In Year 1 a risk has a Pareto distribution with [math]\alpha = 2[/math] and [math]\theta = 3000[/math] . In Year 2 losses inflate by 20%.

An insurance on the risk has a deductible of 600 in each year. [math]P_i[/math], the premium in year [math]i[/math], equals 1.2 times the expected claims.

The risk is reinsured with a deductible that stays the same in each year. [math]R_i[/math], the reinsurance premium in year [math]i[/math], equals 1.1 times the expected reinsured claims.

[math] \frac{R_1}{P_1} = 0.55 [/math]

Calculate [math]\frac{R_2}{P_2}[/math]

• 0.46
• 0.52
• 0.55
• 0.58
• 0.66

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Key: D

For any deductible d and the given severity distribution

[[math]] \begin{aligned} \operatorname{E}[( X − d )_+ ] = \operatorname{E}[ X ) − \operatorname{E}[ X \wedge d ) &= 3000 − 3000 \left(1 - \frac{3000}{3000 +d} \right) \\ &= 3000 \left( \frac{3000}{3000 + d}\right) \\ & = \frac{9,000,000}{3,000 + d} \end{aligned} [[/math]]

So

[[math]] P_1 = 1.2 \frac{9,000,000}{3000 + 600} = 3000[[/math]]

Let r denote the reinsurer’s deductible relative to insured losses. Thus, the reinsurer’s deductible is 600 + r relative to losses. Thus

[[math]] R_1 = 1.1 \left ( \frac{9,000,000}{3000 + 600 + 4}\right ) = 0.55P1 = 0.55(3000) = 1650 \Rightarrow r = 2400 [[/math]]

In Year 2, after 20% inflation, losses will have a Pareto distribution with [math]\alpha = 2 [/math] and [math]\theta = 1.2(3000) = 3600 [/math]. The general formula for expected claims with a deductible of [math]d[/math] is

[[math]] \operatorname{E}[( X − d )_+ ] = 3600 \left( \frac{3600}{3600 + d}\right) = \frac{12,960,000}{3600 + d} [[/math]]

[[math]] P_2 = 1.2 \frac{12,960,000}{3000 + 600} = 3703, R_2 = 1.1 \frac{12,960,000}{3000 + 600 + 2400} = 2160, \frac{R_2}{P_2} = \frac{2160}{3703} = 0.583. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.