Revision as of 18:31, 13 May 2023 by Admin (Created page with "'''Key: A''' Under the initial assumptions, losses, <math>X</math>, have <math>\operatorname{E}(X) = \exp(6 + \frac{1.5^2}{2}) = 1242.65 </math> After 5 years: exponential d...")
Exercise
May 13'23
Answer
Key: A
Under the initial assumptions, losses, [math]X[/math], have [math]\operatorname{E}(X) = \exp(6 + \frac{1.5^2}{2}) = 1242.65 [/math]
After 5 years: exponential distribution with mean 1242.65(1.045 ) = 1511.87 and LER = 0.33, so
[[math]]
0.33 = \frac{\operatorname{E}[Y \wedge d )}{\operatorname{E}(Y)} = \frac{\theta(1-e^{-d/\theta}}{\theta} = 1- e^{-d/1511.87} \Rightarrow d = −1511.87 \ln(1 − 0.33) = 605.47
[[/math]]