Bivariate Normal distribution and correlation Is the CDF of a bivariate normal distribution with mean $(0,0)$ and $\Sigma = ((1,\rho),(\rho,1))$ monotone in the correlation coefficient $\rho$?
 A: Yes.
By definition, the value of the CDF (call it $F_\rho$) at $(s,t)$ is the chance that the first component is less than or equal to $s$ and the second is less than or equal to $t$:
$$F_\rho(s,t) = \frac{1}{2 \pi  \sqrt{1-\rho ^2}}\int_{-\infty}^t \int_{-\infty}^s e^{-\frac{\frac{x^2}{2}-\rho  x y+\frac{y^2}{2}}{1-\rho ^2}} dx dy.$$
Performing the $x$ integration and then differentiating with respect to $\rho$ under the integral sign yields
$$\frac{-1}{2 \pi  \left(1-\rho ^2\right)^{3/2}} \int_{-\infty}^t e^{\frac{s^2-2 \rho  s y+y^2}{2 \left(\rho ^2-1\right)}} (y-\rho  s) dy.$$
This can be integrated directly to produce the PDF 
$$f_\rho(s,t) = \frac{1}{2 \pi  \sqrt{1-\rho ^2}}e^{-\frac{\frac{s^2}{2}-\rho  s t+\frac{t^2}{2}}{1-\rho ^2}}.$$
Because the integrands are so well-behaved, we may reverse the order of integration and differentiation, concluding that for all $(s,t)$,
$$\frac{\partial}{\partial \rho} F_\rho(s,t) = f_\rho(s,t).$$
Because $f_\rho(s,t)\gt 0$ everywhere, $F_\rho$ is strictly monotone in $\rho$ everywhere, QED.
A: The answer would appear to be YES. I am not sure how this is usually proven (since the bivariate Normal cdf does not have a convenient closed form) ... but as a quick thought, I would appeal perhaps to graphical ideas. In particular, consider the contours of the zero mean bivariate Normal as $\rho$ increases, as per:

(source: tri.org.au)
Choose any $(x,y)$ point ... The cdf at $(x,y)$ is the joint integral of the pdf up to $(x,y)$. When $\rho = -1$, the contours of the pdf lie maximally out of alignment with the rectangular area defined by the integral.  When $\rho = 1$, the opposite extreme is attained.
