Separate factorization from linear association

Try

Let X be uniform on (−2,2) and set Y=X². Knowing X determines Y exactly, so the variables are dependent. Yet positive and negative X values balance in a way that cancels linear covariance. Predict which odd moments vanish before calculating.

Predict

Does zero correlation mean there is no relationship? Sketch the parabola and answer from the geometry. Then ask what stronger condition would establish independence: the joint law must factor into its marginals, not merely satisfy one moment identity.

Experiment

Symmetry gives E(X)=0 and E(X³)=0. Because XY=X³, Cov(X,Y)=E(XY)−E(X)E(Y)=0. Both variables have finite positive variance, so their correlation is also zero.

To show dependence numerically, P(Y>1)=P(|X|>1)=1/2. But given X>1, Y>1 with probability one. The conditional probability differs from its marginal. Mark these regions on the parabola: the covariance cancellation does not erase the deterministic relation.

Understand

Independence of random variables means P(X in A,Y in B)=P(X in A)P(Y in B) for all suitable measurable sets A and B. For a joint density, factorization f_(X,Y)=f_X f_Y almost everywhere is an equivalent criterion. Discrete laws use the analogous mass factorization. Support matters: a constant height on a triangle is not a product law on the coordinate supports.

For finite-second-moment variables, covariance is E[(X−EX)(Y−EY)]=E(XY)−E(X)E(Y). Cauchy–Schwarz ensures the product is integrable. Covariance has product units and depends on scale. If both SDs are positive, correlation divides covariance by their product to obtain a unitless value between −1 and one.

Independence implies factorization of suitable product expectations and hence zero covariance. The converse is false, as the square example shows. Zero covariance only constrains one aspect of the joint law. In the special jointly normal family, zero covariance does imply independence, but normal marginals alone do not establish joint normality or that implication.

If either variable is constant almost surely, covariance is zero but correlation is undefined because an SD denominator vanishes. Do not force a displayed zero correlation for a degenerate variable. Similarly, heavy-tailed variables may lack the moments required even to define covariance.

For affine changes, Cov(a+bX,c+dY)=bd Cov(X,Y). Positive rescalings preserve correlation, while one negative rescaling reverses its sign. Shifts do not affect centered quantities. None of these algebraic transformations establishes causality or changes the underlying independence structure when they are invertible.

A sample correlation estimates a population feature under a data-generating model. Finite-sample values, even exactly zero, do not prove independence in the population. Inspect nonlinear patterns, support restrictions and collection design as well as a single coefficient.

Cauchy–Schwarz bounds the magnitude of covariance by the product of standard deviations. This yields the correlation range and provides a consistency check: a proposed covariance larger than that product cannot belong to the stated finite-variance pair.

Apply

Verify the practice covariance and correlation with their units and denominators. Then write a full uncorrelated-but-dependent argument: compute the covariance and separately exhibit a failed conditional-probability equality. Calling a graph curved is useful intuition, but the probability calculation establishes the dependence precisely.

Continue

Next add variables. Expected values add under integrability without independence; variance requires tracking the cross-covariance terms.