Tutorials › Multivariate Statistics (Graduate) › Positive-Definite Tapering for Covariance Estimation

Structured Covariance Models · Tutorial 896 of 975

Positive-Definite Tapering for Covariance Estimation

Understand the conditions that make covariance tapering preserve positive definiteness, and see why simply deleting distant covariances can fail.

Advanced 11 min read

What You'll Learn

  • Define covariance tapering as an entrywise product with a taper matrix.
  • State and prove when tapering preserves positive semidefiniteness and positive definiteness.
  • Verify that the Bartlett triangular taper is positive semidefinite.
  • Work through numerical examples of tapered covariance matrices.
  • Distinguish gradual tapering from hard banding and identify a failure of hard banding.
  • Recognize what positive definiteness guarantees—and what it does not guarantee—about a covariance estimate.

Why Taper a Covariance Estimate?

For measurements indexed by time or location, covariances between nearby observations may be more important, or more reliably estimated, than covariances between distant ones. A structured estimate can reflect this by reducing distant entries and setting sufficiently distant ones to zero. But changing entries one at a time can destroy positive definiteness: the resulting matrix may no longer be a valid covariance matrix. Covariance tapering gives a principled way to modify entries while controlling this risk.

The key is to apply a positive-semidefinite matrix to the estimated covariance entry by entry. Under a simple diagonal condition, this operation preserves positive definiteness. The condition also clarifies why tapering is not the same as retaining entries inside a band and deleting the rest.

Definition and the Positive-Definiteness Guarantee

Definition: For two matrices \(A\) and \(B\) of the same dimensions, their Hadamard product, written \(A\circ B\), is defined by \((A\circ B)_{ij}=A_{ij}B_{ij}\). Given a covariance estimate \(\widehat{\Sigma}\) and a symmetric matrix \(T\), the tapered covariance estimate is \(\widehat{\Sigma}_{\mathrm{tap}}=\widehat{\Sigma}\circ T\). A matrix \(T\) used this way is a taper matrix when it is positive semidefinite; commonly it also has \(T_{ii}=1\), so the variances are unchanged.

A symmetric matrix \(A\) is positive semidefinite if \(x^{\mathsf T}Ax\geq 0\) for every vector \(x\), and positive definite if \(x^{\mathsf T}Ax>0\) for every nonzero vector \(x\). A covariance matrix must be positive semidefinite; positive definiteness is needed for operations such as an ordinary inverse and a nonsingular multivariate normal density.

Theorem: If \(A\) and \(T\) are symmetric positive-semidefinite matrices of the same finite dimension, then \(A\circ T\) is positive semidefinite. If \(A\) is positive definite and \(T_{ii}>0\) for every \(i\), then \(A\circ T\) is positive definite.

This is the Schur product theorem, together with its positive-definite consequence. The diagonal condition matters: a positive-semidefinite taper with a zero diagonal entry has an all-zero corresponding row and column, so its Hadamard product cannot be positive definite.

Proof. Since \(T\) is positive semidefinite, it has a Gram representation \(T_{ij}=u_i^{\mathsf T}u_j\), where \(u_i\in\mathbb{R}^r\) for some finite \(r\). Write \(u_{ik}\) for coordinate \(k\) of \(u_i\), and let \(D_k\) be the diagonal matrix with diagonal entries \(u_{1k},\ldots,u_{pk}\). For any vector \(x\),

$$ x^{\mathsf T}(A\circ T)x =\sum_{i=1}^p\sum_{j=1}^p x_i A_{ij}T_{ij}x_j =\sum_{k=1}^r (D_kx)^{\mathsf T}A(D_kx). $$

If \(A\) is positive semidefinite, every term in the last sum is nonnegative, so \(A\circ T\) is positive semidefinite. Now suppose \(A\) is positive definite, \(T_{ii}>0\) for every \(i\), and \(x\neq 0\). Choose an index \(i\) for which \(x_i\neq 0\). Since \(T_{ii}=u_i^{\mathsf T}u_i=\sum_{k=1}^r u_{ik}^2>0\), there is at least one \(k\) with \(u_{ik}\neq 0\). For that \(k\), \(D_kx\neq 0\), so \((D_kx)^{\mathsf T}A(D_kx)>0\). All other terms are nonnegative. Thus \(x^{\mathsf T}(A\circ T)x>0\), proving positive definiteness. \(\square\)

In particular, if an estimate \(\widehat{\Sigma}\) is positive definite and \(T\) is positive semidefinite with unit diagonal, tapering retains positive definiteness and leaves each estimated variance unchanged. If \(\widehat{\Sigma}\) is only positive semidefinite, the theorem guarantees only that its tapered version is positive semidefinite; it need not become positive definite.

