A company offers a basic life insurance policy to its employees, as well as a supplemental life insurance policy. To purchase the supplemental policy, an employee must first purchase the basic policy. Let [math]X[/math] denote the proportion of employees who purchase the basic policy, and [math]Y[/math] the proportion of employees who purchase the supplemental policy. Let [math]X[/math] and [math]Y[/math] have the joint density function [math]f(x,y) = 2(x+y)[/math] on the region where the density is positive.

Given that 10% of the employees buy the basic policy, calculate the probability that fewer than 5% buy the supplemental policy.

• 0.010
• 0.013
• 0.108
• 0.417
• 0.500

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

Step 1: Identify the Region of Support for the Joint Density Function

The problem states that the joint probability density function is [math]f(x,y) = 2(x+y)[/math] on the region where the density is positive. For this type of problem involving proportions, the region of support is defined by [math]0 \lt y \lt x \lt 1[/math]. This implies:

• [math]X[/math] represents the proportion of employees buying the basic policy, so [math]0 \lt X \lt 1[/math].
• [math]Y[/math] represents the proportion of employees buying the supplemental policy, so [math]0 \lt Y \lt 1[/math].
• To purchase the supplemental policy, an employee must first purchase the basic policy. This implies that the proportion of employees buying the supplemental policy cannot exceed the proportion buying the basic policy, hence [math]Y \lt X[/math].

Step 2: Determine the Marginal Probability Density Function of [math]X[/math]

To find the marginal PDF of [math]X[/math], denoted as [math]f_X(x)[/math], we integrate the joint PDF [math]f(x,y)[/math] with respect to [math]y[/math] over its possible range. From the region of support [math]0 \lt y \lt x \lt 1[/math], [math]y[/math] ranges from [math]0[/math] to [math]x[/math].

[[math]]f_X(x) = \int_0^x 2(x+y) dy[[/math]]

[[math]]f_X(x) = \left[ 2xy + y^2 \right]_0^x[[/math]]

[[math]]f_X(x) = (2x(x) + x^2) - (2x(0) + 0^2)[[/math]]

[[math]]f_X(x) = 2x^2 + x^2 = 3x^2, \quad \text{for } 0 \lt x \lt 1[[/math]]

Step 3: Calculate the Conditional Probability Density Function [math]f(y|x)[/math]

The conditional PDF of [math]Y[/math] given [math]X=x[/math] is defined as the ratio of the joint PDF to the marginal PDF of [math]X[/math]:

[[math]]f(y|x) = \frac{f(x,y)}{f_X(x)}[[/math]]

Substitute the expressions for [math]f(x,y)[/math] and [math]f_X(x)[/math]:

[[math]]f(y|x) = \frac{2(x+y)}{3x^2}[[/math]]

This can be simplified for clarity:

[[math]]f(y|x) = \frac{2}{3}\left(\frac{x}{x^2} + \frac{y}{x^2}\right)[[/math]]

[[math]]f(y|x) = \frac{2}{3}\left(\frac{1}{x} + \frac{y}{x^2}\right), \quad \text{for } 0 \lt y \lt x[[/math]]

Step 4: Evaluate the Conditional PDF for [math]X = 0.10[/math]

We are given that 10% of the employees buy the basic policy, which means [math]X = 0.10[/math]. We substitute this value into the conditional PDF [math]f(y|x)[/math] derived in the previous step:

[[math]]f(y|X=0.10) = \frac{2}{3}\left(\frac{1}{0.10} + \frac{y}{(0.10)^2}\right)[[/math]]

[[math]]f(y|X=0.10) = \frac{2}{3}\left(10 + \frac{y}{0.01}\right)[[/math]]

[[math]]f(y|X=0.10) = \frac{2}{3}(10 + 100y)[[/math]]

The range for [math]y[/math] in this conditional distribution is [math]0 \lt y \lt X[/math], so [math]0 \lt y \lt 0.10[/math].

