A diagnostic test for the presence of a disease has two possible outcomes: 1 for disease present and 0 for disease not present. Let [math]X[/math] denote the disease state (0 or 1) of a patient, and let [math]Y[/math] denote the outcome of the diagnostic test. The joint probability function of [math]X[/math] and [math]Y[/math] is given by:

[[math]] \begin{align*} \operatorname{P}[X = 0, Y = 0] &= 0.800 \\ \operatorname{P}[X = 1, Y = 0] &= 0.050 \\ \operatorname{P}[X = 0, Y = 1] &= 0.025 \\ \operatorname{P}[X = 1, Y = 1] &= 0.125 \\ \end{align*} [[/math]]

Calculate [math]\operatorname{Var}(Y | X = 1) .[/math]

• 0.13
• 0.15
• 0.20
• 0.51
• 0.71

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

Step 1: Understand the Goal and Given Information

The objective is to calculate the conditional variance of [math]Y[/math] given that [math]X=1[/math], which is denoted as [math]\operatorname{Var}(Y | X = 1)[/math]. The joint probability function of [math]X[/math] and [math]Y[/math] is provided:

Joint Probability Function [math]\operatorname{P}[X=x, Y=y][/math]

[math]X[/math]	[math]Y=0[/math]	[math]Y=1[/math]
[math]X=0[/math]	0.800	0.025
[math]X=1[/math]	0.050	0.125

Step 2: Calculate the Marginal Probability [math]\operatorname{P}(X=1)[/math]

To determine conditional probabilities given [math]X=1[/math], we first need the marginal probability [math]\operatorname{P}(X=1)[/math]. This is found by summing the joint probabilities for all possible values of [math]Y[/math] when [math]X=1[/math].

[[math]]\operatorname{P}(X=1) = \operatorname{P}(X=1, Y=0) + \operatorname{P}(X=1, Y=1)[[/math]]

Substituting the given values:

[[math]]\operatorname{P}(X=1) = 0.050 + 0.125 = 0.175[[/math]]

Step 3: Determine the Conditional Probabilities [math]\operatorname{P}(Y=y | X=1)[/math]

We use the definition of conditional probability: [math]\operatorname{P}(A|B) = \frac{\operatorname{P}(A \cap B)}{\operatorname{P}(B)}[/math]. For [math]Y=0[/math]:

[[math]]\operatorname{P}(Y=0 | X=1) = \frac{\operatorname{P}(X=1, Y=0)}{\operatorname{P}(X=1)} = \frac{0.050}{0.175} \approx 0.286[[/math]]

For [math]Y=1[/math]: Since [math]Y[/math] is a binary variable (0 or 1), the sum of its conditional probabilities must be 1.

[[math]]\operatorname{P}(Y=1 | X=1) = 1 - \operatorname{P}(Y=0 | X=1) = 1 - 0.286 = 0.714[[/math]]

Step 4: Calculate the Conditional Expectation [math]\operatorname{E}(Y | X=1)[/math]

The conditional expectation of [math]Y[/math] given [math]X=1[/math] is calculated as:

[[math]]\operatorname{E}(Y | X=1) = \sum_{y \in \{0,1\}} y \cdot \operatorname{P}(Y=y | X=1)[[/math]]

[[math]]\operatorname{E}(Y | X=1) = (0) \cdot \operatorname{P}(Y=0 | X=1) + (1) \cdot \operatorname{P}(Y=1 | X=1)[[/math]]

Substituting the conditional probabilities from Step 3:

[[math]]\operatorname{E}(Y | X=1) = (0) \cdot (0.286) + (1) \cdot (0.714) = 0.714[[/math]]

Step 5: Calculate the Conditional Second Moment [math]\operatorname{E}(Y^2 | X=1)[/math]

The conditional second moment of [math]Y[/math] given [math]X=1[/math] is calculated as:

[[math]]\operatorname{E}(Y^2 | X=1) = \sum_{y \in \{0,1\}} y^2 \cdot \operatorname{P}(Y=y | X=1)[[/math]]

