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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 $$E(X_1+X_2+ \cdots +X_k \mid X_1+X_2+ \cdots +X_n=b)=b \cdot \frac{k}{n} .$$

Srivatsan
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Qiang Li
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1 Answers1

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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,

Patrick Da Silva
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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... – Patrick Da Silva Dec 17 '11 at 07:18
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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 07:20
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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 09:47
  • So is the condition non-negativity not necessary for the condition? – Qiang Li Dec 17 '11 at 16:50
  • 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)? – Did Dec 17 '11 at 18:26
  • @DidierPiau: that is because my brain was not working well last night when it was late. :) – Qiang Li Dec 17 '11 at 21:16
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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 00:16
  • @Didier : I didn't write what I wanted to say, but it was definitively what I wanted to say. Thanks for the correction. – Patrick Da Silva Dec 18 '11 at 03:07
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    Exchangeable instead of iid would also work. – Robert Israel Dec 18 '11 at 03:21
  • @Robert : I've never heard the term exchangeable, what does it stand for? – Patrick Da Silva Dec 18 '11 at 05:58
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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 07:26