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Let $X_1\cdots X_n$ be independent identically distributed random variables and $E(X_1)=0$ and $Var(X_1)<\infty$. Do we have that $E(\frac{X_1}{\sqrt{X_1^2+\cdots+X_n^2}})=0$?

My attempt: $E(\frac{X_1}{\sqrt{X_1^2+\cdots+X_n^2}})=\sum_m E(\frac{X_1}{m}|\sqrt{X_1^2+\cdots+X_n^2}=m)P(\sqrt{X_1^2+\cdots+X_n^2}=m)=\sum_m\frac{1}{m} E(X_1|\sqrt{X_1^2+\cdots+X_n^2}=m)P(\sqrt{X_1^2+\cdots+X_n^2}=m)$

So if $E(X_1|\sqrt{X_1^2+\cdots+X_n^2}=m)=0$ then we're done. But I am not sure if this is correct or not.

Thanks for any hint.

  • 2
    Naturally, $E\left(X_1\mid\sqrt{X_1^2+\cdots+X_n^2}=m\right)$ has no reason to be $0$. Actually the claim does not hold, for a counterexample, try $n=2$, $P(X_1=2)=\frac13$, $P(X_1=-1)=\frac23$. – 2017-02-27
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    @Did Yeah, you are right. Thanks for your counterexample. – 2017-02-27

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