An Introduction to Hidden Markov Model
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This document provides an introduction to Hidden Markov Models (HMMs). It begins by explaining the key differences between Markov Models and HMMs, noting that in HMMs the states are hidden and can only be indirectly observed through observations. It then outlines the main elements of an HMM - the number of states, observations, state transition probabilities, observation probabilities, and initial state distribution. An example HMM is provided. Finally, it briefly introduces three common problems in HMMs - determining the most likely model given observations, determining the most likely state sequence, and determining the model parameters that are most likely to have generated the observations.
An Introduction to Hidden Markov Model
- 1. An Introduction to HMM Browny 2010.07.21
- 2. MM vs. HMM States States Observations
- 3. Markov Model • Given 3 weather states: – {S1, S2, S3} = {rain, cloudy, sunny} Rain Cloudy Sunny Rain 0.4 0.3 0.3 Cloudy 0.2 0.6 0.2 Sunny 0.1 0.1 0.8 • What is the probabilities for next 7 days will be {sun, sun, rain, rain, sun, cloud, sun} ?
- 4. Hidden Markov Model • The states – We don’t understand, Hidden! – But it can be indirectly observed • Example – 北極or赤道(model), Hot/Cold(state), 1/2/3 ice cream(observation)
- 5. Hidden Markov Model • The observation is a probability function of state which is not observable directly Hidden States
- 6. HMM Elements • N, the number of states in the model • M, the number of distinct observation symbols • A, the state transition probability distribution • B, the observation symbol probability distribution in states • π, the initial state distribution λ: model
- 7. Example P(…|C) P(…|H) P(…|Start) P(1|…) 0.7 0.1 P(2|…) 0.2 B: 0.2 Observation P(3|…) 0.1 0.7 P(C|…) 0.8 0.1 0.5 A: π: P(H|…) 0.1 0.8 0.5 Transition initial P(STOP|…) 0.1 0.1 0
- 8. 3 Problems
- 9. 3 Problems 1. 觀察到的現象最符合哪一個模型 P(觀察到的現象|模型) 2. 怎樣的狀態序列最符合觀察到的現 象和已知的模型 P(狀態序列|觀察到的現象, 模型) 3. 怎樣的模型最有可能產生觀察到的 現象 what 模型 maximize P(觀察到的現象| 模型)
- 10. Solution 1 • 已知模型，一觀察序列之產生機率 P(O|λ) R1 R1 R1 S1 S1 S1 R2 R2 R2 S2 R1 S2 R1 S2 R1 R2 R2 R2 S3 R1 S3 R1 S3 R1 R2 R2 R2 1 2 3 t 觀察到 R1 R1 R2 的機率為多少？
- 11. Solution 1 • 考慮一特定的狀態序列 Q = q1, q2 … qT • 產生出一特定觀察序列之機率為 P(O|Q, λ) = P(O1|q1, λ) * P(O2|q2, λ) * … * P(Ot|qt, λ) = bq1(O1) * bq2(O2) * … * bqT(OT)
- 12. Solution 1 • 此一特定序列發生之機率為 P(Q|λ) = πq1 * aq1q2 * aq2q3 * … * aq(T-1)qT • 已知模型，一觀察序列之產生機率 P(O|λ) P(O|λ) = ,q∑ q P(O|Q, λ) * P(Q| λ) q ,..., 1 2 T = q ,q∑ q πq1 * bq1(O1) * aq1q2 bq2(O2)* … * aq(T-1)qT * bqT(OT) ,..., 1 2 T
- 13. Solution 1 狀態的數量) • Complexity (N: 狀態的數量 – 2T*NT ≈ (2T-1)NT muls + NT-1 adds (NT:狀態轉換 組合數) – For N=5 states, T=100 observations, there are order 2*100*5100 ≈ 1072 computations!! • Forward Algorithm – Forward variable αt(i) (給定時間 t 時狀態為 Si 的 條件下，向前 向前局部觀察序列為O1, O2, O3…, Ot 的 向前 機率) at (i ) = P(O1 , O2 ,..., Ot , qt = Si | λ )
- 14. Solution 1 R1 R1 R1 S1 S1 S1 R2 R2 R2 When O1 = R1 S2 R1 S2 R1 S2 R1 R2 R2 R2 S3 R1 S3 R1 S3 R1 R2 R2 R2 1 2 3 t α1 ( i ) = π i bi ( O1 ) 1 ≤ i ≤ N α 2 (1) = α1 (1) a11 + α1 ( 2 ) a21 + α1 ( 3) a31 b1 ( O2 ) α1 (1) = π 1b1 (O1 ) α1 (2) = π 2b2 (O1 ) α 2 ( 2 ) = α1 (1) a12 + α1 ( 2 ) a22 + α1 ( 3) a32 b2 ( O2 ) α1 (3) = π 3b3 (O1 )
- 15. Forward Algorithm • Initialization: α1 (i ) = π i bi (O1 ) 1 ≤ i ≤ N • Induction: N 1 ≤ t ≤ T −1 αt +1 ( j ) = ∑αt ( i ) aij bj ( Ot +1 ) 1 ≤ j ≤ N i=1 • Termination: N P(O | λ ) = ∑ αT (i ) i =1
- 16. Backward Algorithm • Forward Algorithm at (i ) = P(O1 , O2 ,..., Ot , qt = Si | λ ) • Backward Algorithm – 給定時間 t 時狀態為 Si 的條件下，向後 向後局 向後 部觀察序列為 Ot+1, Ot+2, …, OT的機率 βt (i ) = P(Ot +1 , Ot + 2 ,..., OT , qt = Si | λ )
- 17. Backward Algorithm • Initialization βT (i ) = 1 1 ≤ i ≤ N • Induction N t = T −1, T − 2, ...,1 βt (i ) = ∑ aij b j (Ot +1 ) β t +1 ( j ) j =1 1≤ i ≤ N
