When Zero Covariance Fails to Imply Independence

Zero Covariance Is Not Independence

Covariance describes how two variables vary together in a linear sense. A covariance of zero rules out this particular kind of association, but it does not rule out every relationship between the variables. In particular, a nonlinear relationship can be strong even when positive and negative contributions cancel in the covariance calculation.

Independence is a stronger statement about the entire joint distribution: knowing one variable does not change the distribution of the other. We will see that zero covariance does not generally establish independence, and then prove the direction that does hold: independent variables with finite second moments have zero covariance.

Definition: If $X$ and $Y$ have finite second moments, their covariance is $$ \operatorname{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]. $$ They are uncorrelated if $\operatorname{Cov}(X,Y)=0$. They are independent if $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$ for every pair of Borel sets $A,B$. The first condition concerns a moment; the second concerns all such events.

The covariance identity requires finite second moments: by the Cauchy–Schwarz inequality, $E|XY|<\infty$ when $E[X^2]$ and $E[Y^2]$ are finite. A zero covariance is therefore a well-defined statement about the centered product. It is not a general test for independence.

A Symmetry Argument for Nonlinear Dependence

A useful way to produce zero covariance is to make the product inside its expectation odd. Suppose $X$ is symmetric about zero, meaning $X$ and $-X$ have the same distribution. If $g$ is an even function, so that $g(-x)=g(x)$, then $xg(x)$ is odd. Provided the relevant moments exist, its expectation is zero. This cancellation can coexist with a clear nonlinear relationship between $X$ and $g(X)$.

Result: Suppose $X$ is symmetric about zero, $E[X^2]<\infty$, and $Y=g(X)$ has $E[Y^2]<\infty$, where $g$ is a measurable even function. Then $E[X]=0$ and $\operatorname{Cov}(X,Y)=0$.

Proof. Symmetry gives $X$ and $-X$ the same distribution, so $E[X]=E[-X]=-E[X]$, hence $E[X]=0$. The moment assumptions and Cauchy–Schwarz give $E|Xg(X)|<\infty$. Since $g$ is even, the function $h(x)=xg(x)$ is odd. Symmetry therefore gives

$$ E[Xg(X)]=E[h(X)]=E[h(-X)]=-E[h(X)]. $$

Thus $E[Xg(X)]=0$. It follows that

\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=E[Xg(X)]=0.

This proves the result. $\square$

Worked Example: A Normal Variable and Its Square

Let $X\sim N(0,1)$ and set $Y=X^2$. Find the covariance and verify that the variables are nevertheless dependent.

The standard normal distribution is symmetric about zero, and $g(x)=x^2$ is even. The symmetry result applies; directly, $E[X]=0$, $E[Y]=E[X^2]=1$, and

\operatorname{Cov}(X,Y)=E[X^3]-E[X]E[X^2]=0-0\cdot 1=0.

Here $E[X^3]=0$ because the standard normal density is symmetric and $x^3$ is odd. To check dependence, let $A=\{X>1\}$, $B=\{Y>1\}$, and write $q=P(X>1)$. Since $Y=X^2$, the event $B$ is $\{|X|>1\}$. Symmetry gives $P(A)=q$, $P(B)=2q$, and $P(A\cap B)=q$. A standard normal variable has $0<q<1/2$, so $2q^2<q$. Consequently,

P(A\cap B)=q\ne 2q^2=P(A)P(B).

The event probabilities violate the definition of independence. The square is a nonlinear function of $X$, and zero covariance has not removed that dependence.

Two More Counterexamples

The same cancellation is not limited to continuous distributions or to a smooth transformation such as squaring. The next examples make the event-probability check explicit in different settings.

Worked Example: A Three-Point Symmetric Distribution

Let $X$ take each of $-1,0,1$ with probability $1/3$, and let $Y=X^2$. Compute the covariance and test independence.

By symmetry, $E[X]=(-1+0+1)/3=0$. Also, $Y$ equals 1 when $X=-1$ or $X=1$, and equals 0 when $X=0$, so $E[Y]=2/3$. The product $XY=X^3$, giving

E[XY]=\frac{-1+0+1}{3}=0, \qquad \operatorname{Cov}(X,Y)=0-(0)(2/3)=0.

For a direct dependence check, take $A=\{X=0\}$ and $B=\{Y=0\}$. These are the same event, so $P(A\cap B)=1/3$. But

P(A)P(B)=\frac13\cdot\frac13=\frac19\ne\frac13=P(A\cap B).

Thus $X$ and $Y$ are dependent despite being uncorrelated. In this example the dependence is exact: $Y$ is determined by $X$.

