D6 D4 D8

1 2 3

4 5 6

1 2 3

4

1 2 3

5 6 7

4

8

每⼀一次擲骰從三個骰⼦子中挑⼀一個，每⼀一個挑到的機率為1/3 挑到骰⼦子後開始擲，擲完結果記下，並且重覆挑骰再擲可得

1 6 3 5 2

1 6 3 5 2 - observation symbols

1 6 3 5 2 可能分別由 D6 D8 D8 D6 D4 擲出

D6 D8 D8 D6 D4

1 6 3 5 2

hidden state visible state

D6

D4 D8

hidden state transition

D6 D4 D8

D6 1/3 1/3 1/3

D4 1/3 1/3 1/3

D8 1/3 1/3 1/3

1 2 3 4 5 6 7 8

D6 1/6 1/6 1/6 1/6 1/6 1/6 0 0

D4 1/4 1/4 1/4 1/4 0 0 0 0

D8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

Transition matrix

Observation emission matrix

Initial state distribution π = {πi}

Number of states in the model X : N

Number of observation symbols O : M

Transition matrix A = {aij} , where aij represents the transition probability from state i to state j

Observation emission matrix B = {bij} , where bij represent the probability of observation j at state i

幾個骰⼦子 3

觀察到骰出的點數 1-8

D4骰⼦子到D8骰⼦子的機率 1/3

D4骰⼦子出現1的機率 1/4

⾸首次選骰⼦子的機率各 1/3

Three Fundamental Problems And Solutions For HMM

1. The Evaluation Problem - Forward

已知骰⼦子有幾種，每種骰⼦子是什麼，知道擲出來想知道此結果的機率？

D6 D4 D8D6 1/3 1/3 1/3D4 1/3 1/3 1/3D8 1/3 1/3 1/3

1 2 3 4 5 6 7 8

D6 1/6 1/6 1/6 1/6 1/6 1/6 0 0

D4 1/4 1/4 1/4 1/4 0 0 0 0

D8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

求 1 6 3 5 2 的機率？

solution1. The Evaluation Problem - Forward

D6 D4

D6 1/2 1/2

D4 1/2 1/2

1 2 3 4 5 6

D6 1/6 1/6 1/6 1/6 1/6 1/6

D4 1/4 1/4 1/4 1/4 0 0

求 1 2 3的機率？

1 2 3

D6 1/2*1/6(1/2*1/6)*1/2*1/6

＋ (1/2*1/4)*1/2*1/6

(5/288)*1/2*1/6 ＋

(5/192)*1/2*1/6

D4 1/2*1/4(1/2*1/6)*1/2*1/4

＋ (1/2*1/4)*1/2*1/4

(5/288)*1/2*1/4 ＋

(5/192)*1/2*1/4

A : 0.009

solution1. The Evaluation Problem - Forward

HMM parameter μ=(π,A,B)

forward probability : αj(t) when time = t，observation sequence (o1,o2,…,ot)，the probability of hidden state j

when t = 1

[ ]

Three Fundamental Problems And Solutions For HMM

2. The Decoding Problem - Viterbi

已知骰⼦子有幾種，每種骰⼦子是什麼，根據骰⼦子擲出來的結果，想知道擲出來的是哪種骰⼦子？

D6 D4 D8D6 1/3 1/3 1/3D4 1/3 1/3 1/3D8 1/3 1/3 1/3

1 2 3 4 5 6 7 8

D6 1/6 1/6 1/6 1/6 1/6 1/6 0 0

D4 1/4 1/4 1/4 1/4 0 0 0 0

D8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

求1 6 3 5 2結果最⼤大可能的擲骰順序？

maybe D6 D8 D8 D6 D4

solution2. The Decoding Problem - Viterbi

求 1 2 3最⼤大可能的擲骰順序？

1/2*1/4 > 1/2*1/6 D4(1/2*1/6)*1/2*1/6

+ (1/2*1/4)*1/2*1/6

(1/2*1/6)*1/2*1/4 +

(1/2*1/4)*1/2*1/4> D4

(5/288)*1/2*1/6 +

(5/192)*1/2*1/6>

(5/288)*1/2*1/4 +

(5/192)*1/2*1/4D4

A : D4 D4 D4

1 2 3

D6 1/2*1/6(1/2*1/6)*1/2*1/6

＋ (1/2*1/4)*1/2*1/6

(5/288)*1/2*1/6 ＋

(5/192)*1/2*1/6

D4 1/2*1/4(1/2*1/6)*1/2*1/4

＋ (1/2*1/4)*1/2*1/4

(5/288)*1/2*1/4 ＋

(5/192)*1/2*1/4

solution2. The Decoding Problem - Viterbi

when t = 1

Three Fundamental Problems And Solutions For HMM

3. The Learning Problem - Baum-Welch(EM algorithm)

已知骰⼦子有幾種，不知道骰⼦子是什麼，觀察多次骰⼦子擲出來的結果，想反推每種骰⼦子是什麼？

1 6 3 5 2 2 5 7 1 2 4 5 5 6 1 3 4 5 6 8

1 2 3 4 5 6 7 8

D6 ? ? ? ? ? ? ? ?

D4 ? ? ? ? ? ? ? ?

D8 ? ? ? ? ? ? ? ?

D6 D4 D8D6 ？ ？ ？D4 ？ ？ ？D8 ？ ？ ？

solution3. The Learning Problem - Baum-Welch(EM algorithm)

求 1 2 3 4 5 6 7 8

D6 ? ? ? ? ? ? ? ?

D4 ? ? ? ? ? ? ? ?

D8 ? ? ? ? ? ? ? ?

D6 D4 D8D6 ？ ？ ？D4 ？ ？ ？D8 ？ ？ ？

初始挑骰⼦子機率 D6,D4,D8 ?

