lectr16.ppt

1
16. Mean Square Estimation
Given some information that is related to an unknown quantity of
interest, the problem is to obtain a good estimate for the unknown in
terms of the observed data.
Suppose represent a sequence of random
variables about whom one set of observations are available, and Y
represents an unknown random variable. The problem is to obtain a
good estimate for Y in terms of the observations
Let
represent such an estimate for Y.
Note that can be a linear or a nonlinear function of the observation
Clearly
represents the error in the above estimate, and the square of
n
X
X
X ,
,
, 2
1 
.
,
,
, 2
1 n
X
X
X 
)
(
)
,
,
,
(
ˆ 2
1 X
X
X
X
Y n 
 
  (16-1)
)
(

.
,
,
, 2
1 n
X
X
X 
)
(
ˆ
)
( X
Y
Y
Y
X 
 


 (16-2)
2
|
|
PILLAI

2
the error. Since is a random variable, represents the mean
square error. One strategy to obtain a good estimator would be to
minimize the mean square error by varying over all possible forms
of and this procedure gives rise to the Minimization of the
Mean Square Error (MMSE) criterion for estimation. Thus under
MMSE criterion,the estimator is chosen such that the mean
square error is at its minimum.
Next we show that the conditional mean of Y given X is the
best estimator in the above sense.
Theorem1: Under MMSE criterion, the best estimator for the unknown
Y in terms of is given by the conditional mean of Y
gives X. Thus
Proof : Let represent an estimate of Y in terms of
Then the error and the mean square
error is given by
 }
|
|
{ 2

E
),
(

)
(

n
X
X
X ,
,
, 2
1 
}.
|
{
)
(
ˆ X
Y
E
X
Y 
 (16-3)
)
(
ˆ X
Y 

).
,
,
,
( 2
1 n
X
X
X
X 
 ,
ˆ
Y
Y 


}
|
)
(
|
{
}
|
ˆ
|
{
}
|
|
{ 2
2
2
2
X
Y
E
Y
Y
E
E 

 



 (16-4) PILLAI
}
|
|
{ 2

E

3
Since
we can rewrite (16-4) as
where the inner expectation is with respect to Y, and the outer one is
with respect to
Thus
To obtain the best estimator we need to minimize in (16-6)
with respect to In (16-6), since
and the variable appears only in the integrand term, minimization
of the mean square error in (16-6) with respect to is
equivalent to minimization of with respect to
}]
|
{
[
]
[ X
z
E
E
z
E z
X

}]
|
)
(
|
{
[
}
|
)
(
|
{
z
2
z
2
2
X
X
Y
E
E
X
Y
E Y
X 







 



.
X









.
)
(
}
|
)
(
|
{
}]
|
)
(
|
{
[
2
2
2
dx
X
f
X
X
Y
E
X
X
Y
E
E
X



(16-6)
(16-5)
,
 2


.
 ,
0
)
( 
X
fX ,
0
}
|
)
(
|
{ 2

 X
X
Y
E 


2


}
|
)
(
|
{ 2
X
X
Y
E 
 .

PILLAI

4
Since X is fixed at some value, is no longer random,
and hence minimization of is equivalent to
This gives
or
But
since when is a fixed number Using (16-9)
)
(X

}
|
)
(
|
{ 2
X
X
Y
E 

.
0
}
|
)
(
|
{ 2




X
X
Y
E 

(16-7)
0
}
|
)
(
{| 
 X
X
Y
E 
(16-8)
),
(
}
|
)
(
{ X
X
X
E 
  (16-9)
)
(
, X
x
X 
 ).
(x

PILLAI
.
0
}
|
)
(
{
}
|
{ 
 X
X
E
X
Y
E 

5
in (16-8) we get the desired estimator to be
Thus the conditional mean of Y given represents the best
estimator for Y that minimizes the mean square error.
The minimum value of the mean square error is given by
As an example, suppose is the unknown. Then the best
MMSE estimator is given by
Clearly if then indeed is the best estimator for Y
}.
,
,
,
|
{
}
|
{
)
(
ˆ 2
1 n
X
X
X
Y
E
X
Y
E
X
Y 



n
X
X
X ,
,
, 2
1 
.
0
)}
|
{var(
]
}
|
)
|
(
|
{
[
}
|
)
|
(
|
{
)
var(
2
2
2
min






X
Y
E
X
X
Y
E
Y
E
E
X
Y
E
Y
E
X
Y



 



