Linear combination of normal distribution which is not normal

Question

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?

Answer 1

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].