Proving that (X,Y) is not bivariate normal Let $X \sim N(0,1)$ and $Y=X$ if $|X|>c$ and $Y=-X$ if $|X|<c$, for any $c>0$.
I've already proved that $Y \sim N(0,1)$.
How do we prove that $(X,Y)$ is not a bivariate normal? 
I've tried proving that $Cov(X,Y)=0$ and because they are not independent, then we reached our desired conclusion. However, I only get $Cov(X,Y)= 1-E(X^2\mathbb{1}_{|X|<c})$ which is $0$ only for some $c$... and doesn't prove for all values of $c>0$.
Then how can I prove this?
 A: Showing that $Cov(X,Y) = 0$ is not going to lead you anywhere, since if $Cov(X,Y) = 0$, would just indicate that their covariance is 0, and say nothing much about the existence of a joint distribution.
By the definition of a multivariate normal distribution, $(X,Y)$ is bivariate normal if every linear combination of X and Y is Normal. So to show that $(X,Y)$ is not jointly normal, you need to find $a$ and $b$ such that
$$aX + bY \not \sim \text{Normally distributed} $$
So consider $X - Y$.
\begin{align*}
X - Y &= 
\begin{cases}
X - |X| & \text{when} |X| > c\\
X + |X| & \text{when} |X| < c
\end{cases}\\
& = \begin{cases}
0 & \text{ when } |X| > \max\{0, c\}\\
0 & \text{ when } c <0 \text{ and } \min\{0,c\} < |X| < \max\{0, c\}\\
2X & \text{ when } c \geq 0 \text{ and }\min\{0,c\} < |X| < \max\{0, c\}\\
2X & \text {when } |X| < \min\{0, c\}
\end{cases}\\
\end{align*}
Then, trivially, irrespective of the value of $c$,  $P(X - Y = 0) > 0$, and thus $X - Y$ is not normally distributed. Thus, $X$ and $Y$ are not bivariate normal.
A: An interesting observation:

If $(X,Y)$ is a bivariate normal then the support takes every values of $\mathbb{R^2}$. Then $XY$ takes every value in Real line.


Now here $XY=X^2$ if $|X|>c$ and $XY=-X^2$ if $|X|<c$. Then $XY$ do not take any less than $-c^2$, which leads to $(X,Y)$ can not be a bivariate normal.
