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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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    I assume you mean the converse inequality? Since you don't make further requirements, Didier Piau's answer contains a solution.2011-10-10

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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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    Andreas, any comment about your unacceptation of this answer?2011-10-13