lecture20-hmm.md

Lecture 20 Hidden Markov Models (Part II)

High-order HMMs

• 1st-order HMM (i.e. bigram HMM)
• 2nd-order HMM (i.e. trigram HMM)
• 3rd-order HMM
• Hidden States - y
  • The joint probability of the observations and the hidden states in an HMM is given by $P(X=x,Y=y) = C_{y_1}[\prod_{t=1}^T A_{y_t,x_t}][\prod_{t=1}^{T-1} B_{y_t,y_{t+1}}]$
• Observations - x
  • The probability of the observations (marginal probability) in an HMM is given by $P(X=x) = \sum_{y \in Y} p(x=\vec{x},y=\vec{y})$

3 Problems for an HMM

Evaluation

• Compute the probability of a given sequence of observations
• $p(\vec{x}) = \sum_{\vec{y} \in Y_{\vec{x}}} p(\vec{x},\vec{y})$

Viterbi Decoding

• Find the most likely sequence of hidden states, given a sequence of observations
• $\hat{y} = argmax_{\vec{y} \in Y_{\vec{x}}} p(\vec{y}|\vec{x})$

Marginals

• Compute the marginal distribution for a hidden state, given a sequence of observations
• $p(y_t=k|\vec{x}) = \sum_{\vec{y}=y_\vec{x} s.t. y_t=k} p(\vec{y}|\vec{x})$

Examples of 3 Problems

y1	→	y2	→	y3
↓		↓		↓
x1		x2		x3

1. Evaluation: $p(x_1,x_2,x_3)=\sum_{y_1} \sum_{y_2} \sum_{y=3} p(x_1,x_2,x_3,y_1,y_2,y_3)$
2. Viterbi Decoding: $\hat{y_1}, \hat{y_2}, \hat{y_3} = argmax_{\vec{y}} p(y_1,y_2,y_3|x_1,x_2,x_3)$
3. Marginals: $p(y_2=V|x_1,x_2,x_3) = \sum_{y_1} \sum_{y_3} p(y_1,y_2,y_3|x_1,x_2,x_3)$
4. Joint distribution: $p(x_1,x_2,x_3,y_1,y_2,y_3) = p(y_1)p(x_1|y_1)p(x_2|y_2)p(x_3|y_3)p(y_2|y_1)p(y_3|y_2)$
5. $p(\vec{y}|\vec{x}) = \frac{p(\vec{x},\vec{y})}{p(\vec{x})}$
6. For $|\vec{y}|=T$ and $y_t \in {1,\cdots,K}$ there are $K^T$ possible values of $\vec{y}$

Forward-Backward Algorithm

Dataset for Supervised Part-of-Speech (POS) Tagging

Sample1	n	v	p	d	n	$y^{(1)}$
	time	flies	like	an	arrow	$x^{(1)}$
Sample2	n	n	v	d	n	$y^{(2)}$
	time	flies	like	an	arrow	$x^{(2)}$

Brute-Force Algorithm

def eval(vec_x):
    p_x = 0                # p(vec_c)
    for y in all_y(vec_x): # y_x
        p_x += joint(x, y) # p(x,y)
    return p_x

Forward-Backward Algorithm

Forward-Backward Search Space

• Key Idea: the current tag holds all the information for the next tag scoring
• Viterbi Algorithm: Most Probable Assignment
  • Numbers associated with edges and nodes of path
  • Most probable assignment = path with highest product
  • p(v a n) = (1/Z) * product weight of one path
• Marginal probability $p(Y_2=a|X)$ = (1/Z) * total weight of all paths through a
• Find Marginals
  • $\alpha_2(n)$ = total weight of these path prefixes
  • $\beta_2(n)$ = total weight of these path suffixes
  • found by dynamic programming: matrix-vector products
  • Product of $\alpha_2(n)$ and $\beta_2(n)$ gives $ax+ay+az+bx+by+bz+cx+cy+cz$ = total weight of paths, which is more efficient
  • We also need $A(pref., n)$, the opinion of the emission probability at this variable
  • total weight of all paths through n = $\alpha_2(n) A(pref.,n) \beta_2(n)$
  • sum = Z (total weight of all paths) => $p(\vec{x}) = \sum_{\vec{y}} p(\vec{x},\vec{y})$
  • Divided by Z to get marginal probability $p(y_i|\vec{x})$

Forward-Backward Algorithm

• Define:
  • $\alpha_t(k) \simeq p(x_1,\cdots,x_t,y_t=k)$
  • $\beta_t(k) \simeq p(x_{t+1},\cdots,x_T,y_t=k)$
• Assume:
  • $y_0 = START$
  • $y_{T+1} = END$
• Initialize
  • $\alpha_0(START) = 1$ $\alpha_0(k) = 0, \forall k \neq START$
  • $\beta_0(END) = 1$ $\beta_{T+1}(k) = 0, \forall k \neq END$
• (Forward Algo.) For $t = 1,\cdots,T$
  • For $k = 1,\cdots,K$
    • $\alpha_t(k) = p(x_t|y_t=k) \sum_{j=1}^K \alpha_{t-1}(j)p(y_t=k|y_{t-1}=j)$
    • the alphas include the emission probabilities so that we don't multiply them in separately
• (Backward Algo.) For $t = T,\cdots,1$
  • For $k = 1,\cdots,K$
    • $\beta_t(k) = \sum_{j=1}^K p(x_{t+1}|y_{t+1}=j) \beta_{t+1}(j) p(y_{t+1}=j|y_t=k)$
• Compute evaluation $p(\vec{x}) = \alpha_{T+1}(END)$
• Compute marginals $p(y_t=k|\vec{x}) = \frac{\alpha_t(k)\beta_t(k)}{p(\vec{x})}$