Zero Covariance With Nonlinear Dependence

Zero Covariance Is Not Independence

Covariance records a particular kind of association: whether deviations of two variables from their means tend to have the same or opposite signs. It is a useful summary, but it does not capture every relationship. A variable can be completely determined by another variable through a nonlinear rule, while their covariance is zero.

A basic construction makes this precise. Let $X$ have a distribution symmetric about zero, and set $Y=X^2$. Positive and negative values of $X$ cancel in the covariance calculation, even though knowing $X$ determines $Y$. The moment assumptions matter: to define this covariance, a finite third absolute moment suffices, but it does not guarantee that $Y$ has finite variance.

Definition: A random variable $X$ is symmetric about zero if $X$ and $-X$ have the same distribution. Two integrable random variables $X$ and $Y$ have covariance $$ \operatorname{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])] $$ when the centered product is integrable. In particular, $E|X|<\infty$, $E|Y|<\infty$, and $E|XY|<\infty$ suffice; under these conditions, $$ \operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]. $$

The factorization $E[XY]=E[X]E[Y]$ is a covariance calculation, not an independence criterion. Independence is a stronger condition: probabilities of events determined by the two variables must factor. For the construction $Y=X^2$, the relation between the variables is nonlinear, and that relation can be checked explicitly in each model.

Why Symmetry Cancels Odd Moments

The key calculation comes from a general consequence of symmetry. An odd function changes sign when its argument changes sign. If its expectation is integrable, the equal probabilities assigned to $x$ and $-x$ make the positive and negative contributions cancel.

Proposition: If $X$ is symmetric about zero and $E|X|^k<\infty$ for an odd positive integer $k$, then $E[X^k]=0$.

Proof. Symmetry means that $X$ and $-X$ have the same distribution. The moment is integrable by assumption, so equality in distribution gives $$ E[X^k]=E[(-X)^k]=-E[X^k], $$ where the last equality uses that $k$ is odd. Thus $2E[X^k]=0$, and $E[X^k]=0$. $\square$

For the square construction, the first and third moments are the relevant odd moments. A finite third absolute moment also guarantees the integrability needed for the covariance: since $|x|\leq 1+|x|^3$ and $x^2\leq 1+|x|^3$, both $E|X|$ and $E[X^2]$ are finite.

The Covariance Theorem

Theorem: Suppose $X$ is symmetric about zero and $E|X|^3<\infty$. Let $Y=X^2$. Then $E|X|<\infty$, $E|Y|<\infty$, and $E|XY|<\infty$; consequently, $\operatorname{Cov}(X,Y)$ exists and $$ \operatorname{Cov}(X,X^2)=0. $$

Proof. The assumed third absolute moment implies $E|X|<\infty$ and $E[X^2]<\infty$, using $|x|\leq 1+|x|^3$ and $x^2\leq 1+|x|^3$. Since $Y=X^2$, these give $E|Y|=E[X^2]<\infty$. Also, $$ E|XY|=E|X|^3<\infty. $$ Thus the covariance is defined and its product formula applies. By the proposition, symmetry and integrability give $E[X]=0$ and $E[X^3]=0$. Therefore, $$ \operatorname{Cov}(X,X^2) =E[X^3]-E[X]E[X^2] =0-0\cdot E[X^2] =0. $$ This establishes the result. $\square$

The conclusion is about covariance, not the variance of $Y$. In fact, $E[Y^2]=E[X^4]$, and a finite third absolute moment does not guarantee that this fourth moment is finite. The covariance calculation above needs $E|XY|=E|X|^3<\infty$; it does not require a finite variance for $Y$.

Worked Examples

Worked Example: A Four-Point Distribution

Let $X$ assign probability $1/4$ to each of $-2,-1,1,2$, and define $Y=X^2$. The distribution is symmetric, and the possible pairs $(X,Y)$ are $$ (-2,4),\quad (-1,1),\quad (1,1),\quad (2,4), $$ each with probability $1/4$.

First calculate the means: $$ E[X]=\frac{-2-1+1+2}{4}=0,\qquad E[Y]=\frac{4+1+1+4}{4}=\frac{10}{4}=\frac52. $$ Next, $$ E[XY]=E[X^3] =\frac{-8-1+1+8}{4}=0. $$ It follows that $$ \operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y] =0-0\cdot\frac52=0. $$

To check dependence, take $A=\{X=1\}$ and $B=\{Y=1\}$. The joint support shows $A\cap B=\{(X,Y)=(1,1)\}$, so $P(A\cap B)=1/4$. But $P(A)=1/4$, $P(B)=1/2$, and hence $P(A)P(B)=1/8$. Since $1/4\neq1/8$, these variables are dependent despite having zero covariance.

