# 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 $$2/\pi$$.

• Run a few samples of $X_1$ and $X_2$ pairs and plot $Y_1$ against $Y_2$ to see that a zero correlation is not the correct answer – JimB Dec 25 '18 at 7:16
• @JimB Let me check then. I thought $E(X_1X_2\mid X_1X_2>0)$ and $E(X_1X_2\mid X_1X_2<0)$ are equal. – StubbornAtom Dec 25 '18 at 7:18
• Equal in magnitude certainly. But you’re subtracting a negative value so they don’t cancel out. – JimB Dec 25 '18 at 7:26
• @StubbornAtom (edited) Showing they don't have 0 correlation doesn't show that they're not bivariate normal. – Glen_b Dec 25 '18 at 7:28
• @Glen_b That wasn't what I was indicating at. My point was that since I could find the correlation without the joint distribution of $(Y_1,Y_2)$ then maybe I could somehow say that the joint is not bivariate normal without the actual joint distribution. That is possibly less work for the problem at hand. – StubbornAtom Dec 25 '18 at 7:37

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.

• In fact, the joint density must be given by $$p(x,y) = \begin{cases} 2p(x)p(y), & \text{if}~ x > 0, y > 0, \text{or}~ x < 0, y < 0,\\ 0,& \text{otherwise}.\end{cases}$$ as depicted in the third picture in this answer by Moderator cardinal. See this answer for the source. – Dilip Sarwate Dec 25 '18 at 15:02

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.