I was just working through this question.

A compound Poisson risk model is used to model the total claims S experienced by an insurance company over one year, of the form:

$S = X_1 + ... + X_n$

where $ X_i$ represents the size of claim $i$ and $N$ is the total number of claims, following a Poisson distribution. Each claim size $X_i$ is distributed according to $X_i$ ~ $Gamma(α,λ) $

In the 5 preceding years the average number of claims per year has been 14. The average claim size has been 500 and the standard deviation of claim sizes has been 150.

I am supposed to estimate:

a) $α$ and $λ $ for the Gamma distribution

b) The variance and mean for the total number of claims

My attempt at a solution, I am not sure if correct.

a) I used the method of moments.The first moment (i.e. mean) is given by $E(X)= α/λ $, and the second moment (i.e. variance) is given by $ Var(X) = \alpha/\lambda^2 $. Setting E(X) = 500, and Var(X) = $150^2 $, we solve for $ \alpha $ and $ \lambda $:

$500\lambda = \alpha$

Therefore, the variance will be:

$ \alpha/\lambda^2= 150^2$ $ 500\lambda/\lambda^2 = 22500$

$\lambda= 500/22500 = 0.0222$

$\alpha = 11.111 $

b) This part I am completely lost at.


Do we have to use the double expectation theorem here? This would give an expected mean for $S$ of $ 500*14 = 700$.

Still no clue how to calculate the variance though.


The first part looks good.

Do we have to use the double expectation theorem here?

Yes, if I understand you correctly. The law of total expectation, or law of iterated expectation, says in this case that $$ \mathbb{E}[S] = \mathbb{E}[\mathbb{E}( S \mid N) ]. $$ To find the variance, there's an analagous formula called the law of total variance, which says $$ \mathbb{V}[S] = \mathbb{E}[\mathbb{V}( S \mid N) ] + \mathbb{V}[\mathbb{E}( S \mid N) ]. $$ Let me know if that helps.

  • $\begingroup$ Hi, that was helpful! If I'm correct, the 2nd formula will simplify to: V(S) = [E(S)^2]*Var(N) + Var(S)*E(N) I'm confused as to what value Var(N) would take in this formula? $\endgroup$ – user3424575 Dec 20 '18 at 5:01

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