Bayesian Dynamic System Estimation - Using Markov Chain Monte

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Dec 15, 2014 - Bayesian Dynamic System Estimation. Using Markov Chain Monte Carlo Simulation. Brett Ninness, Khoa T. Tran and Christopher M. Kellett.
Bayesian Dynamic System Estimation Using Markov Chain Monte Carlo Simulation

Brett Ninness, Khoa T. Tran and Christopher M. Kellett University of Newcastle, Australia

December 15, 2014

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Motivating Example Compare Prediction Error vs. Conditional Mean estimates

Prediction Error Estimate θˆPE = arg min θ

The Univ. of Newcastle, Australia

N X

2 yt − yˆt (θ)

(1)

t=1

Bayesian Dynamic System Estimation

December 15, 2014

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Motivating Example Compare Prediction Error vs. Conditional Mean estimates

Prediction Error Estimate θˆPE = arg min θ

N X

2 yt − yˆt (θ)

(1)

t=1

Conditional Mean Estimate θˆCM = E{θ | Y } =

Z θ p(θ | Y ) dθ

(2)

where Y = {y1 , y2 , . . . , yN }

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Motivating Example An1output error model

0.8 u G(θ) =

y

θ 1 + θ 2 q − 1 + . . .+ θ 5 q − 4 1 + θ 6 q − 1 + . . .+ θ 9 q − 4

0.6

0.4

To e s t imat e θ = [θ 1 θ 2 . . .θ 9 ] f r om t he input a Z = [u y ] yˆt -out = Gθput (q)udat t

0.2

0 0

yt = yˆt + et

0.2

The Univ. of Newcastle, Australia

0.4

0.6

Bayesian Dynamic System Estimation

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3 / 16

Motivating Example An1output error model

0.8 u G(θ) =

y

θ 1 + θ 2 q − 1 + . . .+ θ 5 q − 4 1 + θ 6 q − 1 + . . .+ θ 9 q − 4

0.6

0.4

To e s t imat e θ = [θ 1 θ 2 . . .θ 9 ] f r om t he input a Z = [u y ] yˆt -out = Gθput (q)udat t

0.2

yt = yˆt + et

Given et ∼ U[−0.3, 0.3] 0 0

0.2

The Univ. of Newcastle, Australia

0.4

0.6

Bayesian Dynamic System Estimation

0.8

December 15, 2014

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3 / 16

Motivating Example An1output error model

0.8 u G(θ) =

y

θ 1 + θ 2 q − 1 + . . .+ θ 5 q − 4 1 + θ 6 q − 1 + . . .+ θ 9 q − 4

0.6

0.4

To e s t imat e θ = [θ 1 θ 2 . . .θ 9 ] f r om t he input a Z = [u y ] yˆt -out = Gθput (q)udat t

0.2

yt = yˆt + et

Given et ∼ U[−0.3, 0.3], the posterior distribution of θ becomes 0 0

0.2

0.4

0.6

0.8

1

p(θ | Y ) ∝ p(θ) p(Y | θ)

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Motivating Example An1output error model

0.8 u G(θ) =

y

θ 1 + θ 2 q − 1 + . . .+ θ 5 q − 4 1 + θ 6 q − 1 + . . .+ θ 9 q − 4

0.6

0.4

To e s t imat e θ = [θ 1 θ 2 . . .θ 9 ] f r om t he input a Z = [u y ] yˆt -out = Gθput (q)udat t

0.2

yt = yˆt + et

Given et ∼ U[−0.3, 0.3], the posterior distribution of θ becomes 0 0

0.2

0.4

 p(Y | θ) =  p(θ | Y ) ∝ H p(θ) H  H

0.6 N Y

0.8

U[−0.3,0.3] (yt − yˆt )

1

(3)

t=1

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Motivating Example Very short and noisy step response 2 yt = G θ( q ) u t + e t G θ( q ) u t

1.5

1

0.5

e ∼ U [−0.3, 0.3]

0

−0.5 0

10

20

30

40

50

Figure : Noise free response (red) vs. simulated measurement (blue) The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Prediction Error Estimate vs. Conditional Mean Estimate Bayesian estimation yields the minimum mean square error estimator

Figure : Estimated frequency responses on the same system using 2 methods Prediction error estimates versus truth (bold)