How a Taper Matrix Can Be Constructed

A common construction begins with measurement locations \(s_1,\ldots,s_p\) and a function \(K_h\) of pairwise separation. The entries are \(T_{ij}=K_h(s_i-s_j)\). To use the theorem, the resulting matrix must be positive semidefinite for the locations under consideration. A function that is nonnegative and decreases with distance is not automatically a valid taper: entrywise values alone do not establish that its matrix is positive semidefinite.

One simple valid taper on equally spaced observations is the Bartlett triangular taper. For integer indices and a positive integer bandwidth \(m\), set

$$ T_{ij}=\left(1-\frac{|i-j|}{m}\right)_+, \qquad (a)_+=\max(a,0). $$

The entries decrease linearly with lag and are zero when \(|i-j|\geq m\). Its positive semidefiniteness follows directly from a Gram representation. For each integer \(i\), define \(g_i(t)=m^{-1/2}\mathbf{1}_{[i,i+m]}(t)\). Then

$$ \int_{\mathbb{R}}g_i(t)g_j(t)\,dt =\frac{(m-|i-j|)_+}{m} =\left(1-\frac{|i-j|}{m}\right)_+ =T_{ij}. $$

Thus \(T\) is a Gram matrix of functions: for any vector \(x\), \(x^{\mathsf T}Tx=\int_{\mathbb{R}}(\sum_i x_i g_i(t))^2\,dt\geq 0\). Also \(T_{ii}=1\). This proves that the Bartlett taper satisfies the conditions for preserving positive definiteness.

Worked Example: Tapering a Three-Variable Covariance Matrix

Consider the covariance matrix and the Bartlett taper with \(m=2\):

$$ \Sigma= \begin{pmatrix} 1&0.8&0.3\\ 0.8&1&0.8\\ 0.3&0.8&1 \end{pmatrix}, \qquad T= \begin{pmatrix} 1&0.5&0\\ 0.5&1&0.5\\ 0&0.5&1 \end{pmatrix}. $$

First check that the starting matrix is positive definite. Its leading principal minors are \(1\), \(1-0.8^2=0.36\), and

$$ \det(\Sigma) =1+2(0.8)(0.8)(0.3)-0.8^2-0.8^2-0.3^2 =1.384-1.37 =0.014. $$

All three leading principal minors are positive, so \(\Sigma\) is positive definite. The taper is positive semidefinite with unit diagonal, as shown by the Bartlett construction. Entrywise multiplication gives

$$ \Sigma\circ T= \begin{pmatrix} 1&0.8(0.5)&0.3(0)\\ 0.8(0.5)&1&0.8(0.5)\\ 0.3(0)&0.8(0.5)&1 \end{pmatrix} = \begin{pmatrix} 1&0.4&0\\ 0.4&1&0.4\\ 0&0.4&1 \end{pmatrix}. $$

Its leading principal minors are \(1\), \(1-0.4^2=0.84\), and \(1-2(0.4^2)=0.68\), all positive. The result is therefore positive definite, as the theorem guarantees. The diagonal variances are unchanged, the lag-one covariances are halved, and the lag-two covariance is set to zero.

Tapering Is Not Hard Banding

Hard banding applies a mask that keeps selected entries and sets the others to zero. For an ordered vector, a bandwidth-\(k\) mask keeps entries with \(|i-j|\leq k\). This operation is also an entrywise product, but its mask is not necessarily positive semidefinite. As a result, it does not have the positive-definiteness guarantee just proved. Tapering instead uses a positive-semidefinite matrix of weights; those weights may decrease gradually to zero, and the matrix property—not merely the appearance of its entries—is what protects the covariance structure.

