HMM-前向后向算法与实现

基本要素
HMM三大问题
概率计算问题

基本要素

状态 \(N\)个
状态序列 \(S = s_1,s_2,...\)
观测序列 \(O=O_1,O_2,...\)
\(\lambda(A,B,\pi)\)
- 状态转移概率 \(A = \{a_{ij}\}\)
- 发射概率 \(B = \{b_{ik}\}\)
- 初始概率分布 \(\pi = \{\pi_i\}\)
观测序列生成过程
- 初始状态
- 选择观测
- 状态转移
- 返回step2

HMM三大问题

概率计算问题（评估问题）

给定观测序列 \(O=O_1O_2...O_T\)，模型 \(\lambda (A,B,\pi)\)，计算 \(P(O|\lambda)\)，即计算观测序列的概率

解码问题

给定观测序列 \(O=O_1O_2...O_T\)，模型 \(\lambda (A,B,\pi)\)，找到对应的状态序列 \(S\)

学习问题

给定观测序列 \(O=O_1O_2...O_T\)，找到模型参数 \(\lambda (A,B,\pi)\)，以最大化 \(P(O|\lambda)\)，

概率计算问题

给定模型 \(\lambda\) 和观测序列 \(O\)，如何计算\(P(O| \lambda)\)？

暴力枚举每一个可能的状态序列 \(S\)

对每一个给定的状态序列

\[P(O|S,\lambda) = \prod^T_{t=1} P(O_t|s_t,\lambda) =\prod^T_{t=1} b_{s_tO_t}
\]
一个状态序列的产生概率

\[P(S|\lambda) = P(s_1)\prod^T_{t=2}P(s_t|s_{t-1})=\pi_1\prod^T_{t=2}a_{s_{t-1}s_t}
\]
联合概率

\[P(O,S|\lambda) = P(S|\lambda)P(O|S,\lambda) =\pi_1\prod^T_{t=2}a_{s_{t-1}s_t}\prod^T_{t=1} b_{s_tO_t}
\]
考虑所有的状态序列

\[P(O|\lambda)=\sum_S\pi_1b_{s_1O_1}\prod^T_{t=2}a_{s_{t-1}s_t}b_{s_tO_t}
\]

\(O\) 可能由任意一个状态得到，所以需要将每个状态的可能性相加。

这样做什么问题？时间复杂度高达 \(O(2TN^T)\)。每个序列需要计算 \(2T\) 次，一共 \(N^T\) 个序列。

前向算法

在时刻 \(t\)，状态为 \(i\) 时，前面的时刻观测到 \(O_1,O_2, ..., O_t\) 的概率，记为 \(\alpha _i(t)\) ：

\[\alpha_{i}(t)=P\left(O_{1}, O_{2}, \ldots O_{t}, s_{t}=i | \lambda\right)
\]

当 \(t=1\) 时，输出为 \(O_1\)，假设有三个状态，\(O_1\) 可能是任意一个状态发出，即

\[P(O_1|\lambda) = \pi_1b_1(O_1)+\pi_2b_2(O_1)+\pi_2b_3(O_1) = \alpha_1(1)+\alpha_2(1)+\alpha_3(1)
\]

当 \(t=2\) 时，输出为 \(O_1O_2\) ，\(O_2\) 可能由任一个状态发出，同时产生 \(O_2\) 对应的状态可以由 \(t=1\) 时刻任意一个状态转移得到。假设 \(O_2\) 由状态 1 发出，如下图

\[P(O_1O_2,s_2=q_1|\lambda) = \pi_1b_1(O_1)a_{11}b_1(O_2)+\pi_2b_2(O_1)a_{21}b_1(O_2)+\pi_2b_3(O_1)a_{31}b_1(O_2) \\
=\bold{\alpha_1(1)}a_{11}b_1(O_2)+\bold{\alpha_2(1)}a_{21}b_1(O_2)+\bold{\alpha_3(1)}a_{31}b_1(O_2) = \bold{\alpha_1(2)}
\]

同理可得 \(\alpha_2(2),\alpha_3(2)\)

\[\bold{\alpha_2(2)} = P(O_1O_2,s_2=q_2|\lambda)
=\bold{\alpha_1(1)}a_{12}b_2(O_2)+\bold{\alpha_2(1)}a_{22}b_2(O_2)+\bold{\alpha_3(1)}a_{32}b_2(O_2)
\\
\bold{\alpha_3(2)} = P(O_1O_2,s_2=q_3|\lambda)
=\bold{\alpha_1(1)}a_{13}b_3(O_2)+\bold{\alpha_2(1)}a_{23}b_3(O_2)+\bold{\alpha_3(1)}a_{33}b_3(O_2)
\]

