I am considering the following question:
Three types of magazines, $A, B,$ and $C$, are for sale in a market. A customer who purchases a magazine will choose $A, B,$ and $C$ with probabilities $0.5$, $0.3$, and $0.2$, respectively. Each day, the total number of customers of the market follows a Poisson distribution with mean $400$. Assume that $25$% of customers will purchase magazines. Further assume that each customer will buy one magazine and the magazines will never be out of stock. Let $X$ and $Y$ be the daily number of sales of magazines $A$ and $B$, respectively. Find (i) $E(X-Y)$ and (ii) $V(X-Y)$.
I am not certain how to proceed with this question, though I believe it would be helpful to use the Law of Total Expectation (i.e., $E(E(U|W)) = E(U)$) and of Total Variance (i.e., $V(U) = V(E(U|W)) + E(V(U|W))$).
I conjecture that $X|N \sim Bin(0.25N, p=0.5), Y|N \sim Bin(0.25N, 0.3)$.
(i) By linearity, $$E(X-Y)= E(E(X|N) - E(Y|N)) = E(0.125N - 0.075N) = 0.05E(N) = 0.05(400) = 20.$$
(ii) Owing to independence of $X, Y$, $V(X-Y) = V(X) + V(Y) = V(E(X|N)) + E(V(X|N)) + V(E(Y|N)) + E(V(Y|N))$, so $$V(X-Y) = V(0.125N) + E(0.25 \times 0.5 \times 0.5N) + V(0.075N) + E(0.25 \times 0.3 \times 0.7N) = 0.02125V(N) + 0.115E(N) = 109(400)/800 = 54.5.$$
I am curious whether these are correct.