Simple nonunifilar binary word generators Sarah Marzen June 1, - - PowerPoint PPT Presentation

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Simple nonunifilar binary word generators Sarah Marzen June 1, 2013 Outline Motivation Results Simple nonunifilar source (SNS) Variation on the SNS A different set of nonunifilar binary word generators (preliminary)

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June 1, 2013

Simple nonunifilar binary word generators

Sarah Marzen

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Outline

✤ Motivation ✤ Results ✤ Simple nonunifilar source (SNS) ✤ Variation on the SNS ✤ A different set of nonunifilar binary word generators (preliminary) ✤ Future work

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Motivation

✤ ɛ-machines are most useful when we have no understanding of the

system-- perfect for biological modeling

✤ Problem: neurobiological data is highly subsampled.

fMRI, EEG, ECOG, electrophysiology Observe behavior

Physics is fun!

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Motivation

✤ These problems can maybe be couched as nonunifilar HMMs.

Asleep Active

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Outline: Results

✤ Simple nonunifilar source (the one studied in class) ✤ Simple nonunifilar source with adjustable transition probabilities ✤ Attempt to extend to continuous case ✤ Binary subsampled HMMs of a particular form, to be described

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SNS

A B 1|1/2 0|1/2 1|1/2 1|1/2

T (0) = ✓

1 2

◆ T (1) = ✓ 1

2 1 2 1 2

◆ π = ✓ 1

2 1 2

◆

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SNS

s0: ...0 s1: ...01 s2: ...012 sn: ...01n sinfty: ...01inf

1|1 0|M10 1|M12 1|M23 1|Mn−1,n 1|1/2 0|1/2 0|Mn,0

... ...

0|M10 Mn−1,n = 1T T (1)n T (0)π 1T T (1)n−1 T (0)π Mn,0 = 1 − Mn,n+1

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SNS

s0: ...0 s1: ...01 s2: ...012 sn: ...01n sinfty: ...01inf

1|1 0|M10 1|M12 1|M23 1|Mn−1,n 1|1/2 0|1/2 0|Mn,0

... ...

0|M10 πn = Mn−1,n πn−1

∞

n=0

πn = 1 πn = 1 4 n + 1 2n

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SNS:

Stat. complexity and entropy rate

Cµ = −

∞

n=0

πn log2 πn = 2.71 bits hµ =

∞

n=0

πnH[Mn,0] = 0.678 bits πn = 1 4 n + 1 2n Mn−1,n = n + 1 2n

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SNS: E from causal shielding

s0: ...0 s1: ...01 s2: ...012 sn: ...01n sinfty: ...01inf

1|1 0|M10 1|M12 1|M23 1|Mn−1,n 1|1/2 0|1/2 0|Mn,0

... ...

0|M10 E =

∞

L=0

hµ(L) hµ ' 0.147 bits hµ(L) = H(L + 1) − H(L) = H[XL+1|RL+1, R0 = µ0] χ = Cµ − E = 2.56 bits

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SNS: Time reversed process?

A B 1|1/2 0|1/2 1|1/2 1|1/2

T (0) = ✓

1 2

◆ T (1) = ✓ 1

2 1 2 1 2

◆ π = ✓ 1

2 1 2

◆ C+

µ = C− µ , χ+ = χ−, Ξ = 0

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SNS v. 2

A B 0|q 1|1 − q 1|1 − p 1|p T (1) = ✓1 − p p 1 − q ◆ , T (0) = ✓0 q ◆ , π = ✓

q p+q p p+q

◆

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SNS v. 2

Same recurrent causal states! s0: ...0 s1: ...01 s2: ...012 sn: ...01n sinfty: ...01inf

1|1 0|M10 1|M12 1|M23 1|Mn−1,n 0|Mn,0

... ...

0|M10 0|q Mn−1,n = 1T T (1)n T (0)π 1T T (1)n−1 T (0)π = p(1 − q)n − (1 − p)nq p(1 − q)n−1 − q(1 − p)n−1

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SNS v. 2

πn = p(1 − q)n − q(1 − p)n p − q × pq p + q

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SNS v. 2

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SNS v. 2: Calculating E from causal shields

s0: ...0 s1: ...01 s2: ...012 sn: ...01n sinfty: ...01inf

1|1 0|M10 1|M12 1|M23 1|Mn−1,n 0|Mn,0

... ...

