HMM-前向后向算法(附python实现)
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 \(\alpha _i(t)\) :
\]
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
\]
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
=\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_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
= \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
#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 \(\beta _i(t)\) :
\]
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
\]
当 \(t=1\) 时,观测为 \(O_{2},O_{3}, …, O_T\)
\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
\]
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
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
\]
\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] //www.cs.sjsu.edu/~stamp/RUA/HMM.pdf
[2]//www.cnblogs.com/fulcra/p/11065474.html
[3] //www.cnblogs.com/sjjsxl/p/6285629.html
[4] //blog.csdn.net/xueyingxue001/article/details/52396494