# 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.

## Some properties

${\displaystyle E[Y]=E[E[Y|N]]=\lambda E[X]}$

${\displaystyle Var[Y]=Var[E[Y|N]]+E[Var[Y|N]]=\lambda \{E^{2}[X]+Var[X]\}}$

The cumulant generating function ${\displaystyle K_{Y}(t)={\mbox{ln}}E[e^{tY}]={\mbox{ln}}E[E[e^{tY}|N]]={\mbox{ln}}E[e^{NK_{X}(t)}]=K_{N}(K_{X}(t))}$

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)
The cumulant generating function treatment above is now in the article. Melcombe (talk) 15:26, 6 August 2008 (UTC)

## Why is ${\displaystyle Y(0)\neq 0}$?

I would have thought that the process should be started in zero? Just thinking in terms of (shudder) actuarial science, a claims process would make very little sense if it started with a claim at time zero? What I'm proposing is to change the definition to the one given on the page for 'Compound Poisson Process'. — Preceding unsigned comment added by Fladnaese (talkcontribs) 18:54, 26 May 2011 (UTC)

Fixed-up for this point. JA(000)Davidson (talk) 08:40, 27 May 2011 (UTC)

I see that a citation is needed for the relationship between the cumulants of the compound Poisson distribution Y, and the moments for the random variables Xi. Back in 1976, I proved this result, that is:

For j > 0, K(j) = lambda * m(j), where:

- K(j) are the cumulants of Y
- m(j) are the moments for the Xi
- lambda is the parameter of the Poisson distribution


I made use of the characteristic function in thís proof. Is my proof of interest as a citation? If so, I can send the reference number and a pdf of the paper I wrote.

David LeCorney (talk) 12:03, 8 June 2012 (UTC)