标签：tor block word each tip 清华 学习 note bin
What‘s HMM?
In problems that have an inherent temporality -- that is, consist of a process that unfolds in time -- we may have states at time t that are influenced directly by a state t-1. Hidden Markov model (HMM) have found greatest use in such problem, for instance speech recognition or gesture recongnition.
HMM has two stochastic process. One is a Markov chain and the other is a "emission" that based on each Markov chain‘s state.
The Markov chain can be decribed as Figure 1
Figure 1
The state space in Markov chain is $\mathbf{W}$
\[
\mathbf{W} = \{w_1, w_2, ..., w_n\}
\]
The state at time t is $w(t)$ and a particular sequence of length T is denoted by $\mathbf{w^T}$.
\[
\mathbf{w^T} = \{w(1), w(2), ..., w(T) \}
\]
where $w(t) = w_i$, $i = 1,2,...,n$
The hidden Markov model can be described as Figure 2.
Figure 2
At every time step t the system is in a state $w(t)$ but now it emits some (visible) symbol $v(t)$. Notice that $v(t)$ can be discrete or continuous, but in the passage, $v(t)$ is assumed to be discrete.
We define a particular sequence of such visible states as
\[\mathbf{V^T} = \{v(1), v(2), ..., v(T)\}\]
where $v(t) = v_i$, $i=1,2,3,...$
We denote the initial distrubution $\pi_i$, transition probabilities $a_{ij}$ among hidden states and for the probability $b_{jk}$ of the emission of a visible state:
\[
\begin{split}
& \pi_i = P(w(1)=w_i) \\
& a_{ij}(t) = P\left(w(t+1)=w_j|w(t)=w_i\right)\\
& b_{jk}(t) = P\left(v(t)=v_k|w(t)=w_j\right)
\end{split}
\]
We have the normalization condition:
\[
\begin{split}
& \sum_{j}a_{ij}(t) = 1\\
& \sum_{k}b_{jk}(t) = 1
\end{split}
\]
Let $\mathbf{A}=[a_{ij}]_{n\times n}$ , $\mathbf{B}=[b_{jk}]_{n\times m}$ and $\mathbf{\pi} = (\pi_i)$, then the HMM can be described as $\mathbf{\lambda} = (\mathbf{A},\mathbf{B},\mathbf{\pi})$
Three Central Issues
When we get an HMM we need to handle three central issues in order to make full use of HMM and have a better understanding of HMM.
① The Evaluation problem
Suppose we have an HMM. Determine the probability that a particular sequence of visible states $\mathbf{V^T}$ generated by that model.
② The Learning problem
Suppose we are given the coarse structure of a model (the number of hidden states and the number of visible states) but not the probability $a_{ij}$, $b_{jk}$ and $\pi_i$. Given a set of training observation of visible symbols, determine these parameters.
③ The Decoding problem
Suppose we have an HMM as well as a set of observation $\mathbf{V^T}$. Determine the most likely sequence of hidden states $\mathbf{w^T}$ that led to those observation.
We will consider each of these problem in turn.
Evaluation problem
Evaluation problem: Given $\mathbf{\lambda} = (\mathbf{A},\mathbf{B},\mathbf{\pi})$ and observation sequence $\mathbf{V^T}$, compute $P(\mathbf{V^T}|\mathbf{\lambda})$.
If we compute $P(\mathbf{V^T}|\mathbf{\lambda})$ directly:
\[
\begin{split}
P(\mathbf{V^T}|\mathbf{\lambda}) &= \sum_{\mathbf{w^T}}P(\mathbf{V^T}|\mathbf{w^T},\mathbf{\lambda})P(\mathbf{w^T}|\mathbf{\lambda})\\
& = \sum_{\mathbf{w^T}}\prod_{t=1}^{T}\pi_i a_{ij}(t) b_{jk}(t)
\end{split}
\]
Analyze the formulation above, $w^T$ totally have $n^T$ situations. Thus, it is an $O(Tn^T)$ calculation, which is quite prohibitive in practice.
A computationaly simpler algorithm -- forward algorithm and backward algorithm -- for the same goal is as follows. Notice that both forward algorithm and backward algorithm can calculate $P(\mathbf{V^T}|\mathbf{\lambda})$.
