Polya's urn model - limit distribution












4












$begingroup$


Let an urn contain w white and b black balls. Draw a ball randomly from the urn and return it together with another ball of the same color. Let $b_n$ be the number of black balls and $w_n$ the number of white balls after the n-th draw-and-replacement. Let $X_n$ be the relative proportion of white balls after the n-th draw-and-replacement.



I start with b=w=1, so the total number of balls after the n-th draw-and-replacement is n+2.
Now I want to find the limit distribution of $X_n$; I already showed that $X_n$ is a martingale and that it converges a.s..
It is



$X_n = dfrac{w_n}{n+2}$ for $n in mathbb{N}_0$.



I've read that the limit distribution is a beta distribution, but I don't know how to get there.

I could write $w_n$ as the sum of $Y_i$ where $Y_i$ is 0, if the i-th ball is black and 1, if the i-th ball is black. Then I'd have



$w_n = 1+sum_{i=1}^{n} Y_i$.



Does this help? How can I proceed?



Thanks! :)






probability-theory probability-distributions convergence stochastic-processes polya-urn-model







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$endgroup$

















    4












    $begingroup$


    Let an urn contain w white and b black balls. Draw a ball randomly from the urn and return it together with another ball of the same color. Let $b_n$ be the number of black balls and $w_n$ the number of white balls after the n-th draw-and-replacement. Let $X_n$ be the relative proportion of white balls after the n-th draw-and-replacement.



    I start with b=w=1, so the total number of balls after the n-th draw-and-replacement is n+2.
    Now I want to find the limit distribution of $X_n$; I already showed that $X_n$ is a martingale and that it converges a.s..
    It is



    $X_n = dfrac{w_n}{n+2}$ for $n in mathbb{N}_0$.



    I've read that the limit distribution is a beta distribution, but I don't know how to get there.

    I could write $w_n$ as the sum of $Y_i$ where $Y_i$ is 0, if the i-th ball is black and 1, if the i-th ball is black. Then I'd have



    $w_n = 1+sum_{i=1}^{n} Y_i$.



    Does this help? How can I proceed?



    Thanks! :)






    probability-theory probability-distributions convergence stochastic-processes polya-urn-model







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Let an urn contain w white and b black balls. Draw a ball randomly from the urn and return it together with another ball of the same color. Let $b_n$ be the number of black balls and $w_n$ the number of white balls after the n-th draw-and-replacement. Let $X_n$ be the relative proportion of white balls after the n-th draw-and-replacement.



      I start with b=w=1, so the total number of balls after the n-th draw-and-replacement is n+2.
      Now I want to find the limit distribution of $X_n$; I already showed that $X_n$ is a martingale and that it converges a.s..
      It is



      $X_n = dfrac{w_n}{n+2}$ for $n in mathbb{N}_0$.



      I've read that the limit distribution is a beta distribution, but I don't know how to get there.

      I could write $w_n$ as the sum of $Y_i$ where $Y_i$ is 0, if the i-th ball is black and 1, if the i-th ball is black. Then I'd have



      $w_n = 1+sum_{i=1}^{n} Y_i$.



      Does this help? How can I proceed?



      Thanks! :)






      probability-theory probability-distributions convergence stochastic-processes polya-urn-model







      share|cite|improve this question











      $endgroup$




      Let an urn contain w white and b black balls. Draw a ball randomly from the urn and return it together with another ball of the same color. Let $b_n$ be the number of black balls and $w_n$ the number of white balls after the n-th draw-and-replacement. Let $X_n$ be the relative proportion of white balls after the n-th draw-and-replacement.



      I start with b=w=1, so the total number of balls after the n-th draw-and-replacement is n+2.
      Now I want to find the limit distribution of $X_n$; I already showed that $X_n$ is a martingale and that it converges a.s..
      It is



      $X_n = dfrac{w_n}{n+2}$ for $n in mathbb{N}_0$.



      I've read that the limit distribution is a beta distribution, but I don't know how to get there.

      I could write $w_n$ as the sum of $Y_i$ where $Y_i$ is 0, if the i-th ball is black and 1, if the i-th ball is black. Then I'd have



      $w_n = 1+sum_{i=1}^{n} Y_i$.



      Does this help? How can I proceed?



      Thanks! :)






      probability-theory probability-distributions convergence stochastic-processes polya-urn-model




      probability-theory probability-distributions convergence stochastic-processes polya-urn-model






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 '16 at 6:29









      BCLC

      1




      1










      asked Jan 29 '15 at 18:04









      Max93Max93

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          $begingroup$

          Refer to this?




          $$M_{Theta}(t) = E[exp(tTheta)]$$



          $$= E[exp(tlim frac{B_n + 1}{n+2})]$$



          $$= E[limexp(t frac{B_n + 1}{n+2})]$$



          $$= lim E[exp(t frac{B_n + 1}{n+2})]$$



          $$= lim frac{1}{n+1}[exp(t frac{1}{n+2}) + exp(t frac{2}{n+2}) + ... + exp(t frac{n+1}{n+2})]$$



          Case 1: $t ne 0$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), text{where} a(n) := e^{frac{t}{n+2}}$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} lim (1-a(n)^{n+1})$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-e^t)$$



          $$= frac{1-e^t}{-t}$$



          $$= frac{e^t-1}{t}$$



          Case 2: $t = 0$



          $$= lim frac{1}{n+1}[exp((0) frac{1}{n+2}) + exp((0) frac{2}{n+2}) + ... + exp((0) frac{n+1}{n+2})]$$



          $$= lim frac{1}{n+1} (1)(n+1) = 1$$



          This is the mgf of $Unif(0,1)$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            @Did Edited. I accounted for $t=0$ this time
            $endgroup$
            – BCLC
            Feb 12 '16 at 22:59











