Question
U, V, follow independent univariate normal with mean zero and variance one. Let X = U and Y = pU+$\sqrt{1-p^2}$V where $|p|$ < 1.
$\Sigma_{XY} = \pmatrix{1 & p \\ p & 1}$
Find the joint distribution of X,Y.
My Work
I note that U and V are independent univariate normal. Define Z = $\pmatrix{U \\ V}$ and W = $\pmatrix{X \\ Y}$. Let W = B * Z where B is a 2x2 constant matrix, then W ~ MVN(B$\mu$, B$\Sigma_{UV}$$\textbf{B}^T$).
\begin{align} \pmatrix{X \\ Y} = \textbf{B}\pmatrix{U \\V} \\ \pmatrix{\textbf{U} \\ p\textbf{U} + \sqrt{1-p^2}\textbf{V}} = \textbf{B}\pmatrix{U \\V} \\ B = \pmatrix{b_{11} & b_{12} \\ b_{21} & b_{22}} \\ \therefore \pmatrix{b_{11}U + b_{12}V \\b_{21}U + b_{22}V} = \pmatrix{\textbf{U} \\ p\textbf{U} + \sqrt{1-p^2}\textbf{V}} \\ \Rightarrow \matrix{b_{11}=1& b_{12}=0\\b_{21} =p & b_{22} = \sqrt{1-p^2}} \\ \textbf{B} = \pmatrix{1&0\\p&\sqrt{1-p^2}}\\ \therefore\textbf{W} =\pmatrix{1 & 0 \\p & \sqrt{1-p^2}}*\pmatrix{U \\V} \end{align}
So we have
B$\mu$ = $\pmatrix{1 & 0 \\p & \sqrt{1-p^2}} * \pmatrix{0 \\ 0} = \pmatrix{0 \\0}$ and
B$\Sigma_{UV}\textbf{B}^T$ = $\Sigma_{XY}$ and therefore W ~ MVN$\pmatrix{\pmatrix{0 \\0}, & \pmatrix{1 & p \\ p & 1}}$
Which can be written explicitly as
$$ f_{XY} = \frac{1}{2\pi * \sqrt{det(\Sigma_{XY})}}*e^{\frac{-1}{2}(\textbf{W} - \mu)^T\Sigma_{XY}^{-1}(\textbf{W} - \mu)} $$
Thank you.