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Given that $X_1,...Xn$ are all identical independent random variables.

$\mathbb{E}(X_1|\sum_{k=1}^{n}X_k)$ = ?

I am unsure how to proceed on this one. I know the default relation: $\mathbb{E}(X|Y)$ = $\mathbb{E}(X*I_{[Y=y]})\over\mathbb{P}(Y=y))$, where I is an indicator function.

Intuitively, I believe the answer should be the sum of the random variables divided by how many random variables or the average of the sum.

Mid
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2 Answers2

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For the late comers for this question. I think clear solution of the problem might be as follows:

Let $Y = \sum_{i=1}^n X_i$, by using the property of conditional expectation $\mathbb{E}[g(Y)|Y] = g(Y)$ (i.e.$\mathbb{E}[Y|Y] = Y$), we can write

$$ Y = \mathbb{E}[X_1 + X_2 + \cdots + X_n |Y] = \mathbb{E}[X_1|Y]+ \mathbb{E}[X_2|Y]+ \cdots + \mathbb{E}[X_n|Y] = n\mathbb{E}[X_1|Y],$$ since $X_i$'s are i.i.d., then $$\mathbb{E}[X_1|Y] = \frac{Y}{n}.$$

S.Meral
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Hint: Since the random variables are exchangeable, all $E[X_i | \sum_{k=1}^n X_k]$ are equal. What is their sum?

Robert Israel
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  • Would it just be the values of each random variable $x_1 + x_2 + .... + x_n$? I assume you were talking about the sum of all expectations. – Mid Mar 26 '15 at 00:50
  • Let the random variable $S_n := \sum_{k=1}^n X_k$, then what is $\sum_{k=1}^n \mathbb E(X_k\mid S_n)$ ? @Mid – Graham Kemp Mar 26 '15 at 00:55
  • I assume $x_1+....+x_n$, because for n = 2, we have (x_1 + x_2)/2 + (x_1 + x_2)/2 = $x_1 + x_2$ – Mid Mar 26 '15 at 01:00
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    @Mid, $\sum_{k=1}^n \mathsf E(X_k\mid S_n) = S_n$. – Graham Kemp Mar 26 '15 at 01:17
  • Okay, so they are still random variables and not actual values in the sum of the expectations? I am not seeing the connection between this and the original problem. – Mid Mar 26 '15 at 01:36
  • If $A$ and $B$ are random variables, $\mathbb E[A|B]$ (assuming it exists) is again a random variable, which is a function of $B$. The original problem asks for $\mathbb E[X_1 | S_n]$. You now know $\sum_{k=1}^n \mathbb E[X_k | S_n] = \mathbb E [ S_n | S_n ] = S_n$. Since the $\mathbb E[X_k | S_n]$ are all equal, each is ... – Robert Israel Mar 26 '15 at 01:57
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    Thanks, I see it now. Each has to be $S_n$/n. – Mid Mar 29 '15 at 04:07
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    Could you explain why $E[S_n|S_n]=S_n$? Also, is each $X_i$ has to be $S_n/n$ if the problem is $E(X_1|\bar{X_n})$, where $\bar{X_n}$ is the average? I think so because the $X_i$ are independent. – Blain Waan Oct 01 '18 at 22:19
  • Definition of conditional expectation. No, the $X_i$ don't have to be equal to $S_n/n$, that's their conditional expectation given $S_n$. – Robert Israel Oct 03 '18 at 05:20