Question

(If needed, use the definitions on the next page to aid you in answering this question) Suppose that Wi and W2 are bivariate normal with mean a (aa2) and covariance matrix Use the following formula 1) ws(/2) to show that the conditional distribution of W1 given W2 w2 is univariate normal In other words, you need to simplify () to the point where you have a form exp 2πσ where μ and σ depend on ai ,a2,bi,b2,p and u2- What are μ and σ2, the conditional mean and variance of WilWw

Answer 1

$The~joint~pdf~of~(W_1,W_2)~is\\ f_{W_1,W_2}(w_1,w_2)=\frac{1}{2\pi b_1b_2\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)} \left(\left(\frac{w_1-a_1}{b_1} \right )^2-2\rho\left(\frac{w_1-a_1}{b_1} \right )\left(\frac{w_2-a_2}{b_2} \right )&plus;\left(\frac{w_2-a_2}{b_2} \right )^2 \right )\right )\\ =\frac{1}{2\pi b_1b_2\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)} \left(\frac{w_1-a_1}{b_1}-\rho\frac{w_2-a_2}{b_2} \right )^2-\frac{1}{2}\left(\frac{w_2-a_2}{b_2} \right )^2\right )\\ =\left(\frac{1}{\sqrt{2\pi} b_1\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)b_1^2} \left(w_1-a_1-\rho\frac{b_1}{b_2}(w_2-a_2) \right )^2\right ) \right )\left(\frac{1}{\sqrt{2\pi}b_2}\exp\left(-\frac{1}{2}\left(\frac{w_2-a_2}{b_2} \right )^2\right ) \right )\\$$The~marginal~pdf~of~W_2~is\\ f_{W_2}(w_2)=\left(\frac{1}{\sqrt{2\pi}b_2}\exp\left(-\frac{1}{2}\left(\frac{w_2-a_2}{b_2} \right )^2\right ) \right )\int_{-\infty}^\infty \left(\frac{1}{\sqrt{2\pi} b_1\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)b_1^2} \left(w_1-a_1-\rho\frac{b_1}{b_2}(w_2-a_2) \right )^2\right ) \right )dw_1\\ =\frac{1}{\sqrt{2\pi}b_2}\exp\left(-\frac{1}{2}\left(\frac{w_2-a_2}{b_2} \right )^2\right )~-\infty<w_2<\infty\\ since~\left(\frac{1}{\sqrt{2\pi} b_1\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)b_1^2} \left(w_1-a_1-\rho\frac{b_1}{b_2}(w_2-a_2) \right )^2\right ) \right )~\\ is~pdf~of~N(a_1&plus;\rho\frac{b_1}{b_2}(w_2-a_2),~(1-\rho^2)b_2^2)\\ so\\~\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} b_1\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)b_1^2} \left(w_1-a_1-\rho\frac{b_1}{b_2}(w_2-a_2) \right )^2\right ) dw_1\\=1\\$$The~conditional~pdf~of~W_1~given~W_2=w_2~is\\ f_{W_1|W_2=w_2}(w_1)=\frac{f_{W_1,W_2}(w_1,w_2)}{f_{W_2}(w_2)}\\ =\frac{=\left(\frac{1}{\sqrt{2\pi} b_1\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)b_1^2} \left(w_1-a_1-\rho\frac{b_1}{b_2}(w_2-a_2) \right )^2\right ) \right )\left(\frac{1}{\sqrt{2\pi}b_2}\exp\left(-\frac{1}{2}\left(\frac{w_2-a_2}{b_2} \right )^2\right ) \right )\\ }{\frac{1}{\sqrt{2\pi}b_2}\exp\left(-\frac{1}{2}\left(\frac{w_2-a_2}{b_2} \right )^2\right )}\\ =\frac{1}{\sqrt{2\pi} b_1\sqrt{1-\rho^2}}\exp\left(-\frac{1}{2(1-\rho^2)b_1^2} \left(w_1-a_1-\rho\frac{b_1}{b_2}(w_2-a_2) \right )^2\right ) ;\\ ~-\infty<w_1<\infty\\ Hence~W_1|W_2=w_2\sim N\left(a_1&plus; \rho\frac{b_1}{b_2}(w_2-a_2) ,~b_1^2(1-\rho^2)\right ).\\ Here,~\mu=a_1&plus; \rho\frac{b_1}{b_2}(w_2-a_2) ,~\sigma^2=b_1^2(1-\rho^2).$