Why Gaussianity Changes the Stationarity Question
Weak stationarity describes a series through its mean and covariance: the mean is constant, and the covariance between two observations depends only on their time separation. Strict stationarity asks for more. It requires the entire joint distribution of any finite collection of observations to remain unchanged when all the time indices are shifted together.
In general, matching means and covariances does not determine a joint distribution. The important exception is a Gaussian series: every finite vector of observations is jointly multivariate normal, and a multivariate normal distribution is determined by its mean vector and covariance matrix. This is why, for Gaussian series with the required moments, weak and strict stationarity are equivalent.
Strict stationarity concerns all finite-dimensional distributions, not just the one-dimensional distributions of individual observations. Weak stationarity concerns only first and second moments. For a Gaussian series, however, those first and second moments contain all the information needed to identify each finite-dimensional distribution.
The Equivalence for Gaussian Series
Suppose \(\{X_t\}\) is a Gaussian series. If it is weakly stationary, then for any times \(t_1,\ldots,t_n\), the vector \((X_{t_1},\ldots,X_{t_n})\) is multivariate normal. Its mean vector is constant across the components, and its covariance matrix is determined by the time differences. Shifting every time by \(r\) leaves both quantities unchanged. The shifted and unshifted vectors therefore have the same distribution.
Conversely, strict stationarity makes every \(X_t\) have the same marginal distribution. If the series has finite second moments, this gives a common mean and variance. The joint distribution of \((X_{t+h},X_t)\) also does not change when the times are shifted, so its covariance depends only on \(h\). Thus strict stationarity implies weak stationarity when the required moments are finite. Gaussian marginals have finite moments, so this condition holds for a Gaussian series.
Why the Gaussian Argument Works
The key fact about a multivariate normal vector \(Z\) is that its distribution is determined by its mean vector \(m\) and covariance matrix \(\Sigma\). In particular, its characteristic function at a vector \(u\) is
Thus, if two multivariate normal vectors have the same mean vector and covariance matrix, their characteristic functions agree for every \(u\), so their distributions agree. To apply this fact to stationarity, define \(Z=(X_{t_1},\ldots,X_{t_n})\) and \(Z_r=(X_{t_1+r},\ldots,X_{t_n+r})\). Under weak stationarity, both have mean vector \((\mu,\ldots,\mu)\). Their covariance entries also agree: the entry in row \(j\), column \(k\) is \(\gamma(t_j-t_k)\) for both vectors, since shifting both times does not change their difference. The two vectors are jointly Gaussian with the same parameters, hence have the same distribution. This holds for every \(n\), set of times, and shift, proving strict stationarity.
The argument uses joint Gaussianity. Knowing only that each \(X_t\) is normally distributed does not tell us that a vector of observations is multivariate normal, so it does not justify identifying that vector’s distribution from its covariance matrix.
Form the observation vector at \(t_1,\ldots,t_n\), and form a second vector at the shifted times \(t_1+r,\ldots,t_n+r\).
Weak stationarity gives the same mean vector for both. Each covariance entry depends on a time difference, which is unchanged by a common shift.
Both vectors are multivariate normal. Equal mean vectors and covariance matrices give equal joint distributions, which is strict stationarity.
Worked Applications
Worked Example: Comparing Two Shifted Gaussian Vectors
Suppose a jointly Gaussian series is weakly stationary with mean \(2\) and autocovariance \(\gamma(h)=9(1/2)^{|h|}\). Compare the distributions of the vectors at times \((0,1,3)\) and \((5,6,8)\), and determine whether the process is strictly stationary.
Method. Compare the two vectors’ mean vectors and covariance matrices, then use the multivariate normal result. The common time shift is \(5\): \((0,1,3)\) becomes \((5,6,8)\).
For each vector, the mean vector is \((2,2,2)^\mathsf{T}\). The pairwise time differences, in absolute value, are \(1\), \(3\), and \(2\). The autocovariances at these lags are
Consequently, the covariance matrix for either vector is
Both vectors are multivariate normal, have the same mean vector, and have the same covariance matrix. Their distributions are therefore identical. Since the same reasoning applies to any finite set of times and any common shift, the Gaussian series is strictly stationary. The common mistake to avoid is comparing only the variances: the off-diagonal covariances are also part of the vector’s distribution.
Worked Example: A Gaussian Series That Fails Stationarity
Let the observations be independent, with \(X_t\sim N(1,4)\) at even times and \(X_t\sim N(1,9)\) at odd times. Determine whether the series is weakly or strictly stationary. Here the second parameter of \(N(\mu,\sigma^2)\) is the variance.
Method. Check the mean and variance first. Independence then gives the covariances at distinct times.
At every time, \(E[X_t]=1\), so the mean is constant. At lag zero, however, the autocovariance is the variance, and it changes with time: \(\operatorname{Cov}(X_0,X_0)=4\), while \(\operatorname{Cov}(X_1,X_1)=9\). Thus the covariance at lag zero is not a single function of lag, and the series is not weakly stationary.
