Talk:Compound Poisson distribution

$E[Y]=E[E[Y|N]]=\backslash lambda\; E[X]$

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

The cumulant generating function

$K\_Y(t)=\backslash mbox\{ln\}\; E[e^\{tY\}]=\backslash mbox\{ln\}\; E[E[e^\{tY\}|N]]=\backslash mbox\{ln\}\; E[e^\{NK\_X(t)\}]=K\_N(K\_X(t))$

: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)

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)

: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)

In the development of the properties section, the notation $E\_N(.)$ 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):

$\backslash varphi\_Y(t)\; =\; \backslash operatorname\{E\}\_Y(e^\{itY\})=\; \backslash operatorname\{E\}\_N\; (\; \backslash left(\backslash operatorname\{E\}\_X\; (e^\{itX\})\; )^N\; \backslash right)=\; \backslash operatorname\{E\}\_N\; ((\backslash varphi\_X(t))^N),\; \backslash ,$

What should be clarified is what justified the second equal sign moving from $E\_Y(.)$ to $E\_N(.)$, 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)