7

Given

$$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right) $$

$Y = Z\sqrt{1-\rho^2} + \rho X$

And

$$ f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right) $$

Show that $P(X>0,Y>0)= \frac{1}{4}+\frac{1}{2\pi}(\arcsin \rho) $

I'm supposed to use the fact that X and Z are independent standard normal random variables, but I don't quite understand how. Any help would be greatly appreciated.

BruceET
  • 52,418
user320559
  • 171
  • Another editor temporarily changed the fraction from $\frac14$ to $\frac12$ - I put it back. Please check that our edits haven't changed the question. – Frentos Mar 08 '16 at 01:12
  • 2
    I just answered a similar question in other site. Just too lazy to retype all the stuffs here. http://www.talkstats.com/showthread.php/64068-Analytic-form-for-the-quadrant-probability-of-a-(standard)-bivariate-t-distribution?p=185687&viewfull=1#post185687 – BGM Mar 08 '16 at 03:12
  • Same question:https://math.stackexchange.com/questions/255368/getting-px0-y0-for-a-bivariate-distribution?rq=1. – StubbornAtom Nov 18 '17 at 15:21

2 Answers2

14

Here's a solution that only uses linear algebra and geometry.

If $\pmatrix{X\\ Y}$ is bivariate normal with mean $\pmatrix{0\\0}$ and covariance matrix $\Sigma=\pmatrix{1&\rho\\\rho&1}$, then $\pmatrix{U\\V}=\Sigma^{-1/2} \pmatrix{X\\Y}$ is bivariate normal with mean $\pmatrix{0\\0}$ and covariance matrix $\pmatrix{1&0\\ 0&1}.$ That is, $U$ and $V$ are independent, standard normal random variables.

The illustration below shows that the probability that $\pmatrix{X\\Y}$ lies in the upper quadrant (in blue), is the same as the probability that $\pmatrix{U\\V}$ lies in the wedge (in orange). Since the distribution of $\pmatrix{U\\V}$ is rotationally invariant, simple geometry gives $\mathbb{P}(X>0,Y>0)={\theta\over 2\pi}.$

enter image description here

With $v=\Sigma^{-1/2}\pmatrix{0\\1}$ and $w=\Sigma^{-1/2}\pmatrix{1\\0},$ we have $\cos(\theta)=\langle v,w\rangle /\|v\| \|w\|.$ But \begin{eqnarray*} \langle v,w\rangle &=& (0\ 1)\,\Sigma^{-1}\pmatrix{1\\0}=-\rho/(1-\rho^2)\\[5pt] \|v\|^2&=&(0\ 1)\,\Sigma^{-1}\pmatrix{0\\1}=1/(1-\rho^2)\\[5pt] \|w\|^2&=&(1\ 0)\,\Sigma^{-1}\pmatrix{1\\0}=1/(1-\rho^2), \end{eqnarray*} so that $\cos(\theta)=-\rho.$ Putting it all together gives $$\mathbb{P}(X>0,Y>0)={\arccos(-\rho)\over 2\pi}.$$

0

First, for the connections among $X, Y,$ and $Z$, please see this page on how to generate two correlated normal random variables. I will leave the verification of means, variances, and correlation to you. Also, I assume $p$ in your (edited) post should be $\rho$, and you should change that.

Here is a brief simulation in R, for the case $\rho = 0.8,$ that might be helpful.

 m = 10^6;  rho = .8;  dlt = sqrt(1 - rho^2)
 x = rnorm(m);  z = rnorm(m)  # indep std norm
 y = dlt*z + rho*x
 mean(x > 0 & y > 0)
 ## 0.397643
 1/4 + 1/(2*pi)*asin(rho)
 ## 0.3975836  # agrees to 3 places
 acos(-rho)/(2*pi)  # appended later in view of @Byron's Answer
 ## 0.3975836

Here is a plot of 30,000 (out of a million) $(X, Y)$ pairs simulated.

enter image description here

Community
  • 1
BruceET
  • 52,418
  • thank you! I thought I was supposed to get the answer above through the use of the formulas. Is this the wrong approach? Otherwise, how could I get to the answer given in the exercise? – user320559 Mar 08 '16 at 02:31
  • Not knowing the context of the problem, I can't say for sure what procedure is required. That's why this is a "Comment" rather than an "Answer." Probably you need to evaluate a double integral of the joint density function of X and Y (over the 'blue' region). In many applications, where $\rho$ is known, a simulated result such as I provided would be OK. My guess is this is intended as an exercise in calculus. Then my simulation is likely useful only to help you visualize the problem. Start by understanding the relationships in the reference. – BruceET Mar 08 '16 at 02:52