# Independence, Mean, Covariance Matrix

Let $$(X, Y)$$ be normal distributed with mean vector $$\mu=c(1,2)$$ and covariance matrix $$\Sigma=\left(\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right) .$$ Define $$Z:=Y-\frac{\Sigma_{1,2}}{\Sigma_{1,1}} X$$ and show that $$(X, Z)$$ are independent and $$(X, Z)$$ is normal distributed. Find the mean $$\mu_{(X, Z)}$$ of $$(X, Z)$$ as well as its covariance matrix $$S$$.

Here's what I've tried so far:

From what I know, $$\Sigma_{1,2}$$ = 1 and $$\Sigma_{1,1} = 2$$. I think that '1,2' just means $$1^{st}$$ row $$2^{nd}$$ column and that's why $$\Sigma_{1,2}$$ = 1 and so on. But I'm not sure how to really solve the question and help would be appreciated. Thank you.

The r.v. $$Z$$ is normally distributed.
1. $$X$$ is independent of $$Z$$:
\begin{aligned} cov(X, Z) &= cov(X, Y - \frac12 X) \\ &=cov(X, Y) - \frac12 cov(X, X) \\ &= 1 - \frac12 \times 2 \\ &=0 \end{aligned}
2. $$(X, Z)$$ is normally distributed since $$X$$ and $$Z$$ are normally distributed and $$X$$ is independent of $$Z$$.
3. $$EX = 1$$, $$EZ = EY-1/2 EX=3/2$$. and the covariance matrix of $$(X, Z)$$ is \begin{aligned} S &= \begin{pmatrix} Var(X) & cov(X, Z) \\ cov(X,Z) & Var(Z) \end{pmatrix} \\ &= \begin{pmatrix} 2 & 0 \\ 0 & Var(Y-\frac12X) \end{pmatrix} \\ &= \begin{pmatrix} 2 & 0 \\ 0 & Var(Y) + \frac14Var(X) - cov(X,Y) \end{pmatrix} \\ &= \begin{pmatrix} 2 & 0 \\ 0 & \frac12 \end{pmatrix} \\ \end{aligned} Hence, $$\begin{pmatrix} X\\ Z\end{pmatrix} \sim N\left(\begin{pmatrix} 1\\ \frac32\end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & \frac12\end{pmatrix}\right)$$
• I just added this comment in case someone thought that always $0$ covariance implies independence Commented Jan 1, 2023 at 11:33