Exercise: Given $p \in (0,1)$ and $(X,Y)$ a pair of RV with distribution density function $$f(x,y)= \frac{1}{2 \pi \sqrt{1-p^2}}\exp \left( -\frac{1}{2(1-p^2)}(x^2-2pxy+y^2) \right) $$ Compute $P(X>0,Y>0)$

My approach: The calculation as it is stated above might be very hard, however since this a guided (not mandatory) exercise I first had to show that in fact $X$ and $Z=(Y-pX)/\sqrt{1-p^2}$ are two independent standard Gaussian RV, which I managed to complete.

My idea was now to use this knowledge and apply it to the above problem. Evidently we have: $$Y >0 \iff Z=\frac{Y-pX}{\sqrt{1-p^2}}> \frac{-pX}{\sqrt{1-p^2}}=:z_X $$ Therefore I obtain $$P(X>0,Y>0)=P(X>0,Z>z_X)=P(X>0)P(Z>z_X) =\frac{1}{2}P(Z>z_X) $$ using the independence of $X,Z$ and that $X$ has standard Gauss distribution. If these steps so far are correct then I am stuck on how to compute $$P(Z>z_X)= P \left( Z> -pX/\sqrt{1-p^2}\right) $$ The resulting integral still seems very complicated to me and I am not sure if there exists a closed form of it, according to some experiments I have done with WolframAlpha it does.

According to WolframAlpha for $p=0.5$ we get $P(X>0,Y>0)=0.33333 $ however this might be a numerical approximation.

Update: As mentioned by Professor @Xi'an since $z_X$ depends on $X$ I cannot use the 'independence' of $X,Z$ in order to write the probabilities as a product. However $$P(X>0,Y>0)=P\left(X>0,Z> -pX/\sqrt{1-p^2}\right) $$ still gives the correct bounds for the integral. The next promising step might be to change into polar coordinates

  • $\begingroup$ You cannot break your probability into a product since $z_X$ depends on $X$. $\endgroup$ – Xi'an Apr 24 '16 at 16:53
  • 1
    $\begingroup$ There is a nice derivation of this (the bivariate Normal orthant prob) at: math.stackexchange.com/questions/1687795/… $\endgroup$ – wolfies Apr 24 '16 at 17:03
  • 1
    $\begingroup$ There is also a derivation in: Stigler (1989), Normal outhant probabilities, American Statistician, 43, p.291 ..., as well as Stuart and Ord, Kendal's Advanced Theory of Stats $\endgroup$ – wolfies Apr 24 '16 at 17:20
  • 1
    $\begingroup$ Perhaps a swtich to polar coordinates, after switching to independent rvs x and z? That still really just leaves the geometry though, as you need to correctly adjust the ranges of integration into polar coordinates. $\endgroup$ – probabilityislogic Apr 24 '16 at 17:25
  • 1
    $\begingroup$ Invariance by rotation of 2D standard normal distribution yields $$\frac14+\frac1{2\pi}\arcsin p.$$ $\endgroup$ – Did Apr 24 '16 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.