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consider the sum of random variables $Y_k = \sum\limits_{j=1}^k X_j $, $X_k$ are i.i.d. Now I want to calculate:

$$E(Y_m| Y_{m+n}=n) \overset{!}{=} m \frac{n}{n+m}$$

How can I get to that?

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  • $\begingroup$ Typo. Sorry:( I edit it. $\endgroup$ – Sarah Mar 30 at 8:13
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    $\begingroup$ Is this just a special case of $E(Y_m \mid Y_{m+n}=c) = c \frac{m}{n+m}$ ? $\endgroup$ – Henry Mar 30 at 8:17
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Since $X_i$'s are i.i.d. it follows that $c=E(X_i|Y_{n+m}=n)$ is independent of $i$. Summing over $i \leq n+m$ we get $(m+n) c=n$ so $c =\frac n {n+m}$. Finally the given expectation is $mc=\frac {mn} {m+n}$.

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  • $\begingroup$ c=E(....) is because of the fact, that E(., X=X(w)) is constant, isn't it? $\endgroup$ – Sarah Mar 30 at 8:28
  • $\begingroup$ If you permute $X_i, 1 \leq i \leq n+m$ then $Y_{n+m}$ does not change. This is the reason the number $EX_i|Y_{n+m}=n)$ does not depend on $i$. @Sarah $\endgroup$ – Kavi Rama Murthy Mar 30 at 8:32
  • $\begingroup$ Why do you sum to n+m? $\endgroup$ – Sarah Mar 30 at 8:33
  • $\begingroup$ @Sarah When you sum to $n+m$ you get something whose value is obviously $n$: $E(\sum\limits_{i=1}^{n}X_i|Y{n+m}=n)=E(Y_{n+m}|Y_{n+m}=n)=n$. $\endgroup$ – Kavi Rama Murthy Mar 30 at 8:35
  • $\begingroup$ Ok then I get the c. Then: $E(Y_m| Y_{n+m}=n) = m E( X_k=X |Y_{n+m}=n)$ $\endgroup$ – Sarah Mar 30 at 8:40
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Note that all permutations of the $X_j$ are equally likely. Thus the desired expectation is invariant if you average over them. If the sum of the first $m+n$ of the $X_j$ is $n$ (or in fact any other value, there’s no reason why it should be linked to the indices), then since each of these $m+n$ occurs in the first $m$ places in a fraction $\frac m{m+n}$ of the permutations, the expectation of the first $m$ is $\frac m{m+n}$ times that value.

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  • $\begingroup$ Thank you for your answer: "Then since each of these m+n occurs in the first m places" $\endgroup$ – Sarah Mar 30 at 8:26
  • $\begingroup$ Can you explain why this holds? So there is also no analytic way to calculate this conditional expected value? $\endgroup$ – Sarah Mar 30 at 8:26
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    $\begingroup$ @Sarah: I'm not sure what you mean by "analytic". This seems rather analytic to me. If you want a more rigorous, formal approach, the one in Kavi Rama Murthy's answer seems to fit the bill. $\endgroup$ – joriki Mar 30 at 8:31
  • $\begingroup$ I do not understand why the m+n occurs in the first m places. $\endgroup$ – Sarah Mar 30 at 8:53
  • $\begingroup$ @Sarah: A random permutation is equally likely to move an $X_j$ to any of the $m+n$ positions. Thus the proportion of permutations that move it to a certain position must be $\frac1{m+n}$, and then the proportion that move it to one of $m$ particular positions must be $\frac m{m+n}$. $\endgroup$ – joriki Mar 30 at 10:00
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Just another solution:

$$E(Y_m| Y_{m+n}=n)=E(\sum_{k=1}^{m} X_k| \sum_{i=1}^{m+n} X_i=n)$$ $$=\sum_{k=1}^{m}E( X_k| \sum_{i=1}^{m+n} X_i=n)=\sum_{i=1}^{m} \frac{n}{n+m}= m \frac{n}{n+m} $$

It is enough to show $E( X_k| \sum_{i=1}^{m+n} X_i=n)=\frac{n}{n+m}$

$$E( \sum_{k=1}^{m+n} X_k| \sum_{i=1}^{m+n} X_i=n)=n$$ $$\Leftrightarrow$$ $$\sum_{k=1}^{m+n}E( X_k| \sum_{i=1}^{m+n} X_i=n)=n$$

$$\Leftrightarrow$$ $$(m+n)E( X_k| \sum_{i=1}^{m+n} X_i=n)=n$$

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    $\begingroup$ Thank you very much. That is the solution I was looking for:) $\endgroup$ – Sarah Mar 30 at 9:13
  • $\begingroup$ You are welcome $\endgroup$ – Masoud Mar 30 at 9:14
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Just another solution

lets $E(X_i)=\mu$. In non-parametric family, $\bar{X}$ is sufficient and complete estimator for $\mu$(that coincide with this example).

$$E(X_k|\sum_{i=1}^{m+n} X_i)=E(X_k|\frac{\sum_{i=1}^{m+n} X_i}{m+n})=E(X_k|\bar{X}_{(m+n)})=g(\bar{X}_{(m+n)})$$

I want to show $g(\bar{X}_{(m+n)})=\bar{X}_{(m+n)}$ almost surely.

$$E(g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)})=E(E(X_k|\bar{X}_{(m+n)}))-\mu=\mu-\mu=0$$ since $\bar{X}_{(m+n)}$ is complete and sufficient and
$g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)}$ is a function of $\bar{X}_{(m+n)}$ so $$P(g(\bar{X}_{(m+n)})-\bar{X}_{(m+n)}=0)=1$$

so $g(\bar{X}_{(m+n)})=\bar{X}_{(m+n)}$ almost surely. so

$$E(X_k|\sum_{i=1}^{m+n} X_i)=\bar{X}_{(m+n)}$$ and

$$E(X_k|\sum_{i=1}^{m+n} X_i=n)=E(X_k|\frac{\sum_{i=1}^{m+n} X_i}{m+n}=\frac{n}{m+n})=\frac{n}{m+n}$$. Finally

$$E(\sum_{k=1}^{m} X_k| \sum_{i=1}^{m+n} X_i=n)=m\frac{n}{m+n}$$

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