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Let the bivariate random variable $A=(A_1,A_2)^T$ have a Gaussian distribution on $\mathbb{R}^2$ with zero mean and covariance matrix be given by

$$\begin{pmatrix} 1 & -0.4\\-0.4 & 1\end{pmatrix}$$.

Let $B$ = $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $C$= $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. Define $X=B^TA,Y=C^TA$. How do I find the covariance of X and Y?

I know that $cov(X,Y) = E(XY)-E(X)E(Y)$. I don't quite understand how to read a covariance matrix.

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  • $\begingroup$ The covariance matrix of a vector $X$ is defined as $\mathbb{E}\left(XX^\top\right) - \mathbb{E}\left(X\right)\mathbb{E}\left(X\right)^\top$. The $(i,j)$th component of it is the covariance between $X_i$ and $X_j$. $\endgroup$ Commented Oct 30, 2021 at 20:40
  • $\begingroup$ Hi, yes I am aware of that property, but I am confused with how I can obtain $cov(X,Y)$, and also what does $B^TA$ and $C^TA$ actually mean? Thanks in advance! @svensvenson $\endgroup$
    – user986741
    Commented Oct 30, 2021 at 20:44

2 Answers 2

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You have $B^\top A = A_1 + 2A_2$ and $C^\top A = 2A_1 + A_2$. Then,

\begin{eqnarray*} Cov\left(X,Y\right) &=& Cov\left(A_1+2A_2,2A_1+A_2\right)\\ &=& 2Cov\left(A_1,A_1\right) + Cov\left(A_1,A_2\right) + 4Cov\left(A_1,A_2\right)+2Cov\left(A_2,A_2\right). \end{eqnarray*}

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  • $\begingroup$ And we can obtain the covariances from the covariance matrix am I right? $\endgroup$
    – user986741
    Commented Oct 30, 2021 at 20:52
  • $\begingroup$ Yes, the $(i,j)$th entry of the covariance matrix is $Cov\left(A_i,A_j\right)$. $\endgroup$ Commented Oct 30, 2021 at 20:55
  • $\begingroup$ @svensvenson then final answer would be $2(1)+1(-0.4)-4(-0.4)+2(1) = 2$? $\endgroup$
    – f.Greening
    Commented Nov 2, 2021 at 12:42
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You can use matrix properties:

$$\begin{split}\text{Cov}(B^TA, C^TA)&=B^T\text{Cov}(A, A)C\\ &=B^T\text{Var}(A)C\end{split}$$

It is 1-dimensional (a scalar).

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