Unable to understand this expression
I am unable to understand the above expression from the 3rd line onwards. Why is P(A|q1...qt-1)P(q1..q-1) = P(A|qt-1)P(q1..q-1)
I am not clear of the subsequent line as well
bayes-theorem
add a comment |
I am unable to understand the above expression from the 3rd line onwards. Why is P(A|q1...qt-1)P(q1..q-1) = P(A|qt-1)P(q1..q-1)
I am not clear of the subsequent line as well
bayes-theorem
Is this from a Markov process?
– gd1035
Nov 29 '18 at 21:59
I think the third line is a typo. I think it shoulbe $P(A|q_{t-1})P(q_1, ... , q_{t-2}) = P(A|q_{t-1})P(q_{t-1}|q_{t-2})P(q_1,....,q_{t-e}) = ..... P(A|q_{t-1})P(q_{t-1}|q_{t-2})....P(q_2|q_1)P(q_1)$.
– fleablood
Nov 29 '18 at 22:02
yes its about Markov process
– user1436508
Nov 29 '18 at 22:15
add a comment |
I am unable to understand the above expression from the 3rd line onwards. Why is P(A|q1...qt-1)P(q1..q-1) = P(A|qt-1)P(q1..q-1)
I am not clear of the subsequent line as well
bayes-theorem
I am unable to understand the above expression from the 3rd line onwards. Why is P(A|q1...qt-1)P(q1..q-1) = P(A|qt-1)P(q1..q-1)
I am not clear of the subsequent line as well
bayes-theorem
bayes-theorem
asked Nov 29 '18 at 21:54
user1436508user1436508
1538
1538
Is this from a Markov process?
– gd1035
Nov 29 '18 at 21:59
I think the third line is a typo. I think it shoulbe $P(A|q_{t-1})P(q_1, ... , q_{t-2}) = P(A|q_{t-1})P(q_{t-1}|q_{t-2})P(q_1,....,q_{t-e}) = ..... P(A|q_{t-1})P(q_{t-1}|q_{t-2})....P(q_2|q_1)P(q_1)$.
– fleablood
Nov 29 '18 at 22:02
yes its about Markov process
– user1436508
Nov 29 '18 at 22:15
add a comment |
Is this from a Markov process?
– gd1035
Nov 29 '18 at 21:59
I think the third line is a typo. I think it shoulbe $P(A|q_{t-1})P(q_1, ... , q_{t-2}) = P(A|q_{t-1})P(q_{t-1}|q_{t-2})P(q_1,....,q_{t-e}) = ..... P(A|q_{t-1})P(q_{t-1}|q_{t-2})....P(q_2|q_1)P(q_1)$.
– fleablood
Nov 29 '18 at 22:02
yes its about Markov process
– user1436508
Nov 29 '18 at 22:15
Is this from a Markov process?
– gd1035
Nov 29 '18 at 21:59
Is this from a Markov process?
– gd1035
Nov 29 '18 at 21:59
I think the third line is a typo. I think it shoulbe $P(A|q_{t-1})P(q_1, ... , q_{t-2}) = P(A|q_{t-1})P(q_{t-1}|q_{t-2})P(q_1,....,q_{t-e}) = ..... P(A|q_{t-1})P(q_{t-1}|q_{t-2})....P(q_2|q_1)P(q_1)$.
– fleablood
Nov 29 '18 at 22:02
I think the third line is a typo. I think it shoulbe $P(A|q_{t-1})P(q_1, ... , q_{t-2}) = P(A|q_{t-1})P(q_{t-1}|q_{t-2})P(q_1,....,q_{t-e}) = ..... P(A|q_{t-1})P(q_{t-1}|q_{t-2})....P(q_2|q_1)P(q_1)$.
– fleablood
Nov 29 '18 at 22:02
yes its about Markov process
– user1436508
Nov 29 '18 at 22:15
yes its about Markov process
– user1436508
Nov 29 '18 at 22:15
add a comment |
1 Answer
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This follows from the Markov property (or the memoryless property) that the conditional probability of a future event only depends on the current state.
See link for more information about this property.
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This follows from the Markov property (or the memoryless property) that the conditional probability of a future event only depends on the current state.
See link for more information about this property.
add a comment |
This follows from the Markov property (or the memoryless property) that the conditional probability of a future event only depends on the current state.
See link for more information about this property.
add a comment |
This follows from the Markov property (or the memoryless property) that the conditional probability of a future event only depends on the current state.
See link for more information about this property.
This follows from the Markov property (or the memoryless property) that the conditional probability of a future event only depends on the current state.
See link for more information about this property.
answered Nov 29 '18 at 23:12
gd1035gd1035
4591210
4591210
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Is this from a Markov process?
– gd1035
Nov 29 '18 at 21:59
I think the third line is a typo. I think it shoulbe $P(A|q_{t-1})P(q_1, ... , q_{t-2}) = P(A|q_{t-1})P(q_{t-1}|q_{t-2})P(q_1,....,q_{t-e}) = ..... P(A|q_{t-1})P(q_{t-1}|q_{t-2})....P(q_2|q_1)P(q_1)$.
– fleablood
Nov 29 '18 at 22:02
yes its about Markov process
– user1436508
Nov 29 '18 at 22:15