state space models
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State Space Models
Let { xt:t T} and { yt:t T} denote two vector valued time series that satisfy the system of equations:
yt = Atxt + vt (The observation equation)
xt = Btxt-1 + ut (The state equation)
The time series { yt:t T} is said to have state-space representation.
Note: { ut:t T} and { vt:t T} denote two vector valued time series that satisfying:
1. E(ut) = E(vt) = 0.
2. E(utusˊ) = E(vtvsˊ) = 0 if t ≠ s.
3. E(ututˊ) = u and E(vtvtˊ) = v.
4. E(utvsˊ) = E(vtusˊ) = 0 for all t and s.
Example: One might be tracking an object with several radar stations. The process {xt:t T} gives the position of the object at time t. The process { yt:t T} denotes the observations at time t made by the several radar stations.
As in the Hidden Markov Model we will be interested in determining position of the object, {xt:t T}, from the observations, {yt:t T} , made by the several radar stations
Example: Many of the models we have considered to date can be thought of a State-Space models
Autoregressive model of order p:
tptpttt uyyyy 2211
Define
Then tty x100
t
t
pt
t
y
y
y
1
1
x
and ttt uBxx 1 State equation
Observation equation
tt
p
u
1
0
0
100
010
1
21
x
Hidden Markov Model: Assume that there are m states. Also that there the observations Yt are discreet and take on n possible values.
Suppose that the m states are denoted by the vectors:
1
0
0
,,
0
1
0
,
0
0
1
21
meee
Suppose that the n possible observations taken at each state are
1
0
0
,,
0
1
0
,
0
0
1
21
nfff
Let
ijmm
itjtij XXP , 1 ee
and
ijnm
itjtij XYP Βef ,
Note
i
im
i
i
itt XXE eΠe
2
1
1
Let
So that
itttt XXEX eu 1
itX eΠ
1 tt XX Π
ttt XX uΠ 1 The State Equation
with
0, 121 ttttt XEXXE uu
Also
Hence
tttttt XXXX uΠuΠ 11
and
tttttttt XXXX uuΠuuΠΠΠ 1111
1111 tttttttttt XXXXXX ΠuuΠΠΠuu
1111 ttttttttt XXXXXEXE ΠΠuuΣu
ΠΠ 11 diagdiag ttt XXXE
where diag(v) = the diagonal matrix with the components of the vector v along the diagonal
then
Since
ttt XX uΠ 1
and
ttt XX uΠ diagdiagdiag 1
11 diagdiag ttt XXXE Π
Thus
ΠΠΠΣu 11 diagdiag tt XX
We have defined
ijnm
itjtij XYP Βef ,
Hence
i
in
i
i
itt XYE eΒe
2
1
Let
tttt XYEY v
tt XY Β
Then
with
The Observation Equation
0v tt XE
ttt XY vΒ
and
ΒΒΒvvΣv ttttt XXXE diagdiag
Hence with these definitions the state sequence of a Hidden Markov Model satisfies:
with
The Observation Equation
0v tt XE
ttt XY vΒ
and ΒΒΒvvΣv ttttt XXXE diagdiag
ttt XX uΠ 1 The State Equation
with 0u tt XE
and ΠΠΠuuΣu 111 diagdiag ttttt XXXE
The observation sequence satisfies:
Kalman Filtering
We are now interested in determining the state vector xt in terms of some or all of the observation vectors y1, y2, y3, … , yT.We will consider finding the “best” linear predictor. We can include a constant term if in addition one of the observations (y0 say) is the vector of 1’s.
We will consider estimation of xt in terms of 1. y1, y2, y3, … , yt-1 (the prediction problem)
2. y1, y2, y3, … , yt (the filtering problem)
3. y1, y2, y3, … , yT (t < T, the smoothing problem)
For any vector x define:
where
sxsxsxs pˆˆˆˆ 21 x
is the best linear predictor of x(i), the ith component of x, based on y0, y1, y2, … , ys.
sx iˆ
The best linear predictor of x(i) is the linear function that of x, based on y0, y1, y2, … , ys that minimizes
2ˆ sxxE ii
Remark: The best predictor is the unique vector of the form:
Where C0, C1, C2, … ,Cs, are selected so that:
sss yCyCyCx 1100ˆ
sis i ,,2,1,0 ˆ yxx
sisE i ,,2,1,0 ˆ i.e. 0yxx
Remark: If x, y1, y2, … ,ys are normally distributed then:
sEs yyyxx ,,,ˆ 21
Cvuv ˆ
Let u and v, be two random vectors than
is the optimal linear predictor of u based on v if
1 vvvuC EE
Remark
State Space Models
Let { xt:t T} and { yt:t T} denote two vector valued time series that satisfy the system of equations:
yt = Atxt + vt (The observation equation)
xt = Btxt-1 + ut (The state equation)
The time series { yt:t T} is said to have state-space representation.
