Chapter 7
State Estimation
What is in this Chapter?
Kalman Filtering plays a central role in parameter estimation. The basic problem is to nd the
best estimate of the state of a plant at time k given (noisy) observation of the past output.
39
40
STATE EST
Probability Theory
1 Preliminaries
2 Probability Spaces
3 Conditioning and Independence
4 Random Variables
5 Distribution and Density Functions
6 Expectation
7 Examples
8 Random Vectors
9 Functions of a Random Variable
10 Statistical Independence
11 Momen
E 231
Solutions # 9
1 A simple calculation reveals that
E [Y ] = 0.25E [2X] = 0
E Y2
= 0.25E 4X 2 = 1
Thus X and Y have identical second order statistics. Typical sample paths of these random
variables are shown below. Notice how dierent they look.
4
Samp
E 231
Solutions # 6
1 There are only four possible events A through D, which can be conveniently depicted in the table
below.
rain
A
C
forecast of rain
forecast of no rain
no rain
B
D
Notice that these events are disjoint, i.e. their pairwise intersection
E 231
Solutions # 7
1 (a) Since X, Y are independent, the joint density function factors as
pXY (x, y) = pX(x) pY (y) =
1
2
2
exp x
2
0
y [2, 3]
else
(b) The conditional density function is easily computed as
pX|Y =y(x|y) =
pXY (x, y)
pY (y)
x2
1
= exp
2
E 231
Assignment # 10
Issued: November 03, 2014
Due: November 10, 2014
1 The Newsboy Problem
A paperboy can buy newspapers at the wholesale price of p per paper.
He sells the newspapers at a retail price of q per paper, with q > p.
The number of customers
E 231
Assignment # 8
Issued: October 27, 2014
Due: November 03, 2014
1 Sample Paths
Let X N (0, 1) and let
Y =
0
with probability 0.75
2X with probability 0.25
(a) Find E [Y ] and E Y 2 .
(b) From part (a) you should see that X, Y have identical second or
E 231
Assignment # 8
Issued: October 15, 2014
Due: October 23, 2014
1 Let X, Y be independent random variables with X N (0, 1) and Y U [2, 3].
(a) Find the joint density function pXY (x, y).
(b) Find the conditional density function pX|Y =y(x|y).
2 Let X,
E 231
Assignment # 6
Issued: October 08, 2014
Due: October 15, 2014
1 According to statistical data, it rains in Berkeley two out of three weekends. Forecasters predict
the weather correctly with probability 0.8 if it rains and with probability 0.7 otherw
E231 HW 4, due Wednesday, October 1, 6:00 PM
1. Suppose A Cmn and rank(A) = k. Let F Cmm be an invertible matrix
that transforms A into it row echelon form (so Aref = F A). Let cfw_c1 , c2 , . . . , ck
be the indices of the unit columns of Aref . Further
HW4
l. (a) Let W 6 CM" be the invertible matrix which transforms A=k into (A*)wf.
The chain of cquivalenccs makes it clear.
{.12 E C" : An: = ()m} {m E C" : :1:'A" =01xm}
{1: E C : :rWlWA = ()lel}
{m E C : :BW'1(A')rcr =01xm}
{WZ I 2 E Cn,z*(A*)rel' = lem
E231 HW 5 (part a), due Wednesday, October 8, 6:00 PM
1. Recall that C is an eigenvalue of A Cnn if there is a v Cn , v = 0n such
that Av = v. We showed that is an eigenvalue if and only if det (In A) = 0.
The characteristic polynomial of A is dened pA (s
E231 HW 5 (part B), due Wednesday, October 8, 6:00 PM
1. Suppose x is a vector-valued function of a scalar real variable, so x(t) Cn for each
t R. Let A Cnn , and the evolution of x is governed by the linear dierential
equation
x(t) = Ax(t)
(1)
We will st
HW3
l.
(a)
(C)
Addition and scalar multiplication in 8 are inherited from R3. The main
axioms (conmmtativity, associativity and distributivity) hold since 5 is a
subset of the set of all n. X 7;, matrices, which we know to be a vector spacc.
For any /1,B
E231 HW 3, due Wednesday, September 24, 6:00 PM
1. Let S nn be dened as
S 33 := A R33 : A = AT
(a) Show that S 33 (with the eld R) is a vector space, with addition and scalar
multiplication as dened from the vector space R33 . Hint: Since we already
know
E231 HW 2, due Wednesday, Sept 17, 6:00 PM
Capital letters denote matrices. If dimensions are not specied, assume that the dimensions are arbitrary, and if several matrices are involved, the dimensions are compatible
among the various matrices.
1. (a) Boo
A.
HWI
1. Here are a few examples that work:
1 l
-1]
. 00
A~[10]
2. Since A is invertible, and B is the common left and right inverse, it must be
that AB = BA = I. Taking transposes of this relation gives BTAT = ATBT =
I. Hence BT is the common left and
HWZ
l.
(a)
(b)
The equation AI: = (l, for
0 1 2 () 0 2 4
0 0 0 1 0 () 1
0 0 0 0 1 1 6
0 0 0 0 0 0 0
A:
has many solutions, and since A is in row-echelon form, determining the
general form of solutions is easy. Since .r2,:r4 and 9:5 each enter only one
equ
E231 HW 1, due Wednesday, September 10, 6:00 PM
Current due-date is next Wednesday. I will extend by a few days, if necessary.
Capital letters denote matrices. If dimensions are not specied, assume that the dimensions are arbitrary, and if several matrice