Probability of reaching a maximum in a random walk












3












$begingroup$


Let's define a random walk in the following way
$$S_0 = 0, S_n = sum_{i=1}^{n} epsilon_i,$$
where:
$$epsilon_i = pm 1.$$
Moreover
$$P(epsilon_i = - 1) = P(epsilon_i = + 1) = frac{1}{2}.$$



Again, let's define a random variable $X_n$ which gives us the position of the first maximum of a random walk of length $2n$.

I am to show that
$$P(X_n = 2k) = P(X_n = 2k + 1) = frac{1}{2}u_{2k}u_{2n-2k}.$$
Here
$$u_{2n} = frac{binom{2n}{n}}{2^{2n}} = P(S_{2n} = 0) = P(S_1 ge 0, S_2 ge0, ldots, S_{2n} ge 0) = P(S_1 neq0, S_2 neq 0, ldots, S_{2n} neq 0).$$
I also know that
$$u_{2k}u_{2n-2k}$$
tells us about the probability that a randomly walking particle is $2k$ ,,moves'' above the $x$ axis and $2n-2k$ below. However I don't know how to combine all those things and give the proof. I would appreciate any hints or tips.






probability-theory stochastic-processes random-walk







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let's define a random walk in the following way
    $$S_0 = 0, S_n = sum_{i=1}^{n} epsilon_i,$$
    where:
    $$epsilon_i = pm 1.$$
    Moreover
    $$P(epsilon_i = - 1) = P(epsilon_i = + 1) = frac{1}{2}.$$



    Again, let's define a random variable $X_n$ which gives us the position of the first maximum of a random walk of length $2n$.

    I am to show that
    $$P(X_n = 2k) = P(X_n = 2k + 1) = frac{1}{2}u_{2k}u_{2n-2k}.$$
    Here
    $$u_{2n} = frac{binom{2n}{n}}{2^{2n}} = P(S_{2n} = 0) = P(S_1 ge 0, S_2 ge0, ldots, S_{2n} ge 0) = P(S_1 neq0, S_2 neq 0, ldots, S_{2n} neq 0).$$
    I also know that
    $$u_{2k}u_{2n-2k}$$
    tells us about the probability that a randomly walking particle is $2k$ ,,moves'' above the $x$ axis and $2n-2k$ below. However I don't know how to combine all those things and give the proof. I would appreciate any hints or tips.






    probability-theory stochastic-processes random-walk







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Let's define a random walk in the following way
      $$S_0 = 0, S_n = sum_{i=1}^{n} epsilon_i,$$
      where:
      $$epsilon_i = pm 1.$$
      Moreover
      $$P(epsilon_i = - 1) = P(epsilon_i = + 1) = frac{1}{2}.$$



      Again, let's define a random variable $X_n$ which gives us the position of the first maximum of a random walk of length $2n$.

      I am to show that
      $$P(X_n = 2k) = P(X_n = 2k + 1) = frac{1}{2}u_{2k}u_{2n-2k}.$$
      Here
      $$u_{2n} = frac{binom{2n}{n}}{2^{2n}} = P(S_{2n} = 0) = P(S_1 ge 0, S_2 ge0, ldots, S_{2n} ge 0) = P(S_1 neq0, S_2 neq 0, ldots, S_{2n} neq 0).$$
      I also know that
      $$u_{2k}u_{2n-2k}$$
      tells us about the probability that a randomly walking particle is $2k$ ,,moves'' above the $x$ axis and $2n-2k$ below. However I don't know how to combine all those things and give the proof. I would appreciate any hints or tips.






      probability-theory stochastic-processes random-walk







      share|cite|improve this question











      $endgroup$




      Let's define a random walk in the following way
      $$S_0 = 0, S_n = sum_{i=1}^{n} epsilon_i,$$
      where:
      $$epsilon_i = pm 1.$$
      Moreover
      $$P(epsilon_i = - 1) = P(epsilon_i = + 1) = frac{1}{2}.$$



      Again, let's define a random variable $X_n$ which gives us the position of the first maximum of a random walk of length $2n$.

