Show that, if $X_{k}\sim\text{Poisson}(\lambda_{k})$ and they are independent for $1 \leq k \leq n$, then \begin{align*} Y = \displaystyle\sum_{k=1}^{n}X_{k}\sim\text{Poisson}\left(\sum_{k=1}^{n}\lambda_{k}\right) \end{align*}
MY SOLUTION
To begin with, I am going determine the moment generating function of $X\sim\text{Poisson}(\lambda)$: \begin{align*} M_{X}(t) = \textbf{E}[e^{tX}] = \sum_{k=0}^{\infty}e^{tk}e^{-\lambda}\frac{\lambda^{k}}{k!} = e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^{t})^{k}}{k!} = e^{-\lambda}e^{\lambda e^{t}} = \exp(-\lambda(1 - e^{t})) \end{align*}
Once these random variables are independent, the moment generating function of its sum is given by the product of each moment generating function: \begin{align*} M_{Y}(t) = \prod_{k=1}^{n}\exp(-\lambda_{k}(1-e^{t})) = \exp\left(-\sum_{k=1}^{n}\lambda_{k}(1-e^{t})\right) \end{align*}
which means that $\displaystyle Y\sim\text{Poisson}\left(\sum_{k=1}^{n}\lambda_{k}\right)$, because random variables that have the same generating function have the same distribution.