Uncorrelated but dependent

Topic: Covariance and the multivariate normal · more questions on this topic
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Let X\sim\operatorname{N}(0,1) and Y=X^2. Show that \operatorname{Cov}(X,Y)=0 although Y is a function of X.

Worked answer (click to show)

The example has to pass two tests: its covariance must vanish, and its joint distribution must still contain dependence. Symmetry settles the first. The deterministic relation settles the second.

\operatorname{Cov}(X,X^2)=\mathbb{E}(X^3)-\mathbb{E}(X)\mathbb{E}(X^2)=0-0=0,

because the third moment of a symmetric distribution about zero vanishes. Yet Y=X^2 is completely determined by X. In particular \mathbb{P}(|X|>1\mid Y\le 1)=0 while \mathbb{P}(|X|>1)>0, so X and Y are not independent. Zero covariance rules out a nonzero linear association. It does not rule out nonlinear dependence.