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Linear Prediction

Module by: Douglas L. Jones. E-mail the author

Recall that for the all-pole design problem, we had the overdetermined set of linear equations: ( h d ⁢00...0 h d ⁢1 h d ⁢0...0 ⋮⋮⋱⋮ h d ⁢N−1 h d ⁢N−2... h d ⁢N−M )⁢ a 1 a 2 ⋮ a M ≃− h d ⁢1 h d ⁢2⋮ h d ⁢N h d 0 0 ... 0 h d 1 h d 0 ... 0 ⋮ ⋮ ⋱ ⋮ h d N 1 h d N 2 ... h d N M a 1 a 2 ⋮ a M h d 1 h d 2 ⋮ h d N with solution a⇀= H d H⁢ H d -1⁢ H d H⁢h⇀d a H d H d -1 H d h d

Let's look more closely at H d H⁢ H d =R H d H d R . r i ​ j r i ​ j is related to the correlation of h d h d with itself: r i ​ j =∑ k =0N−maxij h d ⁢k⁢ h d ⁢k+|i−j| r i ​ j k 0 N i j h d k h d k i j Note also that: H d H⁢h⇀d= r d ⁢1 r d ⁢2 r d ⁢3⋮ r d ⁢M H d h d r d 1 r d 2 r d 3 ⋮ r d M where r d ⁢i=∑ n =0N−i h d ⁢n⁢ h d ⁢n+i r d i n 0 N i h d n h d n i so this takes the form a⇀opt=−(RH⁢r⇀d) a opt R r d , or R⁢a⇀=−r⇀ R a r , where R R is M×M M M , a⇀ a is M×1 M 1 , and r⇀ r is also M×1 M 1 .

Except for the changing endpoints of the sum, r i ​ j ≃r⁢i−j=r⁢j−i r i ​ j r i j r j i . If we tweak the problem slightly to make r i ​ j =r⁢i−j r i ​ j r i j , we get: ( r⁢0r⁢1r⁢2...r⁢M−1 r⁢1r⁢0r⁢1...⋮ r⁢2r⁢1r⁢0...⋮ ⋮⋮⋮⋱⋮ r⁢M−1.........r⁢0 )⁢ a 1 a 2 a 3 ⋮ a M =−r⁢1r⁢2r⁢3⋮r⁢M r 0 r 1 r 2 ... r M 1 r 1 r 0 r 1 ... ⋮ r 2 r 1 r 0 ... ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ r M 1 ... ... ... r 0 a 1 a 2 a 3 ⋮ a M r 1 r 2 r 3 ⋮ r M The matrix R R is Toeplitz (diagonal elements equal), and a⇀ a can be solved for with O⁢M2 O M 2 computations using Levinson's recursion.

Statistical Linear Prediction

Used very often for forecasting (e.g. stock market).

Given a time-series y⁢n y n , assumed to be produced by an auto-regressive (AR) (all-pole) system: y⁢n=−∑ k =1M a k ⁢y⁢n−k+u⁢n y n k 1 M a k y n k u n where u⁢n u n is a white Gaussian noise sequence which is stationary and has zero mean.

To determine the model parameters a k a k minimizing the variance of the prediction error, we seek

min a k a k Ey⁢n+∑ k =1M a k ⁢y⁢n−k2=min a k a k Ey2n+2⁢∑ k =1M a k ⁢y⁢n⁢y⁢n−k+∑ k =1M a k ⁢y⁢n−k⁢∑ l =1M a l ⁢y⁢n−l=min a k a k Ey2n+2⁢∑ k =1M a k ⁢Ey⁢n⁢y⁢n−k+∑ k =1M∑ l =1M a k ⁢ a l ⁢Ey⁢n−k⁢y⁢n−l a k y n k 1 M a k y n k 2 a k y n 2 2 k 1 M a k y n y n k k 1 M a k y n k l 1 M a l y n l a k y n 2 2 k 1 M a k y n y n k k 1 M l 1 M a k a l y n k y n l
(1)

Note:

The mean of y⁢n y n is zero.
ε2=r⁢0+2⁢( r⁢1r⁢2r⁢3...r⁢M )⁢ a 1 a 2 a 3 ⋮ a M +( a 1 a 2 a 3 ... a M )⁢( r⁢0r⁢1r⁢2...r⁢M−1 r⁢1r⁢0r⁢1...⋮ r⁢2r⁢1r⁢0...⋮ ⋮⋮⋮⋱⋮ r⁢M−1.........r⁢0 ) ε 2 r 0 2 r 1 r 2 r 3 ... r M a 1 a 2 a 3 ⋮ a M a 1 a 2 a 3 ... a M r 0 r 1 r 2 ... r M 1 r 1 r 0 r 1 ... ⋮ r 2 r 1 r 0 ... ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ r M 1 ... ... ... r 0
(2)
∂ε2∂ a⇀ =2⁢r⇀+2⁢R⁢a⇀ a ε 2 2 r 2 R a
(3)
Setting Equation 3 equal to zero yields: R⁢a⇀=−r⇀ R a r These are called the Yule-Walker equations. In practice, given samples of a sequence y⁢n y n , we estimate r⁢n r n as r⁢n ^=1N⁢∑ k =0N−ny⁢n⁢y⁢n+k≃Ey⁢k⁢y⁢n+k r n 1 N k 0 N n y n y n k y k y n k which is extremely similar to the deterministic least-squares technique.

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