How do I find the correlation between $Y_1$ and $Y_2$ with matrices?

Question

Question Let $x = [X_1, X_2, X_3, X_4]^T$ be a multivariate and normal random vector and $$ \mu = \begin{bmatrix} 0 \\ -1 \\ 2 \\ 0 \end{bmatrix}, \Sigma =\begin{bmatrix} 4 & 1 &0 & 2 \\ 1 & 4 & 3 & 0 \\ 0 & 3 & 9 & 3 \\ 2 & 0 & 3 & 9 \end{bmatrix}, $$

$Y_1 = X_1 + X_3$, $Y_2 = X_2 - X_4$ and $Y \sim N(A\mu, A \Sigma A^T)$.

How do I find the correlation between $Y_1$ and $Y_2$? $$ A\mu = \begin{bmatrix} 2 \\ -1 \end{bmatrix} $$

$$ A \Sigma A^T = \begin{bmatrix} 13 & -1 \\ -1 & 13 \end{bmatrix} $$

The solution has this

$$\operatorname{Corr}(Y_1, Y_2) = \frac{-1}{\sqrt{13 \times 13}} = -0.08$$

I'm not sure what formula its following.

Answer 1

It is just a matter of making boring but easy calculations:

1. First calculate the covariance (the numerator of your correlation coefficient)

$$Cov(Y_1,Y_2)=E((X_1+X_3)(X_2-X_4)]-E(X_1+X_3)E(X_2-X_4)$$

$$E(X_1X_2)-E(X_1X_4)+E(X_2X_3)-E(X_3X_4)-2(-1)$$

now, for example

$$E(X_1X_2)=Cov(X_1X_2)+E(X_1)E(X_2)=1+0$$

(the values are expressed in the vector of means and in the matrix $\Sigma$, the covariances matrix)

Thus

$$Cov(Y_1,Y_2)=1-2+5-3+2=-1$$

2. The denominator is $\sqrt{V(X_1+X_3)V(X_2-X_4)}$

$$V(X_1+X_3)=V(X_1)+V(X_3)+2Cov(X_1,X_3)=4+9+0=13$$

$$V(X_2-X_4)=V(X_2)+V(X_4)-2Cov(X_2,X_4)=4+9+0=13$$

3. Finally

$$\rho(Y_1,Y_2)=-\frac{1}{13}$$