Deductibles, limits and risk measures

2 free worked examples

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1. Exam PDeductibles, limits and risk measuresExam level

   Losses are exponential with mean 1,000 and a policy has an ordinary deductible of 250. Find the expected payment PER LOSS.

   1. A$750.00$
   2. B$778.80$
   3. C$1,000.00$
   4. D$1,250.00$
   5. E$1,284.03$

   Solution

   1. $E[(X-d{)}_{+}]={\int }_d^{\infty }S(x)dx$, and for an exponential this is $\theta e^{-d/\theta }$.
   2. $=1,000e^{-250/1000}=1,000e^{-0.25}$.
   3. $=1,000(0.778801)$.
   4. $=778.80$. This averages over ALL losses, including the small ones that pay nothing.

   Trap. Answering 750 by subtracting the deductible from the mean, which ignores losses below the deductible.

2. Exam PDeductibles, limits and risk measuresStretch

   Losses are Pareto with $\alpha =3$ and $\theta =1,000$. Find $E[X]$ and comment on the tail.

   1. AMean 333.33; variance infinite
   2. BMean 500.00; variance finite
   3. CMean 500.00; variance infinite
   4. DMean 333.33; variance finite
   5. EMean 1,000.00; variance finite

   Solution

   1. For the two-parameter Pareto, $E[X]=\frac{\theta }{\alpha -1}=\frac{1000}{2}=500$.
   2. The variance exists whenever $\alpha >2$, and here $\alpha =3$.
   3. $\mathrm{Var}(X)=\frac{\alpha {\theta }^2}{(\alpha -1{)}^2(\alpha -2)}=\frac{3\times 10^6}{4}=750,000$.
   4. So the standard deviation is about 866, well above the mean - heavy-tailed, but with both moments finite.

   Trap. Assuming any Pareto has infinite variance; it depends entirely on whether α exceeds 2.