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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 N1 h d N2... h d NM ) 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 Hhd 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 =0Nmaxij h d k h d k+|ij| r i j k 0 N i j h d k h d k i j Note also that: H d Hhd= 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 =0Ni 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 aopt=(RHrd) a opt R r d , or Ra=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 rij=rji r i j r i j r j i . If we tweak the problem slightly to make r i j =rij r i j r i j , we get: ( r0r1r2...rM1 r1r0r1... r2r1r0... rM1.........r0 ) a 1 a 2 a 3 a M =r1r2r3rM 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 OM2 O M 2 computations using Levinson's recursion.

Statistical Linear Prediction

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

Given a time-series yn y n , assumed to be produced by an auto-regressive (AR) (all-pole) system: yn= k =1M a k ynk+un y n k 1 M a k y n k u n where un 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 Eyn+ k =1M a k ynk2=min a k a k Ey2n+2 k =1M a k ynynk+ k =1M a k ynk l =1M a l ynl=min a k a k Ey2n+2 k =1M a k Eynynk+ k =1M l =1M a k a l Eynkynl 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 yn y n is zero.
ε2=r0+2( r1r2r3...rM ) a 1 a 2 a 3 a M +( a 1 a 2 a 3 ... a M )( r0r1r2...rM1 r1r0r1... r2r1r0... rM1.........r0 ) 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 =2r+2Ra a ε 2 2 r 2 R a
(3)
Setting Equation 3 equal to zero yields: Ra=r R a r These are called the Yule-Walker equations. In practice, given samples of a sequence yn y n , we estimate rn r n as rn ^=1N k =0Nnynyn+kEykyn+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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