Worked Example: A Continuous Uniform Variable

Let $X\sim \operatorname{Uniform}(-1,1)$, with density $f(x)=1/2$ for $-1<x<1$, and set $Y=X^2$. By symmetry, $E[X]=0$ and $E[X^3]=0$. Directly, $$ E[Y]=E[X^2]=\frac12\int_{-1}^{1}x^2\,dx =\frac12\left[\frac{x^3}{3}\right]_{-1}^{1} =\frac13. $$ Also $E[XY]=E[X^3]=0$, so $$ \operatorname{Cov}(X,Y)=0-(0)\left(\frac13\right)=0. $$

The joint values lie on the parabola $y=x^2$, for $-1\leq x\leq1$. To verify dependence with events, let $A=\{X>1/2\}$ and $B=\{Y>1/4\}$. The uniform density gives $$ P(A)=\frac{1-1/2}{2}=\frac14,\qquad P(B)=P(|X|>1/2)=\frac{1}{2}. $$ Their intersection is $\{X>1/2\}$, since $X>1/2$ implies $X^2>1/4$. Thus $P(A\cap B)=1/4$, whereas $P(A)P(B)=1/8$. The probabilities do not factor, confirming dependence.

Worked Example: A Standard Normal Variable

Let $X\sim N(0,1)$ and $Y=X^2$. The standard normal distribution is symmetric and has finite absolute moments of every positive integer order. Its mean is $0$, and symmetry gives $E[X^3]=0$. Since $E[X^2]=1$, $$ E[Y]=1,\qquad E[XY]=E[X^3]=0, \qquad \operatorname{Cov}(X,Y)=0-(0)(1)=0. $$

For a direct dependence check, take $A=\{X>1\}$ and $B=\{Y>1\}$. Write $q=P(X>1)$. The normal density is positive on the real line, so $0<q<1/2$. By symmetry, $P(B)=P(|X|>1)=2q$, and $A\subseteq B$, so $P(A\cap B)=q$. If $A$ and $B$ were independent, their intersection would have probability $P(A)P(B)=q(2q)=2q^2$. But $q\neq2q^2$ because $q>0$ and $q<1/2$. Thus these variables are dependent.

How to Read the Construction

The examples show two facts at once. First, symmetry makes the signed contributions to $E[X^3]$ cancel, which forces the covariance to vanish. Second, the square retains information about the magnitude of $X$: values of $X$ with different magnitudes produce different values of $Y$. The joint support in the discrete example consists of four specific points on the curve $y=x^2$; in the continuous examples, the possible pairs also obey that equation.

A support constraint is a helpful way to notice a relationship, but a dependence claim should be verified using the definition. The examples do this by finding events whose joint probability is not the product of their marginal probabilities. This is especially useful when a covariance calculation seems to suggest there is no association: it checks a different, more demanding property.

Common Mistake

Wrong reasoning: “The covariance is zero, so $X$ and $X^2$ must be independent.” This treats zero covariance as if it were a general independence test.
Correction: Zero covariance only says $E[XY]=E[X]E[Y]$, when these expectations exist. Independence requires probability factorization for all relevant pairs of events. In the worked examples, explicit events fail that test.
Moment-condition pitfall: A finite third absolute moment makes the covariance in this construction well-defined, but does not imply $E[X^4]<\infty$. Do not infer that $Y=X^2$ has finite variance from the covariance theorem's assumption.

Key takeaway: For a symmetric $X$ with $E|X|^3<\infty$, the variable $Y=X^2$ has $\operatorname{Cov}(X,Y)=0$. The zero covariance comes from symmetry and does not prevent nonlinear dependence; check independence through event probabilities or the joint law.

Check Your Understanding

Use the moment conditions and probability calculations, not covariance alone, to answer these questions.

Why does $E|X|^3<\infty$ ensure that $E|X|$, $E[X^2]$, and $E|X\cdot X^2|$ are finite?
If $X$ is symmetric about zero and $E|X|^3<\infty$, which two odd moments are used to derive $\operatorname{Cov}(X,X^2)=0$?
For the four-point example, calculate $P(X=-1,Y=1)$ and compare it with $P(X=-1)P(Y=1)$.
In the uniform example, why is $\{X>1/2\}$ contained in $\{X^2>1/4\}$, and what does that imply for their joint probability?
Does a finite third absolute moment guarantee that $Y=X^2$ has finite variance? Explain which moment of $X$ would be needed.