How to justify that $(Y_1,Y_2)$ is not bivariate normal without finding its exact distribution? 
Suppose $X_1$ and $X_2$ are independent $N(0,1)$ variables.
Define $$Y_1=X_1\,\text{sign}(X_2)\quad,\quad Y_2=X_2\,\text{sign}(X_1)$$
I have to show that $(Y_1,Y_2)$ is not bivariate normal and find the correlation between $Y_1$ and $Y_2$.

It is easy to see that both $Y_1$ and $Y_2$ are themselves univariate normal.
And the joint distribution function of $(Y_1,Y_2)$ can be derived to show that it is not jointly normal. But I am wondering if there is an alternate way to see this result, i.e. without explicitly finding the distribution of $(Y_1,Y_2)$ can I justify that the distribution is not jointly normal?
I am asking this because I think I do not require the joint distribution of $(Y_1,Y_2)$ to find the correlation. Since
$$Y_1=\begin{cases}X_1&,\text{ if }X_2>0\\-X_1&,\text{ if }X_2<0\end{cases}\quad\text{ and }\quad Y_2=\begin{cases}X_2&,\text{ if }X_1>0\\-X_2&,\text{ if }X_1<0\end{cases}$$
I have $$Y_1Y_2=\begin{cases}X_1X_2&,\text{ if }X_1X_2>0\\-X_1X_2&,\text{ if }X_1X_2<0\end{cases}$$
so that $$E(Y_1Y_2)=E(|X_1X_2|)=E(|X_1|)E(|X_2|)=\frac{2}{\pi},$$
thus implying that the correlation is $2/\pi$.
 A: 
without explicitly finding the distribution of $(Y_1,Y_2)$ can I justify that the distribution is not jointly normal?

One obvious way would be to see that $Y_1$ and $Y_2$ cannot be opposite in sign, and therefore cannot be bivariate normal. 
Equivalently, note that $Y_1Y_2=\text{sign}(X_1)X_1\,\text{sign}(X_2)X_2$ $=|X_1|\cdot |X_2|$ must be non-negative.
A: To see what happens, let's explicitly find the distribution. 
You could see it as a transformation from the entire plane to the first and third quadrants. 


*

*Transform the first quadrant ($X_1>0, X_2>0$) to itself $Y_1,Y_2 = X_1,X_2$.

*Mirror the third quadrant ($X_1<0, X_2<0$) to the first trough the origin $Y_1,Y_2 = -X_1,-X_2$.

*Mirror the second quadrant ($X_1<0, X_2>0$) along the x-axis to the third quadrant $Y_1,Y_2 = X_1,-X_2$ 

*Mirror the fourth quadrant ($X_1>0,X_2 <0$) along the y-axis to the third quadrant $Y_1,Y_2 = -X_1,X_2$.


Overall you map all four quadrants to only the first and third quadrant. This relates to the answer of Glen_b.