Note: { ut:t T} and { vt:t T} denote two vector valued time series that satisfying:
1. E(ut) = E(vt) = 0.
2. E(utusˊ) = E(vtvsˊ) = 0 if t ≠ s.
3. E(ututˊ) = u and E(vtvtˊ) = v.
4. E(utvsˊ) = E(vtusˊ) = 0 for all t and s.
Let { xt:t T} and { yt:t T} denote two vector valued time series that satisfy the system of equations:
yt = Atxt + vt
xt = Bxt-1 + ut
Let
Kalman Filtering:
stt Es yyyxx ,,,ˆ 21
and
suutt
stu ssE yyyxxxxΣ ,,,ˆˆ 21
Then
1ˆ1ˆ 1 tt tt xΒx
1ˆ1ˆˆ ttt tttttt xAyKxx
111 vΣAΣAAΣK tt
ttttt
ttt
where
One also assumes that the initial vector x0 has mean and covariance matrix an that
μx 0ˆ 0
The covariance matrices are updated
uΣBΣBΣ
11,1
1 ttt
ttt
with
11 tttt
ttt
ttt AΣKΣΣ
ΣΣ 000
Summary: The Kalman equations
uΣBΣBΣ
11,1
1 ttt
ttt1.
11 tttt
ttt
ttt AΣKΣΣ
1ˆ1ˆˆ ttt tttttt xAyKxx
111 vΣAΣAAΣK tt
ttttt
ttt
1ˆ1ˆ 1 tt tt xΒx
2.
3.
4.
5.
μx 0ˆ 0with ΣΣ 0
00and
Now stt Es yyyxx ,,,ˆ 21
hence
Proof:
121 ,,,1ˆ ttt Et yyyxx
1ˆ 1 ttxΒ
1211 ,,, tttE yyyuΒx
1211 ,,, ttE yyyxΒ
Note 121 ,,,1ˆ ttt Et yyyyy 121 ,,, ttttE yyyvxA
1ˆ,,, 121 tE ttttt xAyyyxA
proving (4)
Let 1ˆ tttt yye
1ˆ tttttt xAvxA tttt t vxxA 1ˆ
1ˆ tttt xAy
Let 1ˆ tttt xxd
Given y0, y1, y2, … , yt-1 the best linear predictor of dt using et is:
ttttt EE eeeed 1 tttE eyyyd ,,,, 110 tttE yyyyd ,,,, 110
tttt tt eKxx 1ˆˆHence
tttttttt ttEE vxxAxxed 1ˆ1ˆ
1ˆ ttttt xAyewhere
1ˆˆ tt tt xx
1 ttttt EE eeedKand
Now
ttttt ttE Axxxx 1ˆ1ˆ
t
ttt AΣ 1
(5)
tttttt tEE vxxAee 1ˆ
Also
tttt t vxxA 1ˆ
tttttt ttE AxxxxA 1ˆ1ˆ
tttt tE vxxA 1ˆ
tttttt EtE vvAxxv 1ˆ
111 vΣAΣAAΣK tt
ttttt
tut
hence
vΣAΣA
tt
ttt1
(2)
Thus
1ˆ1ˆ 1 tt tt xΒx
1ˆ1ˆˆ ttt tttttt xAyKxx
111 vΣAΣAAΣK tt
ttttt
ttt
where
101 ,,1ˆ1ˆ
tttttt
tt ttE yyxxxxΣ Also
1ˆ 11 tE ttt xΒuΒx
10 ,,1ˆ tttt t yyxΒuΒx
(4)
(5)
(2)
uΣBΣBΣ
1
1,11 t
ttt
tt
11 tttt
ttt
ttt AΣKΣΣ
The proof that
will be left as an exercise.