Forward algorithm
Define forward variable:
\[
\alpha_t(i) = P(V^t,w(t)=w_i|\lambda)
\]
Then,
\[
\begin{split}
\alpha_1(i) &= P(v(1)=v_k,w(1)=w_i|\lambda) \\
& = P(v(1)=v_k|w(1)=w_i,\lambda)P(w(1)=w_i|\lambda)\\
& = \pi_i b_{ik}(1)
\end{split}
\]
\[
\begin{split}
\alpha_{t+1}(j) &= \left[ \sum_{i=1}^{N}\alpha_t(i)a_{ij}(t) \right] b_{jk}(t+1)\\
1\leq &t\leq T-1, \qquad 1\leq j\leq n
\end{split}
\]
\[
P(\mathbf{V^T}|\mathbf{\lambda}) = \sum_{i=1}^{n}\alpha_T(i)
\]
Backward algorithm
Define backward varible:
\[
\begin{split}
\beta_t(i) &= P(V^T/V^t|w(t)=w_i,\lambda)\\
& = P(v(t+1),v(t+2),...,v(T)|w(t)=w_i,\lambda)
\end{split}
\]
Then,
\[
\begin{split}
\beta_T(i) = 1
\end{split}
\]
\[
\begin{split}
\beta_i(t) &= \sum_{j=1}^{n}a_{ij}b_{jk}(t+1)\beta_j(t+1)\\
1\leq &t\leq T-1, \qquad 1\leq i\leq n
\end{split}
\]
\[
\begin{split}
P(\mathbf{V^T}|\mathbf{\lambda}) & = \sum_{i=1}^{n}\pi_i b_{ik}(1) \beta_1(i)\\
& = \sum_{i=1}^{n}\alpha_1(i) \beta_1(i)
\end{split}
\]
The complexity of both forward and backward algorithm is $O(Tn^2)$, rather than the direct algorithm‘s complexity $O(Tn^T)$ !
Some probability calculation
① Given $\lambda$ and observation $V^T$, the probability of $w(t)=w_i$ is :
\[
\begin{split}
\gamma_t(i) & = P(w(t)=w_i|V^T,\lambda)\\
& = \frac{P(w(t)=w_i,V^T|\lambda)}{P(V^T|\lambda)}\\
& = \frac{\alpha_t(i)\beta_t(i)}{P(V^T|\lambda)}\\
& = \frac{\alpha_t(i)\beta_t(i)}{\sum_{i=1}^{n}\alpha_t(i)\beta_t(i)}
\end{split}
\]
② Given $\lambda$ and observation $V^T$, the probability of $w(t)=w_i\cap w(t+1)=w_j$:
\[
\begin{split}
\xi_t(i,j) &= P(w(t)=w_i,w(t+1)=w_j|V^T,\lambda)\\
& = \frac{P(w(t)=w_i,w(t+1)=w_j,V^T|\lambda)}{P(V^T|\lambda)}\\
& = \frac{P(w(t)=w_i,w(t+1)=w_j,V^T|\lambda)}{\sum_{i=1}^{n}\sum_{j=1}^{n}P(w(t)=w_i,w(t+1)=w_j,V^T|\lambda)}\\
& = \frac{\alpha_t(i)a_{ij}b_{jk}(t+1)\beta_{t+1}(j)}{\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_t(i)a_{ij}b_{jk}(t+1)\beta_{t+1}(j)}
\end{split}
\]
③ We can get some expectation value.
(1) The expectation value of state i in condition of observation $V^T$:
\[
\sum_{t=1}^{T} = \gamma_t(i)
\]
(2) The expectation value of the times that transform from state i to the another state:
\[
\sum_{t=1}^{T-1} = \gamma_t(i)
\]
(3) The expectation value of the times that transform from state i to state j:
\[
\sum_{t=1}^{T-1} = \xi_t(i,j)
\]
Learning problem
Learning problem : Given the observation $V^T$(obvious variable) without $w^T$(hidden variable), compute the parameters of $\lambda = (A,B,\pi)$.
The parameter learining can be completed by EM algorithm(Also called Baum-Welch algorithm).
The log likehood function of the hidden variable and obvious variable:
\[
L = \log P(V^T,w^T|\lambda)
\]
The Q function can be writen as:
\[
\begin{split}
Q(\lambda,\lambda^0) &=\sum_{w^T}\log P(V^T,w^T|\lambda) P(w^T|V^T,\lambda^0) \\
& = \sum_{w^T}\log P(V^T,w^T|\lambda) \frac{P(V^T,w^T|\lambda^0)}{P(V^T|\lambda^0)}
\end{split}
\]
Thus,
\[
\begin{split}
\lambda &= arg\max\limits_{\lambda}Q(\lambda,\lambda^0)\\
& = arg\max\limits_{\lambda} \sum_{w^T}\log P(V^T,w^T|\lambda) \frac{P(V^T,w^T|\lambda^0)}{P(V^T|\lambda^0)}\\
& = arg\max\limits_{\lambda} \sum_{w^T}\log P(V^T,w^T|\lambda) P(V^T,w^T|\lambda^0)
\end{split}
\]
Finnaly, we can get:
\[
\begin{split}
a_{ij} &= \frac{\sum_{t=1}^{T-1}\xi_t(i,j)}{\sum_{t=1}^{T-1}\gamma_t(i)}\\
b_{jk}(t) &= \frac{\sum_{t=1,v(t)=v_k}^{T}\gamma_t(j)}{\sum_{t=1}^{T}\gamma_t(j)}\\
\pi_i & = \gamma_1(i)
\end{split}
\]
Thus, Baum-Welch algorithm can be described as follows:
① Initialize. In terms of n=0, choose $a_{ij}^{(0)}$, $b_{jk}^{(0)}(t)$ and $\pi_i^{(0)}$, and get $\lambda{(0)}=(A^{(0)},B^{(0)},\pi^{(0)})$
② Recursion. for n=1,2,...