- 18. Backward Algorithm R1 R1 R1 S1 S1 S1 R2 R2 R2 When OT = R1 S2 R1 S2 R1 S2 R1 R2 R2 R2 S3 R1 S3 R1 S3 R1 R2 R2 R2 1 2 3 t N β T −1 (1) = ∑ a1 j b j ( OT ) β T ( j ) j =1 = a11b1 ( OT ) + a12 b2 ( OT ) + a13b3 ( OT )
- 19. Solution 2 • 怎樣的狀態序列最能解釋觀察到的現 象和已知的模型 P(狀態序列|觀察到的現象, 模型) • 無精確解，有很多種方式解此問題， 對狀態序列的不同限制有不同的解法 對狀態序列的不同限制
- 20. Solution 2 • 例: Choose the state qt which are individually most likely – γt(i) : the probability of being in state Si at time t, given the observation sequence O, and the model λ P (O | qt = Si , λ ) α t ( i ) βt ( i ) α t ( i ) βt ( i ) γ t (i ) = = = N P (O λ ) P (O λ ) ∑ α t ( i ) βt ( i ) i =1 qt = argmax γ t ( i ) 1 ≤ t ≤ T 1≤i ≤ N
- 21. Viterbi algorithm • The most widely used criteria is to find the “single best state sequence” maxmize P ( Q | O, λ ) ≈ maxmize P ( Q, O | λ ) • A formal technique exists, based on dynamic programming methods, and is called the Viterbi algorithm
- 22. Viterbi algorithm • To find the single best state sequence, Q = {q1, q2, …, qT}, for the given observation sequence O = {O1, O2, …, OT} • δt(i): the best score (highest prob.) along a single path, at time t, which accounts for the first t observations and end in state Si δ t ( i ) = max P q1 q2 ... qt = Si , O1 O2 ... Ot λ 1 q , q ,..., q 2 t −1
- 23. Viterbi algorithm • Initialization - δ1(i) – When t = 1 the most probable path to a state does not sensibly exist – However we use the probability of being in that state given t = 1 and the observable state O1 δ1 ( i ) = π i bi ( O1 ) 1 ≤ i ≤ N ψ (i ) = 0
- 24. Viterbi algorithm • Calculate δt(i) when t > 1 – δt(i) : The most probable path to the state X at time t – This path to X will have to pass through one of the states A, B or C at time (t-1) Most probable path to A: δ t −1 ( A) a AX bX ( Ot )
- 25. Viterbi algorithm • Recursion δ t ( j ) = max δ t −1 ( i ) aij b j ( Ot ) 2≤t ≤T 1≤ i ≤ N ψ t ( j ) = argmax δ t −1 ( i ) aij 1≤ j ≤ N 1≤ i ≤ N • Termination P* = max δ T ( i ) 1≤i ≤ N q = argmax δ T ( i ) * T 1≤i ≤ N
- 26. Viterbi algorithm • Path (state sequence) backtracking qt* = ψ t +1 (qt*+1 ) t = T − 1, T − 2, ..., 1 qT −1 = ψ T (qT ) = argmax δ T −1 ( i ) aiq* * * 1≤i ≤ N T ... ... * * q1 = ψ 2 (q2 )
- 27. Solution 3 • 怎樣的模型 λ = (A, B, π) 最有可能產生 觀察到的現象 what 模型 maximize P(觀察到的現象| 模型) • There is no known analytic solution. We can choose λ = (A, B, π) such that P(O| λ) is locally maximized using an iterative procedure
- 28. Baum-Welch Method • Define ξt(i, j) = P(qt=Si , qt+1=Sj|O, λ) – The probability of being in state Si at time t, and state Sj at time t+1 α t ( i ) aij b j ( Ot +1 ) βt +1 ( j ) ξt ( i, j ) = P (O λ ) α t ( i ) aij b j ( Ot +1 ) βt +1 ( j ) = N N ∑∑ α ( i ) a b ( O ) β ( j ) i =1 j =1 t ij j t +1 t +1
- 29. Baum-Welch Method • γt(i) : the probability of being in state Si at time t, given the observation sequence O, and the model λ α t ( i ) βt ( i ) α ( i ) βt ( i ) γ t (i ) = = N t P (O λ ) ∑ α t ( i ) βt ( i ) • Relate γt(i) to ξt(i, j) i =1 α t ( i ) aij b j ( Ot +1 ) βt +1 ( j ) N ξt ( i, j ) = γ t ( i ) = ∑ ξt ( i, j ) P (O λ ) j =1 α t ( i ) aij b j ( Ot +1 ) βt +1 ( j ) = N N ∑∑ α ( i ) a b ( O ) β ( j ) i =1 j =1 t ij j t +1 t +1
- 30. Baum-Welch Method • The expected number of times that state Si is visited T −1 ∑ γ ( i ) = Expected number of transitions from Si t =1 t • Similarly, the expected number of transitions from state Si to state Sj T −1 ∑ ξ ( i, j ) = Expected number of transitions from S to S t =1 t i j
- 31. Baum-Welch Method • Re-estimation formulas for π, A and B π i = γ1(i) T −1 ∑ξ (i, j) t =1 t expected number of transitions from state Si to S j aij = T −1 = expected number of transitions from state Si ∑γt (i) t =1 T ∑t =1 γ t ( j) s.t. Ot =vk expected number of times in state j and observing symbol vk b j (k) = T = expected number of times in state j ∑γ ( j) t =1 t
- 32. Baum-Welch Method • P(O|λ) > P(O|λ) • Iteratively use λ in place of λ and repeat the re-estimation, we then can improve P(O| λ) until some limiting point is reached