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I'm quite lost with regards to solving this problem. Any help is appreciated. Thanks.

Suppose $X_1, X_2, . . .$ are independent discrete random variables, having the same distribution, and $E[X_i]>0$, for each $i$. Is it true that for any two positive integers $m$ and $n$

$E[\frac{X_1+...+X_m}{X_1+...+X_n}]=\frac{m}{n}$ ?

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marked as duplicate by NCh, GNUSupporter 8964民主女神 地下教會, Community Feb 15 '18 at 2:19

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  • $\begingroup$ Hint. Assuming that $Y_i = \frac{X_1}{X_1+\cdots+X_n}$ is integrable, you can check that $Y_i$’s have all the same distribution and sums up to $1$. $\endgroup$ – Sangchul Lee Feb 15 '18 at 1:05
  • $\begingroup$ The condition $E[X_i]>0$ does not guarantee that the sum in the denominator cannot be zero. And in this case the fraction is undefined. $\endgroup$ – NCh Feb 15 '18 at 1:19
  • $\begingroup$ @NCh Why it doesn't guarantee?! $\endgroup$ – YOUSEFY Feb 15 '18 at 12:47
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    $\begingroup$ @YOUSEFY Please find an expectation of r.v. $X$ with the following pmf: $P(X=0)=1/2=P(X=1)$. $\endgroup$ – NCh Feb 15 '18 at 14:17
  • $\begingroup$ @NCh Professor Chernova, I did not know that you are the one in this account (what a surprise)! I am happy to see you here (I wish I will meet you the institute one day to say Hi). All respect to you! About the problem, I still trying to know how to find E[1/X], it seems it has different properties than E[X] to deal with it (I'm trying)! Best! $\endgroup$ – YOUSEFY Feb 16 '18 at 11:12