Conditional expectation of first roll conditionally on sum of two rolls

Question

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. 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. Can someone please answer this question for me.

Thanks!

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

I get E(X∣Y=y)=y/2 for y≥2. Is that correct?

Answer 1

See this. Additionally/alternatively, Bayes' theorem would let you write $P(X \mid Y)$ in terms of $P(Y \mid X)$, which may make more sense to you.

Answer 2

Eric makes a good suggestion. If you want to proceed using your current method, note that, for instance

$P(X=3,Y=y) = 0$ if $y \leq 3$ and

$P(X=3, Y=y) = P(X=3,\textrm{second die displays a y-3}) = 1/36 \quad$ if $y>3$.

Answer 3

Use symmetry:

• $(X,Y)$ and $(Y-X,Y)$ have the same distribution hence $$E(X\mid Y)=E(Y-X\mid Y).$$
• $X+(Y-X)=Y$ hence $$E(X\mid Y)+E(Y-X\mid Y)=E(Y\mid Y)=Y.$$

Thus, $$E(X\mid Y)=\frac12Y.$$