Find the covariance matrix of a vector of random variables If $X$ and $Y$ are independent and identical gaussian random variables with $\mu=0$ and $\sigma^2=1$, how can I calculate the covariance matrix of the following vector:
$$Z=\begin{pmatrix}
    X+2Y\\
    1-X-Y\\
    2X-Y
  \end{pmatrix}$$ 
I tried to start by writing $E(Z)$, the vector contaning the mean values of $Z$:
$$E(Z)=\begin{pmatrix}
    0\\
    1\\
    0
  \end{pmatrix}$$
And, by definition, I wrote $\operatorname{cov}(Z)=E((Z-E(Z)(Z-E(Z))^T))$, and
$$\operatorname{cov}(Z)=E\begin{pmatrix}
    Z_1^2 & Z_1Z_2 & Z_1Z_3\\
    Z_2Z_1 & Z_2^2 & Z_2Z_3\\
    Z_3Z_1 & Z_3Z_2 & Z_3^2
  \end{pmatrix}$$
with $Z_1=N(0,5), Z_2=N(-1,2), Z_3=N(0,5)$
But I how do I go further? I mean, how do I find the covariance between $Z_1$ and $Z_2$? Can I say they are independent?
 A: $\newcommand{\c}{\operatorname{cov}}$
$$
Z=\begin{bmatrix}
    X+2Y\\
    1-X-Y\\
    2X-Y
  \end{bmatrix}
$$
You can use the fact that covariance is linear in each variable separately, thus:
\begin{align}
\c(X+2Y, 1 - X - Y) = {} & \c(X, 1-X-Y) + 2\c(Y,1-X-Y) \\[10pt]
= {} & \Big( \c(X,1) - \c(X,X) - \c(X,Y)\Big) \\
& {} + \Big(2\c(Y,1) - 2\c(Y,X) - 2\c(Y,Y)\Big) \\[10pt]
= {} & \text{etc.}
\end{align}
You can also use matrix algebra.
$$
\begin{bmatrix} Z_1 \\ Z_2 \\ Z_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}  + \begin{bmatrix} \phantom{-}1 & \phantom{-}2 \\ -1 & -1 \\ \phantom{-}2 & -1 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}
$$
Write this as $Z = MW,$ where $M$ is that $3\times2$ matrix. Then we have
$$
\c(MW) = M \Big( \c W\Big) M^T
$$
and $\c W$ is the $2\times2$ identity matrix.
A: We have that$$\operatorname{cov}(Z)=E\{(Z-E(Z))(Z-E(Z))^T\}=E\left\{\begin{bmatrix}(X+2Y)^2&(X+2Y)(-X-Y)&(X+2Y)(2X-Y)\\(X+2Y)(-X-Y)&(X+Y)^2&(-X-Y)(2X-Y)\\(X+2Y)(2X-Y)&(-X-Y)(2X-Y)&(2X-Y)^2\end{bmatrix}\right\}=\begin{bmatrix}5&-3&0\\-3&2&-1\\0&-1&5\end{bmatrix}$$
