# Calculate $Pr(X \ge 0, Y \ge 0)$ where $X, Y \sim N(0,1)$ and their correlation is 1/2 [duplicate]

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?

• if W is independent of T, you can calculate the joint distribution (because is the product) and parametrize the region – Martín Vacas Vignolo Dec 7 '17 at 2:43
• What would it mean to parametrize the region @vvnitram ? Sorry if that's a dumb question, but I'm pretty new to all this. – strwars Dec 7 '17 at 2:55
• Do you know the transformation of random variable formula for the probability density using jacobian? – Abishanka Saha Dec 7 '17 at 3:13
• Sorry, but I don't @AbishankaSaha – strwars Dec 7 '17 at 3:24
• Already asked before: math.stackexchange.com/questions/255368/…. – 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 $$f_{XY}(x,y)=\frac{1}{2πσXσY\sqrt{1−ρ2}}e^{−\frac{1}{2(1-\rho^2)}[((x−μ_X)/σ_X)^2+((y−μ_Y)/σ_Y)^2−2ρ(x−μ_X)(y−μ_Y)/σ_Xσ_Y]}$$
$$f_{XY}(x,y) = \frac{1}{\sqrt{3}\pi}\int_{0}^{\infty}\int_{0}^{\infty} e^\left(-\frac{2}{3}*(x^2+y^2-xy)\right)dy dx$$
$$P(X\gt0,Y\gt0) = \frac{1}{3}$$