Talk:Compound Poisson distribution
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I assume that E[Y] = λ * E[X]
What is Var[Y] in terms of the distribution of X? Say, if X has a gamma distribution.
[edit] Some properties
E[Y] = E[E[Y | N]] = λE[X]
Var[Y] = Var[E[Y | N]] + E[Var[Y | N]] = λ{E2[X] + Var[X]}
The cumulant generating function ![K_Y(t)=\mbox{ln} E[e^{tY}]=\mbox{ln} E[E[e^{tY}|N]]=\mbox{ln} E[e^{NK_X(t)}]=K_N(K_X(t))](../../../../math/d/c/5/dc5a6855cb8ee88b7f4ad847df264fea.png)
- One could add to the above, that if N has a Poisson distribution with expected value 1, then the moments of X are the cumulants of Y. Michael Hardy 20:39, 23 Apr 2005 (UTC)

