An insurance company’s monthly claims are modeled by a continuous, positive random variable [math]X[/math], whose probability density function is proportional to [math](1 + x)^{- 4}[/math] for [math] x \gt 0 [/math] .

Calculate the company’s expected monthly claims.

• 1/6
• 1/3
• 1/2
• 1
• 3

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

Step 1: Define the Probability Density Function (PDF)

The problem states that the probability density function (PDF) for the monthly claims, denoted by [math]X[/math], is proportional to [math](1 + x)^{-4}[/math] for [math]x \gt 0[/math]. We can express the PDF as:

[[math]]f(x) = \frac{k}{(1+x)^4}, \quad x \gt 0[[/math]]

where [math]k[/math] is a normalization constant that ensures the total probability integrates to 1.

Step 2: Determine the Normalization Constant [math]k[/math]

For [math]f(x)[/math] to be a valid PDF, its integral over its entire domain must equal 1.

[[math]]\int_0^{\infty} f(x) dx = 1[[/math]]

Substituting the expression for [math]f(x)[/math]:

[[math]]\int_0^{\infty} \frac{k}{(1+x)^4} dx = 1[[/math]]

We can pull the constant [math]k[/math] out of the integral:

[[math]]k \int_0^{\infty} (1+x)^{-4} dx = 1[[/math]]

Now, we evaluate the integral. Let [math]u = 1+x[/math], so [math]du = dx[/math]. When [math]x=0[/math], [math]u=1[/math]. When [math]x \to \infty[/math], [math]u \to \infty[/math].

[[math]]k \int_1^{\infty} u^{-4} du = 1[[/math]]

Integrating [math]u^{-4}[/math] with respect to [math]u[/math] gives [math]\frac{u^{-3}}{-3}[/math]:

[[math]]k \left[ \frac{u^{-3}}{-3} \right]_1^{\infty} = 1[[/math]]

[[math]]k \left[ -\frac{1}{3u^3} \right]_1^{\infty} = 1[[/math]]

Evaluating the limits:

[[math]]k \left[ \left( \lim_{u \to \infty} -\frac{1}{3u^3} \right) - \left( -\frac{1}{3(1)^3} \right) \right] = 1[[/math]]

[[math]]k \left[ 0 - \left( -\frac{1}{3} \right) \right] = 1[[/math]]

[[math]]k \left( \frac{1}{3} \right) = 1[[/math]]

Therefore, the normalization constant is:

[[math]]k = 3[[/math]]

The complete PDF is [math]f(x) = \frac{3}{(1+x)^4}[/math].

Step 3: Set up the Expected Value Integral

The expected value of a continuous random variable [math]X[/math] is given by the formula:

[[math]]\operatorname{E}[X] = \int_{-\infty}^{\infty} x f(x) dx[[/math]]

For this problem, the domain of [math]X[/math] is [math](0, \infty)[/math], so we have:

[[math]]\operatorname{E}[X] = \int_0^{\infty} x f(x) dx[[/math]]

Substitute the derived PDF [math]f(x) = \frac{3}{(1+x)^4}[/math]:

[[math]]\operatorname{E}[X] = \int_0^{\infty} x \cdot \frac{3}{(1+x)^4} dx[[/math]]

[[math]]\operatorname{E}[X] = 3 \int_0^{\infty} \frac{x}{(1+x)^4} dx[[/math]]

Step 4: Evaluate the Expected Value Integral

To evaluate this integral, we use a substitution. Let [math]u = 1+x[/math]. This implies [math]x = u-1[/math] and [math]du = dx[/math]. The limits of integration also change:

• When [math]x=0[/math], [math]u = 1+0 = 1[/math].
• When [math]x \to \infty[/math], [math]u \to \infty[/math].

