Problem 12.1 Easy Free 1 variation

Die Size Set by a Die

You roll a fair standard six-sided die, obtaining value $k$ . You then roll a second, fair die with $6+k$ faces labelled $1,2,\dots,6+k$ . Find the expected value shown on the second die.

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $K$ be the first roll and $R$ the second. Given $K=k$ , $R$ is uniform on $\{1,\dots,6+k\}$ , so

\mathbb{E}[R\mid K]=\frac{(6+K)+1}{2}=\frac{7+K}{2}.

By the tower property, and since $\mathbb{E}[K]=\frac{7}{2}$ ,

\mathbb{E}[R]=\mathbb{E}\bigl[\mathbb{E}[R\mid K]\bigr]=\frac{7+\mathbb{E}[K]}{2}=\frac{7+\frac72}{2}=\frac{21}{4}.

\boxed{\mathbb{E}[R]=\tfrac{21}{4}=5.25}

Key idea. The conditional mean is linear in $K$ , so only $\mathbb{E}[K]$ matters, not the full distribution of $K$ .

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 12.1a

Let $\lambda \sim \mathrm{Exp}(1)$ , and given $\lambda$ , let $X \sim \mathrm{Poisson}(\lambda)$ . Compute $\mathbb{E}[X]$ .

Work it out, then compare with the answer.

Solution

Given $\lambda$ , $X\sim\mathrm{Poisson}(\lambda)$ , so $\mathbb{E}[X\mid\lambda]=\lambda$ . By the tower property, and since an $\mathrm{Exp}(1)$ variable has mean $1$ ,

\mathbb{E}[X]=\mathbb{E}\bigl[\mathbb{E}[X\mid\lambda]\bigr]=\mathbb{E}[\lambda]=1.

\boxed{\mathbb{E}[X]=1}

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