properties of Poisson random variables

Proposition 1.

If $X_{\mathrm{1}}\mathrm{,}{X}_{\mathrm{2}}$ are independent Poisson random variables with parameters ${\lambda }_{\mathrm{1}}\mathrm{,}{\lambda }_{\mathrm{2}}$, then $X_{\mathrm{1}}\mathrm{+}{X}_{\mathrm{2}}$ is a Poisson random variable with parameter ${\lambda }_{\mathrm{1}}\mathrm{+}{\lambda }_{\mathrm{2}}$.

Proof.

Let $X:=X_1+X_2$ and $\lambda :={\lambda }_1+{\lambda }_2$, let us calculate the distribution function of $X$:

$F_X(x)$$=$$P(X\le x)=P(X_1+X_2\le x)={\displaystyle \sum _{i=0}^x}P(X_1+X_2=i)$

$=$${\displaystyle \sum _{i=0}^x}{\displaystyle \sum _{j=0}^i}P(X_1=j\text{ and }{X}_2=i-j)={\displaystyle \sum _{i=0}^x}{\displaystyle \sum _{j=0}^i}P(X_1=j)P(X_2=i-j)$

$=$${\displaystyle \sum _{i=0}^x}{\displaystyle \sum _{j=0}^i}{\displaystyle \frac{e^{-{\lambda }_1}{\lambda }_1^j}{j!}}{\displaystyle \frac{e^{-{\lambda }_2}{\lambda }_2^{i-j}}{(i-j)!}}={\displaystyle \sum _{i=0}^x}{\displaystyle \sum _{j=0}^i}{\displaystyle \frac{e^{-\lambda }}{i!}}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right){\lambda }_1^j{\lambda }_2^{i-j}$

$=$${\displaystyle \sum _{i=0}^x}{\displaystyle \frac{e^{-\lambda }}{i!}}{\displaystyle \sum _{j=0}^i}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right){\lambda }_1^j{\lambda }_2^{i-j}={\displaystyle \sum _{i=0}^x}{\displaystyle \frac{e^{-\lambda }}{i!}}{({\lambda }_1+{\lambda }_2)}^i={\displaystyle \sum _{i=0}^x}{\displaystyle \frac{e^{-\lambda }}{i!}}{\lambda }^i.$

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.∎

As a corollary, any sum of independent Poisson random variables is Poisson, with parameter the sum of the parameters from the independent random variables.

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 $X_1,\mathrm{\dots },X_n$ 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 $X$.∎