# 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 $\{x=\pm y\}$. Commented 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. Commented 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
Commented Nov 4, 2015 at 2:08
• @whuber The distribution is concentrated on it, there is nothing random. Is that wrong? Commented Nov 4, 2015 at 9:01
• It's incomplete, because the distribution is not uniform on that line.
– whuber
Commented Nov 4, 2015 at 12:58
• @whuber I see what you mean, I didn't mean to imply that. Thank you. Commented 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. Commented Nov 4, 2015 at 22:20