Revision as of 15:25, 14 May 2023 by Admin (Created page with "Loss amounts have the distribution function <math display = "block"> F(x) = \begin{cases} (x/100)^2, \, 0 \leq x \leq 100 \\ 1, \, x> 100 \end{cases} </math> An insurance...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
ABy Admin
May 14'23

Exercise

Loss amounts have the distribution function

[[math]] F(x) = \begin{cases} (x/100)^2, \, 0 \leq x \leq 100 \\ 1, \, x\gt 100 \end{cases} [[/math]]

An insurance policy pays 80% of the amount of the loss in excess of an ordinary deductible of 20, subject to a maximum payment of 60 per loss.

Calculate the conditional expected claim payment, given that a payment has been made.

  • 37
  • 39
  • 43
  • 47
  • 49

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

ABy Admin
May 14'23

Key: B

Pays 80% of loss over 20, with cap of payment at 60, hence u = 60/0.8 + 20 = 95.

[[math]] \begin{aligned} \operatorname{E}[Y \textrm{per loss}) = \alpha [ \operatorname{E}[ X \wedge 95) − \operatorname{E}[ X \wedge 20)] = 0.8 \left [ \int_0^{95}S(x) dx - \int_0^{20}S(x) dx \right ] \\ = 0.8 \int_{20}^{95}S(x) dx = 0.8 \int_{20}^{95} (1 - \frac{x^2}{10,000} ) dx = 0.8 \left( x - \frac{x^3}{30,000} \right) \Big |_{20}^{95} = 37.35 \\ \operatorname{E}( Y \, \textrm{per payment} ) = \frac{\operatorname{E}(Y \, \textrm{per loss})}{1-F(20)} = \frac{37.35}{0.96} = 38.91 \end{aligned} [[/math]]

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

00