1
$\begingroup$

$\begin{bmatrix}\epsilon_{1}\\ \epsilon_{2}\end{bmatrix}\sim N(\begin{bmatrix}0\\0\end{bmatrix},\begin{bmatrix}1,\rho\\ \rho, 1\end{bmatrix})$. 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! enter image description here

$\endgroup$
  • $\begingroup$ 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 $(x,y)\to (x, y+\rho x ).$ $\endgroup$ – whuber Feb 10 at 3:37
  • $\begingroup$ Thank you very much. This link is indeed very helpful! $\endgroup$ – TD888 Feb 10 at 5:11
1
$\begingroup$

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 $\rho$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – TD888 Feb 10 at 5:06
0
$\begingroup$

According to Formula 26.3.19 of Abramowitz and Stegun's Handbook of Mathematical Functions,

$$\int_0^\infty\int_0^\infty f(x,y;\rho)\;\mathrm dx\;\mathrm dy = \frac 14 + \frac{\arcsin \rho}{2\pi}$$ 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.

$\endgroup$
  • $\begingroup$ Thank you so much! This analytical formula for the integral is awesome, showing exactly what's going on with $\rho$. $\endgroup$ – TD888 Feb 10 at 5:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.