Substituting these into the integral:

[[math]]\operatorname{E}[X] = 3 \int_1^{\infty} \frac{u-1}{u^4} du[[/math]]

Now, we can split the integrand:

[[math]]\operatorname{E}[X] = 3 \int_1^{\infty} \left( \frac{u}{u^4} - \frac{1}{u^4} \right) du[[/math]]

[[math]]\operatorname{E}[X] = 3 \int_1^{\infty} (u^{-3} - u^{-4}) du[[/math]]

Next, we find the antiderivative of each term:

• [math]\int u^{-3} du = \frac{u^{-2}}{-2} = -\frac{1}{2u^2}[/math]
• [math]\int u^{-4} du = \frac{u^{-3}}{-3} = -\frac{1}{3u^3}[/math]

So, the definite integral becomes:

[[math]]\operatorname{E}[X] = 3 \left[ -\frac{1}{2u^2} - \left(-\frac{1}{3u^3}\right) \right]_1^{\infty}[[/math]]

[[math]]\operatorname{E}[X] = 3 \left[ -\frac{1}{2u^2} + \frac{1}{3u^3} \right]_1^{\infty}[[/math]]

Now, we evaluate at the limits:

[[math]]\operatorname{E}[X] = 3 \left[ \left( \lim_{u \to \infty} \left( -\frac{1}{2u^2} + \frac{1}{3u^3} \right) \right) - \left( -\frac{1}{2(1)^2} + \frac{1}{3(1)^3} \right) \right][[/math]]

[[math]]\operatorname{E}[X] = 3 \left[ (0 + 0) - \left( -\frac{1}{2} + \frac{1}{3} \right) \right][[/math]]

[[math]]\operatorname{E}[X] = 3 \left[ - \left( -\frac{3}{6} + \frac{2}{6} \right) \right][[/math]]

[[math]]\operatorname{E}[X] = 3 \left[ - \left( -\frac{1}{6} \right) \right][[/math]]

[[math]]\operatorname{E}[X] = 3 \left( \frac{1}{6} \right)[[/math]]

[[math]]\operatorname{E}[X] = \frac{3}{6} = \frac{1}{2}[[/math]]

Thus, the company's expected monthly claims are [math]\frac{1}{2}[/math].

Summary of Results

The calculated expected monthly claims are [math]\frac{1}{2}[/math].

Key Intermediate and Final Results

Description	Value
Normalization Constant [math]k[/math]	[math]3[/math]
Probability Density Function [math]f(x)[/math]	[math]\frac{3}{(1+x)^4}[/math]
Expected Monthly Claims [math]\operatorname{E}[X][/math]	[math]\frac{1}{2}[/math]

Key Insights

• To define a valid probability density function (PDF) from a proportional relationship, a normalization constant [math]k[/math] must be determined such that the integral of the PDF over its entire domain equals 1.
• The expected value [math]\operatorname{E}[X][/math] for a continuous positive random variable [math]X[/math] is calculated by integrating [math]x \cdot f(x)[/math] over the variable's domain.
• Techniques like substitution (e.g., [math]u = 1+x[/math]) are crucial for simplifying integrals, especially those involving rational functions or powers of binomials. Remember to adjust integration limits accordingly.
• Careful evaluation of definite integrals, including handling limits at infinity, is essential to avoid sign errors and ensure accurate results.

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

Solution: C

The pdf of x is given by [math]f(x) = \frac{k}{(1+x)^4}, \, 0 \lt x \lt \infty [/math]. To find [math]k[/math], note

[[math]] 1 = \int_0^{\infty} \frac{k}{(1+x)^4} dx = - \frac{k}{3} \frac{1}{(1+x)^3} \Big |_{0}^{\infty} = \frac{k}{3} [[/math]]

which implies that [math]k = 3 [/math]. It then follows that

[[math]] \operatorname{E}[x] = \int_0^{\infty} \frac{3x}{(1+x)^4} dx [[/math]]

and substituting [math]u = 1+x, du = dx [/math], we see

[[math]] \begin{align*} \operatorname{E}[x] = \int_1^{\infty} \frac{3(u-1)}{u^4} du &= 3 \int_{1}^{\infty} (u^{-3} - u^{-4}) du \\ &= 3\left[\frac{u^{-2}}{-2} - \frac{u^{-3}}{-3}\right]_0^{\infty} \\ &= 3\left[\frac{1}{2}-\frac{1}{3}\right] = 3/2 - 1 = 1/2. \end{align*} [[/math]]

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