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?

  • 1
    $\begingroup$ What about using $X_1X_2\ge 0$? $\endgroup$ Jan 17, 2019 at 10:51
  • $\begingroup$ What do you mean? $\endgroup$
    – ank13
    Jan 17, 2019 at 11:01
  • $\begingroup$ Note that $X_1X_2=|\xi_1||\xi_2|\ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal. $\endgroup$ Jan 17, 2019 at 12:17

1 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].


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