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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?

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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

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 $\mathrm E(X^{1/n})\geqslant 2^{1/n-1/k}\,\mathrm E(X^{1/k})$.

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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
@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
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

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