Why Zero Cross-Covariance Implies Gaussian Independence

Why This Gaussian Result Is Special

For two random vectors, a zero cross-covariance matrix says that every component of one vector is uncorrelated with every component of the other. It does not, by itself, say that the vectors are independent. The important exception is a jointly Gaussian pair: there, zero cross-covariance is not merely evidence of weak association. It is exactly the condition for independence.

The reason is visible in the joint characteristic function. For a Gaussian vector, its mean and covariance determine the entire characteristic function. When the cross-covariance block vanishes, the exponent separates into a term involving the first vector and a term involving the second. That separation gives the product of their marginal characteristic functions, which characterizes independence.

Definition: Let $X\in\mathbb{R}^p$ and $Y\in\mathbb{R}^q$ have finite second moments. Their cross-covariance matrix is $$ \operatorname{Cov}(X,Y) =E\left[(X-E[X])(Y-E[Y])^\mathsf{T}\right], $$ a $p\times q$ matrix whose $(i,j)$ entry is $\operatorname{Cov}(X_i,Y_j)$. The pair $(X,Y)$ is jointly Gaussian if the concatenated vector $(X^\mathsf{T},Y^\mathsf{T})^\mathsf{T}$ has a multivariate Gaussian distribution. Joint Gaussianity permits a singular covariance matrix.

The Characteristic-Function Argument

For a random vector $Z\in\mathbb{R}^d$, its characteristic function is $\varphi_Z(t)=E[\exp(i t^\mathsf{T}Z)]$, for $t\in\mathbb{R}^d$. If $Z$ is Gaussian with mean $m$ and covariance matrix $\Sigma$, its characteristic function is $$ \varphi_Z(t)=\exp\left(i t^\mathsf{T}m-\frac{1}{2}t^\mathsf{T}\Sigma t\right). $$ This formula also holds when $\Sigma$ is singular. Partition the mean and covariance of the jointly Gaussian vector $(X^\mathsf{T},Y^\mathsf{T})^\mathsf{T}$ as $$ m=\begin{pmatrix}m_X\\m_Y\end{pmatrix}, \qquad \Sigma= \begin{pmatrix} \Sigma_{XX}&\Sigma_{XY}\\ \Sigma_{YX}&\Sigma_{YY} \end{pmatrix}, \qquad \Sigma_{XY}=\operatorname{Cov}(X,Y). $$ For $s\in\mathbb{R}^p$ and $t\in\mathbb{R}^q$, the quadratic term in the joint characteristic function is $$ \begin{pmatrix}s\\t\end{pmatrix}^{\mathsf T} \Sigma \begin{pmatrix}s\\t\end{pmatrix} =s^\mathsf{T}\Sigma_{XX}s +s^\mathsf{T}\Sigma_{XY}t +t^\mathsf{T}\Sigma_{YX}s +t^\mathsf{T}\Sigma_{YY}t. $$ The two middle terms are the cross terms. If $\Sigma_{XY}=0$, both vanish, and the exponent splits into a part depending only on $s$ and a part depending only on $t$.

Theorem: If $X$ and $Y$ are jointly Gaussian, then $X$ and $Y$ are independent if and only if $\operatorname{Cov}(X,Y)=0$. No nonsingularity assumption on their joint covariance matrix is needed.

Proof. First suppose $\operatorname{Cov}(X,Y)=\Sigma_{XY}=0$. Then $\Sigma_{YX}=\Sigma_{XY}^\mathsf{T}=0$. Substituting the block form into the Gaussian characteristic function gives $$ \begin{aligned} \varphi_{(X,Y)}(s,t) &=\exp\left(i s^\mathsf{T}m_X+i t^\mathsf{T}m_Y -\frac{1}{2}s^\mathsf{T}\Sigma_{XX}s -\frac{1}{2}t^\mathsf{T}\Sigma_{YY}t\right)\\ &=\exp\left(i s^\mathsf{T}m_X-\frac{1}{2}s^\mathsf{T}\Sigma_{XX}s\right) \exp\left(i t^\mathsf{T}m_Y-\frac{1}{2}t^\mathsf{T}\Sigma_{YY}t\right)\\ &=\varphi_X(s)\varphi_Y(t). \end{aligned} $$ The last equality follows because the two factors are the marginal characteristic functions of $X$ and $Y$. The characteristic-function criterion for independence states that $X$ and $Y$ are independent exactly when this joint characteristic function factors as $\varphi_X(s)\varphi_Y(t)$ for every $s,t$. Thus $X$ and $Y$ are independent.

