Probability of a Sum of Random Variables I know that $X_i$ are distributed Poisson with parameter $\lambda$. Can anyone explain why the property in the link holds? Thanks.
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 A: This is a property you should know, if  $X_i$ are independent and follow a $\text{Pois}(\lambda_i)$ then
$$X_1+\dotsb+X_n \sim\text{Pois}(\lambda_1+\dotsb+\lambda_n).$$
A: Assuming $\{X_i\}_n$ are independent as well as identically distributed, then we can use induction to demonstrate:

The base case:
$$\begin{align}
\mathsf P(X_1 = t) & = \frac{(1\lambda)^t e^{-1\lambda}}{t!}
\\[2ex]
\mathsf P(X_1+X_2 = t) & = \sum_{s=0}^t \mathsf P(X_1=s)\mathsf P(X_2=t-s)
\\ & =\sum_{s=0}^t \dfrac{\lambda^s e^{-\lambda}}{s!}\dfrac{\lambda^{t-s} e^{-\lambda}}{(t-s)!}
\\ & = \frac{\lambda^t e^{-2\lambda}}{t!}\sum_{s=0}^t \frac {t!}{s!(t-s)!}
\\ & = \frac{2^t\lambda^t e^{-2\lambda}}{t!}\end{align}$$
The iterative step:
$$\begin{align}\mathsf P(\sum_{i=1}^{k+1}X_i = t) & = \sum_{s=0}^t \mathsf P(\sum_{i=1}^{k}X_i =s)\mathsf P(X_k=t-s)
\\ & =\sum_{s=0}^t \dfrac{(k\lambda)^s e^{-k\lambda}}{s!}\dfrac{\lambda^{t-s} e^{-\lambda}}{(t-s)!}
\\ & = \frac{\lambda^t e^{-(k+1)\lambda}}{t!}\sum_{s=0}^t \frac {k^s\,t!}{s!(t-s)!}
\\ & = \frac{(k+1)^t\lambda^t e^{-(k+1)\lambda}}{t!}\end{align}$$
Thus by induction:
$$\mathsf P(\sum_{i=1}^{n}X_i = t) =\frac{(n\lambda)^t e^{-n\lambda}}{t!}$$

Remark: The key step hinges on the Binomial Expansion $(k+1)^t = \sum_{s=0}^t \binom{t}{s}k^s$
