An insurance policy reimburses a loss up to a benefit limit of 10. The policyholder’s loss, [math]Y[/math], follows a distribution with density function:

[[math]] f(y) = \begin{cases} 2y^{-3}, \, y \gt 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected value of the benefit paid under the insurance policy.

• 1.0
• 1.3
• 1.8
• 1.9
• 2.0

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

Step 1: Define the Benefit Payment and Loss Distribution

The insurance policy reimburses a loss, [math]Y[/math], up to a benefit limit of 10. The policyholder's loss, [math]Y[/math], follows a distribution with the density function:

[[math]] f(y) = \begin{cases} 2y^{-3}, \, y \gt 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Let [math]W[/math] denote the benefit paid under the policy. The benefit paid is the minimum of the actual loss [math]Y[/math] and the benefit limit of 10. This can be expressed as:

[[math]] W = \min(Y, 10) = \begin{cases} y, & 1 \lt y \lt 10 \\ 10, & y \geq 10 \end{cases} [[/math]]

Step 2: Set Up the Expected Value Integral

The expected value of the benefit paid, [math]\operatorname{E}[W][/math], is calculated by integrating the product of the benefit amount and the probability density function over the entire range of [math]Y[/math]. Since [math]W[/math] has different definitions for different ranges of [math]Y[/math], we split the integral into two parts:

[[math]] \operatorname{E}[W] = \int_1^{10} W \cdot f(y) \, dy + \int_{10}^{\infty} W \cdot f(y) \, dy [[/math]]

Substitute the definitions of [math]W[/math] and [math]f(y)[/math]:

[[math]] \operatorname{E}[W] = \int_1^{10} y \left(2y^{-3}\right) \, dy + \int_{10}^{\infty} 10 \left(2y^{-3}\right) \, dy [[/math]]

Simplify the integrands:

[[math]] \operatorname{E}[W] = \int_1^{10} 2y^{-2} \, dy + \int_{10}^{\infty} 20y^{-3} \, dy [[/math]]

Step 3: Evaluate Each Integral

We will evaluate each integral separately. First Integral: [math]\int_1^{10} 2y^{-2} \, dy[/math] The antiderivative of [math]2y^{-2}[/math] is [math]2 \frac{y^{-1}}{-1} = -2y^{-1} = -\frac{2}{y}[/math].

[[math]] \int_1^{10} 2y^{-2} \, dy = \left[ -\frac{2}{y} \right]_1^{10} = \left(-\frac{2}{10}\right) - \left(-\frac{2}{1}\right) = -0.2 - (-2) = -0.2 + 2 = 1.8 [[/math]]

Second Integral: [math]\int_{10}^{\infty} 20y^{-3} \, dy[/math] The antiderivative of [math]20y^{-3}[/math] is [math]20 \frac{y^{-2}}{-2} = -10y^{-2} = -\frac{10}{y^2}[/math].

[[math]] \int_{10}^{\infty} 20y^{-3} \, dy = \left[ -\frac{10}{y^2} \right]_{10}^{\infty} = \left(\lim_{y \to \infty} -\frac{10}{y^2}\right) - \left(-\frac{10}{10^2}\right) = 0 - \left(-\frac{10}{100}\right) = 0 - (-0.1) = 0.1 [[/math]]

Step 4: Calculate the Total Expected Benefit

Sum the results from the two integrals to find the total expected value of the benefit paid:

[[math]] \operatorname{E}[W] = 1.8 + 0.1 = 1.9 [[/math]]

Thus, the expected value of the benefit paid under the insurance policy is 1.9.

Key Insights

• When an insurance policy has a benefit limit, the payment function [math]W[/math] is defined as [math]\min(Y, L)[/math], where [math]Y[/math] is the loss and [math]L[/math] is the limit. This means the payment equals the loss if the loss is below the limit, and equals the limit if the loss is at or above the limit.
• Calculating the expected value of a benefit with a limit requires splitting the integral into multiple parts, corresponding to the different definitions of the benefit payment. The integration limits are determined by the policy limit and the domain of the loss distribution.
• For loss distributions defined over an infinite range, improper integrals are often required. Correctly evaluating these involves taking limits as the integration bound approaches infinity.
• Proficiency in basic integration techniques, especially for power functions (e.g., [math]\int x^n \, dx[/math]), is crucial for solving problems involving continuous loss distributions.

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Solution: D

Let [math]W[/math] denote claim payments. Then

[[math]] W = \begin{cases} y, \quad 1 \lt y \lt 10 \\ 10, \quad y \leq 10 \end{cases} [[/math]]

It follows that

[[math]] \operatorname{E}[W] = \int_1^{10} y \frac{2}{y^3} dy + \int_{10}^{\infty} 10 \frac{2}{y^3} dy = -\frac{2}{y} \Big |_1^{10} - \frac{10}{y^2} \Big |_{10}^{\infty} = 1.9. [[/math]]

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