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Let $X_1,...X_n$ be independent, identically distributed and nonnegative random variables, and let $k\le n$. Compute: $$E\left[{\sum_{i=1}^k X_i\over \sum_{i=1}^n X_i}\right].$$ This question has already been asked: Expectation of random variables ratio.

The thing is that my teacher told me that the solution in the link wasn't really a "solution" the correct thing to do is to compute the conditional expectation of $${\sum_{i=1}^k X_i\over \sum_{i=1}^n X_i}$$ given that $\sum_{i=1}^n X_i=m$ where $m$ is a positive integer, but I have no idea how to do this, I would really appreciate if you can help me with this problem.

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You can prove as follows:

$$\begin{align} \mathsf E \Big[ \frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i} \Big] & = \sum _{m=0}^{\infty} \mathsf E \Big[ \frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i} \mid {\sum _{i=1}^n X_i}=m\Big]\mathsf P({\sum _{i=1}^n X_i}=m) \\[1ex] & = \sum _{m=0}^{\infty} \mathsf E \Big[ \frac{\sum _{i=1}^k X_i}{m} \mid {\sum _{i=1}^n X_i}=m\Big]\mathsf P({\sum _{i=1}^n X_i}=m) \\[1ex] & = \sum _{m=0}^{\infty} \frac{1}{m} \sum _{i=1}^{k}\mathsf E \Big[ X_i \mid {\sum _{i=1}^n X_i}=m\Big]\mathsf P({\sum _{i=1}^n X_i}=m) \\[1ex] & = \sum _{m=0}^{\infty} \frac{1}{m} \frac{km}{n} \mathsf P({\sum _{i=1}^n X_i}=m) \\[1ex] & = \frac{k}{n} \sum _{m=0}^{\infty}\mathsf P({\sum _{i=1}^n X_i}=m) \\[1ex] & = \frac{k}{n} \end{align}$$ Note: I have used identical distribution of $X_i$s and this fact that $E \Big[ {\sum _{i=1}^n X_i} \mid {\sum _{i=1}^n X_i}=m\Big] = m$ to conclude that $E \Big[ X_i \mid {\sum _{i=1}^n X_i}=m\Big]= \frac{m}{n}$.

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    $\begingroup$ Note: In fact we do not need $X_i$s to be independent. Identical distribution is enough. $\endgroup$ – Mehdi Jafarnia Jahromi Apr 30 '15 at 16:18
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    $\begingroup$ Actually, $X_i$'s independence is necessary. The $X_i$'s being i.i.d. allows you to prove the $X_i/\Sigma$ being i.d. (but no longer independent). Basically I think what you said in the Note is wrong. See my counter-example at the 2nd half of my answer here $\endgroup$ – antkam Sep 28 at 14:35
  • $\begingroup$ @antkam I agree. Thanks for your counter example. $\endgroup$ – Mehdi Jafarnia Jahromi Sep 29 at 9:30
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$\begin{align} \text{Let }\qquad M &=\sum_{i=1}^n X_i & \text{Define a new random variable} \\[2ex] \mathsf E(\sum_{i=1}^k X_i / \sum_{i=1}^n X_i) & = \mathsf E(\mathsf E(\sum_{i=1}^k X_i / M\mid M)) & \text{Law of Iterated Expectation} \\[1ex] & = \mathsf E(\sum_{i=1}^k \mathsf E(X_i/M\mid M)) & \text{Linearity of Expectation} \\[1ex] & = \mathsf E(\sum_{i=1}^k 1/n) & \{X_i\}\sim\text{ i.i.d.} \\[1ex] & = k/n \end{align}$

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  • $\begingroup$ How exactly does the Xi's being iid justify the penultimate step? $\endgroup$ – BCLC Aug 7 '15 at 23:03
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    $\begingroup$ @BCLC identical means that $ \mathsf E(X_i/M \mid M)=\mathsf E(X_1/M \mid M)$ for all $i\in\{1..n\}$ and so $n \mathsf E(X_1/M\mid M) = \mathsf E(\sum_{i=1}^n X_i/\sum_{i=1}^n X_i \mid \sum_{i=1}^n X_i) = 1$, thus... $\endgroup$ – Graham Kemp Aug 8 '15 at 10:41
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    $\begingroup$ Elegant! But should there be an almost surely there? $E(M/M | M) = M/M E( 1 | M) = M/M = 1$ a.s.? Like if X is $\mathscr{G}$-measurable, $E[X | \mathscr{G}] = X$ a.s.? $\endgroup$ – BCLC Aug 8 '15 at 15:59

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