\[
\begin{split}
a_{ij}^{(n+1)} &= \frac{\sum_{t=1}^{T-1}\xi_t(i,j)}{\sum_{t=1}^{T-1}\gamma_t(i)}\\
b_{jk}^{(n+1)}(t) &= \frac{\sum_{t=1,v(t)=v_k}^{T}\gamma_t(j)}{\sum_{t=1}^{T}\gamma_t(j)}\\
\pi_i^{(n+1)} & = \gamma_1(i)
\end{split}
\]
③ If satisfy the termination condition, algorithm done. Otherwise, go to \ding{173}.
Decoding problem
Decoding problem: Given the observation $V^T$ and model parameters $\lambda = (A,B,\pi)$, calculate the most likely hidden state $w^T$.
To solve decoding problem, we can use Viterbi algorithm, which is a dynamic programming.
Define variable:
\[
\delta_t(i) = \max\limits_{w(1),...,w(t-1)}P(w(t)=w_i,w(t-1),...,w(1),v(t),...,v(1)|\lambda), \qquad i = 1,2,...,n
\]
then,
\[
\delta_{t+1}(j) = \left[\max\limits_{i}\delta_t(i)a_{ij}\right]b_{jk}(t+1)
\]
Define variable:
\[
\psi_t(i) = \arg\max\limits_{1\leq j\leq}[\delta_{t-1}(j)a_{ji}]
\]
So, Vaterbi algorithm can be described as follows:
① Initialization.
\[
\begin{split}
\delta_1(i) &= P(w(1)=w_i,v(1)|\lambda)\\
& = P(v(1)=v_k|w(1)=w_i)P(w(1)=w_i)\\
& = \pi_i b_{ik}(1), \qquad i=1,2,...,n\\
\psi_1(i) &= 0, \qquad i=1,2,...,n
\end{split}
\]
② Recursion. For t=2,3,...,T
\[
\begin{split}
\delta_{t}(i) &= \left[\max\limits_{j}\delta_{t-1}(j)a_{ji}\right]b_{ik}(t), \qquad i=1,2,...,n\\
\psi_t(i) & = \max\limits_{j}\delta_{t-1}(j)a_{ji}, \qquad i=1,2,...,n
\end{split}
\]
③ Termination.
\[
\begin{split}
P^* &= \max\limits_{i}\delta_T(i)\\
i_T^* & = arg\max\limits_{i}\delta_T(i)
\end{split}
\]
④ Backtracking. For t=T-1,...,1
\[
i_t^* = \psi_{t+1}(i_t+1^*)
\]
Finally, we can get the optimal path $I^*=(i^*_1,i^*_2,...,i^*_T)$
Words in the end
At the begining of the new term, I complete this eassy eventually. The purpose that I write this $‘Hidden\quad Markov\quad Model‘$ is to have an understanding of HMM and practice my English writing skill, which is kind of disfluency...I‘m sorry :( ...
The are also some quension after I complete this eassy:
① The relationship between HMM and PGM(Probabilistic Graphical Model). There are some probability calculation in forward algorithm and backward algorithm and they are quite related to PGM.
② In Baum-Welch algorithm, how to use Lagrange method to find the solution?
③ In HMM learning problem, how to deal with multiple sequence when trainning the model?
Reference
[1] 李航.统计学习方法[M].北京:清华大学出版社.
[2] LAWRENCE R. RABINER.A Tutorial on Hidden Markov Models and
Selected Applications in Speech Recognition[J].PROCEEDINGS OF THE IEEE,FEBRUARY 1989
[3] 刘成林.中国科学院大学《模式识别》PPT
标签：tor block word each tip 清华 学习 note bin
原文地址：http://www.cnblogs.com/ghmgm/p/6420099.html