8 6 4 2 0 −2 −4 −6 −8 −6

−4

−2

0

2

4

6

(a) Prediction Error Estimate

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Prediction Error Estimate vs. Conditional Mean Estimate Bayesian estimation yields the minimum mean square error estimator

Figure : Estimated frequency responses on the same system using 2 methods Cond Mean estimates versus truth (bold)

Prediction error estimates versus truth (bold)

8

8

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2

0

0

−2

−2

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−8 −6

−4

−2

0

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(a) Prediction Error Estimate

The Univ. of Newcastle, Australia

6

−8 −6

−4

−2

0

2

4

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(b) Conditional Mean Estimate

Bayesian Dynamic System Estimation

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Computational Challenge of Bayesian Estimation Using numerical integration

0.7 0.6

p(θ | Y )

T he c ondit ional me an e s t imat e Z E {θ | Y } = θ p ( θ | Y ) dθ

0.5 is a mul t i di m e ns i o nal i nt e g ral 0.4 0.3 0.2

U s ing Simps on’s r ule ?

0.1

e v a l s = 30 9 = 1.9683 × 10 1 3

0 −6

−5

−4

The Univ. of Newcastle, Australia

−3

−2

−1

Bayesian Dynamic System Estimation

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1 December 15, 2014

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Monte Carlo Integration Using random sampling

Build a random number generator θk ∼ p(θ | Y )

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Bayesian Dynamic System Estimation

December 15, 2014

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Monte Carlo Integration Using random sampling

Build a random number generator θk ∼ p(θ | Y ), then the central limit theorem gives M 1 X 1 d f (θk ) − E{f (θ)} → N (0, σ 2 ) M M

(4)

k=1

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Monte Carlo Integration Using random sampling

Build a random number generator θk ∼ p(θ | Y ), then the central limit theorem gives M 1 X 1 d f (θk ) − E{f (θ)} → N (0, σ 2 ) M M

(4)

k=1

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Monte Carlo Integration Using random sampling

Build a random number generator θk ∼ p(θ | Y ), then the central limit theorem gives M 1 X 1 d f (θk ) − E{f (θ)} → N (0, σ 2 ) M M

(4)

k=1

However, σ 2 must be finite  σ 2 = σ02 1 + 2

∞ X

  corr f (θ0 ), f (θi ) 

(5)

i=1

where σ02 = var{f (θ0 )} with θ0 ∼ p(θ | Y )

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Monte Carlo Integration Using random sampling

θ

N e ar i.i.d. s ample s

Pe r f e c t c onve r ge nc e

20

20

10

10

0

0

−10

−10

−20

−20

θ

St r ongly c or r e lat e d s ample s

I nac c ur at e H is t ogr am

20

20

10

10

0

0

−10

−10

−20

−20

The Univ. of Newcastle, Australia

Sample his t ogr am Tr ue de ns ity

Bayesian Dynamic System Estimation

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Metropolis Algorithm Random sampling using Markov chains

Let π(θ) be the target distribution we want to draw sample from

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Metropolis Algorithm Random sampling using Markov chains

Let π(θ) be the target distribution we want to draw sample from Take any number generator γ(θk+1 | θk )

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Metropolis Algorithm Random sampling using Markov chains

Let π(θ) be the target distribution we want to draw sample from Take any number generator γ(θk+1 | θk ) and apply an acceptance probability α(θk+1 |θk ) = 1 ∧

The Univ. of Newcastle, Australia

π(θk+1 ) γ(θk |θk+1 ) · π(θk ) γ(θk+1 |θk )

Bayesian Dynamic System Estimation

December 15, 2014

(6)

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Metropolis Algorithm Random sampling using Markov chains

Let π(θ) be the target distribution we want to draw sample from Take any number generator γ(θk+1 | θk ) and apply an acceptance probability α(θk+1 |θk ) = 1 ∧

π(θk+1 ) γ(θk |θk+1 ) · π(θk ) γ(θk+1 |θk )

(6)

If we chose θk+1 − θk ∼ N (0, sk Σk )

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Metropolis Algorithm Random sampling using Markov chains

Let π(θ) be the target distribution we want to draw sample from Take any number generator γ(θk+1 | θk ) and apply an acceptance probability α(θk+1 |θk ) = 1 ∧

π(θk+1 ) π(θk )

·

H   γ(θ ) Hk |θ k+1 H H  γ(θk+1 |θH ) kH 

(6)

If we chose θk+1 − θk ∼ N (0, sk Σk ) then γ(θk+1 |θk ) = γ(θk |θk+1 )

