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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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    Huh? What did your comment mean? lol2011-12-18

1 Answers 1

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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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    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.2011-12-18