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!

  • $\begingroup$ Maybe this will help: math.stackexchange.com/questions/1687795/… $\endgroup$ – user940 Apr 4 '16 at 17:01
  • $\begingroup$ Conventionally one sees the letter $\rho$ rather than the letter $p$ here. Might that be what you intended? $\qquad$ $\endgroup$ – Michael Hardy Apr 4 '16 at 17:03
  • $\begingroup$ @MichaelHardy Yes, but writing $\rho$ takes more keypresses than writing $p$, and since it doesn't really matter how we name $\rho$ here, I thought no one would mind ;) $\endgroup$ – limitIntegral314 Apr 5 '16 at 11:42

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.

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  • $\begingroup$ Yes, I can just do $x = x$ and $z = (y - px)/(\sqrt{1 - p^2})$, compute the inverses, and determine the Jacobian determinant, but what are the new boundaries? $\endgroup$ – limitIntegral314 Apr 5 '16 at 10:21
  • $\begingroup$ The boundaries are where $x=0$ and where $y=0$; thus they are where $x=0$ and $ay+bx=0$. Both are straight lines. $\qquad$ $\endgroup$ – Michael Hardy Apr 5 '16 at 18:50

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