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Weak Stationarity and Spectral Representation · Tutorial 892 of 994

Testing Candidate Covariance Sequences

Use finite-dimensional matrix and quadratic-form checks to test whether a proposed sequence could be a stationary covariance sequence.

Advanced 10 min read

What You'll Learn

  • Form the Toeplitz covariance matrix for a finite block of observations.
  • Derive necessary inequalities from two-point covariance matrices.
  • Use principal minors to test positive semidefiniteness, including singular cases.
  • Find explicit finite quadratic forms that rule out an invalid candidate.
  • Distinguish passing selected finite tests from proving validity at every order.

Why Test a Candidate Sequence?

A proposed covariance sequence may look plausible term by term and still be impossible: the values must work together as covariances of every finite collection of observations. The practical test is to build the covariance matrix for a finite block and check that it is positive semidefinite. This tutorial develops that test through Toeplitz structure, principal minors, and finite quadratic forms. A failed finite test rules out the candidate; passing a few tests does not establish that every required matrix is positive semidefinite.

Let \((X_n)_{n\in\mathbb{Z}}\) be a real-valued weakly stationary process on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(S=\mathbb{R}\). Weak stationarity means that \(\mathbb{E}[X_n]\) is a finite constant independent of \(n\), and that \(\operatorname{Cov}(X_{n+h},X_n)\) depends only on the lag \(h\). Write \(\gamma(h)=\operatorname{Cov}(X_{n+h},X_n)\). For a real-valued process, \(\gamma(-h)=\gamma(h)\). A candidate sequence is a proposed list of real numbers \((\gamma(h))_{h\in\mathbb{Z}}\) with this symmetry.

Definition: For an integer \(m\ge 0\), the order-\(m\) Toeplitz covariance matrix of a candidate sequence is \(T_m=(\gamma(i-j))_{0\le i,j\le m}\). It is positive semidefinite if \(x^{\mathsf T}T_mx\ge 0\) for every real column vector \(x=(x_0,\ldots,x_m)^{\mathsf T}\). The expression \(x^{\mathsf T}T_mx\) is the associated finite quadratic form.

The matrix is called Toeplitz because its entries are constant along each diagonal: the entry depends on \(i-j\), not on \(i\) and \(j\) separately. If the candidate really is a covariance sequence, then \(T_m\) is the covariance matrix of \((X_0,\ldots,X_m)^{\mathsf T}\). Thus it must be positive semidefinite. Indeed, the quadratic form is the variance of a linear combination of observations.

$$ x^{\mathsf T}T_mx =\sum_{i=0}^{m}\sum_{j=0}^{m}x_ix_j\gamma(i-j) =\operatorname{Var}\!\left(\sum_{i=0}^{m}x_iX_i\right)\ge 0. $$

What the Small Matrices Require

The order-zero matrix is just \([\gamma(0)]\), so \(\gamma(0)\ge 0\) is necessary. At order one, the matrix is \(\begin{bmatrix}\gamma(0)&\gamma(1)\\\gamma(1)&\gamma(0)\end{bmatrix}\). Its determinant must be nonnegative, giving \(\gamma(0)^2-\gamma(1)^2\ge 0\). More generally, using the observations at times \(0\) and \(k\) gives the same matrix with \(\gamma(k)\) in the off-diagonal entries. Consequently, every candidate must satisfy

$$ \gamma(0)\ge 0,\qquad |\gamma(k)|\le \gamma(0)\quad\text{for every integer }k. $$

These inequalities are useful screening checks, but they only test two observations at a time. A sequence can satisfy all of them and still fail a larger matrix test. For larger matrices, one systematic method is to inspect principal minors. A principal submatrix is formed by choosing the same set of indices for rows and columns; its determinant is a principal minor. Positive semidefiniteness requires every principal minor to be nonnegative. For a symmetric matrix, this condition is also sufficient, including when some minors are zero.

Theorem (Principal-minor test): A real symmetric finite matrix \(A\) is positive semidefinite if and only if every principal minor of \(A\) is nonnegative.

