An investment account earns an annual interest rate [math]R[/math] that follows a uniform distribution on the interval (0.04, 0.08). The value of a 10,000 initial investment in this account after one year is given by [math]V = 10 000e^R. [/math] Let [math]F[/math] be the cumulative distribution function of [math]V[/math]. Determine [math]F(v)[/math] for values of [math]v[/math] that satisfy [math]0 \lt F(v) \lt 1 [/math].

• [math]\frac{10000e^{v /10000} − 10408}{425}[/math]
• [math]25e^{v/10000} - 0.04[/math]
• [math]\frac{v-10408}{10833-10408}[/math]
• [math]\frac{25}{v}[/math]
• [math]25\left[ \ln(v/10000) - 0.04 \right][/math]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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.