Sum of independent Poisson random variables is a Poisson random variable

Question

Suppose $x_1$ and $x_2$ are independent Poisson random variables with parameters equal to $\lambda_1$ and $\lambda_2$ respectively. Show the sum of $x_1$ and $x_2$ is also a Poisson random variable using the moment generating function.

I know the function for Poisson is $\frac{\lambda^x e^{-\lambda}}{x!}$ and moment for it is $M_x(t) = e^{\lambda(e^t - 1)}$.

So I have to show $M_{x_1 + x_2}(t)$ is of the form $e^{k(e^t - 1)}$. But I don't know where to plug in $\frac{\lambda^{x_1} e^{-\lambda}}{x_1!} + \frac{\lambda^{x_2} e^{-\lambda}}{x_2!}$ into the moment function.

Answer 1

By a basic "law of exponentiation" we have $$e^{\lambda_1(e^t-1)}e^{\lambda_2(e^t-1)}=e^{(\lambda_1+\lambda_2)(e^t-1)}.$$

Answer 2

You have to use the fact that if the random variables are independent, then the MGF of the sum is the product of the MGFS.

$$M_{X_1+X_2}(t) = M_{X_1}(t) M_{X_2}(t).$$