1. CS621: Artificial Intelligence
Pushpak Bhattacharyya CSE Dept., IIT Bombay
Lecture 38-39: Baum Welch Algorithm; HMM training

2. Baum Welch algorithm
Training Hidden Markov Model (not structure learning, i.e., the structure of the HMM is pre-given). This involves:
Learning probability values ONLY
Correspondence with PCFG:
Not learning production rule but probabilities associated with them
Training algorithm for PCFG is called Inside-Outside algorithm

3. Key Intuition a b q r
Given: Training sequence Initialization: Probability values Compute: Pr (state seq | training seq) get expected count of transition compute rule probabilities Approach: Initialize the probabilities and recompute them… EM like approach

4. Building blocks: Probabilities to be usedW1
W2…………… Wn-1 Wn S1 S2 Sn
Sn+1

5. Probabilities to be used, contd…
Exercise 1:- Prove the following:

6. Start of baum-welch algorithmq a a
String = aab aaa aab aaa Sequence of states with respect to input symbols
o/p seq
State seq

7. Calculating probabilities from table
Table of counts
T=#states
A=#alphabet symbols
Now if we have a non-deterministic transitions then multiple state seq possible for the given o/p seq (ref. to previous slide’s feature). Our aim is to find expected count through this.
Src
Dest
O/P
Count q r a 5 b 3 2

8. Interplay Between Two Equationswk
No. of times the transitions sisj occurs in the string

9. Learning probabilitiesq r
a:0.16
b:1.0
Actual (Desired) HMM
a:0.4
b:0.48 q r
a:0.48
b:1.0
Initial guess

10. One run of Baum-Welch algorithm: string ababa
P(path) q r
0.0
0.000
Rounded Total 
0.035
0.01
0.06
0.095
New Probabilities (P) 
(0.01/( )
1.0
0.36
0.581
State sequences
* is considered as starting and ending symbol of the input sequence string
This way through multiple iterations the probability values will converge.

11. Appling Naïve Bayes
Hence multiplying the transition probabilities is valid

12. Discussions Symmetry breaking:
Example: Symmetry breaking leads to no change in initial values
Struck in Local maxima
Label bias problem
Probabilities have to sum to 1.
Values can rise at the cost of fall of values for others.
a:0.5
b:0.25 s
a:0.25 s s s
b:1.0
a:0.5
a:0.5
a:1.0
b:0.5
b:0.5
a:0.25 s
b:0.5 s
b:0.5
Desired
Initialized

13. Computational part
Exercise 2: What is the complexity of calculating the above expression?
Hint: To find this first solve Exercise 1 i.e. understand how probability of given string can be represented as