Worked Example: Zero Covariance With Added Noise

Let $X$ take each of $-1,0,1$ with probability $1/3$. Independently, let $Z$ take $-1$ and $1$ with probability $1/2$ each. Define $Y=|X|+Z$. Show that the covariance is zero but the variables are dependent.

First, $E[X]=0$, $E[|X|]=2/3$, and $E[Z]=0$, so $E[Y]=2/3$. Since $Z$ is independent of $X$,

E[XY]=E[X|X|]+E[XZ] =\frac{-1+0+1}{3}+E[X]E[Z] =0.

Therefore $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0$. To check dependence, let $A=\{X=1\}$ and $B=\{Y=0\}$. If $X=1$, then $Y=1+Z$, so $Y=0$ exactly when $Z=-1$. Hence $P(A\cap B)=(1/3)(1/2)=1/6$. The event $B$ occurs for $(X,Z)=(1,-1)$ or $(-1,-1)$, so $P(B)=1/3$, while $P(A)=1/3$. Thus

P(A\cap B)=\frac16\ne\frac19=P(A)P(B).

The variables are dependent. Here $Y$ is not determined by $X$, because the independent noise $Z$ remains, but its distribution still depends on $X$.

Independence Implies Zero Covariance

The converse direction has a useful regularity condition. If $X$ and $Y$ are independent and have finite second moments, then their product has a finite expectation and its expectation factors. We prove the factorization carefully so that the argument also covers unbounded variables.

Theorem: If $X$ and $Y$ are independent and $E[X^2]<\infty$, $E[Y^2]<\infty$, then $E[XY]=E[X]E[Y]$, and hence $\operatorname{Cov}(X,Y)=0$.

Proof. For each positive integer $m$, define the bounded truncations $X_m=\max(-m,\min(X,m))$ and $Y_m=\max(-m,\min(Y,m))$. Independence of $X$ and $Y$ implies independence of measurable functions of them. For these bounded variables, expectation factors, so $E[X_mY_m]=E[X_m]E[Y_m]$.

As $m$ increases, $X_m\to X$ and $Y_m\to Y$ almost surely. Also $|X_m-X|^2\le X^2$ and $|Y_m-Y|^2\le Y^2$; dominated convergence therefore gives $X_m\to X$ and $Y_m\to Y$ in $L^2$. In particular, $E[X_m]\to E[X]$ and $E[Y_m]\to E[Y]$, since convergence in $L^2$ implies convergence in $L^1$ on a probability space.

The products converge in $L^1$ as well. Indeed, Cauchy–Schwarz gives

E|X_mY_m-XY| \le \{E[(X_m-X)^2]\}^{1/2}\{E[Y_m^2]\}^{1/2} +\{E[X^2]\}^{1/2}\{E[(Y_m-Y)^2]\}^{1/2} \longrightarrow 0.

The second moments of $Y_m$ are bounded by $E[Y^2]$, so both terms tend to zero. Taking limits in $E[X_mY_m]=E[X_m]E[Y_m]$ now yields $E[XY]=E[X]E[Y]$. Substitution into the covariance identity proves $\operatorname{Cov}(X,Y)=0$. $\square$

What Covariance Can and Cannot Tell You

The implication proved above is one-way: under the stated moment conditions, independence guarantees zero covariance, but zero covariance alone does not guarantee independence. The counterexamples show why. Covariance summarizes the average centered product, so positive and negative contributions can cancel even when the joint distribution has a systematic nonlinear pattern.

A common mistake is to say, “The correlation is zero, so the variables are independent.” Correlation is covariance divided by the product of standard deviations, when both variances are positive. A correlation of zero is therefore equivalent to zero covariance in that setting; it does not strengthen the conclusion to independence. The appropriate correction is to describe the pair as uncorrelated and investigate dependence using the joint distribution or relevant event probabilities.

Key takeaway: Independence with finite second moments implies zero covariance. The reverse implication is false in general: symmetry can cancel covariance even when a nonlinear or noisy relationship leaves the variables dependent.

Check Your Understanding

Use calculations or event probabilities to justify each answer.

For $X\sim N(0,1)$ and $Y=X^4$, why does symmetry imply $\operatorname{Cov}(X,Y)=0$?
In the three-point example, which pair of events demonstrates that $X$ and $Y$ are not independent?
In the noisy example, calculate $P(Y=0)$ by listing the possible $(X,Z)$ pairs that produce $Y=0$.
What moment assumptions are used in the theorem that independence implies zero covariance, and where does the proof use them?
State the distinction between “uncorrelated” and “independent” in terms of what each condition says about the joint distribution.