Normal Coordinates Do Not Determine a Joint Law
A collection of random variables can each have a familiar, well-behaved normal distribution while their dependence follows a quite different pattern. Marginal distributions describe one coordinate at a time; they do not, by themselves, describe how coordinates move together. For a Gaussian vector, that joint behavior has a special feature: every linear combination of its coordinates is normal. A single linear combination that is not normal is therefore enough to rule out joint Gaussianity.
This distinction matters in multivariate modeling. Properties often associated with a Gaussian vector—including the familiar relationship between zero covariance and independence—require joint Gaussianity, not merely normality of each coordinate. The examples below make the distinction concrete.
Taking \(a\) to be a coordinate vector shows that every coordinate of a jointly Gaussian vector has a normal marginal distribution. The reverse implication does not follow: checking the marginals checks only the choices of \(a\) that select one coordinate at a time, not all linear combinations.
A Characteristic-Function Criterion
The linear-combination definition has a useful equivalent form. Let \(X\) have finite second moments, let \(\mu=E[X]\), and let \(\Sigma=\operatorname{Cov}(X)\). Its characteristic function is \(\varphi_X(t)=E[\exp(i t^\mathsf{T}X)]\), for \(t\in\mathbb{R}^d\).
Proof. Suppose first that \(X\) is jointly Gaussian. For any \(t\), the scalar \(t^\mathsf{T}X\) is normal with mean \(t^\mathsf{T}\mu\) and variance \(t^\mathsf{T}\Sigma t\), since expectation and variance of a linear combination give those quantities. The characteristic function of a univariate normal variable with mean \(m\) and variance \(v\geq 0\) is \(\exp(iu m-u^2v/2)\). Evaluating the characteristic function of \(X\) at \(t\) therefore gives \[ \varphi_X(t)=E[\exp(i t^\mathsf{T}X)] =\exp\left(i t^\mathsf{T}\mu-\frac{1}{2}t^\mathsf{T}\Sigma t\right). \] This includes \(v=0\), when the scalar is constant.
Conversely, suppose the displayed characteristic-function formula holds. For any fixed \(a\in\mathbb{R}^d\), the characteristic function of \(a^\mathsf{T}X\), evaluated at \(u\in\mathbb{R}\), is \[ E[\exp(iu a^\mathsf{T}X)] =\varphi_X(ua) =\exp\left(iu a^\mathsf{T}\mu-\frac{u^2}{2}a^\mathsf{T}\Sigma a\right). \] This is the characteristic function of a univariate normal distribution with mean \(a^\mathsf{T}\mu\) and variance \(a^\mathsf{T}\Sigma a\). By uniqueness of characteristic functions, \(a^\mathsf{T}X\) has that distribution. This holds for every \(a\), so \(X\) is jointly Gaussian by definition. \(\square\)
The criterion also explains how a failed projection diagnoses a failed joint distribution. For a proposed vector, one need not determine its entire joint density or characteristic function: finding just one coefficient vector \(a\) for which \(a^\mathsf{T}X\) is nonnormal disproves joint Gaussianity.
Counterexamples Built From Sign Changes
Worked Example: Zero-Covariance Normal Marginals
Let \(Z\sim N(0,1)\), and let \(S\) be independent of \(Z\), with \(P(S=1)=P(S=-1)=1/2\). Set \(X=Z\) and \(Y=SZ\). Check the marginals, dependence, covariance, and a projection.
The marginal distribution of \(X\) is standard normal. For \(Y\), conditioning on \(S\) gives \(Y=Z\) with probability \(1/2\) and \(Y=-Z\) with probability \(1/2\). Since \(-Z\sim N(0,1)\), both conditional distributions are standard normal, and hence \(Y\sim N(0,1)\).
The coordinates are dependent: \(P(Y=X)=P(S=1)+P(S=-1,Z=0)=1/2\), because \(P(Z=0)=0\). If \(X\) and \(Y\) were independent standard normal variables, their continuous distributions would give \(P(Y=X)=0\). Their covariance nevertheless vanishes: \[ \operatorname{Cov}(X,Y)=E[XY]=E[SZ^2]=E[S]E[Z^2]=0. \]
Now consider \(X+Y=(1+S)Z\). When \(S=-1\), which has probability \(1/2\), this sum is exactly zero. When \(S=1\), it equals \(2Z\), which is nonzero with probability one. Thus the sum has an atom of probability \(1/2\) at zero and is not constant. A nonconstant normal random variable has no atoms, so \(X+Y\) is not normal. Consequently, \((X,Y)^\mathsf{T}\) is not jointly Gaussian, despite its normal marginals and zero covariance.
