Joe will read $N \sim Pois(\lambda)$ books this year. Each book has a $Pois(\mu)$ number of pages, with book lengths independent of each other and independent of $N$.
a) Find the expected number of book pages that Joe will read next year:
My solution:
Let $X$ be the number of pages read next year. $$E(X) = E(E(X\mid N)) = E(N\mu) = \lambda\mu$$
b) Find the variance of the number of book pages joe will read next year
My solution:
$$Var(X) = E(Var(X\mid N)) + Var(E(X\mid N)) = E(N\mu) + Var(N\mu) = \lambda(\mu + \mu^2)$$
c) For each of the $N$ books, Joe likes it with probability $p$ and dislikes it with probability $1-p$, independently. Find the conditional distribution of how many of the $N$ books Joe likes, given that he dislikes exactly $d$ of the books.
My solution:
$$P(X=k\mid Y=d) = \frac{P(X=k, Y=d)}{P(Y=d)} = \frac{P(N=k+d)}{P(Y=d)} = \frac{\frac{e^{-\lambda}\lambda^{k+d}}{(k+d)!}}{{{\frac{e^{-\lambda q}(\lambda q)^d}{d!}}}}$$
where $Y \sim Pois(\lambda q)$
Could somebody please check that my reasoning is correct ? I'm doing this out of a book that doesn't have solutions and I want to catch any systematic error that I'm making ...
Let $P_i$ be the number of pages in book $i$ and $S_n:=\sum_{i=1}^n P_i$ for nonnegative integers $i$, $n$. It is straightforward to show that $S_n\sim\mathrm{Pois}(n\mu)$. By Wald's identity, the expected total number of pages read is $$ \mathbb E[S_N] = \mathbb E[N]\mathbb E[P_1] = \lambda\mu. $$ To compute the variance of $S_N$, we use the law of total variance: \begin{align} \mathrm{Var}(S_N) &= \mathbb E[\mathrm{Var}(S_N\mid N)] + \mathrm{Var}(\mathbb E[S_N\mid N])\\ &= \mathbb E[N\cdot\mathrm{Var}(P_1)]+\mathrm{Var}(N\cdot\mathbb E[P_1])\\ &= \mu\mathbb E[N] +\mu^2\mathrm{Var}(N)\\ &= \lambda\mu(1+\mu). \end{align} For part c) I am not sure, but I don't believe your solution is correct.