Solution
Both Above Average
Show the problem again
(X, Y) has a bivariate normal distribution with both means 0, both variances 1, and correlation coefficient 1/2. What is the probability that X > 0 and Y > 0? Give a reduced fraction.
Worked solution
The answer is 1/3. Write X = Z₁ and Y = cos(60°) Z₁ + sin(60°) Z₂ with Z₁, Z₂ independent standard normals; this has the right variances and correlation cos 60° = 1/2. So X = u·Z and Y = v·Z where Z = (Z₁, Z₂) and u, v are unit vectors 60° apart. Z is rotationally symmetric. The event X > 0 is the half-plane on u's side, and Y > 0 the half-plane on v's side. Their intersection is a wedge with angle 180° − 60° = 120°, so the probability is 120/360 = 1/3. In general P(X > 0, Y > 0) = 1/4 + arcsin(ρ)/(2π), which gives 1/4 for independent variables and 1/2 for perfectly correlated ones.
Source: Sheppard's orthant formula for the bivariate normal (W. F. Sheppard, 1899); this instance in wide circulation. Statement written for AxiomIQ.