Revision as of 18:19, 13 May 2023 by Admin (Created page with "An individual performs dangerous motorcycle jumps at extreme sports events around the world. The annual cost of repairs to their motorcycle is modeled by a Pareto distributio...")
May 13'23
Exercise
An individual performs dangerous motorcycle jumps at extreme sports events around the world.
The annual cost of repairs to their motorcycle is modeled by a Pareto distribution with [math]\theta = 5000 [/math] and [math] \alpha = 2 [/math].
An insurance policy reimburses motorcycle repair costs subject to the following provisions:
- The annual ordinary deductible is 1000.
- The policyholder pays 20% of repair costs between 1000 and 6000 each year.
- The policyholder pays 100% of the annual repair costs above 6000 until they have paid 10,000 in out-of-pocket repair costs each year.
- The policyholder pays 10% of the remaining repair costs each year.
Calculate the expected annual insurance reimbursement.
- 2300
- 2500
- 2700
- 2900
- 3100
May 13'23
Key: C
Insurance pays 80% of the portion of annual claim between 6,000 and 1,000, and 90% of the portion of annual claims over 14,000.
The 14,000 breakpoint is where the policyholder has paid 10,000:
1000 = deductible
1000 = 20% of costs between 1000 and 6000
8000 = 100% of costs between 14,000 and 6,000
[[math]]
\operatorname{E}(X \wedge x ) = \theta \left( 1 - \frac{\theta}{x + \theta} \right) = \frac{5000x}{x + 5000}
[[/math]]
[math]x[/math] | [math]\operatorname{E}(X \wedge x ) [/math] |
---|---|
1000 | 833.33 |
6000 | 2727.27 |
14000 | 3684.21 |
[math]\infty[/math] | 5000 |
[[math]]
\begin{aligned}
&0.80[ \operatorname{E}[ X \wedge 6000) − \operatorname{E}[ X \wedge 1000)] + 0.90[ \operatorname{E}[ X ) − \operatorname{E}( X \wedge 14000)] \\
&= 0.80[2727, 27 − 833.33] + 0.90[5000 − 3684.21] \\
&= 1515.15 + 1184.21 = 2699.36
\end{aligned}
[[/math]]