Step 5: Calculate the Desired Probability

We need to calculate the probability that fewer than 5% buy the supplemental policy, given [math]X=0.10[/math]. This is [math]P[Y \lt 0.05 | X = 0.10][/math]. We achieve this by integrating the specific conditional PDF [math]f(y|X=0.10)[/math] from [math]0[/math] to [math]0.05[/math].

[[math]]P[Y \lt 0.05 | X = 0.10] = \int_0^{0.05} \frac{2}{3}(10 + 100y) dy[[/math]]

Perform the integration:

[[math]]P[Y \lt 0.05 | X = 0.10] = \left[ \frac{2}{3} \left( 10y + \frac{100y^2}{2} \right) \right]_0^{0.05}[[/math]]

[[math]]P[Y \lt 0.05 | X = 0.10] = \left[ \frac{20}{3}y + \frac{100}{3}y^2 \right]_0^{0.05}[[/math]]

Now, substitute the upper limit [math]y=0.05[/math] (the lower limit [math]y=0[/math] yields 0):

[[math]]P[Y \lt 0.05 | X = 0.10] = \frac{20}{3}(0.05) + \frac{100}{3}(0.05)^2[[/math]]

[[math]]P[Y \lt 0.05 | X = 0.10] = \frac{1}{3} + \frac{100}{3}(0.0025)[[/math]]

[[math]]P[Y \lt 0.05 | X = 0.10] = \frac{1}{3} + \frac{0.25}{3}[[/math]]

To combine these fractions, we can write [math]0.25[/math] as [math]\frac{1}{4}[/math]:

[[math]]P[Y \lt 0.05 | X = 0.10] = \frac{1}{3} + \frac{1/4}{3}[[/math]]

[[math]]P[Y \lt 0.05 | X = 0.10] = \frac{1}{3} + \frac{1}{12}[[/math]]

Finding a common denominator:

[[math]]P[Y \lt 0.05 | X = 0.10] = \frac{4}{12} + \frac{1}{12} = \frac{5}{12}[[/math]]

As a decimal, [math]\frac{5}{12} \approx 0.41666...[/math]. Rounding to three decimal places, the probability is approximately [math]0.417[/math].

Key Insights

• The region of support for the joint density function, often implied by the problem context (e.g., "proportion," "must first purchase"), is critical for defining the integration limits when calculating marginal and conditional distributions.
• To determine a conditional probability [math]P(A|B)[/math] for continuous random variables, one must first compute the marginal probability density function of the conditioning variable.
• The conditional PDF [math]f(y|x)[/math] is found by dividing the joint PDF [math]f(x,y)[/math] by the marginal PDF of [math]X[/math], i.e., [math]f(y|x) = \frac{f(x,y)}{f_X(x)}[/math].
• Once the conditional PDF is established, calculating specific probabilities involves integrating this conditional PDF over the desired range of the dependent variable.

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

Solution: D

We are given that the joint pdf of X and Y is f(x,y) = 2(x+y), 0 < y < x < 1 . Now

[[math]] f_x(x) = \int_0^x (2x + 2y) dy = [ 2xy + y^2]_0^x = 2x^2 + x^2 = 3x^2, 0 \lt x \lt 1 [[/math]]

so

[[math]] f(y|x) = \frac{f(x,y)}{f_x(x)} = \frac{2(x+y)}{3x^2} = \frac{2}{3}(\frac{1}{x} + \frac{y}{x^2}), 0 \lt y \lt x [[/math]]

[[math]] f(y|x = 0.10) = \frac{2}{3} [\frac{1}{0.1} + \frac{y}{0.1}] = \frac{2}{3}[10 + 100y], 0 \lt y \lt 0.10 [[/math]]

[[math]] P[ Y \lt 0.05 | X = 0.10] = \int_0^{0.05} \frac{2}{3}[10 + 100y] dy = [ \frac{20}{3}y + \frac{100}{3}y^2]_0^{0.05} = \frac{1}{3} + \frac{1}{12} = \frac{5}{12} = 0.4167. [[/math]]

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