Expected value of the mean of a Poisson distribution, where the mean has a Gamma distribution

Question

stuck on a question and can't seem to make any progress:

We have an insurance company who expects the number of accidents their policy holders will have is Poisson distributed. The Poisson mean $\Delta$ follows a Gamma distribution with the $\Gamma$(2,1) density function being $f_{\Delta}(\lambda) = \lambda e^{-\lambda} (\lambda > 0)$.

I have been tasked to find the expected value of $\Delta$ for a policy holder having $x$ accidents this year, where $x$ = 0, 1, 2...

So far I'm exploring how to find $E(\Delta | X = x$), but kinda unsure of how to go about this.

Any help would be appreciated.

Answer 1

$E(X)=\alpha \beta,$

where $\alpha$ and $\beta$ are the parameters of the gamma distribution.

I can give a proof, but I think it would be more helpful to look here:

http://www.math.wm.edu/~leemis/chart/UDR/PDFs/Gammapoisson.pdf

I hope I helped!

Answer 2

It is a case of Poisson-Gamma mixture, and we can use the hierarchical model as following.

Let $X$ be the number of accident and its mean $\Delta$.

Now $X|\Delta$ has the distribution $\mathrm {Poisson}$$(\Delta)$ and $\Delta$ has the distribution $\Gamma(2,1)$.

Since $E(X|\Delta)=\Delta$ and $E(\Delta)=2$, the identity $E(X)=E(E(X|\Delta))$ shows that $E(X)=E(\Delta)=2$.