1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $X_{n+1}$)

$\endgroup$
0
$\begingroup$

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 $\lambda = \lambda_1 + \dots + \lambda_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.