Levinson recursion

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Levinson or Levinson-Durbin recursion is a procedure in linear algebra to recursively calculate the solution to an equation involving a Toeplitz matrix. The algorithm runs in Θ(n2) time, which is a strong improvement over Gauss-Jordan elimination, which runs in Θ(n3).

Newer algorithms, called asymptotically fast or sometimes superfast Toeplitz algorithms, can solve in Θ(n logpn) for various p (e.g. p = 2, p = 3). Levinson recursion remains popular for several reasons; for one, it is relatively easy to understand in comparison; for another, it can be faster than a superfast algorithm for small n (usually n < 256 [1]).

The Levinson-Durbin algorithm was proposed first by Norman Levinson in 1947, improved by J. Durbin in 1960, and subsequently improved to 4n2 and then 3n2 multiplications by W. F. Trench and S. Zohar, respectively.

Other methods to process data include Schur decomposition and Cholesky decomposition. In comparison to these, Levinson recursion (particularly Split-Levinson recursion) tends to be faster computationally, but more sensitive to computational inaccuracies like round-off errors.

Contents

[edit] Derivation

[edit] Background

Matrix equations follow the form:

\vec y = \mathbf M \ \vec x

The Levinson-Durbin algorithm may be used for any such equation, so long as M is a known Toeplitz matrix with a nonzero main diagonal. Here \vec y is a known vector (in the list-of-numbers linear algebra sense), and \vec x is an unknown vector of numbers xi yet to be determined.

For the sake of this article, êi is a vector made up entirely of zeroes, except for its i'th place, which holds the value one. Its length will be implicitly determined by the surrounding context. The term N refers to the width of the matrix above -- M is an N×N matrix. Finally, in this article, superscripts refer to an inductive index, whereas subscripts denote indices. For example (and definition), in this article, the matrix Tn is an n×n matrix which copies the upper left n×n block from M -- that is, Tnij = Mij.

Tn is also a Toeplitz matrix; meaning that it can be written as:

\mathbf T^n = \begin{bmatrix} t_0 & t_{-1} & t_{-2} & \dots & t_{-n+1} \\ t_1 & t_0 & t_{-1} & \dots & t_{-n+2} \\ t_2 & t_1 & t_0 & \dots & t_{-n+3} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ t_{n-1}& t_{n-2} & t_{n-3} & \dots & t_0 \\ \end{bmatrix}

[edit] Introductory steps

The algorithm proceeds in two steps. In the first step, two sets of vectors, called the forward and backward vectors, are established. The forward vectors are used to help get the set of backward vectors; then they can be immediately discarded. The backwards vectors are necessary for the second step, where they are used to build the solution desired.

Levinson-Durbin recursion defines the nth "forward vector", denoted \vec f^n, as the vector of length n which satisfies:

\mathbf T^n \vec f^n = \hat e_1

The nth "backward vector" \vec b^n is defined similarly; it is the vector of length n which satisfies:

\mathbf T^n \vec b^n = \hat e_n

An important simplification can occur when M is a symmetric matrix; then the two vectors are related by bni = fnn+1-i -- that is, they are row-reversals of each other. This can save some extra computation in that special case.

[edit] Obtaining the backward vectors

Even if the matrix is not symmetric, then the nth forward and backward vector may be found from the vectors of length n-1 as follows. First, the forward vector may be extended with a zero to obtain:

\mathbf T^n \begin{bmatrix} \vec f^{n-1} \\ 0 \\ \end{bmatrix} = \begin{bmatrix} \ & \ & \ & t_{-n+1} \\ \ & \mathbf T^{n-1} & \ & t_{-n+2} \\ \ & \ & \ & \vdots \\ t_{n-1} & t_{n-2} & \dots & t_0 \\ \end{bmatrix} \begin{bmatrix} \ \\ \vec f^{n-1} \\ \ \\ 0 \\ \ \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ \epsilon_f^n \\ \end{bmatrix}

In going from Tn-1 to Tn, the extra column added to the matrix does not perturb the solution when a zero is used to extend the forward vector. However, the extra row added to the matrix has perturbed the solution; and it has created an unwanted error term εf which occurs in the last place. The above equation gives it the value of:

\epsilon_f^n \ = \ \sum_{i=1}^{n-1} \ M_{ni} \ f_{i}^{n-1} \ = \ \sum_{i=1}^{n-1} \ t_{n-i} \ f_{i}^{n-1}

This error will be returned to shortly and eliminated from the new forward vector; but first, the backwards vector must be extended in a similar (albeit reversed) fashion. For the backwards vector,

\mathbf T^n \begin{bmatrix} 0 \\ \vec b^{n-1} \\ \end{bmatrix} = \begin{bmatrix} t_0 & \dots & t_{-n+2} & t_{-n+1} \\ \vdots & \ & \ & \ \\ t_{n-2} & \ & \mathbf T^{n-1} & \ \\ t_{n-1} & \ & \ & \ \\ \end{bmatrix} \begin{bmatrix} \ \\ 0 \\ \ \\ \vec b^{n-1} \\ \ \\ \end{bmatrix} = \begin{bmatrix} \epsilon_b^n \\ 0 \\ \vdots \\ 0 \\ 1 \\ \end{bmatrix}

Like before, the extra column added to the matrix does not perturb this new backwards vector; but the extra row does. Here we have another unwanted error εb with value:

\epsilon_b^n \ = \ \sum_{i=2}^n \ M_{1i} \ b_{i-1}^{n-1} \ = \ \sum_{i=1}^{n-1} \ t_{-i} \ b_i^{n-1} \

These two error terms can be used to eliminate each other. Using the linearity of matrices,

\mathbf T \left( \alpha \begin{bmatrix} \vec f \\ \ \\ 0 \\ \end{bmatrix} + \beta \begin{bmatrix} 0 \\ \ \\ \vec b \\ \end{bmatrix} \right ) = \alpha \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ \epsilon_f \\ \end{bmatrix} + \beta \begin{bmatrix} \epsilon_b \\ 0 \\ \vdots \\ 0 \\ 1 \\ \end{bmatrix}

If α and β are chosen so that the right hand side yields ê1 or ên, then the quantity in the parentheses will fulfill the definition of the nth forward or backward vector, respectively. With those alpha and beta chosen, the vector sum in the parentheses is simple and yields the desired result.

