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(Created page with "'''Solution: B''' Let Y be the reimbursement. Then, G(115) = P[Y < 115 | X > 20]. For Y to be 115, the costs must be above 120 (up to 120 accounts for a reimbursement of 100)...") |
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\begin{align*} | \begin{align*} | ||
\operatorname{P}[X \leq 150 | X > 20] &= \frac{\operatorname{P}[X \leq 150] - \operatorname{P}[ X \leq 20]}{\operatorname{P}[X \leq 20]} \\ & | \operatorname{P}[X \leq 150 | X > 20] &= \frac{\operatorname{P}[X \leq 150] - \operatorname{P}[ X \leq 20]}{\operatorname{P}[X \leq 20]} \\ & | ||
= \frac{1-e^{150/100}-1 + e^{-20/100}}{1-1 + e^{-20/100}} \\ | = \frac{1-e^{-150/100}-1 + e^{-20/100}}{1-1 + e^{-20/100}} \\ | ||
&= \frac{-e^{-1.5} + e^{-0.2}}{e^{-0.2}} \\ | &= \frac{-e^{-1.5} + e^{-0.2}}{e^{-0.2}} \\ | ||
&= 1-e^{-1.3} \\ | &= 1-e^{-1.3} \\ | ||
Latest revision as of 02:38, 3 July 2024
Solution: B
Let Y be the reimbursement. Then, G(115) = P[Y < 115 | X > 20]. For Y to be 115, the costs must be above 120 (up to 120 accounts for a reimbursement of 100). The extra 15 requires 30 in additional costs. Therefore, we need
[[math]]
\begin{align*}
\operatorname{P}[X \leq 150 | X \gt 20] &= \frac{\operatorname{P}[X \leq 150] - \operatorname{P}[ X \leq 20]}{\operatorname{P}[X \leq 20]} \\ &
= \frac{1-e^{-150/100}-1 + e^{-20/100}}{1-1 + e^{-20/100}} \\
&= \frac{-e^{-1.5} + e^{-0.2}}{e^{-0.2}} \\
&= 1-e^{-1.3} \\
&= 0.727.
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.