Probability that a random walk does not leave a subgraph after $k$ steps
$begingroup$
Suppose I have a connected, finite, graph $G = (V,E)$, and I have some vertex set $U$ such that the subgraph of $G$ induced by the vertices $Y = V setminus U$ is still connected.
Now, suppose I have a random walk $X_n$ on $G$, that follows the transition probabilities induced by the edge weights of $G$. Let $X_0 = u in Y$. I want to know how I can go about calculating $P_u(X_1, dots, X_k in Y)$.
I keep running into something like $prod_{i=1}^{k-1}P_u(X_{i+1} in W | X_{i}in W)$, which I don't know how to simplify either.
graph-theory markov-chains markov-process random-walk
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
$endgroup$
add a comment |
$begingroup$
Suppose I have a connected, finite, graph $G = (V,E)$, and I have some vertex set $U$ such that the subgraph of $G$ induced by the vertices $Y = V setminus U$ is still connected.
Now, suppose I have a random walk $X_n$ on $G$, that follows the transition probabilities induced by the edge weights of $G$. Let $X_0 = u in Y$. I want to know how I can go about calculating $P_u(X_1, dots, X_k in Y)$.
I keep running into something like $prod_{i=1}^{k-1}P_u(X_{i+1} in W | X_{i}in W)$, which I don't know how to simplify either.
graph-theory markov-chains markov-process random-walk
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
$endgroup$
add a comment |
$begingroup$
Suppose I have a connected, finite, graph $G = (V,E)$, and I have some vertex set $U$ such that the subgraph of $G$ induced by the vertices $Y = V setminus U$ is still connected.
Now, suppose I have a random walk $X_n$ on $G$, that follows the transition probabilities induced by the edge weights of $G$. Let $X_0 = u in Y$. I want to know how I can go about calculating $P_u(X_1, dots, X_k in Y)$.
I keep running into something like $prod_{i=1}^{k-1}P_u(X_{i+1} in W | X_{i}in W)$, which I don't know how to simplify either.
graph-theory markov-chains markov-process random-walk
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
$endgroup$
Suppose I have a connected, finite, graph $G = (V,E)$, and I have some vertex set $U$ such that the subgraph of $G$ induced by the vertices $Y = V setminus U$ is still connected.
Now, suppose I have a random walk $X_n$ on $G$, that follows the transition probabilities induced by the edge weights of $G$. Let $X_0 = u in Y$. I want to know how I can go about calculating $P_u(X_1, dots, X_k in Y)$.
I keep running into something like $prod_{i=1}^{k-1}P_u(X_{i+1} in W | X_{i}in W)$, which I don't know how to simplify either.
graph-theory markov-chains markov-process random-walk
graph-theory markov-chains markov-process random-walk
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
asked Dec 1 '18 at 18:05
jackson5jackson5
606512
606512
add a comment |
add a comment |
1 Answer
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$begingroup$
To simplify the problem as much as possible, you can combine all vertices in $U$ into a single vertex $x$. Each edge from $v in Y$ to a vertex in $U$ becomes an edge from $v$ to $x$; if you get parallel edges this way, you should either keep them both, or else combine them by adding the weights.
Next, turn the random walk into an asymmetric Markov chain by making $x$ an absorbing state. This will make it easy to tell when we've left $Y$: if we do, we go to $x$, and then we stay in $x$ forever.
Now, the complementary probability (the probability that you get from your starting vertex $u$ to $x$ within $k$ steps) can be computed by taking powers of the transition matrix of this Markov chain, as usual.
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
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1 Answer
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1 Answer
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$begingroup$
To simplify the problem as much as possible, you can combine all vertices in $U$ into a single vertex $x$. Each edge from $v in Y$ to a vertex in $U$ becomes an edge from $v$ to $x$; if you get parallel edges this way, you should either keep them both, or else combine them by adding the weights.
Next, turn the random walk into an asymmetric Markov chain by making $x$ an absorbing state. This will make it easy to tell when we've left $Y$: if we do, we go to $x$, and then we stay in $x$ forever.
Now, the complementary probability (the probability that you get from your starting vertex $u$ to $x$ within $k$ steps) can be computed by taking powers of the transition matrix of this Markov chain, as usual.
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
$endgroup$
add a comment |
$begingroup$
To simplify the problem as much as possible, you can combine all vertices in $U$ into a single vertex $x$. Each edge from $v in Y$ to a vertex in $U$ becomes an edge from $v$ to $x$; if you get parallel edges this way, you should either keep them both, or else combine them by adding the weights.
Next, turn the random walk into an asymmetric Markov chain by making $x$ an absorbing state. This will make it easy to tell when we've left $Y$: if we do, we go to $x$, and then we stay in $x$ forever.
Now, the complementary probability (the probability that you get from your starting vertex $u$ to $x$ within $k$ steps) can be computed by taking powers of the transition matrix of this Markov chain, as usual.
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
$endgroup$
add a comment |
$begingroup$
To simplify the problem as much as possible, you can combine all vertices in $U$ into a single vertex $x$. Each edge from $v in Y$ to a vertex in $U$ becomes an edge from $v$ to $x$; if you get parallel edges this way, you should either keep them both, or else combine them by adding the weights.
Next, turn the random walk into an asymmetric Markov chain by making $x$ an absorbing state. This will make it easy to tell when we've left $Y$: if we do, we go to $x$, and then we stay in $x$ forever.
Now, the complementary probability (the probability that you get from your starting vertex $u$ to $x$ within $k$ steps) can be computed by taking powers of the transition matrix of this Markov chain, as usual.
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
$endgroup$
To simplify the problem as much as possible, you can combine all vertices in $U$ into a single vertex $x$. Each edge from $v in Y$ to a vertex in $U$ becomes an edge from $v$ to $x$; if you get parallel edges this way, you should either keep them both, or else combine them by adding the weights.
Next, turn the random walk into an asymmetric Markov chain by making $x$ an absorbing state. This will make it easy to tell when we've left $Y$: if we do, we go to $x$, and then we stay in $x$ forever.
Now, the complementary probability (the probability that you get from your starting vertex $u$ to $x$ within $k$ steps) can be computed by taking powers of the transition matrix of this Markov chain, as usual.
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
answered Dec 1 '18 at 18:44
Misha LavrovMisha Lavrov
44.7k556107
44.7k556107
add a comment |
add a comment |
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