# Solved – Proving that (X,Y) is not bivariate normal

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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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.

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