      I am to show that
      $$P(X_n = 2k) = P(X_n = 2k + 1) = frac{1}{2}u_{2k}u_{2n-2k}.$$
      Here
      $$u_{2n} = frac{binom{2n}{n}}{2^{2n}} = P(S_{2n} = 0) = P(S_1 ge 0, S_2 ge0, ldots, S_{2n} ge 0) = P(S_1 neq0, S_2 neq 0, ldots, S_{2n} neq 0).$$
      I also know that
      $$u_{2k}u_{2n-2k}$$
      tells us about the probability that a randomly walking particle is $2k$ ,,moves'' above the $x$ axis and $2n-2k$ below. However I don't know how to combine all those things and give the proof. I would appreciate any hints or tips.






      probability-theory stochastic-processes random-walk




      probability-theory stochastic-processes random-walk






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 18:47









      saz

      79.7k860124




      79.7k860124










      asked Jan 6 at 11:52









      HendrraHendrra

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          1 Answer
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          2












          $begingroup$

          Hints for $mathbb{P}(X_n = 2k)$: For fixed $k in mathbb{N}$ define two auxiliary random walks by



          $$U_j := sum_{i=1}^j epsilon_{2k-i}, qquad 0 leq j leq 2k,$$



          and



          $$V_j := sum_{i=1}^j epsilon_{2k+i}, qquad 0 leq j leq 2n-2k.$$




          1. Check that $(U_j)_{j leq 2k}$ and $(V_j)_{j leq 2n-2k}$ are independent.

          2. Show that $${X_n = 2k} = {U_1 > 0,ldots,U_{2k}>0} cap {V_1leq 0,ldots,V_{2n-2k}leq 0}.$$

          3. Combining Step 1+2 gives $$mathbb{P}(X_n = 2k) = mathbb{P}(U_1 > 0,ldots,U_{2k}>0) mathbb{P}(V_1 leq 0,ldots, V_{2n-2k} leq 0).$$ Use the facts which you mentioned about $u_{2n}$ to prove that the right-hand side equals $frac{1}{2} u_{2k} u_{2n-2k}$.



          Hints for $mathbb{P}(X_n = 2k+1)$: Set $$T_j := sum_{i=2}^{j+1} epsilon_i$$ and denote by $Y_n$ the position of the first maximum of $(T_j)_{j leq 2n}$.




          1. Show that ${X_n =2k+1} = {Y_n = 2k}$.

          2. Deduce from the first part of this problem that $$mathbb{P}(Y_n=2k) = frac{1}{2} u_{2k} u_{2n-k}.$$

          3. Conclude.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            1 Answer
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            active

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            2












            $begingroup$

            Hints for $mathbb{P}(X_n = 2k)$: For fixed $k in mathbb{N}$ define two auxiliary random walks by



            $$U_j := sum_{i=1}^j epsilon_{2k-i}, qquad 0 leq j leq 2k,$$



            and



            $$V_j := sum_{i=1}^j epsilon_{2k+i}, qquad 0 leq j leq 2n-2k.$$




            1. Check that $(U_j)_{j leq 2k}$ and $(V_j)_{j leq 2n-2k}$ are independent.

            2. Show that $${X_n = 2k} = {U_1 > 0,ldots,U_{2k}>0} cap {V_1leq 0,ldots,V_{2n-2k}leq 0}.$$

            3. Combining Step 1+2 gives $$mathbb{P}(X_n = 2k) = mathbb{P}(U_1 > 0,ldots,U_{2k}>0) mathbb{P}(V_1 leq 0,ldots, V_{2n-2k} leq 0).$$ Use the facts which you mentioned about $u_{2n}$ to prove that the right-hand side equals $frac{1}{2} u_{2k} u_{2n-2k}$.



            Hints for $mathbb{P}(X_n = 2k+1)$: Set $$T_j := sum_{i=2}^{j+1} epsilon_i$$ and denote by $Y_n$ the position of the first maximum of $(T_j)_{j leq 2n}$.