Hence(3)
(1)
Example:
What is observe is the time series
tttt uxxx 2211 Suppose we have an AR(2) time series
ttt vxy
{ut|t T} and {vt|t T} are white noise time series with standard deviations u and v.
then
0,
01, 21
1
tt
t
tt
u
x
xuΒx
This model can be expressed as a state-space model by defining:
001 2
121
1
t
t
t
t
t u
x
x
x
x
ttt uΒxx 1or
can be written
The equation:
ttt vxy
tttt
tt vv
x
xy
Ax1
0,1
2vvΣ
Note:
00
02u
uΣ
The Kalman equations
uΣBΣBΣ
11,1
1 ttt
ttt1.
11 tttt
ttt
ttt AΣKΣΣ
tt
ttt
ttss
ss
2212
12111Σ
tt
ttt
ttrr
rr
2212
1211Σ
1ˆ1ˆˆ ttt tttttt xAyKxx
111 vΣAΣAAΣK tt
ttttt
ttt
1ˆ1ˆ 1 tt tt xΒx
2.
3.
4.
5.
Let
The Kalman equations
uΣBΣBΣ
11,1
1 ttt
ttt1.
00
0
0
1
01
2
2
1
122
112
112
11121
2212
1211 utt
tt
tt
tt
rr
rr
ss
ss
1 2 1 1 2 211 11 1 12 1 2 22 2
1 112 11 1 12 2
122 11
2t t t tu
t t t
t t
s r r r
s r r
s r
1
2212
1211
2212
1211
0
101
0
1
vtt
tt
tt
tt
tss
ss
ss
ssK
111 vΣAΣAAΣK tt
ttttt
ttt2.
vt
tv
t
t
vt
t
t
s
ss
s
ss
s
11
12
11
11
1
11
12
11
11 tttt
ttt
ttt AΣKΣΣ
tt
tt
ttt
tt
tt
tt
ss
ss
ss
ss
rr
rr
2212
1211
2212
1211
2212
1211 01K
3.
tt
vt
tv
t
t
tt
tt
ss
s
ss
s
ss
ss1211
211
12
211
11
2212
1211
2
11
2
111111
vt
ttt
s
ssr
211
12111212
vt
tttt
s
sssr
2
11
2
122222
vt
ttt
s
ssr
1ˆ
1ˆ
011ˆ
1ˆ
2
121
1 tx
tx
tx
tx
t
t
t
t
1ˆ1ˆ 1 tt tt xΒx4.
1ˆ1ˆ1ˆ 2211 txtxtx ttt
1ˆ1ˆˆ ttt tttttt xAyKxx5.
1ˆ
1ˆ01
1ˆ
1ˆ
ˆ
ˆ
111 tx
txy
tx
tx
tx
tx
t
ttt
t
t
t
t K
1ˆ
1ˆ
1ˆ
ˆ
ˆ
211
12
211
11
11
txy
s
ss
s
tx
tx
tx
txtt
vt
tv
t
t
t
t
t
t
1ˆ1ˆˆ2
11
11
txys
stxtx tt
vt
t
tt
1ˆ1ˆˆ2
11
1211
txy
s
stxtx tt
vt
t
tt
Now consider finding
These can be found by successive backward recursions for t = T, T – 1, … , 2, 1
Kalman Filtering (smoothing):
Ttt ET yyyxx ,,,ˆ 21
where
suutt
stu ssE yyyxxxxΣ ,,,ˆˆ 21
1ˆˆ1ˆˆ 111 tTtT ttttt xxJxx
1111,11
t
ttt
ttt ΣΒΣJ
The covariance matrices satisfy the recursions
11
11
1,11,1
tt
ttT
tttt
ttT
tt JΣΣJΣΣ
1.
The backward recursions
1ˆˆ1ˆˆ 111 tTtT ttttt xxJxx
1111,11
t
ttt
ttt ΣΒΣJ
11
11
1,11,1
tt
ttT
tttt
ttT
tt JΣΣJΣΣ
2.
3.
tt
ttt
ttss
ss
2212
12111Σ
tt
ttt
ttrr
rr
2212
1211Σ
In the example:
0121
Β
txtx ttt
ttt
tt ˆ and 1ˆ,, 11 ΣΣ
- calculated in forward recursion
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