# Linear combination of normal distribution which is not normal

Let $$\xi_1, \xi_2$$ be i.i.d N(0,1).

Define $$(X_1,X_2)=\begin{cases} (\xi_1, |\xi_2|) \quad \xi_1 \geq 0 \\ (\xi_1,-|\xi_2|) \quad \xi_1 < 0 \end{cases}$$

This means we can rewrite $$X_1=\xi_1$$ and $$X_2=sgn(\xi_1)|\xi_2|$$

I proved that $$X_1$$ and $$X_2$$ are uncorrelated standard normal distributions.

Now I have to prove that they are not bivariate gaussian. I wanted to use the statement that says that $$(X_1,X_2)$$ are distributed as a bivariate gaussian if each linear combination of them is normally distributed. I want to find a counterexample of a linear combination for which this is not true but I cannot. Can someone suggest me one?

• What about using $X_1X_2\ge 0$? Jan 17, 2019 at 10:51
• What do you mean? Jan 17, 2019 at 11:01
• Note that $X_1X_2=|\xi_1||\xi_2|\ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal. Jan 17, 2019 at 12:17

First a correction: how did you conclude that $$X_1$$ and $$X_2$$ are uncorrelated? $$EX_1=0$$ and $$X_1X_2 \geq 0$$, $$EX_1X_2 >0$$ so covariance of $$X_1$$ and $$X_2$$ is strictly positive.
If $$(X_1,X_2)$$ has a bivariate normal distribution then $$E(X_2|X_1)=cX_1+d$$ for some constants $$c$$ and $$d$$. Here $$E(X_2|X_1)=E(X_2|\xi_1)=I_{\xi_1 \geq 0} E|\xi_2|-I_{\xi_1 < 0} E|\xi_2|$$ which is clearly not of the type $$c\xi_1+d$$. [The left side takes three values].