Picking path at random in DAG graph with probability equals to path weight.
$begingroup$
I'm refering to the following paper:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.329.3653&rep=rep1&type=pdf.
There is a lemma states:
Let $G$ be a directed acyclic graph with source $s$ and sink $t$. Let
each edge $e = (v_i, v_j)$ of $G$ have weight $w(e)$ (we also use the
notation $w(e) = w(v_i, v_j)$), and each node have weight $w(vi)$.
Assume without loss of generality that $$sum_{all paths P=(v_1,...,v_{k(P)}) }{ w(v_1)prod_{i=2}^{k(P)}{w(v_{i−1}, v_i)w(v_i) = 1} }$$
It is possible to pick at random a path $P$ consisting of $s = v_1, v_2,...,v_{k(P)} = t$ in time $O(n)$ such that $$Pr(picking P) = w(s)prod_{i=2}^{k(P)} {w(v_{i−1}, v_i)w(v_i) }$$
The aim of the lemma is give a method for sampling in a HMM graph. The proof is by induction on the maximal lenght of a path between s and t.
My questions is:
Why assume that all path weight sum up to 1? and why without loss of generality?
I have thinked that only paths between $s$ and $t$ should sum up to 1. I guess that is related to the proof that is based on maximal leght of a path between $s$ and $t$ but how?
probability graph-theory markov-process sampling-theory hidden-markov-models
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
$endgroup$
add a comment |
$begingroup$
I'm refering to the following paper:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.329.3653&rep=rep1&type=pdf.
There is a lemma states:
Let $G$ be a directed acyclic graph with source $s$ and sink $t$. Let
each edge $e = (v_i, v_j)$ of $G$ have weight $w(e)$ (we also use the
notation $w(e) = w(v_i, v_j)$), and each node have weight $w(vi)$.
Assume without loss of generality that $$sum_{all paths P=(v_1,...,v_{k(P)}) }{ w(v_1)prod_{i=2}^{k(P)}{w(v_{i−1}, v_i)w(v_i) = 1} }$$
It is possible to pick at random a path $P$ consisting of $s = v_1, v_2,...,v_{k(P)} = t$ in time $O(n)$ such that $$Pr(picking P) = w(s)prod_{i=2}^{k(P)} {w(v_{i−1}, v_i)w(v_i) }$$
The aim of the lemma is give a method for sampling in a HMM graph. The proof is by induction on the maximal lenght of a path between s and t.
My questions is:
Why assume that all path weight sum up to 1? and why without loss of generality?
I have thinked that only paths between $s$ and $t$ should sum up to 1. I guess that is related to the proof that is based on maximal leght of a path between $s$ and $t$ but how?
probability graph-theory markov-process sampling-theory hidden-markov-models
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
$endgroup$
$begingroup$
I am not too familiar with graph theory, so only posting this as a comment, but since the $w(v_i)$ are weights, iperhaps they are assuming that an appropriate 'scale-factor' can always be found so that the sum can be assumed to have any value and $1$ just happened to be convenient.
$endgroup$
– Devashish Kaushik
Dec 31 '18 at 11:39
$begingroup$
yes, but why sum for all paths?
$endgroup$
– Pablo Escobar
Dec 31 '18 at 12:24
add a comment |
$begingroup$
I'm refering to the following paper:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.329.3653&rep=rep1&type=pdf.
There is a lemma states:
Let $G$ be a directed acyclic graph with source $s$ and sink $t$. Let
each edge $e = (v_i, v_j)$ of $G$ have weight $w(e)$ (we also use the
notation $w(e) = w(v_i, v_j)$), and each node have weight $w(vi)$.
Assume without loss of generality that $$sum_{all paths P=(v_1,...,v_{k(P)}) }{ w(v_1)prod_{i=2}^{k(P)}{w(v_{i−1}, v_i)w(v_i) = 1} }$$
It is possible to pick at random a path $P$ consisting of $s = v_1, v_2,...,v_{k(P)} = t$ in time $O(n)$ such that $$Pr(picking P) = w(s)prod_{i=2}^{k(P)} {w(v_{i−1}, v_i)w(v_i) }$$
The aim of the lemma is give a method for sampling in a HMM graph. The proof is by induction on the maximal lenght of a path between s and t.
