1
$\begingroup$

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?

  • 0
    If you rule out the estimate $\sqrt{x}\geq \frac{x}{\sqrt{2}}$, then please define when an approximation is not too strong.2011-10-07
  • 0
    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 $02011-10-07
  • 0
    I seems that, in order to get a helpful answer, you should describe what you actually want to achieve.2011-10-07
  • 0
    I want an inequality of the form $\mathbb{E}[x^\frac{1}{r}] < f(\mathbb{E}[x])$ for x arbitrarily distributed between 0 and 2.2011-10-10
  • 0
    I assume you mean the converse inequality? Since you don't make further requirements, Didier Piau's answer contains a solution.2011-10-10

1 Answers 1

2

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

  • 1
    But could there be a different lower bound in terms of $E(X)$?2011-10-07
  • 0
    @Rasmus, as your comment shows, there is. Thanks.2011-10-07
  • 0
    @DidierPiau: Thanks for your answer! But as Rasmus pointed out, and as I tried to say in my question, there are other possible approximations.2011-10-07
  • 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.2011-10-07
  • 0
    You showed that there is no linear function of \mathbb{E}(X)^\frac{1}{n} that lower bounds \mathbb{E}(X^\frac{1}{n}. In my question I am not asking for a linear function of \mathbb{E}(X)^\frac{1}{n} but just any function of \mathbb{E}(X). Maybe I shouldn't have related my question to Jensen's inequality.2011-10-10
  • 1
    This is precisely why (prompted by @Rasmus's comment) I added a lower bound by a nonlinear function of $\mathrm E(X)^{1/n}$.2011-10-10
  • 0
    Andreas, any comment about your unacceptation of this answer?2011-10-13