Finding covariance matrix

Question

We have 2 dimensional normal distribution with PDF given by the formula:

$f(x,y)=\frac{1}{4\sqrt{5}\pi}\exp\left\{-\frac{1}{40}\left(9(x-1)^2+8(x-1)(y+1)+4(y+1)^2\right)\right\}$

Let $U_1=X-Y, \ U_2=X+Y+1, \ U_3=-X$.

Find the covariance matrix of vector $(U_1,U_2,U_3)$.

Firstly, let's rewrite this a bit:

$f(x,y)=\frac{1}{2\pi\sqrt{20}}\exp\left\{-\frac{1}{2\cdot20}\left(9(x-1)^2-2\cdot(-4)(x-1)(y+1)+4(y+1)^2\right)\right\}$

Hence

$\mathbf{C_X}=\begin{bmatrix} 4&-4\\-4&9\end{bmatrix}$

And now I have to find a transformation matrix (or however it's called).

$U_1=X-Y \\ U_2=X+Y+1 \\ U_3=-X$

I am not sure what should I do with this "$+1$".

$A_1=\begin{bmatrix} 1&-1&0\\-1&1&0\\ -1&0&0\end{bmatrix} \\ A_2=\begin{bmatrix} 1&-1&0\\-1&1&1\\ -1&0&0\end{bmatrix}$

And then do multiplication $\mathbf{C_{U_1,U_2,U_3}}=A\mathbf{C_X}A^T$

But which $A$ is the right one?

Edit: of course third raw of $\mathbf{C_X}$ is $(0,0,0)$, just like third column

Answer 1

I think you need to use the following fact (see wiki on affine transformations of multivariate normal): if $\mathbf{x}\sim\mathcal{N}(\mathbf{\mu},\mathbf{\Sigma})$ and $\mathbb{y}=\mathbf{c}+\mathbf{B}\mathbf{x}$, then $\mathbf{y}\sim\mathcal{N}(\mathbf{c}+\mathbf{B}\mathbf{\mu},\mathbf{B}\mathbf{\Sigma}\mathbf{B}')$. ($\mathbf{x}$, $\mathbf{\mu}$ are $n\times 1$, $\mathbf{\Sigma}$ is $n\times n$, $\mathbf{c},\mathbf{y}$ are $m\times 1$, $\mathbf{B}$ is $m\times n$.)

In your application, $\mathbf{\mu}'=\left[\begin{array}{cc}-1&1\end{array}\right]$ and $\mathbf{\Sigma}=\left[\begin{array}{cc}4&-4\\-4&9\end{array}\right]$. And you need $\mathbf{B}=\left[\begin{array}{cc}1&-1\\1&1\\-1&0\end{array}\right]$ and $\mathbf{c}=\left[\begin{array}{c}0\\1\\0\end{array}\right]$. The rest is algebra.