# 91 q proof the quantity argmaxrq mtt1 r q ri

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Unformatted text preview: ∗ ∗ Then we can ﬁnd a state value qt+1 (1) at t + 1 such that (qt (1), qt+1 (1)) is a pair of state values of a Viterbi path using the following procedure: ∗ ∗ qt+1 (1) ∈ argmax mt,t+1 (qt (1), q ). (9.1) q Proof. ∗∗ The quantity argmaxr,q mt,t+1 (r, q ) = {(ri , qi )}i is a set of pairs, each of which is compatible with some Viterbi path. Moreover, any Viterbi path must, at times t and t + 1, have value corresponding to one of the ∗ ∗ pairs. Therefore, there is some i such that qt (1) = ri , and the argmax in ∗ Equation (9.1) chooses the corresponding qi which hence is compatible with some Viterbi path. Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lecture 9 - Feb 6th, 2013 page 9-11 (of 180) k-best without the penalty Island Summary Scratch Computing Viterbi from max-margin - any case We start this process with the singleton max-marginal on the left ∗ q1 (1) ∈ argmax m1 (q ), (9.2) q and then repeat the following recursion, for t = 2 . . . T , as follows ∗ ∗ qt (1) ∈ argmax mt−1,t (qt−1 (1), q ), (9.3) q which, thanks to to Proposition 9.3.1, is guaranteed to be a Viterbi path. Prof. Jeﬀ Bilmes EE596A/Winter 2013/DGMs – Lec...
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## This document was uploaded on 04/05/2014.

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