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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.

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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 $Y_1 + Y_2$.

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  • $\begingroup$ "or uncorrelated, resp." No. $\endgroup$ – Did Nov 14 '14 at 11:00
  • $\begingroup$ Ok, thanks. I edited my answer. I thought that since for $\mathbb{E}[Y_1 Y_2] = \mathbb{E}[Y_1] \mathbb{E}[Y_2]$ it suffices for the two variables to be uncorrelated, it also suffices here. I guess, that was wrong. $\endgroup$ – mathie314 Nov 14 '14 at 11:10
  • $\begingroup$ Independence implies uncorrelated, but sadly the converse is not true in general. Another note: of course you showed this, but it might be useful to explicitly state: $M_{X+Y}(t) = M_X(t)M_Y(t)$. $\endgroup$ – knrumsey Nov 18 '16 at 23:41

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