I am not sure what happens to a bivariate normal distribution when $|\rho| \rightarrow 1$. Is the distribution well defined then? Moreover, when $$ \Phi \left(\frac{x_1}{\sigma_1}, \frac{x_2}{\sigma_2}, \rho \right) = 1 \text{ or } 0?$$ Namely, is it true that if $\Bigl \lvert \frac{x_1}{\sigma_1} \Bigr \lvert \leq M$ and $\Bigl \lvert \frac{x_2}{\sigma_2} \Bigr \lvert \leq L$, where $M, L < \infty$, then $$ \Phi \left(\frac{x_1}{\sigma_1}, \frac{x_2}{\sigma_2}, \rho \right) \neq 1 \text{ and } 0? $$ Can it be stated without any restrictions on $\rho$? If you know any reference, where it is explained well, I would be very grateful.

  • 1
    $\begingroup$ I am not sure what exactly your notation means, but the limit of the joint cumulative probability distribution function is not restricted to only two values $0$ and $1$ but should be a nondecreasing function of both variables like $$F(x,y) = \Phi\left(\min(x, y)\right),$$ being an increasing function of $x$ for $x < y$ that becomes the constant $\Phi(y)$ in the region $x > y$, and similarly for fixed $x$ and varying $y$. Here, $\Phi(\cdot)$ is the univariate standard normal CDF and I have assumed that $X$ and $Y$ are marginally standard normal for simplicity. $\endgroup$ Jul 17 '13 at 11:03
  • $\begingroup$ @DilipSarwate: I didn't get your argument. Let $\Sigma$ be a variance-covariance matrix. If $\Sigma$ is positive definite and $\| \Sigma \| < \infty$ (the norm is smaller than infinity, implying that all elements are smaller than infinity, thus the distribution function is well defined), then $\lim_{x_1 \rightarrow - \infty} \Phi_2(x_1,x_2;\Sigma) = 0$? $\endgroup$
    – Kolibris
    Jul 17 '13 at 11:22
  • $\begingroup$ You are correct. I stated the distribution function when $\rho = 1$ (as does the answer by @TooTone which also gives the result for $\rho = -1$) but the result that TooTone and I stated is not necessarily the limit of $\Phi(x,y;\rho)$ as $\rho$ approaches $1$: $$\lim_{\rho \to 1} \Phi_2(x,y; \rho) ~\text{does not necessarily equal}~ \Phi_1(\min(x,y)).$$ $\endgroup$ Jul 17 '13 at 13:43
  • $\begingroup$ @DilipSarwate Good point. I'm afraid that subtleties like that are beyond my pay-grade! $\endgroup$
    – TooTone
    Jul 17 '13 at 14:04
  • $\begingroup$ @Dilip Nevertheless--understanding your "$\Phi_2$" and "$\Phi_1$" to be the CDFs for arbitrary $\rho$ and $\rho=1$, respectively--the (pointwise) limiting value of the former does equal the latter. $\endgroup$
    – whuber
    Jul 17 '13 at 14:31

Yes, it's well-defined. For convenience and ease of exposition I'm changing your notation to use two standard normal distributed random variables $X$ and $Y$ in place of your $X1$ and $X2$. I.e., $X = (X1 - \mu_1)/\sigma_1$ and $Y = (X2 - \mu_2)/\sigma_2$. To standardize you subtract the mean and then divide by the standard deviation (in your post you only did the latter).

When $\rho=1$, the variables are perfectly correlated (which in the case of a normal distribution means perfectly dependent), so $X=Y$. And when $\rho=-1$, $X=-Y$.

The cumulative distribution function $\Phi(x,y,\rho)$ is defined to be the probability $\rm{Pr}(X\le x \cap Y\le y)$ given the correlation $\rm{Corr}(X,Y)=\rho$. I.e., $X$ has to be $\le x\,$ AND $\;Y$ has to be $\le y$.

So in the case $\rho=1$, $$\begin{eqnarray*} \Phi(x,y) &=& \rm{Pr}(X\le x \cap Y\le y) \\ &=& \rm{Pr}(X\le x \cap X\le y) \\ &=& \rm{Pr}(X\le \rm{min}(x,y)) \\ &=& \Phi_X(\rm{min}(x,y)) \\ &=& \Phi_Y(\rm{min}(x,y)). \\ \end{eqnarray*}$$

Here, $0 < \Phi(x,y) < 1$, so long as $|x|, |y| < \infty$.

