Jointly normal and correlated normal random variables

Question

Is is true that if two normal random variables are correlated, then they are jointly normally distributed? I am not sure how to prove or disprove it.

Answer 1

It is certainly not true. Suppose $X\sim N(0,\sigma^2)$ and $$ Y = \begin{cases} \phantom{-}X & \text{if } -c<X<c, \\ -X & \text{otherwise.} \end{cases} $$ Then $Y\sim N(0,\sigma^2)$. The correlation between $X$ and $Y$ depends on $c$, and for one special value of $c$ it is $0$, and for all others it is not.

(That it is not normally distributed can be seen by observing that its support is constrained to the two lines $y=\pm x$. That $Y$ is normally distributed is a fairly routine exercise.)

Answer 2

No, for an extreme example, take $X \sim N(0, 1)$, and consider the random vector $(X, -X)$, clearly, $X$ and $-X$ are (perfectly) correlated while they are not jointly normally distributed. A simple way to verify this is by noting $X + (-X) \equiv 0$ is not normally distributed. On the other hand, given $(X, Y)$ are jointly normal, then $X + Y$ must also be normally distributed. Contradiction.

Answer 3

For better understanding I would like the to rephrase the question. If two random variables have normal marginal densities are they jointly normally distributed? The answer is NO. This enlightening counterexample appears in the famous book "Counterexamples in Probability and Statistics" by Joseph Romano and Andrew Spiegel.

Let $$g(x,y)=(1/2\pi)\exp\left\{\frac{x^2+y^2}{2}\right\} $$ This is the bivariate Gaussian with mean vector $0$ and var-cov the identity matrix.

Take $$f(x,y)=2g(x,y)\;\text{if}\; xy\geq 0$$ $$f(x,y)=0\;\text{if not}$$

One can verify that $f$ is bivariate density, clearly non-normal, of two correlated variables which are marginally normally distributed.