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Variance and Covariance Structure · Tutorial 358 of 994

Dependent Variables With Zero Covariance

See how covariance can vanish even when one variable determines, or strongly constrains, the other.

Intermediate 9 min read

What You'll Learn

  • Verify zero covariance from joint probabilities or moments.
  • Construct a discrete dependent pair using a symmetric variable and its square.
  • Construct a continuous dependent pair and check dependence with events.
  • Show how independent noise can preserve zero covariance without removing dependence.
  • Distinguish zero covariance from independence using exact probability calculations.

Why Zero Covariance Can Miss Dependence

Covariance measures a particular kind of association: whether deviations of two variables from their means tend to have the same sign or opposite signs in a way that produces a nonzero average product. That is useful, but it does not summarize every possible relationship. A variable may be tightly determined by another variable through a nonlinear rule, while positive and negative contributions to covariance cancel exactly.

The goal here is to build and check such examples. In each one, we will calculate the covariance and separately demonstrate dependence using probabilities of events. Keeping those checks separate matters: a covariance calculation alone cannot establish independence or dependence.

Definition: Random variables \(X\) and \(Y\) are independent if \(P(X\in A,Y\in B)=P(X\in A)P(Y\in B)\) for every pair of Borel sets \(A,B\). They are dependent if this equality fails for at least one pair. If \(E[X^2]\) and \(E[Y^2]\) are finite, their covariance is \(\operatorname{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]\).

The finite-second-moment condition ensures that the covariance is well-defined: by the Cauchy–Schwarz inequality, \(E[|XY|]\leq \sqrt{E[X^2]E[Y^2]}<\infty\). The examples below have finite support or bounded variables, so all required moments exist.

A Symmetry Mechanism

A basic construction pairs a symmetric random variable with its square. Symmetry makes the positive and negative contributions to the covariance cancel, even though squaring preserves information about the magnitude of the original variable.

Theorem: Suppose \(X\) has a distribution symmetric about zero and \(E[|X|^3]<\infty\). Let \(Y=X^2\). Then \(\operatorname{Cov}(X,Y)=0\). If, in addition, \(X^2\) is not almost surely constant, then \(X\) and \(Y\) are dependent.

Here, symmetry about zero means that \(X\) and \(-X\) have the same distribution. To prove the covariance statement, first note that the finite absolute third moment implies \(E[|X|]<\infty\) and \(E[X^2]<\infty\). Symmetry gives \(E[X]=E[-X]=-E[X]\), so \(E[X]=0\). It also gives \(E[X^3]=E[(-X)^3]=-E[X^3]\), hence \(E[X^3]=0\). These expectations exist because \(E[|X|^3]<\infty\). Therefore,

$$ \operatorname{Cov}(X,X^2) =E[X^3]-E[X]E[X^2] =0-0\cdot E[X^2] =0. $$

For the dependence claim, \(Y=X^2\) is a function of \(X\). If \(X\) and \(Y\) were independent, then \(Y\) would be independent of itself: every event determined by \(Y\) is also determined by \(X\). In particular, for each Borel set \(B\), independence would imply \(P(Y\in B)=P(Y\in B)^2\), so that probability would have to be zero or one. A nonconstant \(Y\) cannot have all its event probabilities equal to zero or one. Thus \(Y\) is not independent of \(X\), proving dependence. \(\square\)

Worked Example: A Discrete Variable and Its Square

Let \(X\) have probability \(1/3\) at each of \(-1,0,1\), and set \(Y=X^2\). Calculate the covariance, then verify dependence directly.

The possible pairs \((X,Y)\), each with probability \(1/3\), are \((-1,1)\), \((0,0)\), and \((1,1)\). Thus

$$ E[X]=\frac{-1+0+1}{3}=0,\qquad E[Y]=\frac{1+0+1}{3}=\frac{2}{3},\qquad E[XY]=\frac{-1+0+1}{3}=0. $$

It follows that

$$ \operatorname{Cov}(X,Y) =E[XY]-E[X]E[Y] =0-0\cdot\frac{2}{3} =0. $$

To check dependence, choose \(A=\{X=0\}\) and \(B=\{Y=0\}\). Because \(Y=X^2\), these events are the same event. Hence \(P(X\in A,Y\in B)=1/3\), whereas \(P(X\in A)P(Y\in B)=(1/3)(1/3)=1/9\). Since \(1/3\ne1/9\), the variables are dependent.

Worked Example: A Continuous Variable and Its Square

Let \(X\sim \operatorname{Uniform}(-1,1)\), with density \(f(x)=1/2\) for \(-1<x<1\), and set \(Y=X^2\). Find the covariance and exhibit events that show dependence.

