# Bivariate normal distribution with $|\rho|=1$

I have deduced the bivariate normal density function. However am unaware of what happens when the correlation coefficient $\rho$ tends to 1 and -1?

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 $Y \sim N(\mu_Y,\sigma_Y^2)$

• In measure theoretic terms, when $\rho=\pm 1$, there is a density but only with respect to a measure on the subspace $\{x=\pm y\}$. Nov 4, 2015 at 14:55
• @Xi'an As I said, I didn't want to get into the measure-theoretic version and especially since the OP is asking what happens to the joint density as $\rho$ tends to $+1$ or $-1$ which gets into limits etc. Nov 4, 2015 at 15:01

The Pearson product moment correlation coefficient is a measure of linear dependence. At its extremes, one of the random variables is a linear function of the other with probability one and so there is no randomness.

• How exactly does "the line" manage to "describe" a joint distribution?
– whuber
Nov 4, 2015 at 2:08
• @whuber The distribution is concentrated on it, there is nothing random. Is that wrong? Nov 4, 2015 at 9:01
• It's incomplete, because the distribution is not uniform on that line.
– whuber
Nov 4, 2015 at 12:58
• @whuber I see what you mean, I didn't mean to imply that. Thank you. Nov 4, 2015 at 13:35
• "... the distribution is not uniform on that line." and indeed, since the line is of unbounded length, no distribution can be uniform on that line. Nov 4, 2015 at 22:20