Sum of Poissons is Poisson with adding parameters

Question

I am studying probability and statistics recently, in section multidimensional random variable and its distribution, I encountered a problem:

Let $\pi(\lambda)$ denote the Poisson distribution with parameter $\lambda$ and let $X$ and $Y$ be independent Poisson random variables, $X\sim \pi(\lambda_1), Y\sim \pi(\lambda_2)$. Prove that $X+Y\sim \pi(\lambda_1 + \lambda_2)$.

Answer 1

HINT

Typically such things are done by conditioning. Assume $a,b \in \mathbb{R}$ and let $X \sim \pi(a), Y \sim \pi(b)$ as you suggest. Also let $Z = X+Y$ and note that $$ \begin{split} F_Z(z) &= \mathbb{P}[X+Y \le z] \\ &= \sum_{k=0}^\infty \mathbb{P}[X+Y \le z|Y=k] f_Y(k) \\ &= \sum_{k=0}^\infty \mathbb{P}[X \le z-k] f_Y(k) \\ &= \sum_{k=0}^\infty f_X(z-k) f_Y(k), \end{split} $$ which is called a convolution of $f_X$ and $f_Y$. Can you finish?

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Another approach is to use mgf: $$ M_Z(t) = \mathbb{E}\left[e^{tZ}\right] = \mathbb{E}\left[e^{tX}e^{tY}\right] = M_X(t)M_Y(t), $$ since $X,Y$ are independent. Can you finish this too?