Let [math]T[/math] denote the time in minutes for a customer service representative to respond to 10 telephone inquiries. [math]T[/math] is uniformly distributed on the interval with endpoints 8 minutes and 12 minutes. Let [math]R[/math] denote the average rate, in customers per minute, at which the representative responds to inquiries, and let [math]f(r)[/math] be the density function for [math]R[/math].

Determine [math]f(r)[/math], for [math]\frac{10}{12} \lt r \lt \frac{10}{8}[/math].

• 12/5
• 3-5/2r
• [math]3r-5\ln(r)/2[/math]
• [math]\frac{10}{r^2}[/math]
• [math]\frac{5}{2r^2}[/math]

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

Step 1: Understand the Random Variables and Their Distributions

The random variable [math]T[/math] represents the time in minutes for a customer service representative to respond to 10 telephone inquiries. [math]T[/math] is uniformly distributed on the interval [8, 12] minutes. Its probability density function (PDF) is:

[[math]]f_T(t) = \begin{cases} \frac{1}{12 - 8} = \frac{1}{4} & \text{for } 8 \le t \le 12 \\ 0 & \text{otherwise} \end{cases}[[/math]]

The random variable [math]R[/math] denotes the average rate, in customers per minute, at which the representative responds to inquiries. The relationship between [math]R[/math] and [math]T[/math] is given by:

[[math]]R = \frac{\text{Number of Inquiries}}{\text{Time}} = \frac{10}{T}[[/math]]

To determine the range of [math]R[/math], we use the domain of [math]T[/math]:

• When [math]T = 8[/math] (the minimum time), [math]R = \frac{10}{8}[/math] (the maximum rate).
• When [math]T = 12[/math] (the maximum time), [math]R = \frac{10}{12}[/math] (the minimum rate).

Thus, for [math]T \in [8, 12][/math], the rate [math]R[/math] is in the interval [math]\left[\frac{10}{12}, \frac{10}{8}\right][/math].

Step 2: Derive the Cumulative Distribution Function (CDF) of R

To find the probability density function (PDF) [math]f_R(r)[/math], we first derive the cumulative distribution function (CDF) [math]F_R(r)[/math]. By definition, [math]F_R(r) = P[R \le r][/math]. Substituting the relationship [math]R = \frac{10}{T}[/math]:

[[math]]F_R(r) = P\left[\frac{10}{T} \le r\right][[/math]]

Since [math]T[/math] is always positive (time cannot be negative) and [math]r[/math] is positive within its domain, we can rearrange the inequality. Multiplying both sides by [math]T[/math] and dividing by [math]r[/math] (which is positive, so the inequality sign remains the same) gives:

[[math]]P\left[T \ge \frac{10}{r}\right][[/math]]

This probability can be expressed using the CDF of [math]T[/math], [math]F_T(t)[/math]:

[[math]]P\left[T \ge \frac{10}{r}\right] = 1 - P\left[T \lt \frac{10}{r}\right] = 1 - F_T\left(\frac{10}{r}\right)[[/math]]

Therefore, the CDF of [math]R[/math] is:

[[math]]F_R(r) = 1 - F_T\left(\frac{10}{r}\right)[[/math]]

Step 3: Determine the Probability Density Function (PDF) of R

The PDF [math]f_R(r)[/math] is found by differentiating [math]F_R(r)[/math] with respect to [math]r[/math]:

[[math]]f_R(r) = \frac{d}{dr} F_R(r) = \frac{d}{dr} \left(1 - F_T\left(\frac{10}{r}\right)\right)[[/math]]

Applying the chain rule, where [math]G(t) = F_T(t)[/math] and [math]h(r) = \frac{10}{r}[/math]:

[[math]]\frac{d}{dr} \left(-F_T\left(\frac{10}{r}\right)\right) = - f_T\left(\frac{10}{r}\right) \cdot \frac{d}{dr}\left(\frac{10}{r}\right)[[/math]]

We know that [math]\frac{d}{dr}\left(\frac{10}{r}\right) = -10r^{-2} = -\frac{10}{r^2}[/math]. Substituting this and [math]f_T(t) = \frac{1}{4}[/math] (from Step 1) into the expression:

[[math]]f_R(r) = -\left(\frac{1}{4}\right) \cdot \left(-\frac{10}{r^2}\right) = \frac{10}{4r^2} = \frac{5}{2r^2}[[/math]]

This result is valid for the range [math]\frac{10}{12} \lt r \lt \frac{10}{8}[/math]. Otherwise, [math]f_R(r) = 0[/math].

Key Insights

• The CDF method is a reliable technique for finding the probability density function (PDF) of a transformed random variable [math]R = g(T)[/math]. It involves finding [math]F_R(r) = P[R \le r][/math] and then differentiating [math]F_R(r)[/math] to get [math]f_R(r)[/math].
• When dealing with inverse transformations like [math]R = \frac{10}{T}[/math], it's crucial to correctly manipulate inequalities. For positive [math]T[/math] and [math]r[/math], [math]\frac{10}{T} \le r[/math] implies [math]T \ge \frac{10}{r}[/math].
• The chain rule is essential for differentiating the CDF of the transformed variable: if [math]F_R(r) = 1 - F_T(h(r))[/math], then [math]f_R(r) = -f_T(h(r)) \cdot h'(r)[/math].
• Always establish the correct domain for the transformed random variable [math]R[/math] based on the domain of the original variable [math]T[/math] to fully define the PDF.
• Recall that for a uniform distribution on [math][a, b][/math], its PDF is a constant [math]\frac{1}{b-a}[/math] within the interval.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: E

First note R = 10/T . Then

[[math]] F_R(r) = \operatorname{P}[R ≤ r] = \operatorname{P}[\frac{10}{T} \leq r ] = \operatorname{P}[T \geq \frac{10}{r}] = 1 - F_T(\frac{10}{r}). [[/math]]

Differentiating with respect to

[[math]] \begin{align*} r f_R(r) = F^{'}_R(r) = d/dr (1 - F_T(\frac{10}{r})) = -(\frac{d}{dt}F_T(t)) (\frac{-10}{r^2}) \\ \frac{d}{dt}F_T(t) = f_T(t) = \frac{1}{4} \end{align*} [[/math]]

since [math]T[/math] is uniformly distributed on [8,12]. Therefore

[[math]] f_R(r) = \frac{-1}{4} (\frac{-10}{r^2}) = \frac{5}{2r^2}. [[/math]]

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