To find these coefficients -- specifically, αf, βf, αb, and βb, notice that all the zeroes in the middle of the two vectors above may be disregarded and collapsed, leaving only the equation:

\begin{bmatrix} 1 & \epsilon_b \\ \epsilon_f & 1 \end{bmatrix} \begin{bmatrix} \alpha_f & \alpha_b \\ \beta_f & \beta_b \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

With these solved for (by using the 2x2 matrix inverse formula), the new forward and backward vectors are:

\vec f^n = {1 \over { 1 - \epsilon_b^n \epsilon_f^n }} \begin{bmatrix} \vec f^{n-1} \\ 0 \end{bmatrix} - { \epsilon_f^n \over { 1 - \epsilon_b^n \epsilon_f^n }}\begin{bmatrix} 0 \\ \vec b^{n-1} \end{bmatrix}
\vec b^n = {1 \over { 1 - \epsilon_b^n \epsilon_f^n }} \begin{bmatrix} 0 \\ \vec b^{n-1} \end{bmatrix} - { \epsilon_b^n \over { 1 - \epsilon_b^n \epsilon_f^n }}\begin{bmatrix} \vec f^{n-1} \\ 0 \end{bmatrix}

Performing these vector summations, then, gives the nth forward and backward vectors from the prior ones. All that remains is to find the first of these vectors, and then some quick sums and multiplications give the remaining ones. The first forward and backward vectors are simply:

\vec f^1 = \vec b^1 = \begin{bmatrix}{1 \over M_{11}}\end{bmatrix} = \begin{bmatrix}{1 \over t_0}\end{bmatrix}

[edit] Using the backward vectors

The above steps give the N backward vectors for M. From there, a more arbitrary equation is:

\vec y = \mathbf M \ \vec x

The solution can be built in the same recursive way that the backwards vectors were built. Accordingly, \vec x must be generalized to a sequence \vec x^n, from which \vec x^N = \vec x.

The solution is then built recursively by noticing that if:

\mathbf T^{n-1} \begin{bmatrix} x_1^{n-1} \\ x_2^{n-1} \\ \dots \\ x_{n-1}^{n-1} \\ \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ \dots \\ y_{n-1} \\ \end{bmatrix}

Then, extending with a zero again, and defining an error constant where necessary:

\mathbf T^{n} \begin{bmatrix} x_1^{n-1} \\ x_2^{n-1} \\ \dots \\ x_{n-1}^{n-1} \\ 0 \\ \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ \dots \\ y_{n-1} \\ \epsilon_x^{n-1} \end{bmatrix}

We can then use the nth backward vector to eliminate the error term and replace it with the desired formula as follows:

\mathbf T^{n} \left ( \begin{bmatrix} x_1^{n-1} \\ x_2^{n-1} \\ \dots \\ x_{n-1}^{n-1} \\ 0 \\ \end{bmatrix} + (y_n - \epsilon_x^{n-1}) \ \vec b^n \right ) = \begin{bmatrix} y_1 \\ y_2 \\ \dots \\ y_{n-1} \\ y_n \\ \end{bmatrix}

Extending this method until n = N yields the solution \vec x.

In practice, these steps are often done concurrently with the rest of the procedure, but they form a coherent unit and deserve to be treated as their own step.

[edit] References

Defining sources

  • Levinson, N. (1947). "The Wiener RMS error criterion in filter design and prediction," J. Math. Phys., v. 25, pp. 261-278.
  • Durbin, J. (1960). "The fitting of time series models." Rev. Inst. Int. Stat., v. 28, pp. 233-243.
  • Trench, W. F. (1964). "An algorithm for the inversion of finite Toeplitz matrices." J. Soc. Indust. Appl. Math., v. 12, pp. 515-522.
  • Delsarte, P. and Genin, Y. V. (1986). "The split Levinson algorithm." IEEE Transactions on Acoustics, Speech, and Signal Processing, v. ASSP-34(3), pp. 470–478.

Further work

  • Bunch, J. R. (1985). "Stability of methods for solving Toeplitz systems of equations." SIAM J. Sci. Stat. Comput., v. 6, pp. 349-364. [2]

Summaries

  • Bäckström, T. (2004). "2.2. Levinson-Durbin Recursion." Linear Predictive Modelling of Speech -- Constraints and Line Spectrum Pair Decomposition. Doctoral thesis. Report no. 71 / Helsinki University of Technology, Laboratory of Acoustics and Audio Signal Processing. Espoo, Finland. [3]
  • Claerbout, Jon F. (1976). "Chapter 7 - Waveform Applications of Least-Squares." Fundamentals of Geophysical Data Processing. Palo Alto: Blackwell Scientific Publications. [4]

[edit] See also

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