The case $\rho=-1$ is much more interesting (and I'm not 100% sure I've got this right so would welcome corrections):

$$\begin{eqnarray*} \Phi(x,y) &=& \rm{Pr}(X\le x \cap Y\le y) \\ &=& \rm{Pr}(X\le x \cap -X\le y) \\ &=& \rm{Pr}(X\le x \cap X \ge -y) \\ &=& \rm{Pr}(-y \le X \le x) \\ &=& \Phi_X(x) - \Phi_X(-y) \;\;\;\mbox{(*)}\\ &=& \Phi_X(x) - (1 - \Phi_X(y)) \\ &=& \Phi_X(x) + \Phi_X(y) -1. \end{eqnarray*}$$

Note that the step marked * assumes $-y < x$, or equivalently, $y > -x$. If this doesn't hold, then $\Phi = 0$.

Here, $0 \le \Phi(x,y) < 1$, so long as $|x|, |y| < \infty$. Compared to the case of $\rho=1$, it is now possible to get $\Phi(x,y) = 0$ with finite values of $x$ and $y$. E.g. if $x=1$ and $y=-2$, it's impossible to get both $X\le x$ and $X\ge y$ ($X\le 1$ and $X\ge 2$).

To get some intuition for how the cumulative distribution functions look, I've plotted 3d plots and contour plots for the two cases below.

enter image description here

enter image description here

enter image description here

enter image description here

R code for these plots

grid = expand.grid(x=seq(-3,3,0.05), y=seq(-3,3,0.05))
grid$phi1 = with(grid, pnorm(pmin(x,y)))
grid$phi2 = with(grid, ifelse(-y<x,pnorm(x) + pnorm(y) -1,0))

wireframe(data=grid, phi1 ~ x*y, shade=TRUE, main="X=Y", scales=list(arrows=FALSE))
contourplot(data=grid, phi1 ~ x*y, main="X=Y")
wireframe(data=grid, phi2 ~ x*y, shade=TRUE, main="X=-Y", scales=list(arrows=FALSE))
contourplot(data=grid, phi2 ~ x*y, main="X=-Y", cuts=10)

There are plenty of web pages which cover the bivariate standard normal distribution. Which one you find best is going to be dependent on you. I had a quick search and rather liked the following: http://webee.technion.ac.il/people/adler/lec36.pdf, as it has some nice diagrams on p8 of what happens as $\rho \rightarrow \pm 1$. In the case of $\rho = \pm 1$, plotting $X$ against $Y$ will give you a straight line through the origin, either $y=\pm x$. If you plot this yourself, you should get a good intuition as to why $\rm{min}$ occurs in the formula for $\rho =1$.

  • $\begingroup$ Thank you very much, one problem is solved. I would be very happy if someone could help me with the second part of the question as well. $\endgroup$
    – Kolibris
    Jul 17 '13 at 12:31
  • $\begingroup$ @Kolibris no worries it turned out to be an interesting exercise. Sorry can you clarify exactly what the second part of the question is? $\endgroup$
    – TooTone
    Jul 17 '13 at 12:38
  • 1
    $\begingroup$ +1, the graphics are a very nice feature! How did you draw them? $\endgroup$ Jul 17 '13 at 12:41
  • 1
    $\begingroup$ @COOLSerdash glad you like them! The wireframe plots are nicest I think, as you can see the "ridge" on the $X=Y$ plot, and the triangle of $\Phi=0$ on the $X=-Y" plot. I've added R code (which I checked properly this time!) $\endgroup$
    – TooTone
    Jul 17 '13 at 12:48
  • $\begingroup$ @TooTone: I will reformulate it a bit. Is it true that if I assume that $\| \Sigma \| < \infty$, $|x| < \infty$, $|y| < \infty|$, then $0<\Phi(x,y;\Sigma)<1$? $\endgroup$
    – Kolibris
    Jul 17 '13 at 12:50

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.