Worked Example: A Deterministic Dependence With Normal Marginals
Let \(X\sim N(0,1)\), and define \[ Y=\begin{cases} X, & |X|\leq 1,\\ -X, & |X|>1. \end{cases} \] This construction uses no additional randomness. Verify that \(Y\) is still standard normal and use a projection to test joint Gaussianity.
For any fixed magnitude \(r>0\), a standard normal variable has equally likely signs given \(|X|=r\). The rule defining \(Y\) either keeps both signs or reverses both signs, according to \(r\). In either case, conditional on \(|X|=r\), \(Y\) has probability \(1/2\) at \(r\) and probability \(1/2\) at \(-r\), exactly as \(X\) does. Since \(|Y|=|X|\), averaging these conditional distributions over the distribution of \(|X|\) shows that \(Y\sim N(0,1)\).
But the linear combination \(X+Y\) equals zero whenever \(|X|>1\), an event of positive probability. It equals \(2X\) whenever \(|X|\leq 1\), and is nonzero except on the probability-zero event \(X=0\). Thus \(X+Y\) has a positive atom at zero and also takes nonzero values with positive probability. It cannot be a normal random variable. The coordinates have normal marginals but are not jointly Gaussian; indeed, \(Y\) is a deterministic, nonconstant function of \(X\).
Even an Identity Covariance Matrix Is Not Enough
Worked Example: Three Standard Normal Coordinates With Covariance \(I\)
Let \(Z,W\sim N(0,1)\) be independent, and let \(S\) be an independent random sign with equal probabilities for \(1\) and \(-1\). Define \(X_1=Z\), \(X_2=SZ\), and \(X_3=W\).
The first two coordinates are standard normal by the sign-symmetry argument in the first example, and \(X_3\) is standard normal by construction. Their means are zero. Their variances are one, and \[ \operatorname{Cov}(X_1,X_2)=E[SZ^2]=0. \] Because \(W\) is independent of \((S,Z)\), both covariances involving \(X_3\) are zero. Therefore \(\operatorname{Cov}(X)=I_3\), the \(3\)-by-\(3\) identity matrix.
Consider the projection \(T=X_1+X_2+X_3\). If \(S=-1\), then \(T=W\), so its conditional distribution is \(N(0,1)\). If \(S=1\), then \(T=2Z+W\). Independence gives \(2Z+W\sim N(0,5)\). Thus \(T\) is a mixture, with equal weights, of \(N(0,1)\) and \(N(0,5)\).
A moment calculation confirms that this mixture is not normal. Its mean is zero and its variance is \((1+5)/2=3\). Its fourth moment is \[ E[T^4]=\frac{1}{2}(3\cdot 1^2)+\frac{1}{2}(3\cdot 5^2)=\frac{3}{2}+\frac{75}{2}=39. \] A normal random variable with mean zero and variance \(3\) has fourth moment \(3(3^2)=27\). Since \(39\neq 27\), \(T\) is not normal. Therefore \(X\) is not jointly Gaussian even though every coordinate is standard normal and the covariance matrix is the identity.
How to Use the Linear-Combination Test
The examples share a simple diagnostic. First, verify what the marginals and covariance matrix actually establish. Then choose coefficients that expose dependence and determine the distribution of the resulting scalar projection. If that projection is nonnormal, joint Gaussianity is ruled out. A projection that happens to be normal does not, by itself, prove joint Gaussianity; the definition requires normality for every coefficient vector.
This distinction prevents a common error in multivariate work: treating normal marginals and a covariance matrix as a complete specification of the joint distribution. They do specify the joint law within the class of jointly Gaussian vectors, but outside that class they need not do so. In particular, the assertion that zero covariance implies independence is valid for jointly Gaussian coordinates, not for arbitrary variables with normal marginals. The first example has normal marginals and zero covariance, yet its coordinates are dependent.
Check Your Understanding
For each question, focus on which distributions have actually been established.
- Why does normality of every coordinate follow from joint Gaussianity?
- In the first worked example, what feature of \(X+Y\) shows that it is not normal?
- For the vector in the three-coordinate example, why does an identity covariance matrix not establish joint Gaussianity?
- What must be checked to prove that a vector is jointly Gaussian using the linear-combination definition?
- Does finding one normal linear combination prove joint Gaussianity? Explain.