Normal marginals without joint normality

Topic: Covariance and the multivariate normal · more questions on this topic
‹ Axes of a contour The bivariate conditional ›

Let X\sim\operatorname{N}(0,1) and let W be independent of X with \mathbb{P}(W=1)=\mathbb{P}(W=-1)=\tfrac12. Put Y=WX. Show that Y\sim\operatorname{N}(0,1) and \operatorname{Cov}(X,Y)=0, but that (X,Y) is not bivariate normal. Consider X+Y.

Worked answer (click to show)

Checking the two marginal distributions cannot settle joint normality. Find a linear combination that a jointly Gaussian vector could not produce.

For any y,

\mathbb{P}(Y\le y)=\tfrac12\mathbb{P}(X\le y)+\tfrac12\mathbb{P}(-X\le y)=\Phi(y),

because -X\sim\operatorname{N}(0,1) as well. So Y\sim\operatorname{N}(0,1). The covariance is

\mathbb{E}(XY)=\mathbb{E}(WX^2)=\mathbb{E}(W)\mathbb{E}(X^2)=0.

Now X+Y=(1+W)X equals 2X with probability \tfrac12 and 0 with probability \tfrac12. It therefore has an atom at 0 of mass \tfrac12. A linear combination of a bivariate normal vector is normal. A degenerate normal has no atom unless the combination is identically zero. So (X,Y) cannot be bivariate normal. The pair has normal margins and zero covariance, and it is dependent because |Y|=|X|.