Using the density,

$$ E[X]=\int_{-1}^{1}x\frac{1}{2}\,dx=0,\qquad E[Y]=E[X^2]=\int_{-1}^{1}x^2\frac{1}{2}\,dx =\left[\frac{x^3}{6}\right]_{-1}^{1} =\frac{1}{3}. $$

The mixed moment is

$$ E[XY]=E[X^3]=\int_{-1}^{1}x^3\frac{1}{2}\,dx =\left[\frac{x^4}{8}\right]_{-1}^{1} =0. $$

Therefore, \(\operatorname{Cov}(X,Y)=0-(0)(1/3)=0\). For dependence, take \(A=\{|X|<1/2\}\) and \(B=\{Y<1/4\}\). Since \(Y=X^2\), \(A\) and \(B\) are the same event. The uniform distribution gives \(P(A)=P(B)=1/2\), because the interval \((-1/2,1/2)\) has length 1 within an interval of length 2. Consequently,

$$ P(X\in A,Y\in B)=\frac{1}{2} \ne \frac{1}{2}\cdot\frac{1}{2} =P(X\in A)P(Y\in B). $$

Thus the variables are dependent despite their zero covariance. In fact, the event calculation identifies the relationship directly: knowing whether \(Y<1/4\) tells us exactly whether \(|X|<1/2\).

Adding Noise Without Creating Covariance

Dependence with zero covariance does not require one variable to be an exact function of the other. We can add independent noise to the squared variable. The noise changes the conditional values of \(Y\), but symmetry still cancels the covariance.

Result: Suppose \(X\) and \(\varepsilon\) are independent, have finite second moments, and \(g(X)\) has a finite second moment. Then \(\operatorname{Cov}(X,g(X)+\varepsilon)=\operatorname{Cov}(X,g(X))\). Indeed, bilinearity gives \(\operatorname{Cov}(X,g(X)+\varepsilon)=\operatorname{Cov}(X,g(X))+\operatorname{Cov}(X,\varepsilon)\). Independence and finite moments imply \(E[X\varepsilon]=E[X]E[\varepsilon]\), so \(\operatorname{Cov}(X,\varepsilon)=0\). Substitution proves the result. \(\square\)

Worked Example: Independent Noise and a Dependent Pair

Let \(X\) again take the values \(-1,0,1\) with equal probabilities. Independently, let \(\varepsilon\) take the values \(-1,1\) with probability \(1/2\) each. Define \(Y=X^2+\varepsilon\). Verify zero covariance and then demonstrate dependence.

Independence of \(X\) and \(\varepsilon\) gives \(E[X\varepsilon]=E[X]E[\varepsilon]=0\), since both means are zero. Also, \(E[X^3]=0\), and

$$ E[Y]=E[X^2]+E[\varepsilon]=\frac{2}{3}+0=\frac{2}{3}, \qquad E[XY]=E[X^3]+E[X\varepsilon]=0+0=0. $$

Thus \(\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0-0\cdot(2/3)=0\).

For dependence, choose \(A=\{X=0\}\) and \(B=\{Y=2\}\). The event \(B\) occurs exactly when \(X^2=1\) and \(\varepsilon=1\). Independence gives \(P(B)=P(X^2=1)P(\varepsilon=1)=(2/3)(1/2)=1/3\). But if \(X=0\), then \(Y=\varepsilon\), which cannot equal 2. Therefore \(P(X\in A,Y\in B)=0\), while \(P(X\in A)P(Y\in B)=(1/3)(1/3)=1/9\). These probabilities differ, so \(X\) and \(Y\) are dependent.

What the Covariance Check Does—and Does Not—Say

The calculations in the examples follow a useful order. First check that the moments needed for covariance exist. Then calculate \(E[X]\), \(E[Y]\), and \(E[XY]\), and use their difference to find the covariance. Finally, test dependence separately by finding events whose joint probability fails to factor. A zero result in the covariance step is not a substitute for the event-probability check.

1
Check moments.
Confirm that the variables have finite second moments; then the covariance is well-defined.
2
Compute covariance.
Evaluate \(E[XY]-E[X]E[Y]\), using the joint distribution or a justified representation such as \(Y=g(X)+\varepsilon\).
3
Check dependence independently.
Choose events for \(X\) and \(Y\), and compare their joint probability with the product of their marginal probabilities.

A common mistake is to reason, “The covariance is zero, so the variables are independent.” The conclusion does not follow. Covariance is built from a product of centered values, so positive and negative contributions can cancel. In the square examples, large positive and negative values of \(X\) produce the same \(Y\); the relationship is systematic but nonlinear. The event calculations, not the covariance value, establish the dependence.

The safe interpretation is narrower: zero covariance says that the expected product of the centered variables is zero. It does not say that knowing one variable provides no information about the other. Independence is a stronger requirement, involving factorization for every pair of events, not just a single moment equation.

Key takeaway: Symmetry can make the covariance between \(X\) and \(X^2\) equal zero, even though \(X^2\) is determined by \(X\). Always verify covariance and dependence as separate properties.

Check Your Understanding

Use moment calculations for covariance and event probabilities for dependence.

  1. If \(X\) is symmetric about zero, \(E[|X|^3]<\infty\), and \(Y=X^2\), which moments show that the covariance is zero?
  2. For the discrete pair \(X\in\{-1,0,1\}\) and \(Y=X^2\), give events that violate the independence factorization and calculate both sides.
  3. For \(X\sim\operatorname{Uniform}(-1,1)\) and \(Y=X^2\), why do the events \(\{|X|<1/2\}\) and \(\{Y<1/4\}\) demonstrate dependence?
  4. In the noisy example, why is \(\operatorname{Cov}(X,\varepsilon)=0\), and what event pair demonstrates that \(X\) and \(Y\) are still dependent?
  5. State precisely what zero covariance establishes, and what additional check is needed to establish independence or dependence.

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