Biased random walk

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I have an undirected graph $G$ and when performing a random walk on the graph $G$, I visit a node $u$ with probability proportional to $d_u$ where $d_u$ is degree of node $u$.



I know there exists ways like Metropolis Hastings to redefine edge probabilities such that the probability of visiting a node is 1/|V|.



Given that I have another quality value $q_u$ of every node, is there a way to redefine the edge probabilities so I sample nodes with probability proportional to $q_u$.



I know I can do the sampling of nodes uniformly first and then do rejection sampling but I wanted to know if there is a way to directly do the sampling from the graph by changing transition probabilities online.




graph-theory random-walk stationary-processes




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    I have an undirected graph $G$ and when performing a random walk on the graph $G$, I visit a node $u$ with probability proportional to $d_u$ where $d_u$ is degree of node $u$.



    I know there exists ways like Metropolis Hastings to redefine edge probabilities such that the probability of visiting a node is 1/|V|.



    Given that I have another quality value $q_u$ of every node, is there a way to redefine the edge probabilities so I sample nodes with probability proportional to $q_u$.



    I know I can do the sampling of nodes uniformly first and then do rejection sampling but I wanted to know if there is a way to directly do the sampling from the graph by changing transition probabilities online.




    graph-theory random-walk stationary-processes




    share|cite|improve this question





















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      I have an undirected graph $G$ and when performing a random walk on the graph $G$, I visit a node $u$ with probability proportional to $d_u$ where $d_u$ is degree of node $u$.



      I know there exists ways like Metropolis Hastings to redefine edge probabilities such that the probability of visiting a node is 1/|V|.



      Given that I have another quality value $q_u$ of every node, is there a way to redefine the edge probabilities so I sample nodes with probability proportional to $q_u$.



      I know I can do the sampling of nodes uniformly first and then do rejection sampling but I wanted to know if there is a way to directly do the sampling from the graph by changing transition probabilities online.




      graph-theory random-walk stationary-processes




      share|cite|improve this question











      I have an undirected graph $G$ and when performing a random walk on the graph $G$, I visit a node $u$ with probability proportional to $d_u$ where $d_u$ is degree of node $u$.



      I know there exists ways like Metropolis Hastings to redefine edge probabilities such that the probability of visiting a node is 1/|V|.



      Given that I have another quality value $q_u$ of every node, is there a way to redefine the edge probabilities so I sample nodes with probability proportional to $q_u$.



      I know I can do the sampling of nodes uniformly first and then do rejection sampling but I wanted to know if there is a way to directly do the sampling from the graph by changing transition probabilities online.





      graph-theory random-walk stationary-processes





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      share|cite|improve this question




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      asked Aug 2 at 18:11









      learner

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          Presumably all $q_u > 0$.
          Assuming your graph is connected, and you allow the process to sometimes stay at a vertex rather than jump, it is possible. Let $p_ij$ be the transition probability for moving from $i$ to $j$ ($0$ unless either $i,j$ is an edge or
          $i=j$). Your process will be reversible if for each edge $i,j$, $q_i p_ij = q_j p_ji$. This would be satisfied if $p_ij = q_j$ for all edges $i,j$, but that might make the sum of transition probabilities leaving some nodes $> 1$.
          Therefore if $M = max_i sum_j in N(i) q_j$, we take
          $$eqalignp_ij &= q_j/Mcr p_ii &= 1 - sum_j in N(i) q_j/M$$






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            accepted










            Presumably all $q_u > 0$.
            Assuming your graph is connected, and you allow the process to sometimes stay at a vertex rather than jump, it is possible. Let $p_ij$ be the transition probability for moving from $i$ to $j$ ($0$ unless either $i,j$ is an edge or
            $i=j$). Your process will be reversible if for each edge $i,j$, $q_i p_ij = q_j p_ji$. This would be satisfied if $p_ij = q_j$ for all edges $i,j$, but that might make the sum of transition probabilities leaving some nodes $> 1$.
            Therefore if $M = max_i sum_j in N(i) q_j$, we take
            $$eqalignp_ij &= q_j/Mcr p_ii &= 1 - sum_j in N(i) q_j/M$$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Presumably all $q_u > 0$.
              Assuming your graph is connected, and you allow the process to sometimes stay at a vertex rather than jump, it is possible. Let $p_ij$ be the transition probability for moving from $i$ to $j$ ($0$ unless either $i,j$ is an edge or
              $i=j$). Your process will be reversible if for each edge $i,j$, $q_i p_ij = q_j p_ji$. This would be satisfied if $p_ij = q_j$ for all edges $i,j$, but that might make the sum of transition probabilities leaving some nodes $> 1$.
              Therefore if $M = max_i sum_j in N(i) q_j$, we take
              $$eqalignp_ij &= q_j/Mcr p_ii &= 1 - sum_j in N(i) q_j/M$$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Presumably all $q_u > 0$.
                Assuming your graph is connected, and you allow the process to sometimes stay at a vertex rather than jump, it is possible. Let $p_ij$ be the transition probability for moving from $i$ to $j$ ($0$ unless either $i,j$ is an edge or
                $i=j$). Your process will be reversible if for each edge $i,j$, $q_i p_ij = q_j p_ji$. This would be satisfied if $p_ij = q_j$ for all edges $i,j$, but that might make the sum of transition probabilities leaving some nodes $> 1$.
                Therefore if $M = max_i sum_j in N(i) q_j$, we take
                $$eqalignp_ij &= q_j/Mcr p_ii &= 1 - sum_j in N(i) q_j/M$$






                share|cite|improve this answer













                Presumably all $q_u > 0$.
                Assuming your graph is connected, and you allow the process to sometimes stay at a vertex rather than jump, it is possible. Let $p_ij$ be the transition probability for moving from $i$ to $j$ ($0$ unless either $i,j$ is an edge or
                $i=j$). Your process will be reversible if for each edge $i,j$, $q_i p_ij = q_j p_ji$. This would be satisfied if $p_ij = q_j$ for all edges $i,j$, but that might make the sum of transition probabilities leaving some nodes $> 1$.
                Therefore if $M = max_i sum_j in N(i) q_j$, we take
                $$eqalignp_ij &= q_j/Mcr p_ii &= 1 - sum_j in N(i) q_j/M$$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Aug 2 at 18:38









                Robert Israel

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