Question

Let two variables $X_1$ and $X_2$ are bivariately normally distributed with mean vector component $\mu_1$ and $\mu_2$ and co-variance matrix $\sum$ shown below:

$\sum = \begin{bmatrix} 1 & r\\ r& 1 \end{bmatrix}$.

(a) What is the probability distribution function of joint Gaussian $P(X_1, X_2)$? (Show it with $\mu$ and $\sum$)

(b) What is the eigenvalues of co-variance matrix $\sum$?

(c) Given the condition that the sum of squared values of each eigenvector are equal to 1, what is the eigenvectors of co-variance matrix $\sum$?

please help with all parts!

thank you!

Answer 1

$(a)~The~joint~distribution~of~(X_1,X_2)~is\\ f(x_1,x_2)=\frac{1}{2\pi|\Sigma|^{\frac{1}{2}}}\exp\left(-\frac{1}{2}({\bf x}-\boldsymbol{\mu})^\prime \Sigma^{-1}({\bf x}-\boldsymbol{\mu}) \right ),~-\infty<x_1,x_2<\infty\\ where,~{\bf x}=\begin{pmatrix} x_1\\ x_2 \end{pmatrix},~\boldsymbol{\mu}=\begin{pmatrix} \mu_1\\ \mu_2 \end{pmatrix},~|\Sigma|=1-r^2,~\Sigma^{-1}=\frac{1}{1-r^2}\begin{pmatrix} 1&-r\\ -r&1 \end{pmatrix}\\ ({\bf x}-\boldsymbol{\mu})=\begin{pmatrix} x_1-\mu_1\\ x_2-\mu_2 \end{pmatrix},\\ ~({\bf x}-\boldsymbol{\mu})^\prime\Sigma^{-1}({\bf x}-\boldsymbol{\mu})=\frac{1}{1-r^2}\left((x_1-\mu_1)^2-2r(x_1-\mu_1)(x_2-\mu_2)&plus;(x_2-\mu_2)^2 \right )\\ then~f(x_1,x_2)=\frac{1}{2\pi\sqrt{1-r^2}}\exp\left(-\frac{1}{2(1-r^2)}\left((x_1-\mu_1)^2-2r(x_1-\mu_1)(x_2-\mu_2)&plus;(x_2-\mu_2)^2 \right ) \right ),\\ ~-\infty<x_1,x_2<\infty\\$

$(b)\\ ~|\Sigma-\lambda {\bf I}|=\begin{vmatrix} 1-\lambda &r \\ r & 1-\lambda \end{vmatrix}=(1-\lambda)^2-r^2=0~or,~1-\lambda=\pm r\\ ~or,~\lambda=1-r~or~1&plus;r\\ then~the~eigenvalues~are~1-r~and~1&plus;r.\\ (c)\\ \Sigma {\bf x}=(1-r){\bf x}~or,~\begin{pmatrix} x_1&plus;rx_2\\ rx_1&plus;x_2 \end{pmatrix}=\begin{pmatrix} (1-r)x_1\\ (1-r)x_2 \end{pmatrix}\Rightarrow x_1&plus;rx_2=(1-r)x_1\\ or,~x_2=-x_1....(1)\\ Again~x_1^2&plus;x_2^2=1~or,~2x_2^2=1~(using~(1))~or,~x_2=\pm \frac{1}{\sqrt{2}}\\ Take,~x_2=-\frac{1}{\sqrt{2}}~then~x_1=\frac{1}{\sqrt{2}}.\\ Hence~eigen vector~coorsponding~to~eigenvalue~1-r~is~\begin{pmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{pmatrix}\\$

$\Sigma {\bf x}=(1&plus;r){\bf x}~or,~\begin{pmatrix} x_1&plus;rx_2\\ rx_1&plus;x_2 \end{pmatrix}=\begin{pmatrix} (1&plus;r)x_1\\ (1&plus;r)x_2 \end{pmatrix}\Rightarrow x_1&plus;rx_2=(1&plus;r)x_1\\ or,~x_2=x_1....(1)\\ Again~x_1^2&plus;x_2^2=1~or,~2x_2^2=1~(using~(1))~or,~x_2=\pm \frac{1}{\sqrt{2}}\\ Take,~x_2=\frac{1}{\sqrt{2}}~then~x_1=\frac{1}{\sqrt{2}}.\\ Hence~eigen vector~coorsponding~to~eigenvalue~1&plus;r~is~\begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix}\\$