A manufacturer’s annual losses follow a distribution with density function

[[math]] f(x) = \begin{cases} \frac{2.5(0.6)^{2.5}}{x^{3.5}}, \, x \gt 0.6 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

To cover its losses, the manufacturer purchases an insurance policy with an annual deductible of 2. Calculate the mean of the manufacturer’s annual losses not paid by the insurance policy.

• 0.84
• 0.88
• 0.93
• 0.95
• 1.00

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

Step 1: Understand the Loss Distribution and Deductible

The manufacturer's annual losses, denoted by [math]X[/math], follow a distribution with the given probability density function (PDF):

[[math]] f(x) = \begin{cases} \frac{2.5(0.6)^{2.5}}{x^{3.5}}, & x \gt 0.6 \\ 0, & \text{Otherwise.} \end{cases} [[/math]]

The insurance policy has an annual deductible of [math]D = 2[/math]. This means the manufacturer is responsible for the first 2 units of loss. We need to calculate the mean of the manufacturer’s annual losses not paid by the insurance policy. This is equivalent to calculating the expected retained loss by the manufacturer.

Step 2: Formulate the Expected Retained Loss

The expected amount of loss retained by the manufacturer (the amount not paid by the insurance policy) is calculated as follows:

• If the loss [math]X[/math] is less than or equal to the deductible [math]D[/math] ([math]X \le D[/math]), the manufacturer pays the entire loss [math]X[/math].
• If the loss [math]X[/math] is greater than the deductible [math]D[/math] ([math]X \gt D[/math]), the manufacturer pays only the deductible amount [math]D[/math].

Therefore, the expected retained loss [math]E[R][/math] is given by the integral:

[[math]] E[R] = \int_{\text{support of } X \text{ up to } D} x f(x) \, dx + \int_D^{\infty} D f(x) \, dx [[/math]]

Given that the support of [math]X[/math] starts at [math]0.6[/math] and the deductible [math]D = 2[/math], the formula becomes:

[[math]] E[R] = \int_{0.6}^2 x f(x) \, dx + \int_2^{\infty} 2 f(x) \, dx [[/math]]

Step 3: Substitute the Probability Density Function

Now, we substitute the given PDF [math]f(x) = \frac{2.5(0.6)^{2.5}}{x^{3.5}}[/math] into the integral expression:

[[math]] E[R] = \int_{0.6}^2 x \left( \frac{2.5(0.6)^{2.5}}{x^{3.5}} \right) \, dx + \int_2^{\infty} 2 \left( \frac{2.5(0.6)^{2.5}}{x^{3.5}} \right) \, dx [[/math]]

We can simplify the integrands by combining [math]x[/math] terms and factoring out constants:

[[math]] E[R] = 2.5(0.6)^{2.5} \int_{0.6}^2 x^{1 - 3.5} \, dx + \left( 2 \times 2.5(0.6)^{2.5} \right) \int_2^{\infty} x^{-3.5} \, dx [[/math]]

[[math]] E[R] = 2.5(0.6)^{2.5} \int_{0.6}^2 x^{-2.5} \, dx + 5(0.6)^{2.5} \int_2^{\infty} x^{-3.5} \, dx [[/math]]

Step 4: Evaluate the Integrals

Now we evaluate each integral separately.

• First Integral: [math]2.5(0.6)^{2.5} \int_{0.6}^2 x^{-2.5} \, dx[/math]

The antiderivative of [math]x^{-2.5}[/math] is [math]\frac{x^{-1.5}}{-1.5}[/math].

[[math]] 2.5(0.6)^{2.5} \left[ \frac{x^{-1.5}}{-1.5} \right]_{0.6}^2 = 2.5(0.6)^{2.5} \left( \frac{2^{-1.5}}{-1.5} - \frac{0.6^{-1.5}}{-1.5} \right) [[/math]]

[[math]] = 2.5(0.6)^{2.5} \left( \frac{0.6^{-1.5}}{1.5} - \frac{2^{-1.5}}{1.5} \right) [[/math]]

• Second Integral: [math]5(0.6)^{2.5} \int_2^{\infty} x^{-3.5} \, dx[/math]

The antiderivative of [math]x^{-3.5}[/math] is [math]\frac{x^{-2.5}}{-2.5}[/math].

