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Material Type: Notes; Class: PROBAB MTH,ELEC ENG; Subject: STATISTICS; University: Iowa State University; Term: Unknown 1989;
Typology: Study notes
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Reading: Bertsekas & Tsitsiklis, 4.
EE/STAT 322, #24 1
There are many examples and applications of the estimation problem:
quantity Y that is related to X.
− deterministic (fixed unknown constant)
− probabilistic (a RV, whose PDF/PMF is assumed known)
EE/STAT 322, #24 3
Estimating the value of a signal in noise:
X, W, Y are random variables.
2
W
X = g(Y ) of X
Y could be the signal picked up by a microphone in a classroom with
background noise.
SETUP: There is a RV X whose value we don’t know. There is another
RV Y , whose value we know. Suppose we know the distribution of X|Y.
What is the best guess of X, given that Y = y, in the MMSE sense?
That is, want to find a function
X = g(y) such that we can minimize
J = MSE = E[ (X − g(y))
2
|Y = y]
Any function of y is call an estimator. Some are good, some are bad. The
estimator g(y) that minimizes J is called the MMSE estimator.
The same reasoning as in the previous slide leads to
∂g(y)
= 0 → g(y) = E[X|Y = y]
EE/STAT 322, #24 7
SETUP: Given several measurements Y 1
2
n
, want to find the
MMSE estimate of another RV X.
That is, find a function g(Y 1 , Y 2 ,... , Yn) to minimize
E[(X − g(Y 1
2
n
2
|Y 1
= y 1
2
= y 2
n
= y n
and hence E[(X − g(Y 1
2
n
2
]
The answer is similar: the MMSE estimator is given by
g(Y 1
2
n
1
2
n
Define
X = E[X|Y ] and
X. We have
X|Y = y] = 0, for all y
Because of the iterated expectation, we only need to show E[
X|Y = y] = 0
X|Y = y] = E[X −
X|Y = y]
= E[X|Y = y] − E[
X|Y = y]
= E[X|Y = y] − E[X|Y = y] = 0
EE/STAT 322, #24 9
The nonlinear MMSE estimator based on several measurements performs
well, but
’s
may be difficult to compute (or is in an ugly form)
Solution: Lower our standard and look for something simpler
The equations above can be expressed in a compact form, called Yule-Walker
equations.
[a 1
,... , a n
Y
XY
[a 1
,... , a n
Y + b =
where R Y
T
] is the correlation matrix of vector Y = [Y 1
n
T
,
given by
Y
Y
Y
Y
(n − 1)
Y
Y
Y
(n − 2)
Y
(n − 1) R Y
Y
XY
XY
XY
(n − 1)].
EE/STAT 322, #24 13
X = aY ;
Nonhomogeneous linear estimate of X:
X = aY + b.
X is the LMMSE estimate of X, then E(
X) = E(Y ) = 0, i.e.,
(and Y ) are uncorrelated with (orthogonal to) .
Proof: Consider a homogeneous case, and let
X = aY.
dJ
da
= 0 ⇒E[(X − aY )Y ] = 0 ⇒E[Y ] = 0.
We can also show E[
X] = E[aY ] = 0.
X is estimated from y = [y 1
,... , y n
, and the estimate is
x ˆ = a
y =
n
i=
a i
y i
MSE: J = E[(x − xˆ)
2
] = E[(x −
n
i=
a i
y i
2
].
Using the orthogonality principle,
= x − xˆ ⊥ xˆ.
Also ⊥ y i
i.e. E[y i
] = 0, for i = 1,... , n.
1
y
2
y
2 2
ay
1 1
ay
x
x ˆ
ε
0
11 2 2
x ˆ = ay + ay and x = x ˆ+ε
EE/STAT 322, #24 15
i
where var(X) = σ
2
x
, var( ˆX) = ρ
2
XY
σ
2
x
, and
var() = J = E[(X −
2 ] = (1 − ρ
2
XY
)σ
2
x