Probability that a random walk does not leave a subgraph after $k$ steps












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$begingroup$


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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 1 '18 at 18:05









      jackson5jackson5

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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.






          share|cite|improve this 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.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 18:44









                Misha LavrovMisha Lavrov

                44.7k556107




                44.7k556107






























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