How to Solve Multivariate Probability — Multivariate — discrete joint table (marginal/conditional/covariance) Questions on Exam P
Sample Practice Problem
ID: #ZAQN9
Binary indicators \(X\) and \(Y\) have \(\Pr(X=1,Y=0)=0.110\) and \(\Pr(X=1,Y=1)=0.090\). Calculate the conditional variance of \(Y\) given \(X=1\).
📖 Worked Solution & Strategy
The marginal probability of \(X=1\) is \(\Pr(X=1) = \Pr(X=1, Y=0) + \Pr(X=1, Y=1) = 0.110 + 0.090 = 0.20\). Conditioning on \(X=1\) gives success probability \(p = \Pr(Y=1 \mid X=1) = \frac{\Pr(X=1, Y=1)}{\Pr(X=1)} = \frac{0.090}{0.20} = \frac{9}{20}\). Conditional on \(X=1\), \(Y\) is a Bernoulli random variable with parameter \(p = \frac{9}{20}\). Its conditional variance is \(\operatorname{Var}(Y \mid X=1) = p(1-p) = (\frac{9}{20})(1 - \frac{9}{20}) = \frac{99}{400}\). The evaluated result is \(\frac{99}{400}\), which matches the correct option.