Let [math]X[/math] and [math]Y[/math] be continuous random variables with joint density function

[[math]] f(x,y) = \begin{cases} 24xy, \,\, 0 \lt x \lt 1 \,\, \textrm{and} \,\, 0 \lt y \lt 1-x \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate [math] \operatorname{P}[Y \lt X | X = 1/3][/math]

• 1/27
• 2/27
• 1/4
• 1/3
• 4/9

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

Solution: C

Note that the conditional density function

[[math]] \begin{align*} f(y | x = \frac{1}{3} ) &= \frac{f(1/3,y)}{f_x(1/3)}, \, 0 \lt y \lt \frac{2}{3}, \\ f_x(\frac{1}{3}) &= \int_0^{2/3} 24 (1/3) y dy = \int_0^{2/3} 8y dy = 4y^2 \Big|_0^{2/3} = \frac{16}{9}. \end{align*} [[/math]]

It follows that

[[math]] f(y | x = \frac{1}{3} ) = \frac{9}{16} f(1/3, y) = \frac{9}{2}y, \, 0 \lt y \lt \frac{2}{3} [[/math]]

Consequently,

[[math]] \operatorname{P}[Y \lt X | X = 1/3] = \int_0^{1/3} \frac{9}{2} y dy = \frac{9}{4} y^2 \Big |_0^{1/3} = \frac{1}{4}. [[/math]]

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