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.

Step 1: Understand the Joint Density and Identify the Goal

The problem provides a joint density function for two continuous random variables [math]X[/math] and [math]Y[/math]:

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

Our objective is to calculate the conditional probability [math] \operatorname{P}[Y \lt X | X = 1/3][/math]. To achieve this, we first need to determine the conditional density function [math]f(y | X = 1/3)[/math].

Step 2: Determine the Conditional Density Function Formula and Relevant Values

The conditional density function [math]f(y | x)[/math] is generally defined as:

[[math]] f(y | x) = \frac{f(x,y)}{f_x(x)} [[/math]]

where [math]f_x(x)[/math] is the marginal density function of [math]X[/math] at a given [math]x[/math]. For the given condition [math]X = 1/3[/math], the joint density function becomes:

[[math]] f(1/3, y) = 24(1/3)y = 8y [[/math]]

The support for [math]Y[/math] when [math]X = 1/3[/math] is determined by the condition [math]0 \lt y \lt 1-x[/math], so [math]0 \lt y \lt 1 - 1/3[/math], which simplifies to [math]0 \lt y \lt 2/3[/math].

Step 3: Calculate the Marginal Density [math]f_x(1/3)[/math]

To find the marginal density [math]f_x(1/3)[/math], we integrate [math]f(1/3, y)[/math] with respect to [math]y[/math] over its valid range, which is [math]0 \lt y \lt 2/3[/math]:

[[math]] f_x(1/3) = \int_0^{2/3} f(1/3, y) \, dy = \int_0^{2/3} 8y \, dy [[/math]]

Performing the integration:

[[math]] f_x(1/3) = \left[ 4y^2 \right]_0^{2/3} = 4 \left( \frac{2}{3} \right)^2 - 4(0)^2 = 4 \times \frac{4}{9} = \frac{16}{9} [[/math]]

Step 4: Formulate the Conditional Density [math]f(y | X = 1/3)[/math]

Now we can substitute the joint density [math]f(1/3, y)[/math] and the marginal density [math]f_x(1/3)[/math] into the formula for the conditional density function:

[[math]] f(y | X = 1/3) = \frac{f(1/3, y)}{f_x(1/3)} = \frac{8y}{16/9} [[/math]]

Simplifying the expression:

[[math]] f(y | X = 1/3) = 8y \times \frac{9}{16} = \frac{72y}{16} = \frac{9}{2}y [[/math]]

This conditional density function [math]f(y | X = 1/3)[/math] is valid for [math]0 \lt y \lt 2/3[/math].

Step 5: Calculate the Conditional Probability [math]\operatorname{P}[Y \lt X | X = 1/3][/math]

We are asked to calculate [math]\operatorname{P}[Y \lt X | X = 1/3][/math]. Since [math]X = 1/3[/math] is given, this becomes [math]\operatorname{P}[Y \lt 1/3 | X = 1/3][/math]. To find this probability, we integrate the conditional density function [math]f(y | X = 1/3)[/math] from the lower bound of its support (0) up to [math]1/3[/math]. Note that [math]1/3[/math] is within the valid range of [math]y[/math] for the conditional distribution [math](0, 2/3)[/math].

[[math]] \operatorname{P}[Y \lt 1/3 | X = 1/3] = \int_0^{1/3} f(y | X = 1/3) \, dy = \int_0^{1/3} \frac{9}{2}y \, dy [[/math]]

Performing the integration:

[[math]] \operatorname{P}[Y \lt 1/3 | X = 1/3] = \left[ \frac{9}{2} \frac{y^2}{2} \right]_0^{1/3} = \left[ \frac{9}{4} y^2 \right]_0^{1/3} [[/math]]

[[math]] = \frac{9}{4} \left( \frac{1}{3} \right)^2 - \frac{9}{4} (0)^2 = \frac{9}{4} \times \frac{1}{9} = \frac{1}{4} [[/math]]

Thus, the conditional probability [math] \operatorname{P}[Y \lt X | X = 1/3][/math] is [math]1/4[/math].

Key Insights

• To calculate a conditional probability [math]P(A|B)[/math] for continuous random variables, the first step is to determine the conditional density function [math]f(y|x)[/math].
• The conditional density function is derived by dividing the joint density [math]f(x,y)[/math] by the marginal density [math]f_x(x)[/math], i.e., [math]f(y|x) = \frac{f(x,y)}{f_x(x)}[/math].
• The marginal density [math]f_x(x)[/math] is found by integrating the joint density [math]f(x,y)[/math] with respect to [math]y[/math] over its entire support for a given [math]x[/math].
• Carefully define the support (range) for the variables at each step, especially when calculating marginal and conditional densities, and when setting integration limits for probabilities.
• When calculating a conditional probability like [math]P[Y \lt c | X = x_0][/math], integrate the conditional density [math]f(y|x_0)[/math] from its lower bound up to [math]c[/math], ensuring [math]c[/math] is within the conditional density's support.

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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.