Let [math]X[/math] be a continuous random variable with density function

[[math]] f(x) = \begin{cases} \frac{p-1}{x^p}, \, x \gt 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate the value of [math]p[/math] such that [math]\operatorname{E}(X) = 2 [/math].

• 1
• 2.5
• 3
• 5
• There is no such [math]p[/math]

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

Step 1: Define the Expectation and Set Up the Integral

The expectation of a continuous random variable [math]X[/math] with probability density function [math]f(x)[/math] is given by the formula:

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

Given the density function [math]f(x)[/math] is [math]\frac{p-1}{x^p}[/math] for [math]x \gt 1[/math] and [math]0[/math] otherwise, we set up the integral for [math]\operatorname{E}(X)[/math] as follows:

[[math]]\operatorname{E}(X) = \int_{1}^{\infty} x \cdot \frac{p-1}{x^p} dx[[/math]]

Step 2: Evaluate the Indefinite Integral

First, simplify the integrand and factor out the constant [math](p-1)[/math]:

[[math]]\operatorname{E}(X) = (p-1) \int_{1}^{\infty} x^{1-p} dx[[/math]]

For this integral to converge at infinity, the exponent [math]1-p[/math] must be less than [math]-1[/math]. This implies [math]p \gt 2[/math]. If [math]p \le 2[/math], the expectation would not exist. Now, evaluate the indefinite integral using the power rule:

[[math]]\int x^{1-p} dx = \frac{x^{1-p+1}}{1-p+1} = \frac{x^{2-p}}{2-p}[[/math]]

Step 3: Apply the Limits of Integration

Substitute the limits of integration from [math]1[/math] to [math]\infty[/math] into the evaluated integral:

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

Given that [math]p \gt 2[/math], the exponent [math]2-p[/math] is negative. Therefore, as [math]x \to \infty[/math], [math]x^{2-p} \to 0[/math]. At [math]x = 1[/math], [math]1^{2-p} = 1[/math].

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

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

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

Step 4: Solve for [math]p[/math]

We are given that [math]\operatorname{E}(X) = 2[/math]. Set the derived expression for [math]\operatorname{E}(X)[/math] equal to [math]2[/math]:

[[math]]\frac{p-1}{p-2} = 2[[/math]]

Now, solve the equation for [math]p[/math]:

[[math]]p-1 = 2(p-2)[[/math]]

[[math]]p-1 = 2p-4[[/math]]

[[math]]3 = p[[/math]]

The obtained value [math]p=3[/math] satisfies the condition [math]p \gt 2[/math] for the expectation to exist, which confirms the validity of our solution.

Key Insights

• The expectation of a continuous random variable is calculated by integrating [math]x \cdot f(x)[/math] over the variable's support.
• For improper integrals (integrals with infinite limits) to converge, specific conditions on the integrand's exponent must be met. In this case, for [math]\operatorname{E}(X)[/math] to exist, [math]p[/math] must be greater than [math]2[/math].
• The power rule of integration ([math]\int x^n dx = \frac{x^{n+1}}{n+1}[/math] for [math]n \ne -1[/math]) is fundamental for evaluating integrals of polynomial terms.
• Solving for parameters in a probability distribution often involves algebraic manipulation after setting a known statistical property (like the mean) to its calculated expression.
• Always verify that the obtained parameter value satisfies any necessary conditions for the distribution or its moments to be well-defined.

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Solution: C

[[math]] \begin{align*} \operatorname{E}[X] = \int_{1}^{\infty} x \frac{p-1}{x^p} dx &= (p-1) \int_{1}^{\infty} x^{1-p} dx \\ &= (p-1) \frac{x^{2-p}}{2-p} \Big |_1^{\infty} \\ &= \frac{p-1}{p-2} = 2. \end{align*} [[/math]]

Hence [math]p = 3[/math].

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