# Show the bivariate normal cdf evaluated at (0,0) is increasing in the correlation coefficient

$$\begin{bmatrix}\epsilon_{1}\\ \epsilon_{2}\end{bmatrix}\sim N(\begin{bmatrix}0\\0\end{bmatrix},\begin{bmatrix}1,\rho\\ \rho, 1\end{bmatrix})$$. Show that the joint cdf evaluated at (0,0), i.e., $$F_{\epsilon_{1},\epsilon_{2}}(0,0,\rho)\equiv Pr(\epsilon_{1}\leq 0, \epsilon_{2} \leq0)$$ is monotonically increasing in $$\rho \in (-1,1)$$. Numerical calculation below (using 2000 grid points) shows that this is almost indeed the case. Thanks! • The analysis at stats.stackexchange.com/a/71303/919 should make this result obvious, because as $\rho$ increases, the probability assigned to the first quadrant corrresponds to an ever larger region for the standard bivariate Normal distribution under the area preserving "lifting" transformation $(x,y)\to (x, y+\rho x ).$ – whuber Feb 10 at 3:37
• Thank you very much. This link is indeed very helpful! – JTS365 Feb 10 at 5:11

Take independent standard normal random variables $$X$$ and $$Y$$. Then the joint distribution of $$X$$ and $$\rho X-\sqrt{1-\rho^2}\,Y$$ is the same as joint distribution of $$\epsilon_1$$, $$\epsilon_2$$. This is Cholesky's decomposition. Then $$\mathbb P(\epsilon_1\leq 0, \epsilon_2\leq 0) = \mathbb P\left(X\leq 0, \rho X-\sqrt{1-\rho^2}\,Y \leq 0\right)=\mathbb P\left(X\leq 0, Y\geq \frac{\rho}{\sqrt{1-\rho^2}}X \right)$$ The function $$\frac{\rho}{\sqrt{1-\rho^2}}$$ is monotonly increasing in $$\rho\in[-1,1]$$. Therefore, if $$\rho_1<\rho_2$$ then $$\frac{\rho_1}{\sqrt{1-\rho_1^2}}<\frac{\rho_2}{\sqrt{1-\rho_2^2}}$$. Then for $$X< 0$$,
$$\frac{\rho_1}{\sqrt{1-\rho_1^2}}X > \frac{\rho_2}{\sqrt{1-\rho_2^2}}X$$ and if $$Y$$ is greater the first, it is definitely greater the second. So we have the inclusion of the events: $$\left\{X\leq 0, Y\geq \frac{\rho_1}{\sqrt{1-\rho_1^2}}X \right\} \subset \left\{X\leq 0, Y\geq \frac{\rho_2}{\sqrt{1-\rho_2^2}}X \right\}.$$ This events are equal only when $$X=0$$ which has zero probability. Therefore the probability of the first event is strictly less than the probability of the second, which means that $$\mathbb P(\epsilon_1\leq 0, \epsilon_2\leq 0)$$ is strictly increasing in $$\rho$$.
• Thanks a lot! Your answer is quite elegant, and completely solved my problem. Seems the (0,0) here plays a crucial role, and when evaluated at other points this monotonicity in $\rho$ is not necessarily true. – JTS365 Feb 10 at 5:06
$$\int_0^\infty\int_0^\infty f(x,y;\rho)\;\mathrm dx\;\mathrm dy = \frac 14 + \frac{\arcsin \rho}{2\pi}$$ where $$f(x,y;\rho)$$ is the bivariate normal density of two standard normal random variables with correlation coefficient $$\rho$$. By symmetry, $$\frac 14 + \frac{\arcsin \rho}{2\pi}$$ is also the value of the joint CDF at the origin. Since $$\arcsin\rho$$ increases monotonically from $$-\frac{\pi}{2}$$ to $$+\frac{\pi}{2}$$ as $$\rho$$ increases from $$-1$$ to $$+1$$, this proves the desired result.
• Thank you so much! This analytical formula for the integral is awesome, showing exactly what's going on with $\rho$. – JTS365 Feb 10 at 5:27