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

[[math]] f(x) = \begin{cases} \frac{|x|}{10}, \, -2 \leq x \leq 4 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected value of [math]X[/math].

• 1/5
• 3/5
• 1
• 28/15
• 12/5

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

Step 1: Define the Expected Value Formula

The expected value [math]\operatorname{E}[X][/math] of a continuous random variable [math]X[/math] with a probability density function [math]f(x)[/math] is defined by the integral:

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

Given the density function [math]f(x) = \frac{|x|}{10}[/math] for [math]-2 \leq x \leq 4[/math] and [math]0[/math] otherwise, the integration limits are from [math]-2[/math] to [math]4[/math]. Thus, we need to calculate:

[[math]]\operatorname{E}[X] = \int_{-2}^{4} x \cdot \frac{|x|}{10} \, dx[[/math]]

Step 2: Split the Integral Based on the Absolute Value Function

The absolute value function [math]|x|[/math] changes its definition at [math]x=0[/math]. Therefore, we must split the integral into two parts:

• For [math]-2 \leq x \lt 0[/math], [math]|x| = -x[/math].
• For [math]0 \leq x \leq 4[/math], [math]|x| = x[/math].

Applying this, the expected value integral becomes:

[[math]]\operatorname{E}[X] = \int_{-2}^{0} x \cdot \frac{-x}{10} \, dx + \int_{0}^{4} x \cdot \frac{x}{10} \, dx[[/math]]

This simplifies to:

[[math]]\operatorname{E}[X] = \int_{-2}^{0} \frac{-x^2}{10} \, dx + \int_{0}^{4} \frac{x^2}{10} \, dx[[/math]]

Step 3: Perform the Integration and Evaluate

We now evaluate the definite integrals. The antiderivative of [math]-x^2/10[/math] is [math]-x^3/30[/math], and the antiderivative of [math]x^2/10[/math] is [math]x^3/30[/math].

[[math]]\operatorname{E}[X] = \left[ -\frac{x^3}{30} \right]_{-2}^{0} + \left[ \frac{x^3}{30} \right]_{0}^{4}[[/math]]

[[math]] = \left( -\frac{0^3}{30} - \left(-\frac{(-2)^3}{30}\right) \right) + \left( \frac{4^3}{30} - \frac{0^3}{30} \right)[[/math]]

[[math]] = \left( 0 - \left(-\frac{-8}{30}\right) \right) + \left( \frac{64}{30} - 0 \right)[[/math]]

[[math]] = \left( -\frac{8}{30} \right) + \left( \frac{64}{30} \right)[[/math]]

[[math]] = \frac{-8 + 64}{30} = \frac{56}{30} = \frac{28}{15}[[/math]]

Key Insights

• The expected value of a continuous random variable is found by integrating [math]x \cdot f(x)[/math] over its support.
• When the probability density function involves an absolute value function, like [math]|x|[/math], the integral must be split into sub-intervals based on where the expression inside the absolute value changes sign (typically at [math]x=0[/math]).
• Careful evaluation of definite integrals, especially with negative limits or terms, is essential to avoid sign errors.

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Solution: D

Note that

[[math]] \operatorname{E}[X] = \int_{-2}^0 \frac{-x^2}{10} dx + \int_0^4 \frac{x^2}{10} dx = \frac{-x^3}{30} \Big |_{-2}^0 + \frac{x^3}{30} \Big |_0^4 = \frac{28}{15}. [[/math]]

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