I've been working on this problem and got stuck.

Assume that X and Y have joint normal distribution, that each $X, Y \sim N(0,1)$, and that their correlation is $\frac{1}{2}$. Calculate $Pr(X \ge 0, Y \ge 0)$.

I know that I can rewrite (X,Y) in terms of independent normal variables (W,T) as follows

$$ \begin{bmatrix} X\\ Y\\ \end{bmatrix} = \begin{bmatrix} 1 & 0\\ \rho & \sqrt{1-\rho^2}\\ \end{bmatrix} \begin{bmatrix} X\\ Z\\ \end{bmatrix} $$ where $\rho$ is the correlation. This has left me with $$ X=X\\ Y=\frac{1}{2}X + Z\sqrt{\frac{3}{4}} $$ And so I have $$ Pr(X \ge 0, Y \ge 0) = Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0) $$ So I just realized that the T variable I had before was superfluous, so I removed it. I am wondering from here if it would make sense to put this: $$ Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0)=Pr(X \ge 0, Z\sqrt{\frac{3}{4}} \ge -\frac{1}{2}X) $$ Still not sure how to proceed from here. Maybe simplify to $$ Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0)=Pr(X \ge 0, Z\sqrt{\frac{3}{4}} \ge 0) $$ Since we are looking for $Pr(X \ge 0)$ anyway?

  • $\begingroup$ if W is independent of T, you can calculate the joint distribution (because is the product) and parametrize the region $\endgroup$ – Martín Vacas Vignolo Dec 7 '17 at 2:43
  • $\begingroup$ What would it mean to parametrize the region @vvnitram ? Sorry if that's a dumb question, but I'm pretty new to all this. $\endgroup$ – strwars Dec 7 '17 at 2:55
  • $\begingroup$ Do you know the transformation of random variable formula for the probability density using jacobian? $\endgroup$ – Abishanka Saha Dec 7 '17 at 3:13
  • $\begingroup$ Sorry, but I don't @AbishankaSaha $\endgroup$ – strwars Dec 7 '17 at 3:24
  • $\begingroup$ Already asked before: math.stackexchange.com/questions/255368/…. $\endgroup$ – StubbornAtom Dec 7 '17 at 7:58

Two random variables X and Y are said to have a bivariate normal distribution with parameters $μX, σ2X, μY, σ2Y,$ and $ρ$, if their joint PDF is given by $$f_{XY}(x,y)=\frac{1}{2πσXσY\sqrt{1−ρ2}}e^{−\frac{1}{2(1-\rho^2)}[((x−μ_X)/σ_X)^2+((y−μ_Y)/σ_Y)^2−2ρ(x−μ_X)(y−μ_Y)/σ_Xσ_Y]}$$

$$f_{XY}(x,y) = \frac{1}{\sqrt{3}\pi}\int_{0}^{\infty}\int_{0}^{\infty} e^\left(-\frac{2}{3}*(x^2+y^2-xy)\right)dy dx$$

$$P(X\gt0,Y\gt0) = \frac{1}{3}$$

evaluated by Wolfram

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  • $\begingroup$ So we can just tell it is bivariate normal because they are both normal? Or is joint normal the same as bivariate normal? @SatishRamanathan $\endgroup$ – strwars Dec 7 '17 at 4:44
  • $\begingroup$ If there happens to be a correlation between otherwise independent X and Y, then the joint distribution is given by the above relationship $\endgroup$ – Satish Ramanathan Dec 7 '17 at 5:15
  • $\begingroup$ Look at this link it derives what you are asking for. probabilitycourse.com/chapter5/5_3_2_bivariate_normal_dist.php $\endgroup$ – Satish Ramanathan Dec 7 '17 at 5:34

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