[[math]] 5(0.6)^{2.5} \left[ \frac{x^{-2.5}}{-2.5} \right]_2^{\infty} = 5(0.6)^{2.5} \left( \lim_{b \to \infty} \frac{b^{-2.5}}{-2.5} - \frac{2^{-2.5}}{-2.5} \right) [[/math]]

Since [math]\lim_{b \to \infty} b^{-2.5} = 0[/math]:

[[math]] = 5(0.6)^{2.5} \left( 0 - \frac{2^{-2.5}}{-2.5} \right) = 5(0.6)^{2.5} \left( \frac{2^{-2.5}}{2.5} \right) [[/math]]

[[math]] = 2(0.6)^{2.5} 2^{-2.5} [[/math]]

Step 5: Calculate the Total Expected Retained Loss

Now, we sum the results from both integrals:

[[math]] E[R] = 2.5(0.6)^{2.5} \left( \frac{0.6^{-1.5}}{1.5} - \frac{2^{-1.5}}{1.5} \right) + 2(0.6)^{2.5} 2^{-2.5} [[/math]]

We can factor out [math](0.6)^{2.5}[/math]:

[[math]] E[R] = (0.6)^{2.5} \left[ 2.5 \left( \frac{0.6^{-1.5}}{1.5} - \frac{2^{-1.5}}{1.5} \right) + 2 \times 2^{-2.5} \right] [[/math]]

Substituting the numerical values:

• [math]0.6^{2.5} \approx 0.2788545[/math]
• [math]0.6^{-1.5} \approx 2.1516515[/math]
• [math]2^{-1.5} \approx 0.3535534[/math]
• [math]2^{-2.5} \approx 0.1767767[/math]

[[math]] E[R] \approx 0.2788545 \left[ 2.5 \left( \frac{2.1516515}{1.5} - \frac{0.3535534}{1.5} \right) + 2 \times 0.1767767 \right] [[/math]]

[[math]] E[R] \approx 0.2788545 \left[ 2.5 \left( 1.4344343 - 0.2357023 \right) + 0.3535534 \right] [[/math]]

[[math]] E[R] \approx 0.2788545 \left[ 2.5 \left( 1.198732 \right) + 0.3535534 \right] [[/math]]

[[math]] E[R] \approx 0.2788545 \left[ 2.99683 + 0.3535534 \right] [[/math]]

[[math]] E[R] \approx 0.2788545 \left[ 3.3503834 \right] [[/math]]

[[math]] E[R] \approx 0.9343 [[/math]]

The mean of the manufacturer’s annual losses not paid by the insurance policy is approximately 0.93.

Key Insights

• The "losses not paid by the insurance policy" refers to the amount retained by the insured, which is the loss amount up to the deductible and the deductible amount for losses exceeding it.
• The expected retained loss [math]E[R][/math] for a deductible [math]D[/math] is calculated using two integrals: [math]\int_0^D x f(x) \, dx + \int_D^{\infty} D f(x) \, dx[/math], where the lower limit of the first integral is the start of the loss distribution's support.
• Careful application of integration rules for power functions and evaluation of definite and improper integrals is crucial.
• Constants within the probability density function and the deductible amount must be correctly incorporated into the integration process.

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

Solution: C

The expected payment is

[[math]] \begin{align*} \int_{0.6}^2 x [\frac{2.5(0.6)^{2.5}}{x^{3.5}}] \, dx + \int_2^{\infty} [\frac{2.5(0.6)^{2.5}}{x^{3.5}}] \, dx &= 2.5(0.6)^{3.5} \left( \frac{-x^{1.5}}{1.5} \Big |_{0.6}^{2} + \frac{-x^{2.5}}{2.5} \Big |_2^{\infty} \right) \\ &= 2.5(0.6)^{2.5} \left ( \frac{-2^{-1.5}}{1.5} + \frac{0.6^{-1.5}}{1.5} + 2 \frac{2^{-2.5}}{2.5}\right ) \\ &= 0.9343. \end{align*} [[/math]]

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