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Simple question here (I think). I have to find $E(X\mid Y)(y)$ where $X$ is the value of the first roll and $Y$ is the sum of the two dice. Normally, I wouldn't have any trouble with this sort of problem, but the definitions of the two variables are "flipped," so to speak, making it a little confusing for me. I know that

$E(X|Y)(y) = \sum_x{xP(X\mid Y)}=\frac{\sum_xxP(X=x, Y=y)}{P(Y=y)},$

but this doesn't really get me anywhere. From that summation, I get

$E(X|Y)(y) = \frac{1\cdot P(1, Y=y)}{P(Y=y)} + \frac{2\cdot P(2, Y=y)}{P(Y=y)} + \dots + \frac{6\cdot P(6, Y=y)}{P(Y=y)}.$

Here's where I'm stumped. How can I put $E(X|Y)(y)$ in terms of just $y$?

Thanks!

EDIT: To be clear, I need to find the expected value of the first roll given the sum of the two dice.

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    Sorry for not being clear enough. What @MichaelHardy said is correct. I need to find the expected value of the first roll given the sum of the two dice.2012-11-04

3 Answers 3

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If two rolls are independent, then by symmetry (there is no difference between which one is the first roll and which one is the second), $E(X\mid Y=y)=\frac{y}{2}$ for $y \geq 2$.

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    Doh, why didn't I think of that? Thanks!2012-11-05
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The values of the two dice are identically distributed and (unconditionally) independent so if the sum of them is $Y=y$ then each of their conditional expectations is $\dfrac{y}{2}$, by symmetry.

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    Thanks. Is there any way to arrive at $y/2$ given the work I have in my original post?2012-11-05
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Let us try to work without much in the way of formulas. Find, for example, $E(X\mid Y=6)$.

If $Y=6$, then $X$ can take on the values $1$, $2$, $3$, $4$, $5$. These are equally likely. If that is not clear, one can do a formal computation, for example, of $\Pr(X=2\mid Y=6)$. Now $E(X\mid Y=6)$ is straightforward to calculate. The others are very similar.