excans:4896d55649: Difference between revisions
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(Created page with "'''Solution: E''' We are given that R is uniform on the interval ( 0.04, 0.08 ) and <math>V = 10, 000e^R</math> Therefore, the distribution function of <math>V</math> is give...") |
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\begin{align*} | \begin{align*} | ||
F(v) = \operatorname{P}[V \leq v ] &= \operatorname{P}[10000 e^R \leq v ] = \operatorname{P}[R \leq ln(v) - \ln(10000) ] \\ | F(v) = \operatorname{P}[V \leq v ] &= \operatorname{P}[10000 e^R \leq v ] \\ &= \operatorname{P}[R \leq \ln(v) - \ln(10000) ] \\ | ||
&= \frac{1}{0.04} \int_{0.04}^{\ln(v)-\ln(10000)} dr \\ &= \frac{1}{0.04}r \Big |_{0.04}^{\ln(v)-\ln(10000)} = 25\ln(v) - 25\ln(10000) - 1 \\ | &= \frac{1}{0.04} \int_{0.04}^{\ln(v)-\ln(10000)} dr \\ &= \frac{1}{0.04}r \Big |_{0.04}^{\ln(v)-\ln(10000)}\\ | ||
&= 25\ln(v) - 25\ln(10000) - 1 \\ | |||
&= 25 \left [ \ln(\frac{v}{10000}) - 0.04\right ]. | &= 25 \left [ \ln(\frac{v}{10000}) - 0.04\right ]. | ||
\end{align*} | \end{align*} | ||
Latest revision as of 18:18, 4 May 2023
Solution: E
We are given that R is uniform on the interval ( 0.04, 0.08 ) and [math]V = 10, 000e^R[/math] Therefore, the distribution function of [math]V[/math] is given by
[[math]]
\begin{align*}
F(v) = \operatorname{P}[V \leq v ] &= \operatorname{P}[10000 e^R \leq v ] \\ &= \operatorname{P}[R \leq \ln(v) - \ln(10000) ] \\
&= \frac{1}{0.04} \int_{0.04}^{\ln(v)-\ln(10000)} dr \\ &= \frac{1}{0.04}r \Big |_{0.04}^{\ln(v)-\ln(10000)}\\
&= 25\ln(v) - 25\ln(10000) - 1 \\
&= 25 \left [ \ln(\frac{v}{10000}) - 0.04\right ].
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.