0|M10 0|q

s-1 s-2 s-inf

...

1 1 M−n,0 = 1T T (0) T (1)n−1 π 1T T (1)n−1 π

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SNS v. 2

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SNS v. 2

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SNS v. 2: Time reversed process?

A B 0|q 1|1 − q 1|1 − p 1|p A B 1|1 − p 0|p 1|1 − q 1|q Cµ(p, q) = Cµ(p, q) ⇒ C+

µ = C− µ

⇒ χ+

µ = χ− µ

⇒ Ξ = 0

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SNS v. 2: Attempt at continuous time

d dt ✓p(A, t) p(B, t) ◆ = ✓−kAB kBA kAB −kBA ◆ ✓p(A, t) p(B, t) ◆ ✓p(A, t + ∆t) p(B, t + ∆t) ◆ = ✓1 − kAB∆t kBA∆t kAB∆t 1 − kBA∆t ◆ ✓p(A, t) p(B, t) ◆ ⇒ p = kAB∆t, q = kBA∆t

Continuous time Discretized time

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SNS v. 2

1 2 3 4 5 0.2 0.4 0.6 0.8 1.0 1.2

st P(st)

kAB = 2, kBA = 3

Statistical complexity: differential entropy of this probability distribution?

πt∆t = lim

∆t→0,n∆t=t πn(p = kAB∆t, q = kBA∆t)

πt = kABkBA kAB + kBA kABe−kBAt − kBAe−kABt kAB − kBA

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SNS v. 2: Continuous time stat. comp.

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SNS v. 2

ht = H[kAkB

kAe−kBt − kBe−kAt

Ae−kBt − k2 Be−kAt

∆t]

= kAkB

kAe−kBt − kBe−kAt

Ae−kBt − k2 Be−kAt

∆t 1 log 2 − log2 kAkB

kAe−kBt − kBe−kAt

Ae−kBt − k2 Be−kAt

−kAkB

kAe−kBt − kBe−kAt

Ae−kBt − k2 Be−kAt

∆t log2 ∆t Not sure what to do with these weird factors of time resolution-- they seem to suggest the entropy rate is 0.

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SNS v. 2: Excess entropy in cont. time?

✤ Did not unifilarize the time-reversed epsilon machine, so did not get a

closed form analytic expression for excess entropy

✤ However, if excess entropy is mainly coming from the rule “a 0 must

be followed by a 1” then E ∼

ABk2 BA∆t

(kAB + kBA)3

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SNS v. 2

✤ Excess entropy and statistical complexity capture very different ideas. ✤ E captures how often you are synchronized to internal states ✤ Stat. comp. captures how long-tailed the probability distribution over

causal states is

✤ Going to continuous time maybe introduces an uncountable infinity

f causal states, differential entropies (negative stat. comp.???),

discontinuities in stat. comp. vs. parameters

✤ E captures relaxation of probability distribution over all mixed states

to stationarity

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Last nonunifilar model

A B1 B2 Bn ... Group 0 Group 1 Fully connected, randomly chosen kinetic rates between states

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Last nonunifilar model

s0: ...0 s1: ...01 s2: ...012 sn: ...01n sinfty: ...01inf

1|1 0|M10 1|M12 1|M23 1|Mn−1,n 0|Mn,0

... ...

0|M10

Same recurrent causal states!

Mn−1,n = 1T T (1)n T (0)π 1T T (1)n−1 T (0)π

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Preliminary results

πn = 1T T (1)n T (0)π 1T I − T (1)−1 T (0)π

This n is # of hidden states

hn = H[1T T (1)n+1 T (0)π 1T T (1)n T (0)π ]

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Future directions

✤ Finish up calculating stuff for the last nonunifilar model. ✤ Maybe this has a practical application-- you can estimate the number

f hidden states by knowing the average transition rates and

calculating crypticity? We’ll see.

✤ More nonunifilar models, continuous time, everything.