By <independence> and the <binomial theorem>,
$$
\mathbb P(X+Y=n)
=\sum_{k=0}^n e^{-\lambda}\frac{\lambda^k}{k!}e^{-\mu}\frac{\mu^{n-k}}{(n-k)!}
=e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^n}{n!}.
$$
Therefore \b[$X+Y\sim\operatorname{Poisson}(\lambda+\mu)$].

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 3F / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-2/3f/i/solution