A symmetric return and its magnitude are uncorrelated

Question

A stock's daily return X takes the value -4\%, 0\%, or +4\% with chances 0.25, 0.50, and 0.25. Define Y = X^2, a measure of how big the move was that day regardless of direction. What is the correlation between the return and its size measure?

Accepted Answer

Big-move days have large Y, so Y clearly depends on X. Yet the correlation is 0, because the return is symmetric around zero and Y is an even (direction-blind) function. Show the covariance vanishes: \mathrm{Cov}(X, Y) = E[XY] - E[X]\,E[Y]. The mean of X. The symmetric weights make the up and down moves cancel: E[X] = (-4)(0.25) + (0)(0.50) + (4)(0.25) = -1 + 0 + 1 = 0. So the covariance is just E[XY]. The average of the product. Here XY = X\cdot X^2 = X^3: E[X^3] = (-4)^3(0.25) + 0 + (4)^3(0.25) = (-64)(0.25) + (64)(0.25) = -16 + 16 = 0. Combine. \mathrm{Cov}(X, Y) = 0, \qquad \rho = 0. The …

A symmetric return and its magnitude are uncorrelated

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