 


(16-11)
(16-10)
3
X
Y 
,
3
X
Y  3
ˆ X
Y 
.
}
|
{
}
|
{
ˆ 3
3
X
X
X
E
X
Y
E
Y 

 (16-12)
PILLAI

6
in terms of X. Thus the best estimator can be nonlinear.
Next, we will consider a less trivial example.
Example : Let
where k > 0 is a suitable normalization constant. To determine the best
estimate for Y in terms of X, we need
Thus
Hence the best MMSE estimator is given by


 



otherwise,
0
1
0
,
)
,
(
,
y
x
kxy
y
x
f Y
X
).
|
(
|
x
y
f X
Y
1.
x
0
,
2
)
1
(
2
)
,
(
)
(
2
1
2
1 1
,






  
x
kx
kxy
kxydy
dy
y
x
f
x
f
x
x x
Y
X
X
y
x
1
1
.
1
0
;
1
2
2
/
)
1
(
)
(
)
,
(
)
|
( 2
2
,







 y
x
x
y
x
kx
kxy
x
f
y
x
f
x
y
f
X
Y
X
X
Y
PILLAI
(16-13)

7
Once again the best estimator is nonlinear. In general the best
estimator is difficult to evaluate, and hence next we
will examine the special subclass of best linear estimators.
Best Linear Estimator
In this case the estimator is a linear function of the
observations Thus
where are unknown quantities to be determined. The
mean square error is given by
.
1
)
1
(
3
2
1
1
3
2
1
3
2
)
|
(
}
|
{
)
(
ˆ
2
2
2
3
1
2
3
1 2
1
2
1
1
2
1
2
2
|
x
x
x
x
x
x
y
dy
y
dy
y
dy
x
y
f
y
X
Y
E
X
Y
x
x
x
x x
y
x X
Y




















}
|
{ X
Y
E
(16-14)
Y
ˆ
.
,
,
, 2
1 n
X
X
X 
n
a
a
a ,
,
, 2
1 







n
i
i
i
n
n
l X
a
X
a
X
a
X
a
Y
1
2
2
1
1 .
ˆ  (16-15)
)
ˆ
( l
Y
Y 


PILLAI

8
and under the MMSE criterion should be chosen so
that the mean square error is at its minimum possible
value. Let represent that minimum possible value. Then
To minimize (16-16), we can equate
This gives
But
n
a
a
a ,
,
, 2
1 
}
|
|
{
}
|
ˆ
{|
}
|
|
{ 2
2
2




 i
i
l X
a
Y
E
Y
Y
E
E 
}
|
|
{ 2

E
2
n





n
i
i
i
a
a
a
n X
a
Y
E
n 1
2
,
,
,
2
}.
|
{|
min
2
1 
 (16-17)
,n.
,
,
k
E
ak

2
1
,
0
}
|
{| 2




 (16-18)
.
0
2
|
|
}
|
{|
*
2
2





























k
k
k a
E
a
E
E
a




(16-19)
PILLAI
(16-16)

9
Substituting (16-19) in to (16-18), we get
or the best linear estimator must satisfy
Notice that in (16-21), represents the estimation error
and represents the data. Thus from
(16-21), the error is orthogonal to the data for the
best linear estimator. This is the orthogonality principle.
In other words, in the linear estimator (16-15), the unknown
constants must be selected such that the error
.
)
(
)
(
1
1
k
k
n
i
i
i
k
k
n
i
i
i
k
X
a
X
a
a
Y
a
X
a
Y
a













 




,
0
}
{
2
}
|
{| *
2





k
k
X
E
a
E


.
,
,
2
,
1
,
0
}
{ *
n
k
X
E k 



(16-20)
(16-21)


n
i i
i X
a
Y 1
),
(


n
k
Xk 
1
,
n
k
Xk 
1
,
n
a
a
a ,
,
, 2
1 
PILLAI

10
is orthogonal to every data for the
best linear estimator that minimizes the mean square error.
Interestingly a general form of the orthogonality principle
holds good in the case of nonlinear estimators also.
Nonlinear Orthogonality Rule: Let represent any functional
form of the data and the best estimator for Y given With
we shall show that
implying that
This follows since
 


n
i i
i X
a
Y 1
 n
X
X
X ,
,
, 2
1 
)
(X
h
}
|
{ X
Y
E .
X
}
|
{ X
Y
E
Y
e 

).
(
}
|
{ X
h
X
Y
E
Y
e 


,
0
)}
(
{ 
X
eh
E
.
0
)}
(
{
)}
(
{
]}
|
)
(
[
{
)}
(
{
}
)
(
]
|
[
{
)}
(
{
)}
(
])
|
[
{(
)}
(
{









