# Linear combination of independent random variables that are poisson distributed

Suppose $X_1$ and $X_2$ are independent random variables $X_1$~ Poisson$(\lambda_1)$ and $X_2$~ Poisson$(\lambda_2)$

I want to show $X_1 + X_2$~poisson$(\lambda_1 + \lambda_2)$

I want to then generalize this to a sum of $n$ independent Poisson random variables

An easy way to demonstrate this is by using the property of moment generating functions that says for two independent random variables, $X$ and $Y$, with moment generating functions $M_{X}(t)$ and $M_{Y}(t)$, respectively, the MGF of their sum is given as: $$M_{X+Y}(t) = M_{X}(t) * M_{Y}(t)$$ Since a Poisson random variable with parameter $\lambda$ has MGF $e^{\lambda (e^{t}-1)}$, we can say that if $X\sim Poisson(\lambda_{1})$ and $Y\sim Poisson(\lambda_{2})$, we have: $$M_{X+Y}(t) = e^{\lambda_{1} (e^{t}-1)}*e^{\lambda_{2} (e^{t}-1)} = e^{(\lambda_{1}+\lambda_{2}) (e^{t}-1)}$$ Since a moment generating function uniquely describes a distribution, we can say that this MGF must correspond to $X+Y = Z \sim Poisson(\lambda^{*}= \lambda_{1}+\lambda_{2})$. Using this method, generalizing to the case of a sum of $N$ Poisson random variables is a simple task.

Hint:

First proof $X_1+X_2 ~ Poisson(\lambda_1+\lambda_2)$ (the proof is here if you want to see it).

Second use mathematical induction using what you have proved first to pass from $n$ to $n+1$ (because the sum of the first $n$ is poisson and is added $X_{n+1}$)

The characteristic function of the Poisson distribution is $$\phi_X(t) = \exp(\lambda (e^{it} -1)).$$ If you have $n$ independent Poisson random variables, the characteristic function of the sum $X_1 + \dots + X_n$ calculates to $$\phi_{X_1 + \dots + X_n}(t) = \phi_{X_1}(t) \cdot \ldots \cdot \phi_{X_n}(t) = \exp\left(\sum_{k=1}^{n} \lambda_k (e^{it} -1)\right),$$ which is the characteristic function of a Poisson distribution with parameter $\lambda = \lambda_1 + \dots + \lambda_n$.