# 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

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.
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
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