A variable and its square can have zero correlation

Question

Let X be equally likely to be any of -2, -1, 0, 1, 2. Define Y = X^2, so Y is completely determined by X. What is the correlation between X and Y?

Accepted Answer

The result is a classic surprise: Y is a perfect function of X, yet the correlation is 0. The reason is that correlation only detects straight-line links, and the relationship here is a U-shape, not a line. Correlation is proportional to the covariance, so it is enough to show the covariance is zero: \mathrm{Cov}(X, Y) = E[XY] - E[X]\,E[Y]. The mean of X. The values are symmetric around zero, so they cancel: E[X] = 	frac15(-2 - 1 + 0 + 1 + 2) = 0. Because E[X] = 0, the covariance reduces to just E[XY]. The average of the product. Here XY = X \cdot X^2 = X^3, so we need the average of X^3: E[…

A variable and its square can have zero correlation

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