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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 $y$?


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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Could you say a little more clearly what you are trying to find? You say "the first roll" and "the two dice" but you have not said anything previously about dice or rolls. Perhaps just starting with "Two dice are rolled." would be sufficient. – Matthew Conroy Nov 4 '12 at 23:36
I'd write $\mathbb E(X\mid Y=y)$ for what you appear to mean. – Michael Hardy Nov 4 '12 at 23:36
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. – jrad Nov 4 '12 at 23:38
up vote 2 down vote accepted

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! – jrad Nov 5 '12 at 1:08

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? – jrad Nov 5 '12 at 1:11

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.

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