[[math]]\operatorname{E}(Y^2 | X=1) = (0)^2 \cdot \operatorname{P}(Y=0 | X=1) + (1)^2 \cdot \operatorname{P}(Y=1 | X=1)[[/math]]

Substituting the conditional probabilities:

[[math]]\operatorname{E}(Y^2 | X=1) = (0) \cdot (0.286) + (1) \cdot (0.714) = 0.714[[/math]]

Step 6: Calculate the Conditional Variance [math]\operatorname{Var}(Y | X=1)[/math]

The conditional variance is found using the formula:

[[math]]\operatorname{Var}(Y | X=1) = \operatorname{E}(Y^2 | X=1) - [\operatorname{E}(Y | X=1)]^2[[/math]]

Substituting the values calculated in Step 4 and Step 5:

[[math]]\operatorname{Var}(Y | X=1) = 0.714 - (0.714)^2[[/math]]

[[math]]\operatorname{Var}(Y | X=1) = 0.714 - 0.509796 \approx 0.204204[[/math]]

Rounding to two decimal places, we get:

[[math]]\operatorname{Var}(Y | X=1) \approx 0.20[[/math]]

"'Key Insights"'

• Conditional probabilities are computed by dividing the joint probability of two events by the marginal probability of the conditioning event: [math]\operatorname{P}(A|B) = \frac{\operatorname{P}(A \cap B)}{\operatorname{P}(B)}[/math].
• For a binary random variable (e.g., taking values 0 or 1), its expectation is simply the probability of it taking the value 1. That is, if [math]Z \in \{0,1\}[/math], then [math]\operatorname{E}(Z) = \operatorname{P}(Z=1)[/math].
• Similarly, for a binary random variable [math]Z \in \{0,1\}[/math], its second moment is also equal to the probability of it taking the value 1: [math]\operatorname{E}(Z^2) = (0^2)\operatorname{P}(Z=0) + (1^2)\operatorname{P}(Z=1) = \operatorname{P}(Z=1)[/math].
• The variance of a Bernoulli random variable [math]Z[/math] with [math]\operatorname{P}(Z=1)=p[/math] is given by the formula [math]\operatorname{Var}(Z) = p(1-p)[/math]. In this problem, the conditional distribution of [math]Y[/math] given [math]X=1[/math] is Bernoulli with [math]p = \operatorname{P}(Y=1|X=1) \approx 0.714[/math]. Thus, [math]\operatorname{Var}(Y|X=1) \approx 0.714 \times (1 - 0.714) = 0.714 \times 0.286 \approx 0.204[/math]. This formula provides a direct shortcut for calculating the variance once the conditional probability [math]\operatorname{P}(Y=1|X=1)[/math] is known.

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Solution: C

Note that

[[math]] \begin{align*} \operatorname{P}(Y = 0 | X = 1) &= \frac{\operatorname{P}(X=1, Y= 0)}{\operatorname{P}(X=1)} \\ &= \frac{\operatorname{P}(X=1,Y=0)}{\operatorname{P}(X=1,Y=0) + \operatorname{P}(X=1,Y=1)}\\ &= \frac{0.05}{0.05 + 0.125} = 0.286 \end{align*} [[/math]]

[[math]] \operatorname{P}(Y = 1 | X = 1) = 1 - \operatorname{P}(Y = 0 | X = 1) = 1-0.286 = 0.714. [[/math]]

Therefore

[[math]] \begin{align*} \operatorname{E}(Y | X = 1) = (0) \operatorname{P}(Y = 0 | X = 1) + (1) \operatorname{P}(Y = 1 | X = 1) = (1)(0.714) = 0.714 \\ \operatorname{E}(Y^2 | X = 1) = (0)^2 \operatorname{P}(Y = 0 | X = 1) + (1)^2 \operatorname{P}(Y = 1 | X = 1) = 0.714 \\ \operatorname{Var}(Y | X = 1) = \operatorname{E}(Y^2 | X = 1) – [\operatorname{E}(Y | X = 1)]^2 = 0.714 – (0.714)2 = 0.20 \end{align*} [[/math]]

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