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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.

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1 Answer 1

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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, $$ \begin{pmatrix} X\\ Z\end{pmatrix} \sim N\left(\begin{pmatrix} 1\\ \frac32\end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & \frac12\end{pmatrix}\right) $$
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  • $\begingroup$ Zero covariance in general doesn't imply independance. $\endgroup$
    – mowzorn
    Commented Jan 1, 2023 at 11:25
  • $\begingroup$ @mowzorn Yes, but zero covariance of two normal random variables indicates independence. $\endgroup$
    – Chia
    Commented Jan 1, 2023 at 11:30
  • $\begingroup$ I just added this comment in case someone thought that always $0$ covariance implies independence $\endgroup$
    – mowzorn
    Commented Jan 1, 2023 at 11:33

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