Q155·Section III (Statistics)·155 of 180A conditional variance that is really an independence checkX‾2×1∼MN2 ((μ1μ2),(1ρρ1))\uline{X}_{2\times1}\sim MN_2\!\left( \begin{pmatrix}\mu_1\\\mu_2\end{pmatrix}, \begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}\right)X2×1∼MN2((μ1μ2),(1ρρ1)). Then Var(X1+X2∣X1−X2=Y)\operatorname{Var}(X_1+X_2\mid X_1-X_2=Y)Var(X1+X2∣X1−X2=Y) equals: OptionsA1+ρ1+\rho1+ρBρ(1+ρ)\rho(1+\rho)ρ(1+ρ)C222D2(1+ρ)2(1+\rho)2(1+ρ)Pick an option to check your answer.Show answerShow solution Report a problem with this question← Q154All questionsQ156 →Keyboard: A–D answer · S solution · ←/→ previous/next