excans:13d8516dc1: Difference between revisions
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\begin{align*} | \begin{align*} | ||
F(x) &= \int_0^{x}2e^{-2y} dy = -e^{-2y} \Big |_0^x = 1-e^{-2x} \\ | F(x) &= \int_0^{x}2e^{-2y} dy = -e^{-2y} \Big |_0^x = 1-e^{-2x} \\ | ||
P[X \leq 0.5 | X \leq 1.0] &= \frac{P[X \leq 0.5]}{P[X \leq 1.0]} = \frac{F(0.5)}{F(1.0)} = \frac{1-e^{-1}}{1-e^{-2}} = 0.731. | \operatorname{P}[X \leq 0.5 | X \leq 1.0] &= \frac{\operatorname{P}[X \leq 0.5]}{\operatorname{P}[X \leq 1.0]} = \frac{F(0.5)}{F(1.0)} = \frac{1-e^{-1}}{1-e^{-2}} = 0.731. | ||
\end{align*} | \end{align*} | ||
</math> | </math> | ||
{{soacopyright | 2023}} | {{soacopyright | 2023}} | ||
Latest revision as of 21:54, 8 May 2023
Solution: D
[[math]]
\begin{align*}
F(x) &= \int_0^{x}2e^{-2y} dy = -e^{-2y} \Big |_0^x = 1-e^{-2x} \\
\operatorname{P}[X \leq 0.5 | X \leq 1.0] &= \frac{\operatorname{P}[X \leq 0.5]}{\operatorname{P}[X \leq 1.0]} = \frac{F(0.5)}{F(1.0)} = \frac{1-e^{-1}}{1-e^{-2}} = 0.731.
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.