Worked Example: A Hard-Banding Mask Can Break Positive Definiteness

Take a three-variable equicorrelation matrix with correlation \(0.9\):

$$ \Sigma= \begin{pmatrix} 1&0.9&0.9\\ 0.9&1&0.9\\ 0.9&0.9&1 \end{pmatrix}. $$

Its eigenvalues are \(1-0.9=0.1\) twice and \(1+2(0.9)=2.8\) once, so \(\Sigma\) is positive definite. Now apply the bandwidth-one hard-banding mask

$$ B= \begin{pmatrix} 1&1&0\\ 1&1&1\\ 0&1&1 \end{pmatrix}. $$

The hard-banded result and its determinant are

$$ \Sigma\circ B= \begin{pmatrix} 1&0.9&0\\ 0.9&1&0.9\\ 0&0.9&1 \end{pmatrix}, \qquad \det(\Sigma\circ B)=1-2(0.9^2)=-0.62. $$

A positive-definite matrix must have positive determinant, so the result is not positive definite; in fact, it is indefinite. The mask itself fails the needed condition: \(\det(B)=-1\), so \(B\) is not positive semidefinite. This example shows why retaining the same diagonal and setting only distant covariances to zero does not, by itself, preserve validity.

Worked Example: A Four-Variable Bartlett-Tapered Estimate

Suppose the estimated covariance matrix has the form \(\widehat{\Sigma}_{ij}=0.5^{|i-j|}\) for four equally spaced measurements. It is positive definite: it is the covariance matrix of a four-dimensional Gaussian AR(1) model with marginal variance \(1\) and correlation parameter \(0.5\), whose positive definiteness holds for \(|0.5|<1\). Apply the Bartlett taper with \(m=2\). Its weights are \(1\) at lag zero, \(0.5\) at lag one, and zero at all lags of at least two. The tapered matrix is

$$ \widehat{\Sigma}\circ T= \begin{pmatrix} 1&0.25&0&0\\ 0.25&1&0.25&0\\ 0&0.25&1&0.25\\ 0&0&0.25&1 \end{pmatrix}. $$

For a direct numerical check, let \(D_j\) be the determinant of the leading \(j\)-by-\(j\) principal submatrix, with \(D_0=1\). Expanding the tridiagonal determinants gives \(D_1=1\) and \(D_j=D_{j-1}-0.25^2D_{j-2}\). Therefore \(D_2=1-0.0625=0.9375\), \(D_3=0.9375-0.0625=0.875\), and \(D_4=0.875-0.0625(0.9375)=0.81640625\). All leading principal minors are positive, confirming positive definiteness. This computation agrees with the theorem; the taper reduces the lag-one covariances and removes the more distant ones without changing the variances.

What the Guarantee Does—and Does Not—Say

The preservation result is algebraic. It says that if the input matrix is positive definite and the taper matrix is positive semidefinite with positive diagonal, their Hadamard product is positive definite. It does not say that the tapered estimate is unbiased for the true covariance, that a particular bandwidth is appropriate, or that every taper function with a convenient shape is valid. Those are separate modeling and estimation questions.

Tapering is useful when the data structure makes distant covariances less important or less precisely estimated, and when a sparse estimate is desirable. A taper with compact support can create exact zeros, while positive semidefiniteness preserves the covariance-matrix constraint. However, if the starting estimate is singular, the theorem alone cannot guarantee an invertible result. For example, a sample covariance matrix may be singular when the number of variables is large relative to the number of observations. One must check the starting matrix or use an appropriate regularization before relying on a positive-definite guarantee.

Common mistake: “The taper has nonnegative entries and leaves all variances unchanged, so it must preserve positive definiteness.” Nonnegative entries and a unit diagonal do not imply that the taper matrix is positive semidefinite. The correction is to verify positive semidefiniteness of the entire taper matrix, then check that the input covariance is positive definite and the taper diagonal is strictly positive. Hard-band masks can fail precisely because their matrix is not positive semidefinite.
Key takeaway: Covariance tapering is the Hadamard product \(\widehat{\Sigma}\circ T\). Positive-semidefinite \(T\) preserves positive semidefiniteness, and positive-definite \(\widehat{\Sigma}\) together with strictly positive taper diagonal entries guarantees a positive-definite result. Hard banding has no such guarantee unless its mask is itself positive semidefinite.

Check Your Understanding

Use the definitions and results above to answer the following questions.

  1. What two matrix conditions guarantee that tapering a positive-definite covariance estimate leaves it positive definite?
  2. If both the covariance estimate and taper matrix are only positive semidefinite, what can be concluded about their Hadamard product?
  3. Why does a zero diagonal entry in a positive-semidefinite taper prevent its Hadamard product from being positive definite?
  4. For the Bartlett taper with \(m=3\), what are the weights at lags zero, one, two, and three?
  5. Why does a hard-banding mask with nonnegative entries and unit diagonal still fail to guarantee a valid tapered covariance matrix?
Clinical Biostats