Zero Covariance as an Independence Test
For general random variables, zero covariance rules out linear association but does not rule out dependence. Joint Gaussian vectors are a special case: if two Gaussian subvectors have zero cross-covariance, then they are independent. This is useful because covariance is often much easier to calculate than a joint distribution. The result applies even when the covariance matrix is singular; no density or inverse covariance matrix is required.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). For a random vector \(Z\) with finite second moments, its mean is \(\mathbb{E}[Z]\), and its covariance matrix is \(\operatorname{Cov}(Z)=\mathbb{E}[(Z-\mathbb{E}Z)(Z-\mathbb{E}Z)^{\mathsf T}]\). For subvectors \(X\in\mathbb{R}^p\) and \(Y\in\mathbb{R}^q\), define the cross-covariance matrix by \(\operatorname{Cov}(X,Y)=\mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)^{\mathsf T}]\); its \((a,b)\) entry is \(\operatorname{Cov}(X_a,Y_b)\).
For a random vector \(Z\), its characteristic function is \(\varphi_Z(u)=\mathbb{E}[\exp(iu^{\mathsf T}Z)]\). The standard characteristic-function formula for a Gaussian vector with mean \(m\) and covariance matrix \(\Sigma\) is \(\varphi_Z(u)=\exp(iu^{\mathsf T}m-\tfrac12u^{\mathsf T}\Sigma u)\). This formula holds for singular \(\Sigma\) as well as nonsingular \(\Sigma\). We also use the standard uniqueness theorem for characteristic functions: two probability distributions with the same characteristic function are equal. The uniqueness theorem is a standard result from probability theory; its measure-theoretic proof is not needed here.
The theorem concerns the subvectors as wholes: it says that every event determined by \(X\) is independent of every event determined by \(Y\). It is stronger than saying that each pair of coordinates is uncorrelated. The Gaussian assumption is essential for the implication from zero covariance to independence.
Proof. Let \(m_X=\mathbb{E}X\), \(m_Y=\mathbb{E}Y\), and write the covariance matrix of the concatenated vector in blocks: \(\operatorname{Cov}((X,Y))=\begin{bmatrix}\Sigma_{XX}&\Sigma_{XY}\\\Sigma_{YX}&\Sigma_{YY}\end{bmatrix}\), where \(\Sigma_{XY}=\operatorname{Cov}(X,Y)\) and \(\Sigma_{YX}=\Sigma_{XY}^{\mathsf T}\). For \(s\in\mathbb{R}^p\) and \(t\in\mathbb{R}^q\), the Gaussian characteristic-function formula gives \(\varphi_{X,Y}(s,t)=\exp(i(s^{\mathsf T}m_X+t^{\mathsf T}m_Y)-\tfrac12(s^{\mathsf T}\Sigma_{XX}s+2s^{\mathsf T}\Sigma_{XY}t+t^{\mathsf T}\Sigma_{YY}t))\). If \(\Sigma_{XY}=0\), this expression is the product \(\exp(is^{\mathsf T}m_X-\tfrac12s^{\mathsf T}\Sigma_{XX}s)\exp(it^{\mathsf T}m_Y-\tfrac12t^{\mathsf T}\Sigma_{YY}t)=\varphi_X(s)\varphi_Y(t)\). The joint characteristic function therefore equals the characteristic function of the product of the marginal distributions. By uniqueness of characteristic functions, the joint distribution is that product distribution, which is precisely independence. For the reverse direction, suppose \(X\) and \(Y\) are independent. Each coordinate has a finite second moment because the joint vector is Gaussian. Independence gives \(\mathbb{E}[(X_a-\mathbb{E}X_a)(Y_b-\mathbb{E}Y_b)]=\mathbb{E}[X_a-\mathbb{E}X_a]\mathbb{E}[Y_b-\mathbb{E}Y_b]=0\) for every \(a,b\). Thus every entry of \(\Sigma_{XY}\) is zero. This proves both directions. \(\square\)
The characteristic-function argument also explains why no density assumption appears. In the zero cross-covariance case, the exponent splits into a term involving \(s\) alone and a term involving \(t\) alone. That split remains valid when either marginal, or the full vector, is degenerate.
