New dental and medical plan options will be offered to state employees next year. An actuary uses the following density function to model the joint distribution of the proportion [math]X[/math] of state employees who will choose Dental Option 1 and the proportion [math]Y[/math] who will choose Medical Option 1 under the new plan options:

[[math]] f(x,y) = \begin{cases} 0.50, \,\, 0 \lt x \lt 0.5, 0 \lt y \lt 0.5 \\ 1.25, \,\, 0 \lt x \lt 0.5, 0.5 \lt y \lt 1 \\ 1.50, \,\, 0.5 \lt x \lt 1, 0 \lt y \lt 0.5 \\ 0.75, \,\, 0.5 \lt x \lt 1, 0.5 \lt y \lt 1 \\ \end{cases} [[/math]]

Calculate [math]\operatorname{Var} (Y | X = 0.75)[/math].

• 0.000
• 0.061
• 0.076
• 0.083
• 0.141

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

Step 1: Identify the Relevant Joint Density for [math]X = 0.75[/math]

The given joint density function [math]f(x,y)[/math] is piecewise. We are interested in the conditional variance of [math]Y[/math] given [math]X = 0.75[/math]. Since [math]X = 0.75[/math] falls within the range [math]0.5 \lt x \lt 1[/math], we need to consider the corresponding parts of the density function:

• For [math]0 \lt y \lt 0.5[/math] and [math]0.5 \lt x \lt 1[/math], [math]f(x,y) = 1.50[/math].
• For [math]0.5 \lt y \lt 1[/math] and [math]0.5 \lt x \lt 1[/math], [math]f(x,y) = 0.75[/math].

Thus, for [math]X = 0.75[/math], the joint density function values are:

Joint Density Values for [math]X = 0.75[/math]

Range of [math]Y[/math]	[math]f(0.75, y)[/math]
[math]0 \lt y \lt 0.5[/math]	[math]1.50[/math]
[math]0.5 \lt y \lt 1[/math]	[math]0.75[/math]

Step 2: Calculate the Marginal Density [math]f_X(0.75)[/math]

To find the conditional density [math]f(y | x)[/math], we first need to calculate the marginal density [math]f_X(x)[/math] at [math]x = 0.75[/math], which is given by integrating [math]f(x,y)[/math] over all possible values of [math]y[/math] for the fixed [math]x[/math].

[[math]] f_X(0.75) = \int_0^1 f(0.75,y) dy [[/math]]

Given the piecewise nature of [math]f(0.75,y)[/math], we split the integral:

[[math]] f_X(0.75) = \int_0^{0.5} f(0.75,y) dy + \int_{0.5}^1 f(0.75,y) dy [[/math]]

Substitute the values identified in Step 1:

[[math]] f_X(0.75) = \int_0^{0.5} 1.50 dy + \int_{0.5}^1 0.75 dy [[/math]]

Evaluate the integrals:

[[math]] f_X(0.75) = [1.50y]_0^{0.5} + [0.75y]_{0.5}^1 [[/math]]

[[math]] f_X(0.75) = (1.50 \times 0.5 - 1.50 \times 0) + (0.75 \times 1 - 0.75 \times 0.5) [[/math]]

[[math]] f_X(0.75) = 0.75 + (0.75 - 0.375) [[/math]]

[[math]] f_X(0.75) = 0.75 + 0.375 = 1.125 [[/math]]

Step 3: Determine the Conditional Density [math]f(y | X = 0.75)[/math]

The conditional density function [math]f(y | x)[/math] is defined as [math]f(y | x) = \frac{f(x,y)}{f_X(x)}[/math] for [math]f_X(x) \gt 0[/math]. For [math]X = 0.75[/math], we have [math]f_X(0.75) = 1.125[/math]. Therefore, the conditional density [math]f(y | X = 0.75)[/math] is:

[[math]] f(y | X = 0.75) = \frac{f(0.75, y)}{1.125} [[/math]]

Applying this to the piecewise definition of [math]f(0.75,y)[/math]:

• For [math]0 \lt y \lt 0.5[/math]:

[[math]] f(y | X = 0.75) = \frac{1.50}{1.125} = \frac{1500}{1125} = \frac{4 \times 375}{3 \times 375} = \frac{4}{3} [[/math]]

• For [math]0.5 \lt y \lt 1[/math]:

[[math]] f(y | X = 0.75) = \frac{0.75}{1.125} = \frac{750}{1125} = \frac{2 \times 375}{3 \times 375} = \frac{2}{3} [[/math]]

So, the conditional density function is:

[[math]] f(y | X = 0.75) = \begin{cases} \frac{4}{3}, \,\, 0 \lt y \lt 0.5 \\ \frac{2}{3}, \,\, 0.5 \lt y \lt 1 \\ \end{cases} [[/math]]

Step 4: Calculate the Conditional Expectation [math]E[Y | X = 0.75][/math]

The conditional expectation [math]E[Y | X = 0.75][/math] is calculated by integrating [math]y \cdot f(y | X = 0.75)[/math] over the range of [math]y[/math]:

[[math]] E[Y | X = 0.75] = \int_0^1 y \cdot f(y | X = 0.75) dy [[/math]]

Split the integral based on the piecewise definition of [math]f(y | X = 0.75)[/math]:

[[math]] E[Y | X = 0.75] = \int_0^{0.5} y \left(\frac{4}{3}\right) dy + \int_{0.5}^1 y \left(\frac{2}{3}\right) dy [[/math]]

[[math]] E[Y | X = 0.75] = \frac{4}{3} \int_0^{0.5} y dy + \frac{2}{3} \int_{0.5}^1 y dy [[/math]]