Conversely, suppose $X$ and $Y$ are independent. Their Gaussian distributions have finite second moments. For each component pair $i,j$, independence gives $$ \operatorname{Cov}(X_i,Y_j) =E[(X_i-E[X_i])(Y_j-E[Y_j])] =E[X_i-E[X_i]]E[Y_j-E[Y_j]] =0. $$ Every entry of $\operatorname{Cov}(X,Y)$ is therefore zero. This proves both directions. $\square$

Worked Examples

Worked Example: Factoring a Gaussian Characteristic Function

Suppose $X=(X_1,X_2)^\mathsf{T}$ and $Y$ are jointly Gaussian with means $E[X]=(1,-2)^\mathsf{T}$, $E[Y]=3$, and covariance blocks $$ \operatorname{Cov}(X)= \begin{pmatrix}4&1\\1&9\end{pmatrix}, \qquad \operatorname{Var}(Y)=4, \qquad \operatorname{Cov}(X,Y)= \begin{pmatrix}0\\0\end{pmatrix}. $$ Determine whether $X$ and $Y$ are independent by examining the joint characteristic function.

Let $s=(s_1,s_2)^\mathsf{T}$, and let $t$ be the scalar argument for $Y$. The mean term is $i(s_1-2s_2+3t)$. The covariance quadratic form is $$ s^\mathsf{T}\operatorname{Cov}(X)s+4t^2 =4s_1^2+2s_1s_2+9s_2^2+4t^2. $$ There is no cross term because $\operatorname{Cov}(X,Y)=0$. Hence $$ \begin{aligned} \varphi_{(X,Y)}(s,t) &=\exp\left(i(s_1-2s_2+3t) -\frac12(4s_1^2+2s_1s_2+9s_2^2+4t^2)\right)\\ &=\exp\left(i(s_1-2s_2)-\frac12(4s_1^2+2s_1s_2+9s_2^2)\right) \exp\left(i3t-\frac12(4t^2)\right). \end{aligned} $$ The first factor is $\varphi_X(s)$, and the second is $\varphi_Y(t)$. The joint characteristic function factors for all $s,t$, so $X$ and $Y$ are independent.

Worked Example: A Singular Gaussian Pair

Let $Z$ and $W$ be independent $N(0,1)$ random variables. Define $X=(Z,Z)^\mathsf{T}$ and $Y=(W,-W)^\mathsf{T}$. These vectors each have a singular covariance matrix. Verify their cross-covariance and characteristic-function factorization.

Since $Z$ and $W$ are independent and have mean zero, each entry of the cross-covariance matrix is zero: $$ \operatorname{Cov}(X,Y) = \begin{pmatrix} E[ZW]&E[-ZW]\\ E[ZW]&E[-ZW] \end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}. $$ For $s=(s_1,s_2)^\mathsf{T}$ and $t=(t_1,t_2)^\mathsf{T}$, we have $s^\mathsf{T}X=(s_1+s_2)Z$ and $t^\mathsf{T}Y=(t_1-t_2)W$. Independence of $Z$ and $W$, together with the standard normal characteristic function, gives $$ \varphi_{(X,Y)}(s,t) =\exp\left(-\frac12(s_1+s_2)^2\right) \exp\left(-\frac12(t_1-t_2)^2\right) =\varphi_X(s)\varphi_Y(t). $$ Thus the vectors are independent. The covariance of $X$ is $\begin{pmatrix}1&1\\1&1\end{pmatrix}$, which is singular; the covariance of $Y$ is $\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$, also singular. This example confirms that the theorem does not require invertible covariance matrices.

