# Measure Theory: How to compute the conditional expectation of max of dice tosses?

Consider a dice with $$f$$ faces and let $$(X_n)_{1 \le n \le N}$$ be the outcomes of the tosses. For $$1 \le n \le N$$ we set $$\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$$ and additionally $$\mathcal{F}_0 = \{\emptyset, \Omega \}$$. Consider the RVs $$Z_0 := 0$$ and $$Z_n := \max_{1 \le k \le n} X_k \quad (1 \le n \le N)$$.

Compute the conditional expectation $$\mathbb{E}(Z_n \mid \mathcal{F}_{n-1})$$ for $$(1 \le n \le N)$$.

I understand that intuitively $$Z_n$$ describes precisely the maximum number yielded by the $$n$$th dice toss when we know the results of the previous $$n-1$$ tosses. I furthermore suppose that we can model this setting as a measure space $$(\Omega, \mathcal{A}, \mathbb{P})$$ via $$\Omega := \{1,\ldots,f\}$$ and consider $$X_1,\ldots,X_N$$ as uniformly distributed independent canonical RVs, i.e. $$X_i(\omega) := \omega$$. For the $$\sigma$$-Algebra $$\mathcal{A}$$ I suppose it is best to set $$\mathcal{A} = \mathcal{P}(\Omega)$$. However, I do not see how to transfer my "continuous" definition (see below) of condtional expectation $$\mathbb{E}(X \mid \mathcal{F})$$ over to this discrete case.

Let $$(\Omega, \mathcal{A}, \mathbb{P})$$ be a probability space and let $$X$$ be a real RV with $$E(\lvert X \rvert) < \infty$$. $$\mathbb{E}(X \mid \mathcal{F})$$ is uniquely defined as a RV via the conditions

1. $$\mathbb{E}(X \mid \mathcal{F})$$ is $$\mathcal{F}$$ measurable,

2. $$\mathbb{E}(X \mid \mathcal{F}) \in L^1(\mathbb{P})$$

3. $$\int_A \mathbb{E}(X \mid \mathcal{F}) d \mathbb{P} = \int_A X d \mathbb{P}$$ for all $$A \in \mathcal{F}$$.

• That's a very useful and important exercise. What is $A$ in the definition of the conditional expectation? Oct 4, 2022 at 22:31
• By the way, $Z_n$ is simply the maximum (not an expectation, it is the actual, random, outcome), exactly as defined: $Z_n=\max_kX_k$. Oct 4, 2022 at 22:34
• Follow-up: In this example we can explicitly write down $\sigma(X_1,\dots,X_n)$. What is it? Oct 4, 2022 at 22:54
• The elements $E$ of $\mathcal F_n$ are subsets $E\subseteq\Omega$. One example of an event is $E=\{\omega\in\Omega:X_1(\omega)=1\}$, which is sometimes written as $E=\{X_1=1\}$. The random variable $X_1:\Omega\rightarrow\mathbb R$ is not an element of $\Omega$. I'll give you a hint: We have $\mathcal F_1=\{\{X_1\in E\}:E\subseteq\{1,\dots,f\}\}$. Can you explain why? Oct 4, 2022 at 23:16
• The $n=2$, $f=4$ case is here: math.stackexchange.com/q/2755966/215011 Oct 5, 2022 at 20:42

$$Z_n = \max(\underbrace{\max_{k \leq n-1} X_{k}}_{Z_{n-1}}, X_n)$$
$$Z_{n-1}$$ is $$\mathcal{F}_{n-1}$$-measurable and $$X_n$$ and $$\mathcal{F}_{n-1}$$ are independent. So in $$E[Z_n\vert \mathcal{F}_{n-1}] = E[\max(Z_{n-1}, X_n)\vert \mathcal{F}_{n-1}]$$ we can treat $$Z_{n-1}$$ as a constant and integrate out $$X_n$$ using its unconditional distribution.
$$E[\max(Z_{n-1}, X_n)\vert \mathcal{F}_{n-1}] = \frac{1}{f}\sum_{i=1}^f\max(Z_{n-1},i)$$
The quantity $$E[Z_n\mid \mathcal{F}_{n-1}]$$ answers the question, "Given the results of the first $$n-1$$ die rolls, what is the expected value of the maximum of the first $$n$$ rolls?" It should be further clear that the only relevant information from the first $$n-1$$ rolls we need is the previous maximum, $$Z_{n-1}$$.
There are two things that can occur. If $$X_n\le Z_{n-1}$$, then the maximum is unchanged, i.e. $$Z_n=Z_{n-1}$$. Otherwise, $$X_n>Z_{n-1}$$. Conditional on $$X_n>Z_{n-1}$$, the variable $$X_n$$ is uniformly distributed over the interval $$\{Z_{n-1}+1,Z_{n-1}+2,\dots,f\}.$$ The expectation of such a uniform variable is $$\frac{(Z_{n-1}+1)+f}{2}$$ Therefore, \begin{align}\newcommand{\F}{\mathcal F} E[Z_n\mid \F_{n-1}]= \underbrace{\frac{Z_{n-1}}{f}}_ {\substack{\text{Probability }} \\ \text{that X_{n}\le Z_{n-1}}} \cdot Z_{n-1}+ \underbrace{\left(1-\frac{Z_{n-1}}{f}\right)}_{\substack{\text{Probability }} \\ \text{that X_{n}> Z_{n-1}}} \frac{(Z_{n-1}+1)+f}{2} \end{align}