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.

Step 1: Identify Given Variables and Distributions

The problem provides information about an investment account and its interest rate, [math]R[/math], as well as the value of an initial investment, [math]V[/math].

• Interest Rate ([math]R[/math]): [math]R[/math] follows a uniform distribution on the interval [math](0.04, 0.08)[/math]. The probability density function (PDF) for [math]R[/math] is:

[[math]]f_R(r) = \frac{1}{b-a} = \frac{1}{0.08 - 0.04} = \frac{1}{0.04} = 25 \quad \text{for } 0.04 \lt r \lt 0.08[[/math]]

• Investment Value ([math]V[/math]): The value of a $10,000 initial investment after one year is given by [math]V = 10 000e^R[/math].
• Objective: Determine the cumulative distribution function (CDF) of [math]V[/math], denoted as [math]F(v)[/math], for values of [math]v[/math] that satisfy [math]0 \lt F(v) \lt 1[/math].

Step 2: Express the CDF of [math]V[/math] in Terms of [math]R[/math]

The cumulative distribution function (CDF) of [math]V[/math], denoted as [math]F(v)[/math], is defined as the probability that [math]V[/math] takes on a value less than or equal to [math]v[/math]. We substitute the given expression for [math]V[/math] and rearrange the inequality to isolate [math]R[/math].

[[math]] \begin{align*} F(v) &= \operatorname{P}[V \leq v] \\ &= \operatorname{P}[10000 e^R \leq v] \end{align*} [[/math]]

To isolate [math]R[/math], we divide by 10,000 and then take the natural logarithm of both sides:

[[math]] \begin{align*} e^R &\leq \frac{v}{10000} \\ R &\leq \ln\left(\frac{v}{10000}\right) \\ R &\leq \ln(v) - \ln(10000) \end{align*} [[/math]]

So, the CDF of [math]V[/math] can be expressed as:

[[math]] F(v) = \operatorname{P}\left[R \leq \ln(v) - \ln(10000)\right] [[/math]]

Step 3: Integrate the PDF of [math]R[/math]

Since [math]R[/math] follows a uniform distribution on [math](0.04, 0.08)[/math] with PDF [math]f_R(r) = 25[/math] for [math]0.04 \lt r \lt 0.08[/math], the probability [math]\operatorname{P}[R \leq x][/math] for a given value [math]x[/math] is found by integrating the PDF from the lower bound of the distribution up to [math]x[/math]. In our case, [math]x = \ln(v) - \ln(10000)[/math]. For [math]0 \lt F(v) \lt 1[/math], this value of [math]x[/math] must fall within the open interval [math](0.04, 0.08)[/math]. The integral for [math]F(v)[/math] is:

[[math]] F(v) = \int_{0.04}^{\ln(v)-\ln(10000)} f_R(r) \, dr = \int_{0.04}^{\ln(v)-\ln(10000)} 25 \, dr [[/math]]

Step 4: Evaluate the Integral and Simplify the Expression

Now, we evaluate the definite integral from Step 3:

[[math]] \begin{align*} F(v) &= \int_{0.04}^{\ln(v)-\ln(10000)} 25 \, dr \\ &= 25r \Big |_{0.04}^{\ln(v)-\ln(10000)} \\ &= 25 \left( \left[ \ln(v) - \ln(10000) \right] - 0.04 \right) \\ &= 25 \left[ \ln\left(\frac{v}{10000}\right) - 0.04 \right] \end{align*} [[/math]]

This gives us the cumulative distribution function [math]F(v)[/math] for [math]V[/math]. To satisfy the condition [math]0 \lt F(v) \lt 1[/math], the argument of the logarithm, [math]\ln(v/10000)[/math], must be within the range of [math]R[/math], i.e., [math]0.04 \lt \ln(v/10000) \lt 0.08[/math]. This implies that [math]v[/math] must be in the interval [math](10000e^{0.04}, 10000e^{0.08})[/math]. Numerically, this corresponds to [math]v \in (10408.1077, 10832.8707)[/math].

Key Insights

• The cumulative distribution function (CDF) [math]F_Y(y)[/math] of a transformed random variable [math]Y = g(X)[/math] can be found by expressing [math]\operatorname{P}[Y \leq y][/math] as [math]\operatorname{P}[g(X) \leq y][/math] and then solving for [math]X \leq g^{-1}(y)[/math].
• For a uniformly distributed random variable [math]X \sim U(a, b)[/math], its PDF is [math]f_X(x) = \frac{1}{b-a}[/math] for [math]a \lt x \lt b[/math]. The CDF is found by integrating the PDF from [math]a[/math] to [math]x[/math]: [math]F_X(x) = \int_a^x f_X(t) dt = \frac{x-a}{b-a}[/math] for [math]a \lt x \lt b[/math].
• When dealing with exponential transformations like [math]V = Ke^R[/math], the natural logarithm is the inverse function, i.e., [math]R = \ln(V/K)[/math]. This is a crucial step in transforming the inequality to find the CDF.
• Logarithm properties, such as [math]\ln(a/b) = \ln(a) - \ln(b)[/math], are often useful for simplifying expressions involving logarithms.
• The condition [math]0 \lt F(v) \lt 1[/math] implies that the argument of the CDF must fall within the open interval of the underlying random variable's range. For [math]R \sim U(a, b)[/math], this means [math]a \lt \ln(v/10000) \lt b[/math].

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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.