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.

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.