Worked Example: Two Scalar Gaussian Variables
Let \(X,Y\) be jointly Gaussian with means \(\mathbb{E}X=1\), \(\mathbb{E}Y=-2\), variances \(\operatorname{Var}(X)=4\), \(\operatorname{Var}(Y)=9\), and covariance \(\operatorname{Cov}(X,Y)=0\). Determine whether they are independent.
The covariance matrix is \(\begin{bmatrix}4&0\\0&9\end{bmatrix}\). For real \(s,t\), substitute the means and covariance entries into the Gaussian characteristic-function formula: \(\varphi_{X,Y}(s,t)=\exp(i(s-2t)-\tfrac12(4s^2+9t^2))\). Separating the terms gives \(\varphi_{X,Y}(s,t)=\exp(is-2s^2)\exp(-2it-\tfrac92t^2)\). The first factor is \(\varphi_X(s)\), since \(is\mathbb{E}X-\tfrac12s^2\operatorname{Var}(X)=is-2s^2\). The second is \(\varphi_Y(t)\), since \(it\mathbb{E}Y-\tfrac12t^2\operatorname{Var}(Y)=-2it-\tfrac92t^2\). Hence the joint characteristic function factors into the marginal characteristic functions, and the theorem gives independence.
Check: the cross-covariance matrix has its only entry equal to \(0\), as assumed. The two factors reproduce the specified means and variances, so the factorization uses the stated parameters.
Worked Example: A Gaussian Vector and a Gaussian Subvector
Suppose \((X_1,X_2,Y)\) is jointly Gaussian with mean \((1,-1,2)\) and covariance matrix \(\begin{bmatrix}2&1&0\\1&2&0\\0&0&3\end{bmatrix}\). Show that \(X=(X_1,X_2)\) and \(Y\) are independent.
The block of cross-covariances between \(X\) and \(Y\) is the column \(\begin{bmatrix}0\\0\end{bmatrix}\), so it is zero. To see the characteristic-function split directly, take \(s=(s_1,s_2)^{\mathsf T}\) and \(t\in\mathbb{R}\). The mean term is \(s_1-s_2+2t\), and the covariance quadratic form is \(2s_1^2+2s_1s_2+2s_2^2+3t^2\). Therefore \(\varphi_{X,Y}(s,t)=\exp(i(s_1-s_2)-\tfrac12(2s_1^2+2s_1s_2+2s_2^2))\exp(2it-\tfrac32t^2)\). The first factor depends only on \(s\), and the second only on \(t\); they are the marginal characteristic functions of \(X\) and \(Y\). Thus the subvectors are independent.
Check: the displayed covariance matrix is positive semidefinite: for real \(a,b,c\), its quadratic form is \(a^2+b^2+(a+b)^2+3c^2\), which is nonnegative. Its two cross-covariance entries are both zero, matching the condition used in the theorem.
Worked Example: A Degenerate Gaussian Pair
Let \(Z\) be standard normal, and set \(X=(Z,2Z)\) and \(Y=0\). The pair is jointly Gaussian because each linear combination of its coordinates is a constant multiple of \(Z\). Its cross-covariance entries are \(\operatorname{Cov}(Z,0)=0\) and \(\operatorname{Cov}(2Z,0)=0\). Therefore \(X\) and \(Y\) are independent by the theorem, despite the singular covariance matrix.
In fact, \(Y\) is constant. For any event \(A\) determined by \(X\) and any event \(B\) determined by \(Y\), \(B\) has probability \(0\) or \(1\). If \(\mathbb{P}(B)=0\), then \(\mathbb{P}(A\cap B)=0=\mathbb{P}(A)\mathbb{P}(B)\); if \(\mathbb{P}(B)=1\), then \(\mathbb{P}(A\cap B)=\mathbb{P}(A)=\mathbb{P}(A)\mathbb{P}(B)\). This verifies independence directly.