Proof. Suppose first that \(A\) is positive semidefinite. Every principal submatrix \(A_I\), indexed by a set \(I\), is positive semidefinite: for any vector on \(I\), extend it by zeros outside \(I\) and use the quadratic-form inequality for \(A\). By the spectral theorem, the eigenvalues of the symmetric matrix \(A_I\) are nonnegative. Its determinant, the product of those eigenvalues, is therefore nonnegative. This proves necessity.

For sufficiency, suppose every principal minor of \(A\) is nonnegative. For \(\varepsilon>0\), consider \(A+\varepsilon I\). For any index set \(I\), expansion of the determinant by multilinearity gives \(\det(A_I+\varepsilon I)=\sum_{J\subseteq I}\varepsilon^{|I|-|J|}\det(A_J)\), where the determinant for the empty set is defined to be \(1\). Each term is nonnegative, and the term for \(J=\varnothing\) is \(\varepsilon^{|I|}>0\). Thus every leading principal minor of \(A+\varepsilon I\) is positive.

To see why this makes \(A+\varepsilon I\) positive definite, use induction on its size. The size-one case follows from its positive leading minor. At a larger size, write the matrix in block form \(\begin{bmatrix}B&b\\b^{\mathsf T}&c\end{bmatrix}\), where \(B\) is the leading block of one smaller size. By induction, \(B\) is positive definite and invertible. The determinant identity gives \(c-b^{\mathsf T}B^{-1}b=\det(A+\varepsilon I)/\det(B)>0\), with the determinants here taken for the full matrix and for \(B\). For any \(u\) and scalar \(v\), its quadratic form is \((u+B^{-1}bv)^{\mathsf T}B(u+B^{-1}bv)+(c-b^{\mathsf T}B^{-1}b)v^2\). This is positive for every nonzero \((u,v)\), so \(A+\varepsilon I\) is positive definite. Finally, for every \(x\), \(x^{\mathsf T}Ax=\lim_{\varepsilon\downarrow0}x^{\mathsf T}(A+\varepsilon I)x\ge0\). Therefore \(A\) is positive semidefinite. \(\square\)

For covariance testing, the theorem means that one can test a finite Toeplitz matrix by checking all of its principal minors, not just the leading ones. Strictly positive leading principal minors are a useful test for positive definiteness, but covariance matrices are allowed to be singular. When a minor is zero, the nonnegative-minor criterion still applies.

Worked Tests

Worked Example: A Candidate That Passes Through Order Two

Consider the even candidate with \(\gamma(0)=2\), \(\gamma(1)=\gamma(-1)=1\), and \(\gamma(2)=\gamma(-2)=0\). Test the Toeplitz matrix through order two. The matrix is

$$ T_2=\begin{bmatrix} 2&1&0\\ 1&2&1\\ 0&1&2 \end{bmatrix}. $$

Its one-by-one principal minors are all \(2\). The two-by-two principal minors are \(2\cdot2-1\cdot1=3\) for indices \(\{0,1\}\), \(2\cdot2-0\cdot0=4\) for indices \(\{0,2\}\), and \(2\cdot2-1\cdot1=3\) for indices \(\{1,2\}\). The full determinant is \(2(2\cdot2-1\cdot1)-1(1\cdot2-1\cdot0)=6-2=4\). Every principal minor is positive, so this matrix is positive definite and, in particular, positive semidefinite.

Check: the two-point bounds at these lags are \(|\gamma(1)|=1\le2=\gamma(0)\) and \(|\gamma(2)|=0\le2=\gamma(0)\), consistent with the matrix calculation. This verifies only the matrix through order two; it does not by itself test all larger orders.

Worked Example: Pairwise Bounds Pass, but the Three-Point Test Fails

Now take \(\gamma(0)=1\), \(\gamma(1)=\gamma(-1)=\tfrac34\), and \(\gamma(2)=\gamma(-2)=0\). The two-point bounds hold: \(\tfrac34\le1\) and \(0\le1\). But the order-two matrix is

$$ T_2=\begin{bmatrix} 1&\tfrac34&0\\ \tfrac34&1&\tfrac34\\ 0&\tfrac34&1 \end{bmatrix}. $$

Its one-by-one principal minors are \(1\). Its two-by-two principal minors are \(1-\tfrac{9}{16}=\tfrac7{16}\), \(1\), and \(\tfrac7{16}\), all nonnegative. However, its determinant is \(1-\tfrac9{16}-\tfrac9{16}=-\tfrac18\), so the principal-minor test rules it out.

