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.

Step 1: Determine the Marginal Density Function of X

The marginal density function of [math]X[/math], denoted as [math]f_X(x)[/math], is found by integrating the joint density function [math]f(x,y)[/math] with respect to [math]y[/math] over its defined range.

[[math]] f_X(x) = \int_x^{x+1} 2x \, dy [[/math]]

Performing the integration, we treat [math]2x[/math] as a constant with respect to [math]y[/math]:

[[math]] f_X(x) = \left[ 2xy \right]_x^{x+1} [[/math]]

Substituting the limits of integration:

[[math]] f_X(x) = 2x(x+1) - 2x(x) [[/math]]

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

Thus, the marginal density function of [math]X[/math] is:

[[math]] f_X(x) = 2x, \quad 0 \lt x \lt 1 [[/math]]

Step 2: Determine the Conditional Density Function of Y given X=x

The conditional density function of [math]Y[/math] given [math]X=x[/math], denoted as [math]f_{Y|X}(y|x)[/math], is calculated using the formula [math]f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)}[/math]. We are given the joint density function [math]f(x,y) = 2x[/math] for [math]0 \lt x \lt 1, x \lt y \lt x+1[/math], and from Step 1, we found the marginal density function [math]f_X(x) = 2x[/math] for [math]0 \lt x \lt 1[/math]. Substituting these into the formula:

[[math]] f_{Y|X}(y|x) = \frac{2x}{2x} = 1 [[/math]]

Therefore, the conditional density function is:

[[math]] f_{Y|X}(y|x) = \begin{cases} 1, & x \lt y \lt x+1 \\ 0, & \text{otherwise} \end{cases} [[/math]]

This indicates that [math]Y[/math] given [math]X=x[/math] follows a uniform distribution over the interval [math](x, x+1)[/math].

Step 3: Calculate the Conditional Expectation of Y given X=x, [math]E[Y|X=x][/math]

For a random variable uniformly distributed over an interval [math](a,b)[/math], the expectation is [math]\frac{a+b}{2}[/math]. In this case, [math]a=x[/math] and [math]b=x+1[/math]. Alternatively, we can calculate it by integration:

[[math]] E[Y|X=x] = \int_x^{x+1} y \cdot f_{Y|X}(y|x) \, dy = \int_x^{x+1} y \cdot 1 \, dy [[/math]]

Performing the integration:

[[math]] E[Y|X=x] = \left[ \frac{1}{2}y^2 \right]_x^{x+1} [[/math]]

Substituting the limits:

[[math]] E[Y|X=x] = \frac{1}{2}(x+1)^2 - \frac{1}{2}x^2 [[/math]]

Expanding and simplifying the expression:

[[math]] E[Y|X=x] = \frac{1}{2}(x^2 + 2x + 1) - \frac{1}{2}x^2 [[/math]]

[[math]] E[Y|X=x] = \frac{1}{2}x^2 + x + \frac{1}{2} - \frac{1}{2}x^2 [[/math]]

[[math]] E[Y|X=x] = x + \frac{1}{2} [[/math]]

Step 4: Calculate the Conditional Second Moment of Y given X=x, [math]E[Y^2|X=x][/math]

The conditional second moment is calculated by integrating [math]y^2[/math] multiplied by the conditional density function:

[[math]] E[Y^2|X=x] = \int_x^{x+1} y^2 \cdot f_{Y|X}(y|x) \, dy = \int_x^{x+1} y^2 \cdot 1 \, dy [[/math]]

Performing the integration:

[[math]] E[Y^2|X=x] = \left[ \frac{1}{3}y^3 \right]_x^{x+1} [[/math]]

Substituting the limits:

[[math]] E[Y^2|X=x] = \frac{1}{3}(x+1)^3 - \frac{1}{3}x^3 [[/math]]

Expanding [math](x+1)^3[/math] and simplifying:

[[math]] E[Y^2|X=x] = \frac{1}{3}(x^3 + 3x^2 + 3x + 1) - \frac{1}{3}x^3 [[/math]]

[[math]] E[Y^2|X=x] = \frac{1}{3}x^3 + x^2 + x + \frac{1}{3} - \frac{1}{3}x^3 [[/math]]

[[math]] E[Y^2|X=x] = x^2 + x + \frac{1}{3} [[/math]]

Step 5: Calculate the Conditional Variance of Y given X=x

The conditional variance [math]\operatorname{Var}[Y|X=x][/math] is given by the formula:

[[math]] \operatorname{Var}[Y|X=x] = E[Y^2|X=x] - (E[Y|X=x])^2 [[/math]]

Substitute the results obtained from Step 3 ([math]E[Y|X=x] = x + \frac{1}{2}[/math]) and Step 4 ([math]E[Y^2|X=x] = x^2 + x + \frac{1}{3}[/math]):

[[math]] \operatorname{Var}[Y|X=x] = \left( x^2 + x + \frac{1}{3} \right) - \left( x + \frac{1}{2} \right)^2 [[/math]]

First, expand the squared term:

[[math]] \left( x + \frac{1}{2} \right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{2} + \left( \frac{1}{2} \right)^2 = x^2 + x + \frac{1}{4} [[/math]]

Now substitute this back into the variance formula:

[[math]] \operatorname{Var}[Y|X=x] = x^2 + x + \frac{1}{3} - (x^2 + x + \frac{1}{4}) [[/math]]

Distribute the negative sign:

[[math]] \operatorname{Var}[Y|X=x] = x^2 + x + \frac{1}{3} - x^2 - x - \frac{1}{4} [[/math]]

The [math]x^2[/math] and [math]x[/math] terms cancel out:

[[math]] \operatorname{Var}[Y|X=x] = \frac{1}{3} - \frac{1}{4} [[/math]]

Find a common denominator (12) and subtract the fractions:

[[math]] \operatorname{Var}[Y|X=x] = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} [[/math]]

Key Insights

• Calculating conditional variance requires determining both the conditional expectation [math]E[Y|X=x][/math] and the conditional second moment [math]E[Y^2|X=x][/math] of the random variable.
• The fundamental formula for conditional variance is [math]\operatorname{Var}[Y|X] = E[Y^2|X] - (E[Y|X])^2[/math].
• To find the conditional density function [math]f_{Y|X}(y|x)[/math], one must first compute the marginal density function [math]f_X(x)[/math] and then apply the definition [math]f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)}[/math].
• Recognizing a conditional distribution as a standard probability distribution (e.g., uniform) can significantly simplify calculations. For a uniform distribution [math]U(a,b)[/math], the variance is given by [math]\frac{(b-a)^2}{12}[/math]. In this problem, [math]Y|X=x \sim U(x, x+1)[/math], so [math]a=x[/math] and [math]b=x+1[/math]. Thus, [math]\operatorname{Var}[Y|X=x] = \frac{((x+1)-x)^2}{12} = \frac{1^2}{12} = \frac{1}{12}[/math], which directly confirms the result obtained through integration.
• Even if the limits of integration for the conditional distribution depend on [math]x[/math], the conditional variance can still be a constant if the length of the interval of the conditional distribution is fixed. In this case, the interval length for [math]Y|X=x[/math] is [math](x+1) - x = 1[/math], which is constant.

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

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.