Is the CDF of a bivariate normal distribution with mean $(0,0)$ and $\Sigma = ((1,\rho),(\rho,1))$ monotone in the correlation coefficient $\rho$?
Distributions – How Bivariate Normal Distribution Relates to Correlation
distributionsmultivariate analysisnormal distribution
Related Solutions
\begin{align} y &= \Phi(x_1,x_2,\rho) = \int_{-\infty}^{x_1}\left[\int_{-\infty}^{x_2} \phi(a,b,\rho)\,\mathrm db\right]\,\mathrm da\\ \frac{\partial y}{\partial x_1} &= \frac{\partial}{\partial x_1}\Phi(x_1,x_2,\rho) = \frac{\partial}{\partial x_1}\int_{-\infty}^{x_1} \left[\int_{-\infty}^{x_2} \phi(a,b,\rho)\,\mathrm db\right]\,\mathrm da\\ &= \int_{-\infty}^{x_2} \phi(x_1,b,\rho)\,\mathrm db \end{align} via the rule for differentiating under the integral sign. Similarly, $$\frac{\partial y}{\partial x_2} = \int_{-\infty}^{x_1} \phi(a,x_2,\rho)\,\mathrm da.$$ If you don't recall the rule for differentiating integrals, see for example, the comments following this answer on math.SE.
The derivative with respect to $\rho$ is straightforward to find but messy in its details. We have that $$\phi(x_1,x_2,\rho)=\frac{1}{2\pi\sqrt{1-\rho^2}} \exp\left[-\frac{x^2 -2\rho xy + y^2}{2(1-\rho^2)}\right]$$ whose partial derivative with respect to $\rho$ is left to the OP to find. If $g(x_1,x_2,\rho)$ denotes this partial derivative, then $$\frac{\partial}{\partial \rho}\Phi(x_1,x_2,\rho) = \frac{\partial}{\partial \rho}\int_{-\infty}^{x_1} \left[\int_{-\infty}^{x_2} \phi(a,b,\rho)\,\mathrm db\right]\,\mathrm da = \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} g(a,b,\rho)\,\mathrm db\,\mathrm da$$
A simple-minded (that is, non-measure-theoretic) version of the answer is as follows.
If random variables $X$ and $Y$ are such that
every point $(x,y)$ in a region $\mathcal A$ of the plane is a possible realization of $(X,Y)$
The area of $\mathcal A$ is greater than $0$
and
- $P\{(X,Y) \in \mathcal A\} = 1$
then $X$ and $Y$ are said to be jointly continuous random variables, and their probabilistic behavior can be determined from their joint density function $f_{X,Y}(x,y)$ whose support is $\mathcal A$. Note that $X$ and $Y$ are also (marginally) continuous random variables.
If $(X,Y)$ have a bivariate normal distribution, then they are marginally normal random variables too. In particular, $X$ and $Y$ are continuous random variables.
But $X$ and $Y$ are jointly continuous (and thus enjoy the bivariate normal joint density function that you have found or been told about) only if their (Pearson) correlation coefficient $\rho \in (-1,1)$. When $\rho = \pm 1$, $X$ and $Y$ are not jointly continuous and they don't have a joint density function. They do, however, continue to enjoy the properties stated in the highlighted paragraph above. that is, they are still said to have a bivariate normal distribution (even though they don't have the bivariate normal density), and they are individually normal random variables (and hence continuous). Note that in this case, all realizations of $(X,Y)$ lie on the straight line $$y = \mu_Y + \frac{\sigma_Y}{\sigma_X}(x-\mu_X)$$ passing through $(\mu_X,\mu_Y)$. Note that the straight line has zero area. Since $Y = \mu_Y + \frac{\sigma_Y}{\sigma_X}(X-\mu_X)$, any questions about the probabilistic behavior of $(X,Y)$ can be translated into a question about the probabilistic behavior of $X$ alone and answered based on the knowledge that $X \sim N(\mu_X,\sigma_X^2)$. Since it is also true that $X = \mu_X + \frac{\sigma_X}{\sigma_Y}(Y-\mu_Y)$, contrary-minded folks might prefer to translate the question about $(X,Y)$ that has been asked into a question about the probabilistic behavior of $Y$ alone and answer it based on the knowledge that $Y \sim N(\mu_Y,\sigma_Y^2)$
Best Answer
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.