The stock prices of two companies at the end of any given year are modeled with random variables [math]X[/math] and [math]Y[/math] that follow a distribution with joint density function

[[math]] f(x,y) = \begin{cases} 2x, \,\, 0 \lt x \lt 1, x \lt y \lt x +1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Determine the conditional variance of [math]Y[/math] given that [math]X = x[/math].

• 1/12
• 7/6
• [math]x + 1/2 [/math]
• [math]x^2 - 1/6[/math]
• [math]x^2 +x + 1/3[/math]

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

Solution: A

Let [math]f_1(x)[/math] denote the marginal density function of X. Then

[[math]] f_1(x) = \int_x^{x+1}2x dy = 2xy \Big |_x^{x+1} = 2x(x+1-x) = 2x, \, 0 \lt x \lt 1. [[/math]]

Consequently,

[[math]] f(y |x) = \frac{f(x,y)}{f_1(x)} = \begin{cases} 1, \, x \lt y \lt x+ 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Consequently,

[[math]] f(y | x) = \frac{f(x,y)}{f_1(x)} = \begin{cases} 1, \, x \lt y \lt x+ 1 \\ 0, \, \textrm{otherwise}\end{cases} [[/math]]

[[math]] \begin{align*} E[Y | X ] = \int_x^{x+1} ydy = \frac{1}{2}y^2 \Big |_{x}^{x+1} ydy &= \frac{1}{2}y^2 \Big |_x^{x+1}\\ &= \frac{1}{2}(x+1)^2 - \frac{1}{2}x^2 \\ &= \frac{1}{2}x^2 + x+ \frac{1}{2} - \frac{1}{2}x^2\\ &= x + \frac{1}{2} \end{align*} [[/math]]

[[math]] \begin{align*} E[Y^2 | X ] = \int_x^{x+1}y^2 dy \frac{1}{3}y^3 \Big |_{x}^{x+1} = \frac{1}{3}(x+1)^3- \frac{1}{3}x^3 \\ = \frac{1}{3}x^3 + x^2 + x+ \frac{1}{3} - \frac{1}{3}x^3 = x^2 + x + \frac{1}{3} \end{align*} [[/math]]

[[math]] \begin{align*} \operatorname{Var}[Y | X] &= E[Y^2 | X] - \{ E[Y | X]\}^2 = x^2 + x + \frac{1}{3} - (x + \frac{1}{2})^2 \\ &= x^2 + x + \frac{1}{3} - x^2 -x - \frac{1}{4} \\ &= \frac{1}{12} \end{align*} [[/math]]

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