Compute expected value from two dimensional normal distribution

Question

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.

Answer 1

The term in the exponent of $f(x,y)$ can be written as

$$\begin{split}-\frac 18\left[4x^2+8x(y+3)+6(y+3)^2\right] &= -\frac 12 \left[x^2+2x(y+3)+\frac 32(y+3)^2\right]\\ &=-\frac 12\begin{pmatrix}x\\ y+3\end{pmatrix}^T\begin{pmatrix}1&1\\1&3/2\end{pmatrix}\begin{pmatrix}x\\y+3\end{pmatrix}\end{split}$$

Thus $\Sigma^{-1}=\begin{pmatrix}1&1\\1&3/2\end{pmatrix}$ and the covariance matrix is $\Sigma=\frac 1{3/2-1}\begin{pmatrix}3/2&-1\\-1&1\end{pmatrix}=\begin{pmatrix}3&-2\\-2&2\end{pmatrix}$

Besides this, your calculations look correct, so just redo the parts that need changing.