My questions is:
Why assume that all path weight sum up to 1? and why without loss of generality?
I have thinked that only paths between $s$ and $t$ should sum up to 1. I guess that is related to the proof that is based on maximal leght of a path between $s$ and $t$ but how?
probability graph-theory markov-process sampling-theory hidden-markov-models
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
$endgroup$
I'm refering to the following paper:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.329.3653&rep=rep1&type=pdf.
There is a lemma states:
Let $G$ be a directed acyclic graph with source $s$ and sink $t$. Let
each edge $e = (v_i, v_j)$ of $G$ have weight $w(e)$ (we also use the
notation $w(e) = w(v_i, v_j)$), and each node have weight $w(vi)$.
Assume without loss of generality that $$sum_{all paths P=(v_1,...,v_{k(P)}) }{ w(v_1)prod_{i=2}^{k(P)}{w(v_{i−1}, v_i)w(v_i) = 1} }$$
It is possible to pick at random a path $P$ consisting of $s = v_1, v_2,...,v_{k(P)} = t$ in time $O(n)$ such that $$Pr(picking P) = w(s)prod_{i=2}^{k(P)} {w(v_{i−1}, v_i)w(v_i) }$$
The aim of the lemma is give a method for sampling in a HMM graph. The proof is by induction on the maximal lenght of a path between s and t.
My questions is:
Why assume that all path weight sum up to 1? and why without loss of generality?
I have thinked that only paths between $s$ and $t$ should sum up to 1. I guess that is related to the proof that is based on maximal leght of a path between $s$ and $t$ but how?
probability graph-theory markov-process sampling-theory hidden-markov-models
probability graph-theory markov-process sampling-theory hidden-markov-models
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
asked Dec 31 '18 at 10:24
Pablo EscobarPablo Escobar
1
1
$begingroup$
I am not too familiar with graph theory, so only posting this as a comment, but since the $w(v_i)$ are weights, iperhaps they are assuming that an appropriate 'scale-factor' can always be found so that the sum can be assumed to have any value and $1$ just happened to be convenient.
$endgroup$
– Devashish Kaushik
Dec 31 '18 at 11:39
$begingroup$
yes, but why sum for all paths?
$endgroup$
– Pablo Escobar
Dec 31 '18 at 12:24
add a comment |
$begingroup$
I am not too familiar with graph theory, so only posting this as a comment, but since the $w(v_i)$ are weights, iperhaps they are assuming that an appropriate 'scale-factor' can always be found so that the sum can be assumed to have any value and $1$ just happened to be convenient.
$endgroup$
– Devashish Kaushik
Dec 31 '18 at 11:39
$begingroup$
yes, but why sum for all paths?
$endgroup$
– Pablo Escobar
Dec 31 '18 at 12:24
$begingroup$
I am not too familiar with graph theory, so only posting this as a comment, but since the $w(v_i)$ are weights, iperhaps they are assuming that an appropriate 'scale-factor' can always be found so that the sum can be assumed to have any value and $1$ just happened to be convenient.
$endgroup$
– Devashish Kaushik
Dec 31 '18 at 11:39
$begingroup$
I am not too familiar with graph theory, so only posting this as a comment, but since the $w(v_i)$ are weights, iperhaps they are assuming that an appropriate 'scale-factor' can always be found so that the sum can be assumed to have any value and $1$ just happened to be convenient.
$endgroup$
– Devashish Kaushik
Dec 31 '18 at 11:39
$begingroup$
yes, but why sum for all paths?
$endgroup$
– Pablo Escobar
Dec 31 '18 at 12:24
$begingroup$
yes, but why sum for all paths?
$endgroup$
– Pablo Escobar
Dec 31 '18 at 12:24
add a comment |
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$begingroup$
I am not too familiar with graph theory, so only posting this as a comment, but since the $w(v_i)$ are weights, iperhaps they are assuming that an appropriate 'scale-factor' can always be found so that the sum can be assumed to have any value and $1$ just happened to be convenient.
$endgroup$
– Devashish Kaushik
Dec 31 '18 at 11:39
$begingroup$
yes, but why sum for all paths?
$endgroup$
– Pablo Escobar
Dec 31 '18 at 12:24