Check: \(\operatorname{Var}(Z)=1\), so the covariance matrix of \((Z,2Z,0)\) is \(\begin{bmatrix}1&2&0\\2&4&0\\0&0&0\end{bmatrix}\). It is singular, but the zero cross-covariance condition still holds.
Why Zero Covariance Is Not Enough in General
For arbitrary random variables, covariance records only an average product of centered values. Positive and negative contributions can cancel even when one variable contains substantial information about the other. The following example gives an exact calculation of that cancellation.
Worked Example: Dependent Variables with Zero Covariance
Let \(U\) be uniform on \((-1,1)\), and define \(V=U^2\). Calculate their covariance and test whether they are independent.
The density of \(U\) is \(1/2\) on \((-1,1)\). Thus \(\mathbb{E}U=\tfrac12\int_{-1}^1u\,du=0\), \(\mathbb{E}V=\mathbb{E}[U^2]=\tfrac12\int_{-1}^1u^2\,du=\tfrac12[\tfrac{u^3}{3}]_{-1}^1=\tfrac13\), and \(\mathbb{E}[UV]=\mathbb{E}[U^3]=\tfrac12\int_{-1}^1u^3\,du=0\). Consequently, \(\operatorname{Cov}(U,V)=\mathbb{E}[UV]-\mathbb{E}U\,\mathbb{E}V=0-0\cdot\tfrac13=0\).
Now use the strict events \(A=\{U>\tfrac12\}\) and \(B=\{V>\tfrac14\}\). Since \(U\) is uniform on an interval of length \(2\), \(\mathbb{P}(A)=\tfrac12(1-\tfrac12)=\tfrac14\). The event \(B\) is \(\{|U|>\tfrac12\}\), whose two intervals have total length \(1\), so \(\mathbb{P}(B)=\tfrac12\). Their intersection is \(A\), because \(U>\tfrac12\) implies \(U^2>\tfrac14\); hence \(\mathbb{P}(A\cap B)=\tfrac14\). But \(\mathbb{P}(A)\mathbb{P}(B)=\tfrac14\cdot\tfrac12=\tfrac18\), not \(\tfrac14\). Therefore \(U\) and \(V\) are dependent.
Check: the zero covariance calculation and the event calculation answer different questions. The covariance is zero, while the strict inequality \(\mathbb{P}(A\cap B)\ne\mathbb{P}(A)\mathbb{P}(B)\) disproves independence.
This example isolates the role of Gaussianity. In a jointly Gaussian vector, the characteristic function is determined by the mean and covariance, and a block-diagonal covariance matrix makes that characteristic function factor. For general distributions, zero cross-covariance does not force such a factorization. Independence always implies zero cross-covariance when the second moments exist, but the converse needs the joint Gaussian assumption.
Common mistake. A tempting but incorrect argument is: “\(\operatorname{Cov}(U,V)=0\), therefore \(U\) and \(V\) are independent.” The uniform-square example shows the error concretely: it has covariance \(0\), yet \(\mathbb{P}(A\cap B)=\tfrac14\) while \(\mathbb{P}(A)\mathbb{P}(B)=\tfrac18\). The correction is to check joint Gaussianity before using zero covariance as an independence test. If joint Gaussianity is known, the theorem applies; if it is not, zero covariance alone is insufficient.
Check Your Understanding
Use the definitions and characteristic-function factorization to answer each question.
- For jointly Gaussian \(X\in\mathbb{R}^2\) and \(Y\in\mathbb{R}^3\), how many cross-covariance entries must be zero to conclude independence?
- Why does the zero-cross-covariance theorem still apply when the joint Gaussian covariance matrix is singular?
- If \(X\) and \(Y\) are independent and have finite second moments but are not jointly Gaussian, what can be concluded about their cross-covariance?
- For \(U\) uniform on \((-1,1)\) and \(V=U^2\), verify that \(\{U>\tfrac12\}\cap\{V>\tfrac14\}=\{U>\tfrac12\}\). Why does this show dependence?
- What additional assumption is needed before zero covariance can be used to conclude independence?