May 06'23

Exercise

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.

May 06'23

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.

00