所以

\[P(O_1O_2|\lambda) =P(O_1O_2,s_2=q_1|\lambda)+ P(O_1O_2,s_2=q_2|\lambda) +P(O_1O_2,s_2=q_3|\lambda)\\
= \alpha_1(2)+\alpha_2(2)+\alpha_3(2)
\]

所以前向算法过程如下：

step1：初始化 \(\alpha_i(1)= \pi_i*b_i(O_1)\)

step2：计算 \(\alpha_i(t) = (\sum^{N}_{j=1} \alpha_j(t-1)a_{ji})b_i(O_{t})\)

step3：\(P(O|\lambda) = \sum^N_{i=1}\alpha_i(T)\)

相比暴力法，时间复杂度降低了吗？

当前时刻有 \(N\) 个状态，每个状态可能由前一时刻 \(N\) 个状态中的任意一个转移得到，所以单个时刻的时间复杂度为 \(O(N^2)\),总时间复杂度为 \(O(TN^2)\)

代码实现

例子：

假设从三个袋子 {1,2,3}中取出 4 个球 O={red,white,red,white}，模型参数\(\lambda = (A,B,\pi)\) 如下，计算序列O出现的概率

#状态 1 2 3
A = [[0.5,0.2,0.3],
	 [0.3,0.5,0.2],
	 [0.2,0.3,0.5]]

pi = [0.2,0.4,0.4]

# red white
B = [[0.5,0.5],
	 [0.4,0.6],
	 [0.7,0.3]]

step1：初始化 \(\alpha_i(1)= \pi_i*b_i(O_1)\)

step2：计算 \(\alpha_i(t) = (\sum^{N}_{j=1} \alpha_j(t-1)a_{ji})b_i(O_{t})\)

step3：\(P(O|\lambda) = \sum^N_{i=1}\alpha_i( T)\)

#前向算法
def hmm_forward(A,B,pi,O):
    T = len(O)
    N = len(A[0])
    #step1 初始化
    alpha = [[0]*T for _ in range(N)]
    for i in range(N):
        alpha[i][0] = pi[i]*B[i][O[0]]

    #step2 计算alpha(t)
    for t in range(1,T):
        for i in range(N):
            temp = 0
            for j in range(N):
                temp += alpha[j][t-1]*A[j][i]
            alpha[i][t] = temp*B[i][O[t]]

    #step3
    proba = 0
    for i in range(N):
        proba += alpha[i][-1]
    return proba,alpha

A = [[0.5,0.2,0.3],[0.3,0.5,0.2],[0.2,0.3,0.5]]
B = [[0.5,0.5],[0.4,0.6],[0.7,0.3]]
pi = [0.2,0.4,0.4]
O = [0,1,0,1]
hmm_forward(A,B,pi,O)  #结果为 0.06009

结果

后向算法

在时刻 \(t\)，状态为 \(i\) 时，观测到 \(O_{t+1},O_{t+2}, ..., O_T\) 的概率，记为 \(\beta _i(t)\) ：

\[\beta_{i}(t)=P\left(O_{t+1},O_{t+2}, ..., O_T | s_{t}=i, \lambda\right)
\]

当 \(t=T\) 时，由于 \(T\) 时刻之后为空，没有观测，所以 \(\beta_i(t)=1\)

当 \(t = T-1\) 时，观测 \(O_T\) ，\(O_T\) 可能由任意一个状态产生

\[\beta_i(T-1) = P(O_T|s_{t}=i,\lambda) = a_{i1}b_1(O_T)\beta_1(T)+a_{i2}b_2(O_T)\beta_2(T)+a_{i3}b_3(O_T)\beta_3(T)
\]

