Let $X$ be the value of the first die and $Y$ the sum of the two dice. Find $E(X / Y)$

Question

Consider an experiment of rolling two dice. Let $X$ be the value of the first die and $Y$ the sum of the two dice. Find $E(X / Y)$, ie, obtain the value of $E(x/y) (y)$ for all $y$

Good evening, I could solve the problem. I have no idea how to start. You could give me some tips to solve it, please. Or some bibliographic references with similar exercises?

Answer 1

Assuming you mean $\mathrm{E}(X|Y)$, then since the distribution of each die is identical and independent, if we are given that their sum is $y$, the expected value of each die would be the same: $y/2$.

If you did mean $\mathrm{E}(X/Y)$, then in light of the preceding, this would be $\frac12$.

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Explicit Calculation of $\boldsymbol{\mathrm{E}(X|Y)}$

Given that the sum of the dice is $y$, the possibilities for $x$ are from $y-6$ to $6$ with equal probability of each. That is, the probability that the first die is $x\in\mathbb{Z}$ is $$ \frac1{13-y}[y-6\le x\le6] $$ where the brackets are Iverson brackets.

Thus, the expected value would be $$ \begin{align} \sum_{x=y-6}^6\frac{x}{13-y} &=\frac{\frac126(6+1)-\frac12(y-7)(y-6)}{13-y}\\ &=\frac{21-21+\frac{13}2y-\frac12y^2}{13-y}\\[9pt] &=\frac y2 \end{align} $$ This nicely agrees with the symmetry argument given above.

Answer 2

We show how to handle the problem for one value of $y$, say $y=9$.

Given that $Y=9$, $X$ takes on values $3$ to $6$ with equal probabilities. Thus $$E(X|Y=9)=\frac{3+4+5+6}{4}.$$

One value of $y$ done, $10$ more to do.

Remark: The symmetry argument of robjohn is much better.