Random vector (X, Y) has the two-dimensional normal distribution with the density $ f(x,y)=\frac{1}{2\pi\sqrt{2}}\exp\left\{ -\frac{1}{8} \left[ 4x^2 +8x(y+3)+6(y+3)^2 \right] \right\} $
Compute $\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ]$
Find the covariance matrix of the random vector $(U, V) = (-2X+Y, X-Y)$
Solution:
$det \; C = 2$
$ f(x,y)=\frac{1}{2\pi\sqrt{2}}\exp\left\{ -\frac{1}{4} \left[ 2x^2 +4x(y+3)+3(y+3)^2 \right] \right\} $
$ m = \begin{pmatrix} 0\\ -3 \end{pmatrix} = \begin{pmatrix} \mathbb{E} X\\ \mathbb{E} Y \end{pmatrix} $
$ C = \begin{pmatrix} 3 & 2\\ 2 & 2 \end{pmatrix} = \begin{pmatrix} Var X & Cov(X, Y)\\ Cov(X, Y) & Var Y \end{pmatrix} $
$ \mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ] = \mathbb{E}[XY] - 2\mathbb{E}X - 3\mathbb{E}[Y^2] + 1 $
$ Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}X\mathbb{E}Y \rightarrow \mathbb{E}[XY]=2 $
$ Var Y = \mathbb{E}(Y^2)-(\mathbb{E}Y)^2 \rightarrow (\mathbb{E}Y)^2 = 11 $
$ \mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ] =2-2\cdot0-3\cdot11+1=-30 $
$ A \cdot \begin{pmatrix} X\\ Y \end{pmatrix} = \begin{pmatrix} -2X & Y\\ X & -Y \end{pmatrix} \rightarrow A = \begin{pmatrix} -2 & 1\\ 1 & -1 \end{pmatrix} $
$ C_{UV} = A \cdot C \cdot A^T = \begin{pmatrix} 6 & -2\\ 2 & 1 \end{pmatrix} $
I am asking for verification of the solution and corrections if needed.