Why the Argument Fails for General Distributions

For a general pair, the joint characteristic function need not have a quadratic exponent. Knowing the cross-covariance gives information about second moments, including the mixed second-order behavior at the origin, but it does not determine the characteristic function at all argument values. Independence requires factorization for every $s,t$, a much stronger condition than vanishing cross-covariance.

Worked Example: Zero Covariance Without Independence

Let $X$ be uniform on $(-1,1)$, and define $Y=X^2-\frac13$. Show that $X$ and $Y$ have zero covariance but are dependent.

Symmetry gives $E[X]=0$ and $E[X^3]=0$. Also, $$ E[Y]=E[X^2]-\frac13=\frac13-\frac13=0. $$ Consequently, $$ \operatorname{Cov}(X,Y) =E[XY]-E[X]E[Y] =E[X^3-\tfrac13X] =0. $$ To check dependence, consider $A=\{|X|<1/2\}$ and $B=\{Y<-1/12\}$. The definition of $Y$ gives $$ Y<-\frac{1}{12} \quad\Longleftrightarrow\quad X^2-\frac13<-\frac{1}{12} \quad\Longleftrightarrow\quad |X|<\frac12. $$ Thus $A=B$. Since $X$ is uniform on an interval of length $2$, $P(A)=P(B)=1/2$, and $P(A\cap B)=1/2$, not $P(A)P(B)=1/4$. The variables are dependent. In particular, their joint characteristic function cannot factor as the product of their marginal characteristic functions: if it did, the characteristic-function criterion would imply independence, contradicting the event calculation.

Common pitfall: “The cross-covariance is zero, so the characteristic function factors.” This skips the essential hypothesis. Zero cross-covariance removes the mixed terms from the Gaussian quadratic exponent; a general joint characteristic function has no such quadratic form. Factorization must be established for all arguments, not inferred from a covariance calculation.

A related caution is that Gaussian marginals alone are not enough. Let $X\sim N(0,1)$, let $S$ be an independent random sign taking $1$ and $-1$ with equal probability, and set $Y=SX$. Both $X$ and $Y$ have standard normal marginal distributions, and $$ \operatorname{Cov}(X,Y)=E[SX^2]=E[S]E[X^2]=0. $$ But $|Y|=|X|$ with probability one, so the two variables are dependent. Their pair is not jointly Gaussian. The theorem concerns joint Gaussianity, not merely normality of each marginal.

Why the Equivalence Matters

Under a joint Gaussian model, testing whether the cross-covariance block is zero is also testing independence between the two vector blocks. That interpretation is stronger than saying there is no linear association: it rules out every form of dependence between the blocks within the model. This is why the Gaussian assumption matters when interpreting cross-covariance tests.

Outside that model, a zero cross-covariance result has a narrower meaning. It says that the componentwise linear covariances vanish; it does not rule out deterministic, nonlinear, or other distributional dependence. The characteristic-function argument pinpoints the distinction: Gaussianity lets the covariance encode the whole joint characteristic function, while for general distributions covariance does not.

Check Your Understanding

Use the block characteristic-function argument and the distinction between joint and marginal Gaussianity.

For jointly Gaussian $X\in\mathbb{R}^p$ and $Y\in\mathbb{R}^q$, identify the terms in the covariance quadratic form that disappear when $\operatorname{Cov}(X,Y)=0$.
Why does factorization of the joint characteristic function imply independence?
Does the zero-cross-covariance equivalence require the joint covariance matrix to be nonsingular? Explain.
In the uniform-variable example, verify that $P(A\cap B)\ne P(A)P(B)$ for the specified events.
Why do Gaussian marginal distributions, by themselves, not permit the Gaussian independence result to be applied?