Proposition 1   If $X_1,X_2$ are independent Poisson random variables with parameters $\lambda_1,\lambda_2$ , then $X_1+X_2$ is a Poisson random variable with parameter $\lambda_1+\lambda_2$ .

Proof. Let $X:=X_1+X_2$ and $\lambda:=\lambda_1+\lambda_2$ , let us calculate the distribution function of $X$ : \begin{eqnarray*} F_X(x) &=& P(X\le x) = P(X_1+X_2\le x) = \sum_{i=0}^x P(X_1+X_2=i) \\ &=& \sum_{i=0}^x \sum_{j=0}^i P(X_1=j \mbox{ and } X_2=i-j) = \sum_{i=0}^x \sum_{j=0}^i P(X_1=j)P(X_2=i-j) \\ &=& \sum_{i=0}^x \sum_{j=0}^i \frac{e^{-\lambda_1} \lambda_1^j}{j!} \frac{e^{-\lambda_2} \lambda_2^{i-j}}{(i-j)!} = \sum_{i=0}^x \sum_{j=0}^i \frac{e^{-\lambda}}{i!} \binom{i}{j} \lambda_1^j \lambda_2^{i-j} \\ &=& \sum_{i=0}^x \frac{e^{-\lambda}}{i!} \sum_{j=0}^i \binom{i}{j} \lambda_1^j \lambda_2^{i-j} = \sum_{i=0}^x \frac{e^{-\lambda}}{i!} (\lambda_1+\lambda_2)^i = \sum_{i=0}^x \frac{e^{-\lambda}}{i!} \lambda^i. \end{eqnarray*}As a result, $X$ is a Poisson random variable with parameter $\lambda$ . Notice that in the fifth equation, we used the assumption that $X_1$ and $X_2$ are independent.

Proposition 2   A Poisson random variable is infinitely divisible.

Proof. Let $X$ be a Poisson random variable with parameter $\lambda$ . Let $n$ be any positive integer. Let $X_1,\ldots, X_n$ be independent identically distributed Poisson random variables with parameter $\frac{\lambda}{n}$ . Then the sum of these random variables is easily seen to be Poisson, with parameter $\lambda$ , and is therefore identically distributed as $X$ .