Let $X sim N(0,1)$ and $Y=X$ if $|X|>c$ and $Y=-X$ if $|X|<c$, for any $c>0$.

I've already proved that $Y sim N(0,1)$.

How do we prove that $(X,Y)$ is not a bivariate normal?

I've tried proving that $Cov(X,Y)=0$ and because they are not independent, then we reached our desired conclusion. However, I only get $Cov(X,Y)= 1-E(X^2mathbb{1}_{|X|<c})$ which is $0$ only for some $c$… and doesn't prove for all values of $c>0$.

Then how can I prove this?

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#### Best Answer

Showing that $Cov(X,Y) = 0$ is not going to lead you anywhere, since if $Cov(X,Y) = 0$, would just indicate that their covariance is 0, and say nothing much about the existence of a joint distribution.

By the definition of a multivariate normal distribution, $(X,Y)$ is bivariate normal if every linear combination of X and Y is Normal. So to show that $(X,Y)$ is not jointly normal, you need to find $a$ and $b$ such that $$aX + bY not sim text{Normally distributed} $$

So consider $X – Y$.

begin{align*} X – Y &= begin{cases} X – |X| & text{when} |X| > c\ X + |X| & text{when} |X| < c end{cases}\ & = begin{cases} 0 & text{ when } |X| > max{0, c}\ 0 & text{ when } c <0 text{ and } min{0,c} < |X| < max{0, c}\ 2X & text{ when } c geq 0 text{ and }min{0,c} < |X| < max{0, c}\ 2X & text {when } |X| < min{0, c} end{cases}\ end{align*}

Then, trivially, irrespective of the value of $c$, $P(X – Y = 0) > 0$, and thus $X – Y$ is not normally distributed. Thus, $X$ and $Y$ are not bivariate normal.