X
Yh
E
X
Yh
E
X
X
Yh
E
E
X
Yh
E
X
h
X
Y
E
E
X
Yh
E
X
h
X
Y
E
Y
E
X
eh
E
PILLAI
(16-22)

11
Thus in the nonlinear version of the orthogonality rule the error is
orthogonal to any functional form of the data.
The orthogonality principle in (16-20) can be used to obtain
the unknowns in the linear case.
For example suppose n = 2, and we need to estimate Y in
terms of linearly. Thus
From (16-20), the orthogonality rule gives
Thus
or
n
a
a
a ,
,
, 2
1 
2
1 and X
X
2
2
1
1
ˆ X
a
X
a
Yl 

0
}
)
{(
}
X
{
0
}
)
{(
}
X
{
*
2
2
2
1
1
*
2
*
1
2
2
1
1
*
1








X
X
a
X
a
Y
E
E
X
X
a
X
a
Y
E
E


}
{
}
|
{|
}
{
}
{
}
{
}
|
{|
*
2
2
2
2
1
*
2
1
*
1
2
*
1
2
1
2
1
YX
E
a
X
E
a
X
X
E
YX
E
a
X
X
E
a
X
E




PILLAI

12
(16-23) can be solved to obtain in terms of the cross-
correlations.
The minimum value of the mean square error in (16-17) is given by
But using (16-21), the second term in (16-24) is zero, since the error is
orthogonal to the data where are chosen to be
optimum. Thus the minimum value of the mean square error is given
by
2
1 and a
a























}
{
}
{
}
|
{|
}
{
}
{
}
|
{|
*
2
*
1
2
1
2
2
*
2
1
*
1
2
2
1
YX
E
YX
E
a
a
X
E
X
X
E
X
X
E
X
E
(16-23)
2
n

n
a
a
a ,
,
, 2
1 
,
i
X
}.
{
min
}
{
min
}
)
(
{
min
}
{
min
}
|
{|
min
*
1
,
,
,
*
,
,
,
1
*
,
,
,
*
,
,
,
2
,
,
,
2
2
1
2
1
2
1
2
1
2
1
l
n
i
i
a
a
a
a
a
a
n
i
i
i
a
a
a
a
a
a
a
a
a
n
X
E
a
Y
E
X
a
Y
E
E
E
n
n
n
n
n





















(16-24)
PILLAI

13
where are the optimum values from (16-21).
Since the linear estimate in (16-15) is only a special case of
the general estimator in (16-1), the best linear estimator that
satisfies (16-20) cannot be superior to the best nonlinear estimator
Often the best linear estimator will be inferior to the best
estimator in (16-3).
This raises the following question. Are there situations in
which the best estimator in (16-3) also turns out to be linear ? In
those situations it is enough to use (16-21) and obtain the best
linear estimators, since they also represent the best global estimators.
Such is the case if Y and are distributed as jointly Gaussian
We summarize this in the next theorem and prove that result.
Theorem2: If and Y are jointly Gaussian zero
n
a
a
a ,
,
, 2
1 
)
(X

}.
|
{ X
Y
E
}
{
}
|
{|
}
)
{(
}
{
1
*
2
1
*
*
2









n
i
i
i
n
i
i
i
n
Y
X
E
a
Y
E
Y
X
a
Y
E
Y
E 

(16-25)
n
X
X
X ,
,
, 2
1 
n
X
X
X ,
,
, 2
1 
PILLAI

14
mean random variables, then the best estimate for Y in terms of
is always linear.
Proof : Let
represent the best (possibly nonlinear) estimate of Y, and
the best linear estimate of Y. Then from (16-21)
is orthogonal to the data Thus
Also from (16-28),
n
X
X
X ,
,
, 2
1 
}
|
{
)
,
,
,
(
ˆ 2
1 X
Y
E
X
X
X
Y n 
 
 (16-26)



n
i
i
i
l X
a
Y
1
ˆ (16-27)
.
1
, n
k
Xk 

(16-28)
.
1
,
0
}
X
{ *
k n
k
E 


 (16-29)
.
0
}
{
}
{
}
{
1


 

n
i
i
i X
E
a
Y
E
E  (16-30)
PILLAI
1
n
l i i
i
Y Y Y a X


    


