# Conditional expectation of iid nonnegative random variables

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,

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

• Hint: $E[Y_i|\sum Y_i=y]=\frac yn$, then add. – lulu Sep 4 '15 at 22:07
• That might have been too terse. To expand: $$\sum E[Y_i|\sum Y_i=y]=E[\sum Y_i|\sum Y_i=y]=y$$ – lulu Sep 4 '15 at 22:10

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 $$E\left[Y_1+...+Y_k\;|\;Y_1+Y_2+...+Y_n=y\right]=kE=\frac {ky}{n}$$
• 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. – hjung Sep 4 '15 at 22:42
• 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. – lulu Sep 5 '15 at 2:05