# If $X_i$ are iid, finding $E(X_1 + X_2 + \cdots + X_k \mid X_1 + X_2+ \cdots +X_n=b)$

I just wonder if anybody can help me to prove the following identity:

Given a series of i.i.d. non-negative random variables $X_1, X_2, ..., X_n$, then $$E(X_1+X_2+ \cdots +X_k \mid X_1+X_2+ \cdots +X_n=b)=b \cdot \frac{k}{n} .$$

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19 minutes.  – Did Dec 17 '11 at 9:49
Huh? What did your comment mean? lol – Patrick Da Silva Dec 18 '11 at 8:14

## 1 Answer

You can reduce yourself to the case where $k = 1$ because the expectation is a linear operator.

Since $X_i$'s are i.i.d., $$\mathbb E(X_i \, | \, X_1 + \dots + X_n = b )$$ does not depend on $i$ (as long as $1 \le i \le n$). Thus $$n \, \mathbb E \left( X_i \, \left| \sum_{i=1}^n X_i = b \right. \right) = \sum_{i=1}^n \, \mathbb E \left(X_i \, \left| \, \sum_{i=1}^n X_i = b \right. \right) = \mathbb E \left( \sum_{i=1}^n X_i \, \left| \, \sum_{i=1}^n X_i = b \right. \right) = b$$ so that $$\mathbb E \left( X_i \, \left| \sum_{i=1}^n X_i = b \right. \right) = \frac bn.$$ Your case can then be solved by linearity of expectation.

Hope that helps,

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Well, I was unsure myself, but don't remember seeing a "the". Irrespective of the correct usage, I like your answer, +1. :-) – Srivatsan Dec 17 '11 at 7:20

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Since $X_i$'s are independent is (not necessary and) not enough. Independent and identically distributed is enough. – Did Dec 17 '11 at 9:47

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Really? And somebody is forcing you to accept answers (1) which you do not understand, (2) 19 minutes after you submitted the question, and (3) while your brain is not working? Wow... – Did Dec 18 '11 at 0:16

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Exchangeable instead of iid would also work. – Robert Israel Dec 18 '11 at 3:21

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Patrick: A random vector $(X_k)_{1\leqslant k\leqslant n}$ is exchangeable when the distribution of $(X_{\sigma(k)})_{1\leqslant k\leqslant n}$ does not depend on the permutation $\sigma$. Every i.i.d. sequence is exchangeable. If $(Y_k)_k$ is i.i.d. and independent on $Z$ and $X_k=\Phi(Y_k,Z)$, then $(X_k)_k$ is exchangeable. If $(Y_k)_k$ is i.i.d. and $S$ is their sum, then $(Y_k)_k$ conditionally on $[S=s]$ is exchangeable. And so on. There is a LLN for exchangeable sequence where one converges to a (possibly non degenerate) tail random variable... This is a nice subject, if you ask me. – Did Dec 18 '11 at 7:26
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