HMM automatic vowel detection

[MarcoP]

Marco,

from the figures you have shown, with the white and red characters against a black background, I conclude that the algorithm really goes ahead by assigning each character in the text to one of the possible states.
When doing this, it tries to maximise the consistency of having each character always being generated by the same state, but also to alternate as much as possible.

Hi Rene,
the red/white plots are something I produce on the basis of the B matrix.

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This is the B matrix for Koen's example:
   state:1  0.00088 0.31915
 * state:0  0.00040 0.00000
 2 state:0  0.00403 0.00000
 F state:1  0.00000 0.00031
 S state:1  0.00000 0.00031
 T state:0  0.00040 0.00000
 Z state:0  0.00081 0.00000
 a state:1  0.03163 0.19249
 b state:0  0.00685 0.00000
 c state:0  0.00103 0.00013
 d state:0  0.04113 0.00000
 e state:1  0.00980 0.03370
 f state:0  0.00250 0.00117
 g state:0  0.04032 0.00000
 h state:1  0.00000 0.00092
 i state:1  0.00000 0.21475
 k state:1  0.01406 0.01693
 l state:0  0.01935 0.00000
 m state:0  0.06855 0.00000
 n state:0  0.11573 0.00000
 o state:1  0.00000 0.18095
 p state:0  0.00901 0.00205
 q state:1  0.00672 0.02806
 r state:0  0.26380 0.00054
 s state:0  0.25686 0.00000
 t state:0  0.06613 0.00000
 u state:1  0.00572 0.00854
 z state:0  0.03427 0.00000

I assign the state:0/1 information on the basis of the probability for each character to be higher in state 0 or 1.
For instance:
 a state:1  0.03163 0.19249
0.03163<0.19249 I count 'a' as state:1

 b state:0  0.00685 0.00000
0.00685>0.00000 I count 'b' as state:0


I highlighted in red all the characters for which P0(char)<P1(char).
From the B graphical representation, it should be clear that in most cases the model has a clear preference for generating each character in one of the two states. But in this example, 'k' results ambiguous:
 k state:1  0.01406 0.01693
(the probabilities for state:0 and state:1 are very close).