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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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    19 minutes. $ $2011-12-17
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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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    @Srivatsan : Speaking of your edit, its funny that you noticed this because I actually hesitated to put it there. XD It won't change the quality of the answer so I don't wanna argue about that...2011-12-17
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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. :-)2011-12-17
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    *Since $X_i$'s are independent* is (not necessary and) not enough. *Independent and identically distributed* is enough.2011-12-17
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    So is the condition non-negativity not necessary for the condition?2011-12-17
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    Of course it is not. But something escapes me: how come you accepted this answer if (1) you might find the crucial step difficult to grasp because its justification (that I mentioned in my comment) is lacking (but maybe you do not?) and (2) you still have basic questions about the answer (as witnessed by your last comment)?2011-12-17
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    @DidierPiau: that is because my brain was not working well last night when it was late. :)2011-12-17
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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...2011-12-18
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    @Didier : I didn't write what I wanted to say, but it was definitively what I wanted to say. Thanks for the correction.2011-12-18
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    *Exchangeable* instead of iid would also work.2011-12-18
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    @Robert : I've never heard the term *exchangeable*, what does it stand for?2011-12-18
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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