Revision as of 13:29, 14 May 2023 by Admin (Created page with "For an insurance: #The number of losses per year has a Poisson distribution with <math>\lambda = 10 </math> . #Loss amounts are uniformly distributed on (0, 10). #Loss amount...")
ABy Admin
May 14'23
Exercise
For an insurance:
- The number of losses per year has a Poisson distribution with [math]\lambda = 10 [/math] .
- Loss amounts are uniformly distributed on (0, 10).
- Loss amounts and the number of losses are mutually independent.
- There is an ordinary deductible of 4 per loss.
Calculate the variance of aggregate payments in a year.
- 36
- 48
- 72
- 96
- 120
ABy Admin
May 14'23
Key: C
Since loss amounts are uniform on (0, 10), 40% of losses are below the deductible (4), and 60% are above. Thus, claims occur at a Poisson rate [math]\lambda^* = 0.6(10) = 6 . [/math]
Since loss amounts were uniform on (0, 10), claims are uniform on (0, 6).
Let N = number of claims; X = claim amount; S = aggregate claims.
[[math]]
\begin{aligned}
& \operatorname{E}[ N ] = \operatorname{E}[ N ] = \lambda^* = 6\\
& \operatorname{E}[ X ] = (6 − 0) / 2 = 3\\
&\operatorname{E}[ X ] = (6 − 0)2 /12 = 3 \\
&\operatorname{E}[ S ] = \operatorname{E}[ N ]\operatorname{E}[ X ] + \operatorname{E}[ X ] 2\operatorname{E}[ N ] = 6(3) + 32 (6) = 72
\end{aligned}
[[/math]]