[[math]] E[Y | X = 0.75] = \frac{4}{3} \left[\frac{y^2}{2}\right]_0^{0.5} + \frac{2}{3} \left[\frac{y^2}{2}\right]_{0.5}^1 [[/math]]

[[math]] E[Y | X = 0.75] = \frac{4}{3} \left(\frac{(0.5)^2}{2} - 0\right) + \frac{2}{3} \left(\frac{1^2}{2} - \frac{(0.5)^2}{2}\right) [[/math]]

[[math]] E[Y | X = 0.75] = \frac{4}{3} \left(\frac{0.25}{2}\right) + \frac{2}{3} \left(\frac{1}{2} - \frac{0.25}{2}\right) [[/math]]

[[math]] E[Y | X = 0.75] = \frac{4}{3} (0.125) + \frac{2}{3} (0.5 - 0.125) [[/math]]

[[math]] E[Y | X = 0.75] = \frac{4}{3} (0.125) + \frac{2}{3} (0.375) [[/math]]

[[math]] E[Y | X = 0.75] = \frac{0.5}{3} + \frac{0.75}{3} = \frac{1.25}{3} = \frac{5}{12} [[/math]]

Step 5: Calculate the Conditional Second Moment [math]E[Y^2 | X = 0.75][/math]

The conditional second moment [math]E[Y^2 | X = 0.75][/math] is calculated by integrating [math]y^2 \cdot f(y | X = 0.75)[/math] over the range of [math]y[/math]:

[[math]] E[Y^2 | X = 0.75] = \int_0^1 y^2 \cdot f(y | X = 0.75) dy [[/math]]

Split the integral:

[[math]] E[Y^2 | X = 0.75] = \int_0^{0.5} y^2 \left(\frac{4}{3}\right) dy + \int_{0.5}^1 y^2 \left(\frac{2}{3}\right) dy [[/math]]

[[math]] E[Y^2 | X = 0.75] = \frac{4}{3} \int_0^{0.5} y^2 dy + \frac{2}{3} \int_{0.5}^1 y^2 dy [[/math]]

[[math]] E[Y^2 | X = 0.75] = \frac{4}{3} \left[\frac{y^3}{3}\right]_0^{0.5} + \frac{2}{3} \left[\frac{y^3}{3}\right]_{0.5}^1 [[/math]]

[[math]] E[Y^2 | X = 0.75] = \frac{4}{3} \left(\frac{(0.5)^3}{3} - 0\right) + \frac{2}{3} \left(\frac{1^3}{3} - \frac{(0.5)^3}{3}\right) [[/math]]

[[math]] E[Y^2 | X = 0.75] = \frac{4}{3} \left(\frac{0.125}{3}\right) + \frac{2}{3} \left(\frac{1}{3} - \frac{0.125}{3}\right) [[/math]]

[[math]] E[Y^2 | X = 0.75] = \frac{4}{9} (0.125) + \frac{2}{9} (1 - 0.125) [[/math]]

[[math]] E[Y^2 | X = 0.75] = \frac{0.5}{9} + \frac{2}{9} (0.875) [[/math]]

[[math]] E[Y^2 | X = 0.75] = \frac{0.5}{9} + \frac{1.75}{9} = \frac{2.25}{9} = \frac{9/4}{9} = \frac{1}{4} [[/math]]

Step 6: Calculate the Conditional Variance [math]\operatorname{Var}(Y | X = 0.75)[/math]

The conditional variance is calculated using the formula:

[[math]] \operatorname{Var}(Y | X = 0.75) = E[Y^2 | X = 0.75] - (E[Y | X = 0.75])^2 [[/math]]

From Step 4, [math]E[Y | X = 0.75] = \frac{5}{12}[/math]. From Step 5, [math]E[Y^2 | X = 0.75] = \frac{1}{4}[/math]. Substitute these values into the variance formula:

[[math]] \operatorname{Var}(Y | X = 0.75) = \frac{1}{4} - \left(\frac{5}{12}\right)^2 [[/math]]

[[math]] \operatorname{Var}(Y | X = 0.75) = \frac{1}{4} - \frac{25}{144} [[/math]]

To subtract these fractions, find a common denominator, which is 144.

[[math]] \operatorname{Var}(Y | X = 0.75) = \frac{36}{144} - \frac{25}{144} [[/math]]

[[math]] \operatorname{Var}(Y | X = 0.75) = \frac{11}{144} [[/math]]

Convert the fraction to a decimal:

[[math]] \operatorname{Var}(Y | X = 0.75) \approx 0.076388... [[/math]]

Rounding to three decimal places, this is [math]0.076[/math].

Key Insights

• When dealing with piecewise joint density functions for conditional probabilities, correctly identify the relevant function segments based on the given condition.
• The conditional probability density function [math]f(Y|X=x)[/math] is found by dividing the joint density [math]f(x,y)[/math] by the marginal density [math]f_X(x)[/math], where [math]f_X(x) = \int f(x,y) dy[/math].
• For piecewise conditional densities, calculations of expected values and second moments involve splitting the integral across the relevant intervals where the density function changes definition.
• The variance of a conditional distribution is calculated using the standard formula [math]\operatorname{Var}(Y|X=x) = E[Y^2|X=x] - (E[Y|X=x])^2[/math], just like unconditional variance, but using conditional expectations.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: C

[[math]] f(y | x = 0.75) = \frac{f(0.75,y)}{\int_0^1 f(0.75,y) dy} = \frac{f(0.75,y)}{1.125}. [[/math]]

Thus,

[[math]] f(y | x= 0.75) = \begin{cases} 4/3, \, 0 \lt y \lt 0.5 \\ 2/3, \, 0.5 \lt y \lt 1 \end{cases} [[/math]]

which leads to [math]\operatorname{Var}(Y | X = 0.75 ) = 11/144 = 0.076 [/math].

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