当 \(t=1\) 时，观测为 \(O_{2},O_{3}, ..., O_T\)

\[\begin{aligned}
\beta_1(1)
&= P(O_{2},O_{3}, ..., O_T|s_1=1,\lambda)\\
&=a_{11}b_1(O_2)\beta_1(2)+a_{12}b_2(O_2)\beta_2(2)+a_{13}b_3(O_2)\beta_3(2)
\\
\quad
\\
\beta_2(1)
&= P(O_{2},O_{3}, ..., O_T|s_1=2,\lambda)\\
&=a_{21}b_1(O_2)\beta_1(2)+a_{22}b_2(O_2)\beta_2(2)+a_{23}b_3(O_2)\beta_3(2)
\\
\quad
\\
\beta_3(1)
&=P(O_{2},O_{3}, ..., O_T|s_1=3,\lambda)\\
&=a_{31}b_1(O_2)\beta_1(2)+a_{32}b_2(O_2)\beta_2(2)+a_{33}b_3(O_2)\beta_3(2)
\end{aligned}
\]

所以

\[P(O_{2},O_{3}, ..., O_T|\lambda) = \beta_1(1)+\beta_2(1)+\beta_3(1)
\]

后向算法过程如下：

step1：初始化 \(\beta_i(T)=1\)

step2：计算 \(\beta_i(t) = \sum^N_{j=1}a_{ij}b_j(O_{t+1})\beta_j(t+1)\)

step3：\(P(O|\lambda) = \sum^N_{i=1}\pi_ib_i(O_1)\beta_i(1)\)

时间复杂度 \(O(N^2T)\)

代码实现

还是上面的例子

#后向算法
def hmm_backward(A,B,pi,O):
    T = len(O)
    N = len(A[0])
    #step1 初始化
    beta = [[0]*T for _ in range(N)]
    for i in range(N):
        beta[i][-1] = 1

    #step2 计算beta(t)
    for t in reversed(range(T-1)):
        for i in range(N):
            for j in range(N):
                beta[i][t]  += A[i][j]*B[j][O[t+1]]*beta[j][t+1]

    #step3
    proba = 0
    for i in range(N):
        proba += pi[i]*B[i][O[0]]*beta[i][0]
    return proba,beta

A = [[0.5,0.2,0.3],[0.3,0.5,0.2],[0.2,0.3,0.5]]
B = [[0.5,0.5],[0.4,0.6],[0.7,0.3]]
pi = [0.2,0.4,0.4]
O = [0,1,0,1]
hmm_backward(A,B,pi,O)  #结果为 0.06009

结果

前向-后向算法

回顾前向、后向变量：

\(a_i(t)\) 时刻 \(t\)，状态为 \(i\) ，观测序列为 \(O_1,O_2, ..., O_t\) 的概率
\(\beta_i(t)\) 时刻 \(t\)，状态为 \(i\) ，观测序列为 \(O_{t+1},O_{t+2}, ..., O_T\) 的概率

\[\begin{aligned}
P(O,s_t=i|\lambda)
&= P(O_1,O_2, ..., O_T,s_t=i|\lambda)\\
&= P(O_1,O_2, ..., O_t,s_t=i,O_{t+1},O_{t+2}, ..., O_T|\lambda)\\
&= P(O_1,O_2, ..., O_t,s_t=i|\lambda)*P(O_{t+1},O_{t+2}, ..., O_T|O_1,O_2, ..., O_t,s_t=i,\lambda) \\
&= P(O_1,O_2, ..., O_t,s_t=i|\lambda)*P(O_{t+1},O_{t+2}, ..., O_T,s_t=i|\lambda)\\
&= a_i(t)*\beta_i(t)
\end{aligned}
\]

即在给定的状态序列中，\(t\) 时刻状态为 \(i\) 的概率。

使用前后向算法可以计算隐状态，记 \(\gamma_i(t) = P(s_t=i|O,\lambda)\) 表示时刻 \(t\) 位于隐状态 \(i\) 的概率

\[P\left(s_{t}=i, O | \lambda\right)=\alpha_{i}(t) \beta_{i}(t)
\]

\[\begin{aligned}
\gamma_{i}(t)
&=P\left(s_{t}={i} | O, \lambda\right)=\frac{P\left(s_{t}={i}, O | \lambda\right)}{P(O | \lambda)} \\
&=\frac{\alpha_{i}(t) \beta_{i}(t)}{P(O | \lambda)}=\frac{\alpha_{i}(t) \beta_{i}(t)}{\sum_{i=1}^{N} \alpha_{i}(t) \beta_{i}(t)}
\end{aligned}
\]

references：

[1] https://www.cs.sjsu.edu/~stamp/RUA/HMM.pdf

[2]https://www.cnblogs.com/fulcra/p/11065474.html

[3] https://www.cnblogs.com/sjjsxl/p/6285629.html

[4] https://blog.csdn.net/xueyingxue001/article/details/52396494