15
Using (16-29)-(16-30), we get
From (16-31), we obtain that and are zero mean uncorrelated
random variables for But itself represents a Gaussian
random variable, since from (16-28) it represents a linear combination
of a set of jointly Gaussian random variables. Thus and X are
jointly Gaussian and uncorrelated random variables. As a result, and
X are independent random variables. Thus from their independence
But from (16-30), and hence from (16-32)
Substituting (16-28) into (16-33), we get
.
1 n
k 

 k
X


.
1
,
0
}
{
}
{
}
{ *
*
n
k
X
E
E
X
E k
k 


 
 (16-31)

}.
{
}
|
{ 
 E
X
E  (16-32)
,
0
}
{ 

E
.
0
}
|
{ 
X
E  (16-33)
0
}
|
{
}
|
{
1


 

X
X
a
Y
E
X
E
n
i
i
i

PILLAI

16
or
From (16-26), represents the best possible estimator,
and from (16-28), represents the best linear estimator.
Thus the best linear estimator is also the best possible overall estimator
in the Gaussian case.
Next we turn our attention to prediction problems using linear
estimators.
Linear Prediction
Suppose are known and is unknown.
Thus and this represents a one-step prediction problem.
If the unknown is then it represents a k-step ahead prediction
problem. Returning back to the one-step predictor, let
represent the best linear predictor. Then
.
}
|
{
}
|
{
1
1
l
n
i
i
i
n
i
i
i Y
X
a
X
X
a
E
X
Y
E 

 



(16-34)
)
(
}
|
{ x
X
Y
E 

 
n
i i
i X
a
1
n
X
X
X ,
,
, 2
1  1

n
X
,
1

 n
X
Y
,
k
n
X 
1
ˆ 
n
X
PILLAI

17
where the error
is orthogonal to the data, i.e.,
Using (16-36) in (16-37), we get
Suppose represents the sample of a wide sense stationary
i
X
(16-35)
,
1
,
ˆ
1
1
1
1
2
2
1
1
1
1
1
1





















n
n
i
i
i
n
n
n
n
i
i
i
n
n
n
n
a
X
a
X
X
a
X
a
X
a
X
a
X
X
X


(16-36)
.
1
,
0
}
{ *
n
k
X
E k
n 


 (16-37)







1
1
*
*
.
1
,
0
}
{
}
{
n
i
k
i
i
k
n n
k
X
X
E
a
X
E  (16-38)
PILLAI
1
1
ˆ = ,
n
n i i
i
X a X


 


18
stochastic process so that
Thus (16-38) becomes
Expanding (16-40) for we get the following set of
linear equations.
Similarly using (16-25), the minimum mean square error is given by
*
*
)
(
}
{ i
k
k
i
k
i r
r
k
i
R
X
X
E 
 



)
(t
X
(16-39)
.
1
,
1
,
0
}
{ 1
1
1
*
n
k
a
r
a
X
E n
n
i
k
i
i
k
n 



 




 (16-40)
,
,
,
2
,
1 n
k 

.
0
2
0
1
0
1
0
*
3
3
*
2
2
*
1
1
1
2
1
3
0
2
*
1
1
1
2
3
1
2
0
1
n
k
r
r
a
r
a
r
a
r
a
k
r
r
a
r
a
r
a
r
a
k
r
r
a
r
a
r
a
r
a
n
n
n
n
n
n
n
n
n
n


































(16-41)
PILLAI

19
The n equations in (16-41) together with (16-42) can be represented as
Let
.
}
)
{(
}
{
}
{
}
|
{|
0
1
*
2
3
*
1
2
*
1
1
1
*
1
1
1
*
1
*
1
*
2
2
r
r
a
r
a
r
a
r
a
r
a
X
X
a
E
X
E
Y
E
E
n
n
n
n
n
i
i
n
i
n
i
n
i
i
n
n
n
n




























(16-42)
.
0
0
0
0
1 2
n
3
2
1
0
*
1
*
1
*
1
0
*
2
*
1
2
0
*
1
*
2
1
1
0
*
1
2
1
0











































































n
n
n
n
n
n
n
n
a
a
a
a
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
(16-43)
PILLAI

