Why Zero Covariance Is Not Enough
Covariance describes a particular kind of relationship: whether deviations of two variables from their means tend to have a signed product. It is useful, but it does not summarize every possible form of dependence. A variable can determine another variable exactly through a nonlinear rule, even when their covariance is zero.
A standard example is a symmetric random variable \(X\) and its square \(Y=X^2\). Knowing \(X\) determines \(Y\), so the variables are dependent whenever \(X^2\) is not constant. Yet symmetry can make their covariance zero. This distinction matters whenever a statistical argument uses zero covariance as evidence about a relationship between variables.
For \(Y=X^2\), having finite fourth moment, \(E[X^4]<\infty\), is a sufficient condition for both variables to have finite second moments. It also ensures the expectations used below exist. The examples will satisfy this condition. The earlier material on covariance establishes that independence implies zero covariance when the required moments exist; the converse is the point to examine here.
How Symmetry Makes the Covariance Vanish
A random variable is symmetric about zero when \(X\) and \(-X\) have the same distribution. For such a variable, the expectation of an integrable odd function of \(X\) is zero. In particular, when the moments exist, \(E[X]=0\) and \(E[X^3]=0\). In contrast, \(X^2\) is an even function, so its expectation need not be zero.
The covariance claim follows by expanding the centered product, or equivalently using the shortcut formula \(\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]\). Here \(XY=X^3\) and \(E[Y]=E[X^2]\). Thus both \(E[X^3]\) and \(E[X]\) are zero by symmetry, while \(E[X^2]\) is finite. The dependence claim follows because \(Y=X^2\) is determined by \(X\). More formally, if \(X\) and \(Y\) were independent, then \(Y\), a function of \(X\), would be independent of itself. An event \(B\) defined using \(Y\) would then have to satisfy \(P(Y\in B)=P(Y\in B)^2\), so its probability would be zero or one. That is impossible for a nonconstant \(Y\), which has at least one event with probability strictly between zero and one.
This is a short derivation of why the covariance vanishes: the product \(X\cdot X^2\) is the odd function \(X^3\), and the mean of \(X\) is also zero. The square can still vary substantially. Zero covariance therefore does not mean that \(X\) provides no information about \(X^2\).
Worked Examples
Worked Example: A Three-Point Symmetric Distribution
Let \(X\) take the values \(-2,0,2\), each with probability \(1/3\), and define \(Y=X^2\). Calculate the means and covariance, then verify that \(X\) and \(Y\) are dependent.
Method and moments. The probabilities sum to \(1/3+1/3+1/3=1\). Pairing the equal-probability values \(2\) and \(-2\) gives \(E[X]=(-2)(1/3)+0(1/3)+2(1/3)=0\). For the square, \(E[X^2]=4(1/3)+0(1/3)+4(1/3)=8/3\). Since \(Y=X^2\), this also gives \(E[Y]=8/3\).
Covariance. The product \(XY=X^3\), so \(E[XY]=(-8)(1/3)+0(1/3)+8(1/3)=0\). Therefore \(\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0-(0)(8/3)=0\).
Dependence check. Consider the event \(A=\{X=0\}\) and the event \(B=\{Y=0\}\). Since \(Y=X^2\), these events occur together, so \(P(A\cap B)=1/3\). But \(P(A)P(B)=(1/3)(1/3)=1/9\), which is not \(1/3\). The event-factorization condition fails, so \(X\) and \(Y\) are dependent.
Check and common mistake avoided. The computed covariance is zero, while the event calculation proves dependence. It would be incorrect to conclude independence merely from the zero covariance.
Worked Example: A Continuous Uniform Variable on a Symmetric Interval
Suppose \(X\sim\operatorname{Uniform}(-3,3)\), with density \(f_X(x)=1/6\) for \(-3<x<3\), and let \(Y=X^2\). Find \(E[X]\), \(E[Y]\), and \(\operatorname{Cov}(X,Y)\). Then use events to establish dependence.
Means and moments. By integration, \(E[X]=(1/6)\int_{-3}^{3}x\,dx=0\), because the integrand is odd on a symmetric interval. For the square, \(E[Y]=E[X^2]=(1/6)\int_{-3}^{3}x^2\,dx\). An antiderivative of \(x^2\) is \(x^3/3\), so \(E[X^2]=(1/6)[x^3/3]_{-3}^{3}=(1/6)(9-(-9))=18/6=3\).
Covariance. Here \(XY=X^3\), and \(E[XY]=(1/6)\int_{-3}^{3}x^3\,dx=0\), since \(x^3\) is odd. It follows that \(\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0-(0)(3)=0\).
Dependence check. Let \(A=\{X>1.5\}\) and \(B=\{Y>2.25\}\). The interval for \(A\) is \((1.5,3)\), of length \(1.5\), so \(P(A)=1.5/6=1/4\). The event \(B\) means \(|X|>1.5\). Its two intervals have total length \(1.5+1.5=3\), so \(P(B)=3/6=1/2\). Their intersection is \(\{X>1.5\}\), so \(P(A\cap B)=1/4\). If the events were independent, their joint probability would be \(P(A)P(B)=(1/4)(1/2)=1/8\). Since \(1/4\ne1/8\), the variables are dependent.