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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1

Metropolis Algorithm Random sampling using Markov chains

Let π(θ) be the target distribution we want to draw sample from Take any number generator γ(θk+1 | θk ) and apply an acceptance probability α(θk+1 |θk ) = 1 ∧

π(θk+1 ) π(θk )

·

H   γ(θ ) Hk |θ k+1 H H  γ(θk+1 |θH ) kH 

(6)

If we chose θk+1 − θk ∼ N (0, sk Σk ) then γ(θk+1 |θk ) = γ(θk |θk+1 ) N ar r ow pr op os al

Tar ge t de ns ity W ide pr op os al 0

1

θ2

The Univ. of Newcastle, Australia

3

4

5

Bayesian Dynamic System Estimation

December 15, 2014

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1

Metropolis Algorithm Random sampling using Markov chains

Let π(θ) be the target distribution we want to draw sample from Take any number generator γ(θk+1 | θk ) and apply an acceptance probability α(θk+1 |θk ) = 1 ∧

π(θk+1 ) π(θk )

·

H   γ(θ ) Hk |θ k+1 H H  γ(θk+1 |θH ) kH 

(6)

If we chose θk+1 − θk ∼ N (0, sk Σk ) then γ(θk+1 |θk ) = γ(θk |θk+1 ) 10

N ar r ow pr op os al

Tar ge t

θ2

0

−10

P r op os al

Tar ge t de ns ity W ide pr op os al 0

1

θ2

The Univ. of Newcastle, Australia

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4

θ1 5 −20 −15

−10

−5

Bayesian Dynamic System Estimation

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5

10

December 15, 2014

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Adaptive Metropolis Sampling Tuning sk and Σk

Optimise sk using Robbins–Monro approximation α(θk+1 | θk ) − α∗ log sk+1 = log sk + k

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

(7)

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Adaptive Metropolis Sampling Tuning sk and Σk

Optimise sk using Robbins–Monro approximation α(θk+1 | θk ) − α∗ log sk+1 = log sk + k Calculate the sample variance Σk by the recursive equation  T k −1 1  Σk+1 = Σk + θk+1 − θ¯k θk+1 − θ¯k k k +1

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

(7)

(8)

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Adaptive Metropolis Sampling Tuning sk and Σk

Optimise sk using Robbins–Monro approximation α(θk+1 | θk ) − α∗ log sk+1 = log sk + k Calculate the sample variance Σk by the recursive equation  T k −1 1  Σk+1 = Σk + θk+1 − θ¯k θk+1 − θ¯k k k +1

(7)

(8)

10 α∗ ∈ [0.2, 0.4] 5

0

−5

The Univ. of Newcastle, −10 Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Slice Sampling Transform π into a uniform distribution on a “slice”

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

(9)

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ)φ(h | θ) dh

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ) φ(h | θ) dh

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ) φ(h | θ) dh = π(θ)

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

(10)

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ) φ(h | θ) dh = π(θ)

(10)

If φ(h | θ) = U[0, π(θ)]

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ) φ(h | θ) dh = π(θ)

(10)

If φ(h | θ) = U[0, π(θ)] = 1/π(θ), ∀ 0 ≤ h ≤ π(θ)

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ) φ(h | θ) dh = π(θ)

(10)

If φ(h | θ) = U[0, π(θ)] = 1/π(θ), ∀ 0 ≤ h ≤ π(θ) then φ(h, θ) = π(θ)/π(θ) = 1

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

(11)

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ) φ(h | θ) dh = π(θ)

(10)

If φ(h | θ) = U[0, π(θ)] = 1/π(θ), ∀ 0 ≤ h ≤ π(θ) then φ(h, θ) = 1Υ , Υ = {(h, θ) | 0 ≤ h ≤ π(θ)}

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

(11)

December 15, 2014

11 / 16

Slice Sampling Transform π into a uniform distribution on a “slice”

Introduce h ∼ φ(h | θ) and form φ(h, θ) = π(θ)φ(h | θ)

(9)

The marginal density of θ ∼ φ is Z φ(θ) = π(θ) φ(h | θ) dh = π(θ)

(10)

If φ(h | θ) = U[0, π(θ)] = 1/π(θ), ∀ 0 ≤ h ≤ π(θ) then φ(h, θ) = 1Υ , Υ = {(h, θ) | 0 ≤ h ≤ π(θ)}

(11)

