The distribution of [math]Y[/math], given [math]X[/math] , is uniform on the interval [0, [math]X[/math]]. The marginal density of [math]X[/math] is

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

Determine the conditional density of [math]X[/math] , given [math]Y = y[/math] where positive.

• 1
• 2
• 2x
• 1/y
• 1/(1-y)

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

Solution: E

The support of (X,Y) is 0 < y < x < 1.

[[math]] f_{X,Y}(x.y) = f(y |x) f_X(x) = 2 \quad \textrm{on that support.} [[/math]]

It is clear geometrically (a flat joint density over the triangular region 0 < y < x < 1) that when [math]Y = y [/math] we have [math]X \sim U(y,1) [/math] so that

[[math]] f(x |y ) = \frac{1}{1-y} \quad \textrm{for} \quad y \lt x \lt 1. [[/math]]

By computation:

[[math]] f_Y(y) = \int_y^1 2dx = 2-2y \Rightarrow f(x | y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{2}{2-2y} = \frac{1}{1-y} \quad \textrm{for} \quad y \lt x \lt 1 [[/math]]

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