Notes on computational linguistics.pdf - UCLA Department of ...

Stabler - Lx 185/209 2003
octave:21> gplot [1:10] result title "x",\
> result using 1:3 title "y", result using 1:4 title "z"
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
path of a Markov chain
0
1 2 3 4 5 6 7 8 9 10
(67) Notice that the initial distribution and transition matrix can be represented by a finite state machine
with no vocabulary and no final states:
0.7
0.2
0.2
0.7
s
1
0.2
s
2
0.1
s
3
0.1
(68) Notice that no Markov chain can be such that after a sequence of states acccc there is a probability of
0.8 that the next symbol will be an a, thatis,
P(a|acccc) = 0.8
when it is also the case that
P(b|bcccc) = 0.8
This follows from the requirement mentioned in (61) that in each row i, the sum of the transition
probabilities from that state
qj∈ΩX P(qj|qi) = 1, and so we cannot have both P(b|c) = 0.8andP(a|c) =
0.8.
(69) Chomsky (1963, p337) observes that the Markovian property that we see in state sequences does not
always hold in regular languages. For example, the following finite state machine, to which we have
141
0.7
1
0.1
x
y
z