Generalized Polya's Urn












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In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.



Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?






probability statistics stochastic-processes random-variables random







share|cite|improve this question









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    0












    $begingroup$


    In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.



    Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?






    probability statistics stochastic-processes random-variables random







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.



      Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?






      probability statistics stochastic-processes random-variables random







      share|cite|improve this question









      $endgroup$




      In Polya's urn, we have $b$ black balls and $w$ white balls at time $t$. At time $t+1$, we have $b+1$ black balls with probability $frac{b}{w+b}$ and $w+1$ white balls with probability $frac{w}{w+b}$.



      Is there a way to generalize the probability function? Can you construct a generalized Polya's urn such that the probability of $b+1$ black and $w+1$ white balls at $t+1$ is $f(b, w)$, where $f: {mathbb{N}^+}^2 rightarrow {0, 1}^2$?






      probability statistics stochastic-processes random-variables random




      probability statistics stochastic-processes random-variables random






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 5 '18 at 23:57









      ux74bn1ux74bn1

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      163






















          1 Answer
          1






          active

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          0












          $begingroup$

          It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent





          • $f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls


          • $1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls


          and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead



          A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
            $endgroup$
            – ux74bn1
            Dec 6 '18 at 16:10










          • $begingroup$
            @ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
            $endgroup$
            – Henry
            Dec 6 '18 at 19:34











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent





          • $f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls


          • $1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls


          and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead



          A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
            $endgroup$
            – ux74bn1
            Dec 6 '18 at 16:10










          • $begingroup$
            @ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
            $endgroup$
            – Henry
            Dec 6 '18 at 19:34
















          0












          $begingroup$

          It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent





          • $f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls


          • $1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls


          and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead



          A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
            $endgroup$
            – ux74bn1
            Dec 6 '18 at 16:10










          • $begingroup$
            @ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
            $endgroup$
            – Henry
            Dec 6 '18 at 19:34














          0












          0








          0





          $begingroup$

          It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent





          • $f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls


          • $1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls


          and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead



          A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis






          share|cite|improve this answer









          $endgroup$



          It is possible to use any function $f:mathbb N^+ times mathbb N^+ to [0,1]$ to represent





          • $f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b+1$ black balls and $w$ white balls


          • $1-f(b,w)$ as the probability of moving from $b$ black balls and $w$ white balls to $b$ black balls and $w+1$ white balls


          and given the starting position, then to find the distribution of the number or proportion of black balls or of white balls a given number of steps later. With the original Polya urn you have $f(b,w)=frac{b}{b+w}$ and $1-f(b,w)=frac{w}{b+w}$, but any other function can be used instead



          A further generalisation could be to allow transitions to other states, not just adding a single ball with each step, or to have more than two colours, though these might require a more complicated function or a switch to a Markov-chain type of analysis







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 12:02









          HenryHenry

          99.5k479165




          99.5k479165












          • $begingroup$
            What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
            $endgroup$
            – ux74bn1
            Dec 6 '18 at 16:10










          • $begingroup$
            @ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
            $endgroup$
            – Henry
            Dec 6 '18 at 19:34


















          • $begingroup$
            What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
            $endgroup$
            – ux74bn1
            Dec 6 '18 at 16:10










          • $begingroup$
            @ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
            $endgroup$
            – Henry
            Dec 6 '18 at 19:34
















          $begingroup$
          What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
          $endgroup$
          – ux74bn1
          Dec 6 '18 at 16:10




          $begingroup$
          What if you also generalize the updating of the number of balls at each round? Rather than increment by 1, you use a function g() which takes in the number of balls of each color at $t$, and outputs the number at $t+1$.
          $endgroup$
          – ux74bn1
          Dec 6 '18 at 16:10












          $begingroup$
          @ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
          $endgroup$
          – Henry
          Dec 6 '18 at 19:34




          $begingroup$
          @ux74bn1 - Yes, though you might then find a Markov chain a more useful approach
          $endgroup$
          – Henry
          Dec 6 '18 at 19:34


















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