5.2 The Compound Poisson
Result 5.2.1. If \(N\sim \) POI\((\lambda )\) then \[E(S) = \lambda \mu ,\qquad \Var (S) = \lambda \,E\left (X^{2}\right ),\qquad M_S(t) = \exp \left \{\lambda \left [M_X(t)-1\right ]\right \}.\]
Proof. For a Poisson variable \(E(N)=\Var (N)=\lambda \) and \(G_N(t)=e^{\lambda (t-1)}\). Substituting into the theorem, \[\Var (S) = \lambda \sigma ^{2} + \lambda \mu ^{2} = \lambda \left (\sigma ^{2}+\mu ^{2}\right ) = \lambda \,E\left (X^{2}\right ),\] and \(M_S(t) = G_N\left (M_X(t)\right ) = \exp \left \{\lambda \left [M_X(t)-1\right ]\right \}\). □
Note. The variance collapsing to \(\lambda E\left (X^{2}\right )\) is a genuine simplification, and it is peculiar to the Poisson: it happens because the mean and variance of a Poisson variable coincide, so the two terms combine into the second moment. It is worth remembering as a check — if a compound Poisson calculation does not reduce to \(\lambda E\left (X^{2}\right )\), something has gone wrong.
Example 5.2.2. Let \(S = X_1+\cdots +X_N\) where \(N\sim \) POI\((15)\) and each \(X_i\) has the GAM\((\alpha ,\beta )\) distribution with shape \(\alpha \) and scale \(\beta \), independently of \(N\). Find \(E(S)\) and \(\Var (S)\).
Solution. For the gamma, \(\mu = \alpha \beta \) and \(\sigma ^{2} = \alpha \beta ^{2}\), so \[E\left (X^{2}\right ) = \sigma ^{2}+\mu ^{2} = \alpha \beta ^{2} + \alpha ^{2}\beta ^{2} = \alpha (\alpha +1)\beta ^{2}.\] Hence \[E(S) = 15\,\alpha \beta ,\qquad \Var (S) = 15\,\alpha (\alpha +1)\beta ^{2}.\] As a check, the general formula gives \(15\alpha \beta ^{2} + 15\alpha ^{2}\beta ^{2}\), which is the same.
Example 5.2.3. An insurer expects \(200\) claims a year, the number following a Poisson distribution. Individual claims have mean \(\$1{,}500\) and standard deviation \(\$2{,}000\). Find the mean and standard deviation of the annual total.
Solution. Here \(\lambda = 200\), \(\mu = 1500\) and \(\sigma = 2000\), so \[E(S) = 200(1500) = \$300{,}000 ,\] \[E\left (X^{2}\right ) = 2000^{2} + 1500^{2} = 4{,}000{,}000 + 2{,}250{,}000 = 6{,}250{,}000 ,\] \[\Var (S) = 200\times 6{,}250{,}000 = 1{,}250{,}000{,}000 , \qquad \text {sd}(S) = \$35{,}355 .\]
The standard deviation is about \(12\%\) of the mean, and it is worth seeing where it comes from: the claim-size term contributes \(200\times 2000^{2}\) and the claim-number term \(200\times 1500^{2}\), so uncertainty about how many claims occur accounts for over a third of the total variance. An insurer that modelled the claim count as fixed at \(200\) would understate the standard deviation by more than a fifth.
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