Conditional expectation of maximum of dice roll

Question

I am trying to understand better the concept of conditional expectation, and I am facing the following exercise:

I have two independent dice $X,Y$ with four faces s.t. $$P(X=i)=1/4=P(Y=i) , i\in\{1,2,3,4\}$$ What is the conditional expectation of the maximum of the two, given the value of one of them?

First of all, what is it asking me to compute? $\mathbb{E}(max(X,Y)|Y)$ or $\mathbb{E}(max(X,Y)|Y=y)$

I am kind of lost here: $$\mathbb{E}(max(X,Y)|Y=y)= \sum_{i=1}^4 i\cdot P(\{max = i\}|\{Y=y\})$$

because the value of $P(\{max = i\}|\{Y=y\})$ changes according to $i$ and to $y$ so I can not find a generic form of the value. For instance, is it true that $P(\{max=1\}|\{Y=y\}) = \begin{cases} 1/4 &\text{if} &y=1 \\ 0 &else \end{cases}\\ P(\{max=2\}|\{Y=y\}) = \begin{cases} 1/4 &\text{if} &y=1 \\ 1/2 &\text{if} &y=2 \\ 0 &else \end{cases}$

or am I doing something wrong?

Answer 1

It is possible to find the answer directly with conditional expectations, avoiding conditional probabilities.

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For a fixed $j\in\{1,2,3,4\}$ we find:

$\begin{aligned}\mathsf{E}\left[\max\left(X,Y\right)\mid Y=j\right] & =\mathsf{E}\max\left(X,j\right)\\ & =\frac{1}{4}\sum_{i=1}^{4}\max\left(i,j\right)\\ & =\frac{1}{4}\left[\sum_{i=1}^{j}\max\left(i,j\right)+\sum_{i=j+1}^{4}\max\left(i,j\right)\right]\\ & =\frac{1}{4}\left[\sum_{i=1}^{j}j+\sum_{i=j+1}^{4}i\right]\\ & =\frac{1}{4}\left[j^{2}+\frac{1}{2}\left(4-j\right)\left(5+j\right)\right]\\ & =\frac{1}{8}\left[j^{2}-j+20\right] \end{aligned} $

This allows the conclusion that: $$\mathsf{E}\left[\max\left(X,Y\right)\mid Y\right]=\frac{1}{8}\left[Y^{2}-Y+20\right]$$

Answer 2

You fix $y$ first. The key is to compare $X$ with $y$ and divide it into three cases. Think of all possibilities when $\{Y = y\}$. $$\{(X, Y) = (x, y) \mid x = 1, \dots, y\} = \{\max(X,Y) = y\} \cap \{Y = y\}.$$

When $y = 4$, the answer is obviously four, so assume $y \in \{1,2,3\}$ for the rest.

\begin{align} & E[\max(X,Y) \mid Y=y] \\ &= \sum_{i=1}^4 i \, P(\max(X,Y) = i \mid Y = y) \\ &= \sum_{i=y}^4 i \, P(\max(X,Y) = i \mid Y = y) \\ &= y \, P(\max(X,Y) = y \mid Y = y) + \sum_{i=y+1}^4 i \, P(\max(X,Y) = i \mid Y = y) \\ &= y \cdot \frac{y}{4} + \sum_{i=y+1}^4 i \, P(X = i \mid Y = y) \\ &= \frac{y^2}{4} + \sum_{i=y+1}^4 i \, P(X = i) \\ &= \frac{y^2}{4} + \frac14 \sum_{i=y+1}^4 i \\ &= \frac{y^2}{4} + \frac14 \frac{((y+1)+4)(4-(y+1)+1)}{2} \\ &= \frac{2y^2 - y^2 - y + 20}{8} \\ &= \frac{y^2 - y + 20}{8} \end{align}

Observe that when $y = 4$, $\dfrac{y^2 - y + 20}{8} = 4$, so the above polynomial is the desired answer.