The joint probability density for [math]X[/math] and [math]Y[/math] is

[[math]] f(x,y) = \begin{cases} 2e^{-(x + 2y)}, \,\, x \gt 0, y \gt 0\\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the variance of [math]Y[/math] given that [math]X \gt 3 [/math] and [math]Y \gt 3 [/math].

• 0.25
• 0.50
• 1.00
• 3.25
• 3.50

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

Step 1: Identify Independence and Marginal Distributions

The joint probability density function for [math]X[/math] and [math]Y[/math] is given by:

[[math]]f(x,y) = \begin{cases} 2e^{-(x + 2y)}, \,\, x \gt 0, y \gt 0\\ 0, \, \textrm{Otherwise.}\end{cases}[[/math]]

We can factorize [math]f(x,y)[/math] into a product of two functions, one depending only on [math]x[/math] and the other only on [math]y[/math]:

[[math]]f(x,y) = e^{-x} \cdot 2e^{-2y}[[/math]]

This factorization holds for the specified domain [math]x \gt 0[/math] and [math]y \gt 0[/math]. Since the joint PDF can be expressed as a product of two functions, each solely dependent on one variable, and the support is a rectangular region, [math]X[/math] and [math]Y[/math] are independent random variables. The marginal probability density functions are:

• [math]f_X(x) = e^{-x}[/math] for [math]x \gt 0[/math], which indicates that [math]X[/math] follows an exponential distribution with rate parameter [math]\lambda_X = 1[/math].
• [math]f_Y(y) = 2e^{-2y}[/math] for [math]y \gt 0[/math], which indicates that [math]Y[/math] follows an exponential distribution with rate parameter [math]\lambda_Y = 2[/math].

Step 2: Address the Conditional Information

We are asked to calculate [math]Var(Y | X \gt 3, Y \gt 3)[/math]. Impact of X > 3: Since [math]X[/math] and [math]Y[/math] are independent (as established in Step 1), any information about [math]X[/math] (such as [math]X \gt 3[/math]) does not affect the distribution or moments of [math]Y[/math]. Therefore, the condition [math]X \gt 3[/math] can be ignored. The problem simplifies to finding [math]Var(Y | Y \gt 3)[/math]. Impact of Y > 3: The exponential distribution possesses the memoryless property. This property states that for an exponential random variable [math]Y[/math] with rate parameter [math]\lambda[/math], the conditional distribution of [math]Y - t[/math] given [math]Y \gt t[/math] is the same as the unconditional distribution of [math]Y[/math]. In our case, with [math]t=3[/math], the distribution of [math]Y | Y \gt 3[/math] is equivalent to the distribution of [math]3 + Y'[/math], where [math]Y'[/math] is an exponential random variable with the same rate parameter [math]\lambda_Y = 2[/math] as [math]Y[/math]. We want to calculate [math]Var(Y | Y \gt 3)[/math]. Using the relationship derived from the memoryless property:

[[math]]Var(Y | Y \gt 3) = Var(3 + Y')[[/math]]

A fundamental property of variance is that adding a constant to a random variable does not change its variance: [math]Var(c + Z) = Var(Z)[/math] for any constant [math]c[/math] and random variable [math]Z[/math]. Applying this property:

[[math]]Var(3 + Y') = Var(Y')[[/math]]

Since [math]Y'[/math] has the same distribution as [math]Y[/math] (i.e., [math]Y' \sim \text{Exp}(\lambda_Y = 2)[/math]), it follows that [math]Var(Y') = Var(Y)[/math]. Therefore, we conclude that [math]Var(Y | X \gt 3, Y \gt 3) = Var(Y)[/math].

Step 3: Calculate the Variance of Y

From Step 1, we established that [math]Y[/math] follows an exponential distribution with rate parameter [math]\lambda_Y = 2[/math]. For an exponential distribution with parameter [math]\lambda[/math], the variance is given by the formula [math]1/\lambda^2[/math]. Using this formula for [math]Y[/math]:

[[math]]Var(Y) = \frac{1}{\lambda_Y^2} = \frac{1}{2^2} = \frac{1}{4} = 0.25[[/math]]

Thus, the variance of [math]Y[/math] given that [math]X \gt 3[/math] and [math]Y \gt 3[/math] is [math]0.25[/math].

Key Insights

• Independence from Joint PDF: If a joint probability density function [math]f(x,y)[/math] can be factored into [math]f_X(x)f_Y(y)[/math] over a rectangular support, then the random variables [math]X[/math] and [math]Y[/math] are independent.
• Conditional Probabilities with Independence: When random variables are independent, a condition on one variable does not affect the distribution or moments of the other variable.
• Memoryless Property of Exponential Distribution: The exponential distribution is memoryless. This means [math]Var(Y | Y \gt t) = Var(Y)[/math] for an exponential random variable [math]Y[/math] and any [math]t \gt 0[/math]. More generally, the distribution of [math]Y - t[/math] given [math]Y \gt t[/math] is the same as the unconditional distribution of [math]Y[/math].
• Variance of a Constant Shift: Adding or subtracting a constant to a random variable does not change its variance. That is, [math]Var(Y + c) = Var(Y)[/math] for any constant [math]c[/math].
• Moments of Exponential Distribution: For an exponential distribution with rate parameter [math]\lambda[/math], the mean is [math]1/\lambda[/math] and the variance is [math]1/\lambda^2[/math].

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Solution: A

Because [math]f(x,y)[/math] can be written as [math]f(x) f(y) = e^{− x} 2e^{−2y}[/math] and the support of [math]f(x,y)[/math] is a cross product, [math]X[/math] and [math]Y[/math] are independent. Thus, the condition on [math]X[/math] can be ignored and it suffices to just consider [math]f(y) = 2e^{−2y}.[/math]

Because of the memoryless property of the exponential distribution, the conditional density of Y is the same as the unconditional density of [math]Y+3.[/math]

Because a location shift does not affect the variance, the conditional variance of [math]Y[/math] is equal to the unconditional variance of [math]Y[/math]. Because the mean of [math]Y[/math] is 0.5 and the variance of an exponential distribution is always equal to the square of its mean, the requested variance is 0.25.

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