A random vector (X, Y) is normally distributed with the following joint density:
$$ f^{X,Y}(x,y) = \frac{e^{\frac{1}{2} \left(-5 \text{x$$}{}^2-2 \text{x$$} \text{y$$}-\text{y$$}{} ^2\right)}}{\pi } $$
I need to calculate the joint density
$$ f^{U,V} $$
where
$$ U = X + Y $$
and
$$ V = 2X $$
Am I not mistaken that the location of my new distribution is going to be the following vector?
$$ \begin{bmatrix} \mathbb{E} [X + Y]\\ \mathbb{E} [2X] \end{bmatrix} $$
I've calculated the marginal pdfs for x and y:
$$ f^{X}(x,y) = \sqrt{\frac{2}{\pi }} e^{-2x^2} $$ $$ f^{Y}(x,y) = \sqrt{\frac{2}{5 \pi }} e^{-\frac{2 y^2}{5}} $$
and obtained the expected values for x and y: $$ \mathbb{E} [X] = \int_{}^{} x * f^{X}(x,y) dx = 0 $$
$$ \mathbb{E} [Y] = \int_{}^{} y * f^{Y}(x,y) dy = 0 $$
Now since x and y are normally distributed the location of x + y is going to be 0 as well.
Does it make any sense? How do I get the covariance matrix? Thanks a lot for any hints!
Following @stochastic's comment, I was able to figure it out. This is how it goes:
$$ f^{X,Y}(x,y) = \frac{e^{\frac{1}{2} \left(-5 \text{x$$}{}^2-2 \text{x$$} \text{y$$}-\text{y$$}{} ^2\right)}}{\pi } $$
$$ U = X + Y $$
$$ V = 2X $$
Our transformation function looks then like this:
$$ T(\begin{bmatrix} x \\ y \\ \end{bmatrix}) = \begin{bmatrix} u \\ v \\ \end{bmatrix} = \begin{bmatrix} x + y \\ 2x \\ \end{bmatrix} $$
We need its inverse:
$$ T^{-1}(\begin{bmatrix} x \\ y \\ \end{bmatrix}) = T(\begin{bmatrix} \frac{1}{2}v \\ u - \frac{1}{2}v \\ \end{bmatrix}) $$
and the jacobian:
$$ DT^{-1} = \begin{bmatrix} 0 & \frac{1}{2}\\ 1 & -\frac{1}{2} \\ \end{bmatrix} $$
The determinant of said jacobian is:
$$ |DT^{-1}| = -\frac{1}{2} $$
From here we're ready to obtain the transformed joint density by substituting for u and v, and normalizing by the absolute value of the jacobian: $$ f^{U,V}(u,v) = f^{X,Y}(\frac{1}{2}v, u - \frac{1}{2}v) * |-\frac{1}{2}| = \frac{1}{\pi} e^{-\frac{u^{2}}{2} -\frac{v^{2}}{2} } $$