achieve. –  Rasmus Oct 7 '11 at 11:59
    
I want an inequality of the form $\mathbb{E}[x^\frac{1}{r}] < f(\mathbb{E}[x])$ for x arbitrarily distributed between 0 and 2. –  Andreas Mueller Oct 10 '11 at 12:30
    
I assume you mean the converse inequality? Since you don't make further requirements, Didier Piau's answer contains a solution. –  Rasmus Oct 10 '11 at 13:34
add comment

1 Answer 1

If $n>1$, there cannot exist a positive $c_n$ such that $\mathrm E(X^{1/n})\geqslant c_n\mathrm E(X)^{1/n}$ for every $[0,2]$ valued random variable $X$. To see this assume that $X=2$ with probability $p$ and $X=0$ with probability $1-p$. Then one asks that $p2^{1/n}\geqslant c_n(2p)^{1/n}$, hence $c_n\leqslant p^{1-1/n}$. When $p\to0^+$, one gets $c_n\leqslant0$ as soon as $n>1$.

On the other hand, since $X\leqslant2$ almost surely, $X^{1/n}\geqslant2^{-1+1/n}X$ almost surely, hence $\mathrm E(X^{1/n})\geqslant 2^{-1+1/n}\mathrm E(X)$. Likewise, for every positive $k\leqslant n$, $X^{1/n}\geqslant2^{1/n-1/k}\,X^{1/k}$ almost surely, hence $\mathrm E(X^{1/n})\geqslant 2^{1/n-1/k}\,\mathrm E(X^{1/k})$.

share|improve this answer
1  
But could there be a different lower bound in terms of $E(X)$? –  Rasmus Oct 7 '11 at 7:56
    
@Rasmus, as your comment shows, there is. Thanks. –  Did Oct 7 '11 at 8:08
    
@DidierPiau: Thanks for your answer! But as Rasmus pointed out, and as I tried to say in my question, there are other possible approximations. –  Andreas Mueller Oct 7 '11 at 10:28
1  
@Andreas, Surely I am too weak at mind-reading but I have access to what you wrote in your question, and nothing else! I believe my answer answers THAT. If you want an answer to a different problem, the proper way to proceed is to post another question on MSE. –  Did Oct 7 '11 at 12:27
1  
This is precisely why (prompted by @Rasmus's comment) I added a lower bound by a nonlinear function of $\mathrm E(X)^{1/n}$. –  Did Oct 10 '11 at 14:02
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.