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.

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.