Sampling from π(θ) ⇔ uniform sampling from Υ The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Slice Sampling Transform π into a uniform distribution on a “slice” 0.8

h π ( θ k)

θ k + 1 ∼ U [S ( h k ) ]

0.6 hk 0.4

Υ = {( h , θ ) | 0 ≤ h ≤ π ( θ ) }

0.2

0 −1

0

1

θk

2

3

4

5

Slice Sampling Transform π into a uniform distribution on a “slice” 0.8

h π ( θ k)

θ k + 1 ∼ U [S ( h k ) ]

0.6 hk 0.4

Υ = {( h , θ ) | 0 ≤ h ≤ π ( θ ) }

0.2

0 −1

0

1

θk

2

3

4

5

hk ∼ ϕ(h | θk ) = U[0, π(θk )]

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Slice Sampling Transform π into a uniform distribution on a “slice” 0.8

h π ( θ k)

θ k + 1 ∼ U [S ( h k ) ]

0.6 hk 0.4

Υ = {( h , θ ) | 0 ≤ h ≤ π ( θ ) }

0.2

0 −1

0

1

θk

2

3

4

5

hk ∼ ϕ(h | θk ) = U[0, π(θk )] θk+1 ∼ ϕ(θ | hk ) = U[S(hk )]

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

12 / 16

Slice Sampling Transform π into a uniform distribution on a “slice” 0.8

h π ( θ k)

θ k + 1 ∼ U [S ( h k ) ]

0.6 hk 0.4

Υ = {( h , θ ) | 0 ≤ h ≤ π ( θ ) }

0.2

0 −1

0

1

θk

2

3

4

5

hk ∼ ϕ(h | θk ) = U[0, π(θk )] θk+1 ∼ ϕ(θ | hk ) = U[S(hk )] S(hk ) = {θ | π(θ) ≥ hk } The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Slice Sampling Uniform sampling on line segments 10 8

C ar t e s ian dir e c t ions

6 4 2 0

S ( h k)

−2 −4 −6

Eige n-dir e c t ions

−8

Σk → ek

The−10 Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Simulation Study Compare Slice sampling vs. Adaptive Metropolis

Same dynamics yt = Gθ (q)ut + et , et ∼ U[−0.3, 0.3]

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Simulation Study Compare Slice sampling vs. Adaptive Metropolis

Same dynamics yt = Gθ (q)ut + et , et ∼ U[−0.3, 0.3]

The test models include Chebyshev low–pass filters with order 1–5.

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Simulation Study Compare Slice sampling vs. Adaptive Metropolis

Same dynamics yt = Gθ (q)ut + et , et ∼ U[−0.3, 0.3]

The test models include Chebyshev low–pass filters with order 1–5. Performance indicator is an average squared error kˆ pM (θ) −

π(θ)k2

nθ Z h i2 1 X = p ˆM (θ(i) ) − π(θ(i) ) dθ(i) nθ

(12)

i=1

at different sample sizes M

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Simulation Study Compare Slice sampling vs. Adaptive Metropolis

Same dynamics yt = Gθ (q)ut + et , et ∼ U[−0.3, 0.3]

The test models include Chebyshev low–pass filters with order 1–5. Performance indicator is an average squared error kˆ pM (θ) −

π(θ)k2

nθ Z h i2 1 X = p ˆM (θ(i) ) − π(θ(i) ) dθ(i) nθ

(12)

i=1

at different sample sizes M The true target density is approximated with 2 × 109 samples.

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Simulation Study Convergence rate of Adaptive Metropolis and Slice Sampling

Or de r Or de r Or de r Or de r Or de r −3

1 2 3 4 5

−3

10

kp (.) − π(.)k s e

10

−4

−4

10

10

6

7

10

10 I t e r at i on s

8

10

(a) Adaptive Metropolis The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

6

7

10

10 I t e r at i on s

(b) Slice Sampling December 15, 2014

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Questions & Feedbacks

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Further Research Metropolis within Slice Sampling

π(θ )

Ste p 1: Se le c t a random slic e h ∼ U [0, π(θ )]

Ste p 2: Prop ose a ne w sample θ ′ ∼ N (θ , sΣ )

N (θ , sΣ ) h

Ste p 3: Ke e p θ ′ if π(θ ′ ) ≥ h Othe rw ise , ke e p old sample .

s=

3

2 log h + n θ log (2π) + log|Σ | n θ χ 2n θ(β )

