Why does $P(Q_t = q | X_{0:L} = i_{0:L}) = P(Q_t = q, X_{0:L} = i_{0:L})$?
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This is a derivation of an equation used to maximize the posterior probability that $Q_m = i_m$ given a model and a sequence of observations.
$Q_m$ is a RV which maps to some $q in S$, the state space.
$X_m$ is a RV which maps to some $i in Sigma$, the emission space.
$Theta$ parameterizes the HMM.
probability probability-theory markov-chains markov-process
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
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add a comment |
$begingroup$
This is a derivation of an equation used to maximize the posterior probability that $Q_m = i_m$ given a model and a sequence of observations.
$Q_m$ is a RV which maps to some $q in S$, the state space.
$X_m$ is a RV which maps to some $i in Sigma$, the emission space.
$Theta$ parameterizes the HMM.
probability probability-theory markov-chains markov-process
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
$endgroup$
add a comment |
$begingroup$
This is a derivation of an equation used to maximize the posterior probability that $Q_m = i_m$ given a model and a sequence of observations.
$Q_m$ is a RV which maps to some $q in S$, the state space.
$X_m$ is a RV which maps to some $i in Sigma$, the emission space.
$Theta$ parameterizes the HMM.
probability probability-theory markov-chains markov-process
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
$endgroup$
This is a derivation of an equation used to maximize the posterior probability that $Q_m = i_m$ given a model and a sequence of observations.
$Q_m$ is a RV which maps to some $q in S$, the state space.
$X_m$ is a RV which maps to some $i in Sigma$, the emission space.
$Theta$ parameterizes the HMM.
probability probability-theory markov-chains markov-process
probability probability-theory markov-chains markov-process
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
asked Dec 9 '18 at 21:48
greedIsGoodAhagreedIsGoodAha
676
676
add a comment |
add a comment |
1 Answer
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$begingroup$
The text has an error.
It should read,
$P_theta(Q_t = k | X_{0:L} = i_{0:L}) = frac{ P_theta(Q_t, X_{0:L} = i_{0:L})} {P_theta(X_{0:L} = i_{0:L})}$
that is, the probability that the $t$th state is $k$ given the observation set is the probability that the $t$th state is $k$ and the observation set is observed, dividing out the probability that the observation set is observed. This follows from Bayes' Law.
The probability of observing the observation set is independent from a change in $q$, so in the context of finding the argmax of $Q_t = q in S$ the term can be discarded.
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
$endgroup$
add a comment |
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$begingroup$
The text has an error.
It should read,
$P_theta(Q_t = k | X_{0:L} = i_{0:L}) = frac{ P_theta(Q_t, X_{0:L} = i_{0:L})} {P_theta(X_{0:L} = i_{0:L})}$
that is, the probability that the $t$th state is $k$ given the observation set is the probability that the $t$th state is $k$ and the observation set is observed, dividing out the probability that the observation set is observed. This follows from Bayes' Law.
The probability of observing the observation set is independent from a change in $q$, so in the context of finding the argmax of $Q_t = q in S$ the term can be discarded.
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
$endgroup$
add a comment |
$begingroup$
The text has an error.
It should read,
$P_theta(Q_t = k | X_{0:L} = i_{0:L}) = frac{ P_theta(Q_t, X_{0:L} = i_{0:L})} {P_theta(X_{0:L} = i_{0:L})}$
that is, the probability that the $t$th state is $k$ given the observation set is the probability that the $t$th state is $k$ and the observation set is observed, dividing out the probability that the observation set is observed. This follows from Bayes' Law.
The probability of observing the observation set is independent from a change in $q$, so in the context of finding the argmax of $Q_t = q in S$ the term can be discarded.
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
$endgroup$
add a comment |
$begingroup$
The text has an error.
It should read,
$P_theta(Q_t = k | X_{0:L} = i_{0:L}) = frac{ P_theta(Q_t, X_{0:L} = i_{0:L})} {P_theta(X_{0:L} = i_{0:L})}$
that is, the probability that the $t$th state is $k$ given the observation set is the probability that the $t$th state is $k$ and the observation set is observed, dividing out the probability that the observation set is observed. This follows from Bayes' Law.
The probability of observing the observation set is independent from a change in $q$, so in the context of finding the argmax of $Q_t = q in S$ the term can be discarded.
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
$endgroup$
The text has an error.
It should read,
$P_theta(Q_t = k | X_{0:L} = i_{0:L}) = frac{ P_theta(Q_t, X_{0:L} = i_{0:L})} {P_theta(X_{0:L} = i_{0:L})}$
that is, the probability that the $t$th state is $k$ given the observation set is the probability that the $t$th state is $k$ and the observation set is observed, dividing out the probability that the observation set is observed. This follows from Bayes' Law.
The probability of observing the observation set is independent from a change in $q$, so in the context of finding the argmax of $Q_t = q in S$ the term can be discarded.
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
answered Dec 9 '18 at 23:42
greedIsGoodAhagreedIsGoodAha
676
676
add a comment |
add a comment |
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