Baum-WelchA1

B1

π1

假設 已知1 6 3 5 2 2 5 7 1 2 4 5 5 6 1 3 4 5 6 8

A2

B2

π2

updateAn

Bn

πn

收斂

A B

π

solution3. The Learning Problem - Baum-Welch(EM algorithm)

forward probability

backward probability

when t = T-1

when t = T

solution3. The Learning Problem - Baum-Welch(EM algorithm)

γj(t) : when time = t，the probability of hidden state j

HMM parameter μ=(π,A,B) and observation sequence O

ξij(t) : the probability of when time = t，the probability of hidden state i,when time = t+1，the probability of hidden state j

solution3. The Learning Problem - Baum-Welch(EM algorithm)

update HMM parameter

Initial state distribution π = {πi}

Number of states in the model: N

Number of observation symbols: M

Transition matrix A = {aij} , where aij represents the transition probability from state i to state j

Observation emission matrix B = {bj(Ot)} , where bj(Ot) represent the probability of observing Ot at state j

幾個骰⼦子 3

觀察到骰出的點數 1-8

D4骰⼦子到D8骰⼦子的機率 1/3

D4骰⼦子出現1的機率 1/4

⾸首次選骰⼦子的機率各 1/3

The Evaluation Problem - Forward

The Decoding Problem - Viterbi

The Learning Problem - Baum-Welch

已知骰⼦子有幾種，每種骰⼦子是什麼，想知道擲出來想知道此結果的機率？

已知骰⼦子有幾種，每種骰⼦子是什麼，根據骰⼦子擲出來的結果，想知道擲出來的是哪種骰⼦子？

已知骰⼦子有幾種，不知道骰⼦子是什麼，觀察多次骰⼦子擲出來的結果，想反推每種骰⼦子是什麼？

S

O

Discrete

Discrete

S

O

Discrete

Continuous

S

O

Continuous

Continuous

D6 D8

1 6

Mixture of Multinomials

Mixture of Gaussians

Factor Analysis

HMM HMM LDS

The Decoding Problem - Viterbi

中⽂文分詞 (jieba)

states set : {B:begin, M:middle, E:end, S:single}

我天天去實驗室 SBESBME我/天天/去/實驗室 S/BE/S/BME

initiaStateB 0.15M 0.005E 0.005S 0.84

B M E SB 0 0.3 0.7 0M 0 0.2 0.8 0E 0.4 0 0 0.6S 0.5 0 0 0.5

國 ⺠民 我 挺 柱 …B -8 -7 -6 -8 -7 …M -10 -9 -8 -10 -7 …E -8 -8 -6 -9 -7 …S -9 -8 -5 -7 -8 …

經驗法則，⽤用現有收集詞庫統計詞頻即可得知

S

O

Discrete

Discrete

數字太⼩小經過log處理

The Learning Problem - Baum-WelchS

O

Discrete

Continuous Gaussian HMM of stock data

mean μ

Gaussian Normal Distribution

variance σ2

probability density function :

S1

S2 S3

S1-diff(μ, σ2)

S1-volume(μ, σ2)

S2-diff(μ, σ2)S2-volume(μ, σ2)

S3-diff(μ, σ2)

S3-volume(μ, σ2)

S = 3 O = 2

Gaussian = 3*2 = 6

如果每⼀一個S所噴出O的結果符合 gaussian normal distribution

diff = close_v(today) - close_v(yesterday) volume = volume(today) O =[diff, volume]

S1 : rise S2：decline S3：calm

S

O

Discrete

Continuous

Gaussian HMM of stock data

diff = close_v(today) - close_v(yesterday) volume = volume(today) O =[diff, volume]

S1 S2 S3S1 ？ ？ ？S2 ？ ？ ？S3 ？ ？ ？

S1 : rise S2：decline S3：calm

diff volume10/1 -0.2 7000010/2ㄉ

0.1 6900010/3 0.1 85000… … …

diff volumeS1 S1-diff(μ, σ2) S1-volume(μ, σ2)

S2 S2-diff(μ, σ2) S2-volume(μ, σ2)

S3 S3-diff(μ, σ2) S3-volume(μ, σ2)

input : numbers of state & O out put : A,B

predict next status

A B

O

⽤用問題3解出A,B，就可以⽤用問題2 去decode，求出預測結果

BUT!

如果每⼀一個S所噴出O的結果符合 gaussian normal distribution

Gaussian Mixture Model GMM

S -> O ⼀一個狀態S噴出的O對應⼀一個Gaussian

S -> k -> O ⼀一個狀態S有k個管道，每個管道 噴出的O對應⼀一個Gaussian

12 3ck × N( v , μ k , σ2k )

N( v , μ , σ2 )

多個Gaussian組成⼀一個GMM

⼀一個狀態S噴出的O 常態分佈符合

不符合

Gaussian HMM

GMM HMM

S1

S2 S3

S1-diff(μ, σ2)

S1-volume(μ, σ2)

S2-volume(μ, σ2)

S3-volume(μ, σ2)

S2-diff(μ, σ2)

S3-diff(μ, σ2)

每⼀一個S所噴出O的結果符合GMM

g ng b nbg 0.6 0.1 0.2 0.1ng 0.2 ? 0.3 0.2b ? ?nb ? ?

定義 statusgamer的層級

diff wifiS1 S1-diff(μ, σ2) 0.2

S2 S2-diff(μ, σ2) 0.4

S3 S3-diff(μ, σ2) 0.6

DiscreteContinuous收集 O 觀察特徵

決定模型

S

O

Discrete

Continuous

MultinomialHMM

S

O

Discrete

Continuous

GaussianHMM GMMHMMn_mix (注意 overfitting)

boring的層級