Let $X$ and $Y$ have the bivariate normal density function,
$$ f(x, y) = \frac{1}{2 \pi \sqrt{1 - p^2}} \exp \left\{ - \frac{1}{2(1 - p^2)} (x^2 - 2pxy + y^2) \right\} $$
for fixed $p \in (-1, 1)$. Let $Z = (Y - pX)/\sqrt{1 - p^2}$. I have already proven that $X$ and $Z$ are independent $N(0, 1)$ variables. Now I want to determine $\mathbb{P}(X > 0, Y > 0)$. I know that this can be written as $$ \mathbb{P} (X > 0, Z > -pX / \sqrt{1 - p^2} ) $$ I thought about using the Jacobian variable change thing, with $$ u = x^2 + z^2, v = x / (x^2 + z^2) $$ such that the inverses are given by $$ z = uv, ~~~ x = \sqrt{u - u^2v^2} $$ (note that $x > 0$, so the $-\sqrt{\phantom{x}}$ solution of $x$ is irrelevant). However, the (double) integral I then get is an integral I still cannot compute. Also, I don't know to where $u$ and $v$ map the set $$ \{(x, z) \in \mathbb{R}^2 : x > 0, z > -px / \sqrt{1 - p^2} \} $$ which I need to know in order to know the boundaries for the new (double) integral.
Is this even the right approach? If yes, how should I proceed? If no, what is the right approach?
Thanks in advance!
It seems to me your tranformation, begin non-linear, is needlessly complicated. You already have a linear tranformation: $$ \begin{bmatrix} x \\ y \end{bmatrix} \mapsto \begin{bmatrix} x \\ z \end{bmatrix} = \begin{bmatrix} x \\ ay+bx \end{bmatrix} $$ The Jacobian determinant for this is a constant.