We have two random variables $Y_1\sim Poisson(\lambda_1)$ and $Y_2\sim Poisson(\lambda_2)$.
And I know that the moment-generating function $m(t)$ for a random variable with parameter $\lambda$ is $$m(t)=e^{\lambda(e^t -1)}$$
How may I use the moment-generating functions to determine the distribution of $Y_1+Y_2$? Thank you very much.
Best Answer
The moment-generating function $m_X(t)$ of the random variable $X$ is actually $\mathbb{E} [e^{tX}]$.
This means: The moment-generating function of $Y_1 + Y_2$ is
$\mathbb{E} [e^{t(Y_1 + Y_2)}] = \mathbb{E} [e^{tY_1} e^{tY_2}]$.
If $Y_1$ and $Y_2$ are independent , then $\mathbb{E} [e^{tY_1} e^{tY_2}] = \mathbb{E} [e^{tY_1}]\mathbb{E} [e^{tY_2}]$.
Now you plug the moment-generating functions of $Y_1$ and $Y_2$ (which you already know) into this and you get the moment-generating function of $Y_1 + Y_2$. What is left is to deduce the actual distribution of $Y_1 + Y_2$.