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