The failure can also be seen directly in a finite quadratic form. Take \(x=(1,-\tfrac32,1)^{\mathsf T}\). The diagonal contribution to \(x^{\mathsf T}T_2x\) is \(1+\tfrac94+1=\tfrac{17}{4}\). The off-diagonal contribution is \(2\cdot\tfrac34\bigl(-\tfrac32-\tfrac32\bigr)=-\tfrac92\). Thus \(x^{\mathsf T}T_2x=\tfrac{17}{4}-\tfrac{18}{4}=-\tfrac14<0\). A covariance matrix cannot have a negative quadratic form, so this candidate is impossible.

Worked Example: The Zero-Variance Case

Suppose a candidate has \(\gamma(0)=0\) but \(\gamma(1)=\gamma(-1)=\tfrac12\). It satisfies the simple nonnegativity check \(\gamma(0)\ge0\), but its order-one matrix is

$$ T_1=\begin{bmatrix} 0&\tfrac12\\ \tfrac12&0 \end{bmatrix}. $$

The determinant is \(0\cdot0-(\tfrac12)^2=-\tfrac14\), which is negative. Equivalently, the vector \(x=(1,-1)^{\mathsf T}\) gives \(x^{\mathsf T}T_1x=0-\tfrac12-\tfrac12+0=-1\). Either calculation rejects the candidate.

This also illustrates why the case \(\gamma(0)=0\) cannot be treated as if the two-point bound were informative only when \(\gamma(0)>0\). The determinant inequality for the order-one matrix gives \(\gamma(0)^2-\gamma(k)^2\ge0\). If \(\gamma(0)=0\), it forces \(\gamma(k)=0\) for every tested lag \(k\).

A Reliable Testing Procedure

For a candidate sequence, work from small matrices upward, using exact arithmetic where possible. At each order, write the full Toeplitz matrix before calculating; this reduces errors in the lag indices. Check the simple constraints \(\gamma(0)\ge0\) and \(|\gamma(k)|\le\gamma(0)\), then compute principal minors or exhibit a vector whose quadratic form is negative. A single negative principal minor or a single negative quadratic form is conclusive: the candidate cannot be a covariance sequence.

Do not interpret a successful finite check too broadly. Passing the order-one test says only that every two-point matrix of the tested form passes. Passing through order \(m\) checks blocks of at most \(m+1\) consecutive observations; it does not establish positive semidefiniteness at every order. A proposed sequence is not certified by a few reassuring determinants. Conversely, a negative test at any finite order settles the question immediately.

Common mistake. A tempting but incorrect shortcut is to check only \(|\gamma(k)|\le\gamma(0)\) and declare the candidate valid. In the second example, the checked values are \(\gamma(0)=1\), \(|\gamma(1)|=\tfrac34\le1\), and \(|\gamma(2)|=0\le1\), yet the order-two determinant is \(-\tfrac18\) and the explicit quadratic form is \(-\tfrac14\). The correction is to test joint consistency using the full finite covariance matrix, not just separate pairs of values.

Check Your Understanding

For each proposed test, distinguish a necessary check from a conclusion about all orders.

  1. Write the order-two Toeplitz matrix for \(\gamma(0)=3\), \(\gamma(1)=1\), and \(\gamma(2)=-1\). Compute its determinant.
  2. Why does a negative principal minor rule out a candidate covariance sequence?
  3. For \(\gamma(0)=1\), \(\gamma(1)=\tfrac12\), and \(\gamma(2)=1\), which two-point bound already fails?
  4. If \(\gamma(0)=0\), what does the order-one determinant condition require of \(\gamma(k)\)?
  5. What does it mean if all principal minors of \(T_2\) are nonnegative? What does it not establish about \(T_3\) and higher orders?