20
Notice that is Hermitian Toeplitz and positive definite. Using
(16-44), the unknowns in (16-43) can be represented as
Let
n
T
.
0
*
1
*
1
*
1
1
0
*
1
2
1
0

















r
r
r
r
r
r
r
r
r
r
r
r
T
n
n
n
n
n




(16-44)


























































1
2
2
n
1
3
2
1
of
column
Last
0
0
0
0
1
n
n
n
n T
T
a
a
a
a



 (16-45)
PILLAI

21
Then from (16-45),
Thus
.
1
,
1
2
,
1
1
,
1
1
,
2
22
21
1
,
1
12
11
1






















n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
T
T
T
T
T
T
T
T
T
T




.
1
1
,
1
1
,
2
1
,
1
2
2
1



































n
n
n
n
n
n
n
n
n
T
T
T
a
a
a

 
,
0
1
1
,
1
2

 
 n
n
n
n
T

(16-46)
(16-47)
(16-48)
PILLAI

22
and
Eq. (16-49) represents the best linear predictor coefficients, and they
can be evaluated from the last column of in (16-45). Using these,
The best one-step ahead predictor in (16-35) taken the form
and from (16-48), the minimum mean square error is given by the
(n +1, n +1) entry of
From (16-36), since the one-step linear prediction error
.
1
1
,
1
1
,
2
1
,
1
1
,
1
2
1



































n
n
n
n
n
n
n
n
n
n
n T
T
T
T
a
a
a


(16-49)
n
T
.
1

n
T
.
)
(
1
ˆ
1
1
,
1
,
1
1 




 









n
i
i
n
i
n
n
n
n
n X
T
T
X (16-50)
,
1
1
1
1
1 X
a
X
a
X
a
X n
n
n
n
n
n 



 

 
 (16-51)
PILLAI

23
we can represent (16-51) formally as follows
Thus, let
them from the above figure, we also have the representation
The filter
represents an AR(n) filter, and this shows that linear prediction leads
to an auto regressive (AR) model.
n
n
n
n
n z
a
z
a
z
a
X 





 



 1 1
2
1
1
1 
,
1
)
( 1
2
1
1 n
n
n
n z
a
z
a
z
a
z
A 







  (16-52)
.
)
(
1
1


 n
n
n X
z
A

n
n
n
n z
a
z
a
z
a
z
A
z
H 









1
2
1
1
1
1
)
(
1
)
(
 (16-53)
PILLAI

24
The polynomial in (16-52)-(16-53) can be simplified using
(16-43)-(16-44). To see this, we rewrite as
To simplify (16-54), we can make use of the following matrix identity
)
(z
An
)
(z
An























































2
n
1
1
)
1
(
2
1
1
)
1
(
1
2
1
)
1
(
2
1
0
0
0
]
1
,
,
,
,
[
1
]
1
,
,
,
,
[
1
)
(






n
n
n
n
n
n
n
n
n
n
n
T
z
z
z
a
a
a
z
z
z
z
a
z
a
z
a
z
a
z
A
(16-54)
PILLAI
.
0
0 1 












 







B
CA
D
C
A
I
AB
I
D
C
B
A
(16-55)

25
Taking determinants, we get
In particular if we get
Using (16-57) in (16-54), with
.
1
B
CA
D
A
D
C
B
A 

 (16-56)
,
0

D
.
0
)
1
(
1
C
B
A
A
B
CA
n



(16-57)












 



2
1
)
1
(
0
,
],
1
,
,
,
,
[
n
n
n
n
B
T
A
z
z
z
C



PILLAI

26
we get
Referring back to (16-43), using Cramer’s rule to solve for
we get
.
1
|
|
0
1
,
,
,
0
0
0
|
|
)
1
(
)
(
1
)
1
(
1
0
*
2
*
1
1
1
0
*
1
2
1
0
2
1
2












z
z
z
r
r
r
r
r
r
r
r
r
r
r
r
T
z
z
T
T
z
A
n
n
n
n
n
n
n
n
n
n
n
n
n
n






 

(16-58)
),
1
(
1 

n
a
PILLAI
1
|
|
|
|
|
|
1
2
0
1
1
0
2
1 

 



n
n
n
n
n
n
n
n
T
T
T
r
r
r
r
a 





27
or
Thus the polynomial (16-58) reduces to
The polynomial in (16-53) can be alternatively represented as
in (16-60), and in fact represents a stable
.
0
|
|
|
|
1
2



n
n
n
T
T

.
1
1
|
|
1
)
(
1
2
1
1
1
)
1
(
1
0
*
2
*
1
1
1
0
*
1
2
1
0
1
n
n
n
n
n
n
n
n
n
n
n
z
a
z
a
z
a
z
z
z
r
r
r
r
r
r
r
r
r
r
r
r
T
z
A
