Check and common mistake avoided. The interval lengths give probabilities under the stated uniform density, and the joint event is only the positive interval, not both tails. This verifies dependence without relying only on the fact that \(Y\) is a function of \(X\).
Worked Example: Unequal Probabilities on Symmetric Support
Let \(X\) have the following distribution: \(P(X=-3)=P(X=3)=1/6\), and \(P(X=-1)=P(X=1)=1/3\). Set \(Y=X^2\). Calculate the covariance and use a joint-probability check to determine whether the variables are independent.
Check the distribution and means. The probabilities sum to \(1/6+1/6+1/3+1/3=1\). The equal probabilities at opposite values give \(E[X]=(-3)(1/6)+(3)(1/6)+(-1)(1/3)+(1)(1/3)=0\). The square has value \(9\) with total probability \(1/3\), and value \(1\) with total probability \(2/3\). Therefore \(E[Y]=E[X^2]=9(1/3)+1(2/3)=3+2/3=11/3\).
Covariance. The product \(XY=X^3\). Pairing opposite values, \(E[XY]=(-27)(1/6)+(27)(1/6)+(-1)(1/3)+(1)(1/3)=0\). Thus \(\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0-(0)(11/3)=0\).
Dependence check. Take \(A=\{X=3\}\) and \(B=\{Y=9\}\). Then \(P(A\cap B)=1/6\), \(P(A)=1/6\), and \(P(B)=1/3\). If independent, the joint probability would be \((1/6)(1/3)=1/18\). Since \(1/6\ne1/18\), the variables are dependent.
Check and common mistake avoided. The distribution is not uniform on its four support points, but it remains symmetric: each value and its negative have equal probabilities. That symmetry, rather than equal probability at every support point, is what makes the odd moments cancel.
Common Exam Trap: Confusing Odd and Even Moments
For \(X\sim\operatorname{Uniform}(-3,3)\), the third moment is zero, but the second moment is not. The distinction follows directly from the integrals:
A mistaken calculation might use the cancellation of the odd moment to claim \(E[X^2]=0\), or confuse the two integrals and use zero as the mean of \(Y=X^2\). The correction is to identify the power before integrating: \(x^3\) is odd and cancels across the interval, while \(x^2\) is even and has expectation \(3\). The covariance is zero because it uses \(E[X^3]\) and \(E[X]\), not because \(E[X^2]\) is zero.
The more consequential trap is a logical one: “the covariance is zero, therefore the variables are independent.” The examples show why that implication fails. Covariance tests the expected centered product; independence requires factorization for every pair of events. A nonlinear relationship can be invisible to the covariance calculation.
Check Your Understanding
For each problem, calculate the requested moments or probabilities before deciding whether zero covariance implies independence.
- Let \(X\) take \(-1,0,1\) with equal probabilities and set \(Y=X^2\). Find \(E[X]\), \(E[Y]\), and \(\operatorname{Cov}(X,Y)\); then compare \(P(X=0,Y=0)\) with \(P(X=0)P(Y=0)\). (Answer: \(E[X]=0\), \(E[Y]=2/3\), covariance \(0\); the joint probability is \(1/3\) and the product is \(1/9\), so the variables are dependent.)
- Let \(X\sim\operatorname{Uniform}(-3,3)\) and \(Y=X^2\). For \(A=\{X>1.5\}\) and \(B=\{Y>2.25\}\), calculate \(P(A)\), \(P(B)\), and \(P(A\cap B)\), and state whether the events are independent. (Answer: \(P(A)=1/4\), \(P(B)=1/2\), \(P(A\cap B)=1/4\); the product is \(1/8\), so they are not independent.)
- Let \(X\) take \(-2,2\) with probabilities \(0.3,0.3\), and \(-1,1\) with probabilities \(0.2,0.2\). Set \(Y=X^2\). Calculate \(E[Y]\) and the covariance, then compare \(P(X=2,Y=4)\) with \(P(X=2)P(Y=4)\). (Answer: \(E[Y]=2.8\), covariance \(0\); the joint probability is \(0.3\) and the product is \(0.18\), so the variables are dependent.)
- Suppose \(X\) is symmetric about zero, \(E[X^4]<\infty\), and \(E[X^2]=7\). Let \(Y=X^2\), and suppose \(P(X^2=1)=0.4\). Find the covariance and use the event \(\{Y=1\}\) to assess whether \(X\) and \(Y\) are independent. (Answer: covariance \(0\); since \(P(Y=1)=0.4\) is neither zero nor one and \(Y\) is a function of \(X\), the variables are dependent.)
- Let \(X\) be symmetric about zero with finite fourth moment and suppose \(P(X^2=0)=0.2\) and \(P(X^2=4)=0.8\). For \(Y=X^2\), calculate \(E[Y]\) and \(\operatorname{Cov}(X,Y)\), then test independence using the event \(\{Y=0\}\). (Answer: \(E[Y]=3.2\), covariance \(0\); \(Y\) is nonconstant and a function of \(X\), so \(X\) and \(Y\) are dependent.)