A probabilistic model of a sequence in which hidden states form a Markov chain (each state depends only on its predecessor) and each hidden state emits one observable, given by initial, transition and emission probabilities.
Markov property:hi depends only on hi−1, not on earlier states or observations. The hidden states form a bigram model (N-Gram Model with n=2) over H.
Output independence:oi depends only on hi.
With an END state. For POS-Tagging a special hidden state END is appended after the last position, with learned transition probabilities Pr(END∣h). The likelihood gets the extra factor Pr(END∣hn), which encodes which states are likely at the end of a sequence (exam WS 19/20, Task 5.4).
Decoding. Given o1,…,on, the most likely hidden sequence is
argh1,…,hnmaxPr(h1,…,hn,o1,…,on),
which the Viterbi Algorithm computes in O(n⋅∣H∣2) instead of trying all ∣H∣n sequences. Maximizing the joint probability is the same as maximizing Pr(h1,…,hn∣o1,…,on), because Pr(o1,…,on) does not depend on the hi.
Training for POS tagging on a tagged corpus by relative frequencies: Pr(t′∣t)=#(tt′)/#(t) (tag t′ follows tag t), Pr(w∣t)=#(w tagged t)/#(t), Pr(t)= fraction of sentences that start with t.
Posterior of a single state.Pr(hi=h∣o1,…,on)=∑sequences with hi=hPr(h,o)/∑all sequencesPr(h,o). For short sequences this is a direct sum over all ∣H∣n hidden sequences (exam WS 17/18, Task 6.1).
Zero probabilities are decisive. If an observable can only be emitted by one state (emission probability 0 for all others), every positive-probability sequence must use that state at that position. This can determine the most likely sequence without any computation (exam WS 17/18, Task 6.2).
Brute force over all ∣H∣n sequences with the likelihood formula is correct but exponential in n (exam WS 17/18, Task 6.4).