(16-59)
(16-60)
PILLAI
)
(z
An
)
(
~
)
(
1
)
( n
AR
z
A
z
H
n


28
AR filter of order n, whose input error signal is white noise of
constant spectral height equal to and output is
It can be shown that has all its zeros in provided
thus establishing stability.
Linear prediction Error
From (16-59), the mean square error using n samples is given
by
Suppose one more sample from the past is available to evaluate
( i.e., are available). Proceeding as above
the new coefficients and the mean square error can be determined.
From (16-59)-(16-61),
n

|
|
/
|
| 1

n
n T
T .
1

n
X
1

n
X
.
0
|
|
|
|
1
2



n
n
n
T
T
 (16-61)
0
1
1 ,
,
,
, X
X
X
X n
n 

2
1

n

PILLAI
.
|
|
|
| 1
2
1
n
n
n
T
T 
 
 (16-62)
)
(z
An 1
|
| 
z
0
|
| 
n
T

29
Using another matrix identity it is easy to show that
Since we must have or for every n.
From (16-63), we have
or
since Thus the mean square error decreases as more
and more samples are used from the past in the linear predictor.
In general from (16-64), the mean square errors for the one-step
predictor form a monotonic nonincreasing sequence
).
|
|
1
(
|
|
|
|
|
| 2
1
1
2
1 

 
 n
n
n
n s
T
T
T (16-63)
,
0
|
| 
k
T 0
)
|
|
1
( 2
1 
 
n
s 1
|
| 1 

n
s
)
|
|
1
(
|
|
|
|
|
|
|
| 2
1
1
1
2
2
1






n
n
n
n
n
s
T
T
T
T
n
n








,
)
|
|
1
( 2
2
1
2
2
1 n
n
n
n s 

 

 

(16-64)
PILLAI
.
1
)
|
|
1
( 2
1 
 
n
s

30
whose limiting value
Clearly, corresponds to the irreducible error in linear
prediction using the entire past samples, and it is related to the power
spectrum of the underlying process through the relation
where represents the power spectrum of
For any finite power process, we have
and since Thus
2
2
2
1
2

 


 


 
 k
n
n (16-65)
.
0
2



)
(nT
X
).
(nT
X
( ) 0
XX
S  
2 1
exp ln ( ) 0.
2
XX
S d


  


 
 
 
 
 
 (16-66)
( ) ,
XX
S d


 


 

PILLAI
0
2



ln ( ) ( ) .
XX XX
S d S d
 
 
   
 
 
  
 
( ( ) 0), ln ( ) ( ).
XX XX XX
S S S
  
 
(16-67)

31
Moreover, if the power spectrum is strictly positive at every
Frequency, i.e.,
then from (16-66)
and hence
i.e., For processes that satisfy the strict positivity condition in
(16-68) almost everywhere in the interval the final
minimum mean square error is strictly positive (see (16-70)).
i.e., Such processes are not completely predictable even using
their entire set of past samples, or they are inherently stochastic,
( ) 0, in - ,
XX
S    
  
ln ( ) .
XX
S d


 


 

2 1
exp ln ( ) 0
2
XX
S d e


  

 
 
 
  
 
 

(16-68)
(16-69)
(16-70)
),
,
( 


PILLAI

32
since the next output contains information that is not contained in
the past samples. Such processes are known as regular stochastic
processes, and their power spectrum is strictly positive.
)
(
XX
S




Power Spectrum of a regular stochastic Process
PILLAI
Conversely, if a process has the following power spectrum,
such that in then from (16-70),
( ) 0
XX
S   2
1 

 
 .
0
2




2

1


 
)
(
XX
S

33
Such processes are completely predictable from their past data
samples. In particular
is completely predictable from its past samples, since consists
of line spectrum.
in (16-71) is a shape deterministic stochastic process.
 

k
k
k
k t
a
nT
X )
cos(
)
( 
 (16-71)
( )
XX
S 
)
(nT
X

1
 k


)
(
XX
S
PILLAI

lectr16.ppt

More Related Content

What's hot (20)

Similar to lectr16.ppt (20)

Recently uploaded (20)

lectr16.ppt