Revision as of 11:01, 7 May 2023 by Admin (Created page with "'''Solution: C''' Let Y represent the payment made to the policyholder for a loss subject to a deductible D. That is <math display = "block"> Y = \begin{cases} 0, \quad 0...")
Exercise
ABy Admin
May 07'23
Answer
Solution: C
Let Y represent the payment made to the policyholder for a loss subject to a deductible D. That is
[[math]]
Y = \begin{cases}
0, \quad 0 \leq X \leq D \\
x-D, \quad D \lt x \leq 1
\end{cases}
[[/math]]
Then since [math]\operatorname{E}[X] = 500 [/math], we want to choose D so that
[[math]]
\frac{500}{4} = \int_{D}^{1000} \frac{1}{1000} (x-D) dx = \frac{1}{1000} \frac{(x-D)^2}{2} \Big |_D^{1000} = \frac{(1000-D)^2}{2000}
[[/math]]
which implies that [math]D=500[/math] (or [math]D =1500 [/math] which is extraneous).