4

θ The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

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Simulations Quality of convergence

F SS 0.8

0.6

0.6

Pr obability

Pr obability

AM 0.8

0.4

0.2

0 −3

0.4

0.2

−2.5

−2

−1.5

−1 a1

−0.5

0

0.5

0 −3

1

−2.5

(a) Metropolis Sampling

−2

−1.5

−1 a1

−0.5

0

0.5

1

(b) Slice Sampling A MSS

Pr obability

0.8

0.6

0.4

0.2

0 −3

−2.5

−2

−1.5

−1 a1

−0.5

0

0.5

1

(c) Metropolis + Slice Sampling The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Simulations Speed of computation

C P U T im e ( s ) Or de r Or de r Or de r Or de r Or de r

377 ( s ) 1 2 3 4 5

76.1 ( s ) 38.8 ( s ) Me t r op olis The Univ. of Newcastle, Australia

Me t r op olis + Slic e Bayesian Dynamic System Estimation

Slic e Sam pling December 15, 2014

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Further research Robust MCMC simulation techniques

Is there a common framework for the whole range of MCMC algorithms available? How much do we gain from combining multiple MCMC strategies in one computing solution? How do we best utilise parallel computing in MCMC simulation?

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Further research Simulated Annealing

Te mp e r e d de ns ity Tar ge t de ns ity

0.15

0.1

0.05 β = 0.01

0 −60

−40

The Univ. of Newcastle, Australia

−20

0

20

Bayesian Dynamic System Estimation

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80

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Further research Simulated Annealing

Te mp e r e d de ns ity Tar ge t de ns ity

0.15

0.1

0.05 β = 0.027826

0 −60

−40

The Univ. of Newcastle, Australia

−20

0

20

Bayesian Dynamic System Estimation

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80

December 15, 2014

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Further research Simulated Annealing

Te mp e r e d de ns ity Tar ge t de ns ity

0.15

0.1

0.05 β = 0.077426

0 −60

−40

The Univ. of Newcastle, Australia

−20

0

20

Bayesian Dynamic System Estimation

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Further research Simulated Annealing

Te mp e r e d de ns ity Tar ge t de ns ity

0.15

0.1

0.05 β = 0.21544

0 −60

−40

The Univ. of Newcastle, Australia

−20

0

20

Bayesian Dynamic System Estimation

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80

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Further research Simulated Annealing

Te mp e r e d de ns ity Tar ge t de ns ity

0.15

0.1

0.05 β = 0.59948

0 −60

−40

The Univ. of Newcastle, Australia

−20

0

20

Bayesian Dynamic System Estimation

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80

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Further research Simulated Annealing

Te mp e r e d de ns ity Tar ge t de ns ity

0.15

0.1

0.05 β =1

0 −60

−40

The Univ. of Newcastle, Australia

−20

0

20

Bayesian Dynamic System Estimation

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December 15, 2014

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Further research Simulated Tempering

β β β β β

0.15

= = = = =

0.01 0.031623 0.1 0.31623 1

0.1

0.05

0 −60

−40

The Univ. of Newcastle, Australia

−20

0

20

Bayesian Dynamic System Estimation

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Further research Direction Sampling

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

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Limitations The 1 % chance of failure

Fail to converge to the target distribution for approx. 1 % of the cases No indication of such failure is given beside visual inspection of the raw data The target distribution is evidently multimodal

The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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Limitations The 1 % chance of failure

AM

MSS

ǫe

0.25

ǫe

0.25

0.2

0.2 2

4 6 I t e r at ion A MSS

8

10

2

5

x 10

8

4 6 I t e r at ion

8

10 5

x 10

ǫe

0.25

ǫe

0.25

4 6 I t e r at ion F SS

0.2

0.2 2

4 6 I t e r at ion

The Univ. of Newcastle, Australia

8

10 5

x 10

Bayesian Dynamic System Estimation

2

10 5

x 10

December 15, 2014

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Slice Sampling Uniform sampling on line segments

π ( θ k) I k = {θ k + r e k } ∩ S ( h k )

St e p 1: Se le c t a dir e c t ion e k t o de fine t he int e val I k hk St e p 2: Me asur e t he w idt h of t he int e r val I k St e p 3: Sample unif or mly f r om t he int e r val θ ∼ U [I k ] θk The Univ. of Newcastle, Australia

Bayesian Dynamic System Estimation

December 15, 2014

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