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I am studying Ross's book, stochastic processes. There is the following lemma:


Let $Y_1, Y_2, ... , Y_n$ be iid nonnegative random variable. Then,

$E[Y_1+ \cdots +Y_k | Y_1+\cdots+Y_n=y] = \frac{k}{n} \cdot y, \quad k=1,\cdots,n$


But, I really can't understand why this lemma can be established. Could you please help me?

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    $\begingroup$ Hint: $E[Y_i|\sum Y_i=y]=\frac yn$, then add. $\endgroup$ – lulu Sep 4 '15 at 22:07
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    $\begingroup$ That might have been too terse. To expand: $$\sum E[Y_i|\sum Y_i=y]=E[\sum Y_i|\sum Y_i=y]=y$$ $\endgroup$ – lulu Sep 4 '15 at 22:10
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Informally, as all the $Y_i$ are identically distributed, the fact that they sum to $y$ implies that their (conditional) expectation must be $\frac yn$. To see this formally, note that we certainly have equality of all the $E[Y_i\;|\;Y_1+...+Y_n=y]$, call the common value $E$. Then

$$nE=E\left[Y_1\;|\;Y_1+Y_2+...+Y_n=y\right]+...+E\left[Y_n\;|\;Y_1+Y_2+...+Y_n=y\right]=E\left[Y_1+...+Y_n\;|\;Y_1+...+Y_n=y\right]=y$$

Thus $E=\frac yn$.

But then $$E\left[Y_1+...+Y_k\;|\;Y_1+Y_2+...+Y_n=y\right]=kE=\frac {ky}{n}$$

As desired.

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  • $\begingroup$ Thanks for your answer! From your hints, I have some questions. 1) By $E[ \sum Y_i | \sum Y_i = y] = y$, can I argue $E[ x | x = y] = x \cdot p(x|x=y) |_{x=y} = y \cdot 1 = y$ ? 2) I think the nonnegative part in the lemma can be ignored, because it doesn't have an impact on$E[ \sum Y_i | \sum Y_i = y] = y$ and $E[Y_i|\sum Y_i=y]=\frac{y}{n}$. Am I right? Thanks again for your help. $\endgroup$ – hjung Sep 4 '15 at 22:42
  • $\begingroup$ Well, I;m not sure what $x$ is, but yes. $E[x|x=y]=y$. Indeed, if $x=y$ then we are absolutely sure that $x$ is $y$, so of course it's conditional expectation is $y$. That's exactly what your argument says. $\endgroup$ – lulu Sep 5 '15 at 2:05

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