A joint table where the covariance turns out to be zero

Question

Let X = 1 if a website visitor clicks an ad and Y = 1 if they are on a mobile device (each 0 otherwise). The joint chances are: Y=0 Y=1 --------- X=0 0.28 0.42 X=1 0.12 0.18 What is the covariance of X and Y?

Accepted Answer

Apply the definition: \mathrm{Cov}(X, Y) = E[XY] - E[X]\,E[Y]. The two averages. E[X] = 0.12 + 0.18 = 0.30, \qquad E[Y] = 0.42 + 0.18 = 0.60. The product term. E[XY] = P(X=1, Y=1) = 0.18. Combine. \mathrm{Cov}(X, Y) = 0.18 - (0.30)(0.60) = 0.18 - 0.18 = 0. The covariance is exactly zero. That is no coincidence here: check any cell against the product of its row and column chances, for example P(X=1, Y=1) = 0.18 = 0.30 	imes 0.60. Every cell equals its row chance times its column chance, which is the definition of X and Y being independent. Independent variables always have zero covariance. T…

A joint table where the covariance turns out to be zero

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	$Y=0$	$Y=1$
$X=0$	0.28	0.42
$X=1$	0.12	0.18