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          $begingroup$

          Refer to this?




          $$M_{Theta}(t) = E[exp(tTheta)]$$



          $$= E[exp(tlim frac{B_n + 1}{n+2})]$$



          $$= E[limexp(t frac{B_n + 1}{n+2})]$$



          $$= lim E[exp(t frac{B_n + 1}{n+2})]$$



          $$= lim frac{1}{n+1}[exp(t frac{1}{n+2}) + exp(t frac{2}{n+2}) + ... + exp(t frac{n+1}{n+2})]$$



          Case 1: $t ne 0$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), text{where} a(n) := e^{frac{t}{n+2}}$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} lim (1-a(n)^{n+1})$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-e^t)$$



          $$= frac{1-e^t}{-t}$$



          $$= frac{e^t-1}{t}$$



          Case 2: $t = 0$



          $$= lim frac{1}{n+1}[exp((0) frac{1}{n+2}) + exp((0) frac{2}{n+2}) + ... + exp((0) frac{n+1}{n+2})]$$



          $$= lim frac{1}{n+1} (1)(n+1) = 1$$



          This is the mgf of $Unif(0,1)$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            @Did Edited. I accounted for $t=0$ this time
            $endgroup$
            – BCLC
            Feb 12 '16 at 22:59
















          0












          $begingroup$

          Refer to this?




          $$M_{Theta}(t) = E[exp(tTheta)]$$



          $$= E[exp(tlim frac{B_n + 1}{n+2})]$$



          $$= E[limexp(t frac{B_n + 1}{n+2})]$$



          $$= lim E[exp(t frac{B_n + 1}{n+2})]$$



          $$= lim frac{1}{n+1}[exp(t frac{1}{n+2}) + exp(t frac{2}{n+2}) + ... + exp(t frac{n+1}{n+2})]$$



          Case 1: $t ne 0$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), text{where} a(n) := e^{frac{t}{n+2}}$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} lim (1-a(n)^{n+1})$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-e^t)$$



          $$= frac{1-e^t}{-t}$$



          $$= frac{e^t-1}{t}$$



          Case 2: $t = 0$



          $$= lim frac{1}{n+1}[exp((0) frac{1}{n+2}) + exp((0) frac{2}{n+2}) + ... + exp((0) frac{n+1}{n+2})]$$



          $$= lim frac{1}{n+1} (1)(n+1) = 1$$



          This is the mgf of $Unif(0,1)$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            @Did Edited. I accounted for $t=0$ this time
            $endgroup$
            – BCLC
            Feb 12 '16 at 22:59














          0












          0








          0





          $begingroup$

          Refer to this?




          $$M_{Theta}(t) = E[exp(tTheta)]$$



          $$= E[exp(tlim frac{B_n + 1}{n+2})]$$



          $$= E[limexp(t frac{B_n + 1}{n+2})]$$



          $$= lim E[exp(t frac{B_n + 1}{n+2})]$$



          $$= lim frac{1}{n+1}[exp(t frac{1}{n+2}) + exp(t frac{2}{n+2}) + ... + exp(t frac{n+1}{n+2})]$$



          Case 1: $t ne 0$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), text{where} a(n) := e^{frac{t}{n+2}}$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} lim (1-a(n)^{n+1})$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-e^t)$$



          $$= frac{1-e^t}{-t}$$



          $$= frac{e^t-1}{t}$$



          Case 2: $t = 0$



          $$= lim frac{1}{n+1}[exp((0) frac{1}{n+2}) + exp((0) frac{2}{n+2}) + ... + exp((0) frac{n+1}{n+2})]$$



          $$= lim frac{1}{n+1} (1)(n+1) = 1$$



          This is the mgf of $Unif(0,1)$






          share|cite|improve this answer











          $endgroup$



          Refer to this?




          $$M_{Theta}(t) = E[exp(tTheta)]$$



          $$= E[exp(tlim frac{B_n + 1}{n+2})]$$



          $$= E[limexp(t frac{B_n + 1}{n+2})]$$



          $$= lim E[exp(t frac{B_n + 1}{n+2})]$$



          $$= lim frac{1}{n+1}[exp(t frac{1}{n+2}) + exp(t frac{2}{n+2}) + ... + exp(t frac{n+1}{n+2})]$$



          Case 1: $t ne 0$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), text{where} a(n) := e^{frac{t}{n+2}}$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} lim (1-a(n)^{n+1})$$



          $$= lim frac{a(n)}{(n+1)(1-a(n))} (1-e^t)$$



          $$= frac{1-e^t}{-t}$$



          $$= frac{e^t-1}{t}$$



          Case 2: $t = 0$



          $$= lim frac{1}{n+1}[exp((0) frac{1}{n+2}) + exp((0) frac{2}{n+2}) + ... + exp((0) frac{n+1}{n+2})]$$



          $$= lim frac{1}{n+1} (1)(n+1) = 1$$



          This is the mgf of $Unif(0,1)$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 12 '16 at 22:59

























          answered Feb 6 '16 at 7:58









          BCLCBCLC

          1




          1








          • 2




            $begingroup$
            @Did Edited. I accounted for $t=0$ this time
            $endgroup$
            – BCLC
            Feb 12 '16 at 22:59














          • 2




            $begingroup$
            @Did Edited. I accounted for $t=0$ this time
            $endgroup$
            – BCLC
            Feb 12 '16 at 22:59








          2




          2




          $begingroup$
          @Did Edited. I accounted for $t=0$ this time
          $endgroup$
          – BCLC
          Feb 12 '16 at 22:59




          $begingroup$
          @Did Edited. I accounted for $t=0$ this time
          $endgroup$
          – BCLC
          Feb 12 '16 at 22:59


















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