It is also not strictly stationary: \(X_0\sim N(1,4)\) and \(X_1\sim N(1,9)\) have different distributions. This is a jointly Gaussian series because its observations are independent normal variables, but Gaussianity does not by itself make a series stationary. The common mistake is to see that the means agree and infer stationarity; the variance must also be time-invariant.
Worked Example: Gaussian Marginals Do Not Guarantee Strict Stationarity
For each integer \(k\), let \(A_k\) and \(B_k\) be independent random signs, each equal to \(1\) or \(-1\) with probability \(1/2\). Take all these signs to be independent across blocks. Within block \(k\), define the signs at times \(3k+1,3k+2,3k+3\) as \(A_k,B_k,A_kB_k\), respectively. Let \(M_t\) be independent copies of the absolute value of a standard normal variable, independent of all the signs, and set \(X_t=\varepsilon_tM_t\), where \(\varepsilon_t\) is the sign assigned to time \(t\). Check the marginal distributions, weak stationarity, and strict stationarity.
Method. Calculate the first two moments and compare one triple’s joint distribution with that of a shifted triple. This example avoids the mistaken assumption that normal marginals alone make a series Gaussian.
Each sign \(\varepsilon_t\) takes the values \(1\) and \(-1\) equally often. Multiplying an independent absolute standard normal variable by such a sign gives a standard normal variable. Thus \(X_t\sim N(0,1)\) for every \(t\), so \(E[X_t]=0\) and \(\operatorname{Var}(X_t)=1\).
Any two distinct signs are independent: within a block, each pair among \(A_k,B_k,A_kB_k\) is independent, and signs in different blocks are independent. The magnitudes are independent as well. Therefore distinct observations are uncorrelated, so \(\operatorname{Cov}(X_{t+h},X_t)=0\) for every nonzero \(h\), regardless of \(t\). The series is weakly stationary, with \(\gamma(0)=1\) and \(\gamma(h)=0\) for \(h\ne0\).
Now compare the triples at times \((1,2,3)\) and \((2,3,4)\). For the first triple, the product of its signs is \(A_0B_0(A_0B_0)=1\). Since each magnitude is positive with probability one, \(P(X_1X_2X_3>0)=1\). For the shifted triple, the product of its signs is \(B_0(A_0B_0)A_1=A_0A_1\). The independent signs \(A_0\) and \(A_1\) have a product equally likely to be positive or negative, so \(P(X_2X_3X_4>0)=1/2\). These triple distributions differ, and the series is not strictly stationary. The example is weakly but not strictly stationary because its finite vectors are not jointly Gaussian, even though every individual observation is normal.
Common Exam Trap
A frequent error is to reason, “The mean is constant and the covariance depends only on lag, so the series is strictly stationary,” without checking joint Gaussianity. Those moment conditions establish weak stationarity. They establish strict stationarity only when the finite-dimensional distributions are determined by those moments, as they are for a jointly Gaussian series.
The reverse implication also needs care. Strict stationarity alone does not guarantee finite means or variances for an arbitrary series. To infer weak stationarity, the required moments must exist. For a Gaussian series, finite second moments are part of the setting, so this qualification is satisfied.
Check Your Understanding
For each problem, decide which stationarity conclusion is justified and identify the condition that matters.
- A jointly Gaussian series has \(E[X_t]=3\) and \(\gamma(h)=4(0.6)^{|h|}\). State whether it is weakly stationary, strictly stationary, both, or neither. (Answer: Both; weak stationarity and joint Gaussianity imply strict stationarity.)
- A jointly Gaussian series has \(X_t\sim N(0,2)\) for even \(t\) and \(X_t\sim N(0,5)\) for odd \(t\). Does it satisfy either form of stationarity? (Answer: Neither; its variance and marginal distribution change with time.)
- A strictly stationary series has finite second moments. Which weak-stationarity conditions follow, and why? (Answer: Both; its common marginal distribution gives a constant finite mean and variance, and shifted pairs have the same covariance, so covariance depends only on lag.)
- A series has normal marginals, mean zero, variance one, and zero covariance at every nonzero lag. Can strict stationarity be concluded from these facts alone? (Answer: No; Gaussian marginals do not establish joint Gaussianity, and first and second moments alone need not determine higher-dimensional distributions.)
- A weakly stationary jointly Gaussian series has \(E[X_t]=1\), \(\gamma(0)=9\), and \(\gamma(2)=2\). Compare the means and covariance between observations two time units apart in vectors \((X_4,X_6)\) and \((X_{10},X_{12})\), then state the implication. (Answer: Both mean vectors are \((1,1)^\mathsf{T}\), and both covariance matrices are \(\begin{pmatrix}9&2\\2&9\end{pmatrix}\); the two vectors have the same distribution, as required by strict stationarity.)