Talk:Compound Poisson distribution

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.

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

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

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

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 (talk • contribs) 18:54, 26 May 2011 (UTC) [reply ]

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

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) [reply ]

In the development of the properties section, the notation ${\displaystyle E_{N}(.)}${\displaystyle E_{N}(.)} appears which indicates with which distribution the expectation is to be calculated.

I suggest to continue to use this explicit notation throughout the demonstration, for clarity. Would this be right (my expertise is in construction):

${\displaystyle \varphi _{Y}(t)=\operatorname {E} _{Y}(e^{itY})=\operatorname {E} _{N}(\left(\operatorname {E} _{X}(e^{itX}))^{N}\right)=\operatorname {E} _{N}((\varphi _{X}(t))^{N}),,円}${\displaystyle \varphi _{Y}(t)=\operatorname {E} _{Y}(e^{itY})=\operatorname {E} _{N}(\left(\operatorname {E} _{X}(e^{itX}))^{N}\right)=\operatorname {E} _{N}((\varphi _{X}(t))^{N}),,円}

What should be clarified is what justified the second equal sign moving from ${\displaystyle E_{Y}(.)}${\displaystyle E_{Y}(.)} to ${\displaystyle E_{N}(.)}${\displaystyle E_{N}(.)}, which I believe is the law of total expectation.

Also, this sentence could be clarified: "and hence, using the probability-generating function of the Poisson distribution," Does it mean that you use the PGF to manipulate the previous equation to get the result? That I don’t see how, maybe is should be a bit more explicit. I do get to the result by applying de definition of E(.) and after identifying terms get the result.

P.S. this could/should be done also in pages such as « law of total variance » etc.

Scharleb (talk) 22:41, 18 December 2020 (UTC) [reply ]

• Start-Class mathematics articles
• Low-priority mathematics articles
• Start-Class Statistics articles
• Low-importance Statistics articles
• WikiProject Statistics articles