            1. Show that ${X_n =2k+1} = {Y_n = 2k}$.

            2. Deduce from the first part of this problem that $$mathbb{P}(Y_n=2k) = frac{1}{2} u_{2k} u_{2n-k}.$$

            3. Conclude.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Hints for $mathbb{P}(X_n = 2k)$: For fixed $k in mathbb{N}$ define two auxiliary random walks by



              $$U_j := sum_{i=1}^j epsilon_{2k-i}, qquad 0 leq j leq 2k,$$



              and



              $$V_j := sum_{i=1}^j epsilon_{2k+i}, qquad 0 leq j leq 2n-2k.$$




              1. Check that $(U_j)_{j leq 2k}$ and $(V_j)_{j leq 2n-2k}$ are independent.

              2. Show that $${X_n = 2k} = {U_1 > 0,ldots,U_{2k}>0} cap {V_1leq 0,ldots,V_{2n-2k}leq 0}.$$

              3. Combining Step 1+2 gives $$mathbb{P}(X_n = 2k) = mathbb{P}(U_1 > 0,ldots,U_{2k}>0) mathbb{P}(V_1 leq 0,ldots, V_{2n-2k} leq 0).$$ Use the facts which you mentioned about $u_{2n}$ to prove that the right-hand side equals $frac{1}{2} u_{2k} u_{2n-2k}$.



              Hints for $mathbb{P}(X_n = 2k+1)$: Set $$T_j := sum_{i=2}^{j+1} epsilon_i$$ and denote by $Y_n$ the position of the first maximum of $(T_j)_{j leq 2n}$.




              1. Show that ${X_n =2k+1} = {Y_n = 2k}$.

              2. Deduce from the first part of this problem that $$mathbb{P}(Y_n=2k) = frac{1}{2} u_{2k} u_{2n-k}.$$

              3. Conclude.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Hints for $mathbb{P}(X_n = 2k)$: For fixed $k in mathbb{N}$ define two auxiliary random walks by



                $$U_j := sum_{i=1}^j epsilon_{2k-i}, qquad 0 leq j leq 2k,$$



                and



                $$V_j := sum_{i=1}^j epsilon_{2k+i}, qquad 0 leq j leq 2n-2k.$$




                1. Check that $(U_j)_{j leq 2k}$ and $(V_j)_{j leq 2n-2k}$ are independent.

                2. Show that $${X_n = 2k} = {U_1 > 0,ldots,U_{2k}>0} cap {V_1leq 0,ldots,V_{2n-2k}leq 0}.$$

                3. Combining Step 1+2 gives $$mathbb{P}(X_n = 2k) = mathbb{P}(U_1 > 0,ldots,U_{2k}>0) mathbb{P}(V_1 leq 0,ldots, V_{2n-2k} leq 0).$$ Use the facts which you mentioned about $u_{2n}$ to prove that the right-hand side equals $frac{1}{2} u_{2k} u_{2n-2k}$.



                Hints for $mathbb{P}(X_n = 2k+1)$: Set $$T_j := sum_{i=2}^{j+1} epsilon_i$$ and denote by $Y_n$ the position of the first maximum of $(T_j)_{j leq 2n}$.




                1. Show that ${X_n =2k+1} = {Y_n = 2k}$.

                2. Deduce from the first part of this problem that $$mathbb{P}(Y_n=2k) = frac{1}{2} u_{2k} u_{2n-k}.$$

                3. Conclude.






                share|cite|improve this answer











                $endgroup$



                Hints for $mathbb{P}(X_n = 2k)$: For fixed $k in mathbb{N}$ define two auxiliary random walks by



                $$U_j := sum_{i=1}^j epsilon_{2k-i}, qquad 0 leq j leq 2k,$$



                and



                $$V_j := sum_{i=1}^j epsilon_{2k+i}, qquad 0 leq j leq 2n-2k.$$




                1. Check that $(U_j)_{j leq 2k}$ and $(V_j)_{j leq 2n-2k}$ are independent.

                2. Show that $${X_n = 2k} = {U_1 > 0,ldots,U_{2k}>0} cap {V_1leq 0,ldots,V_{2n-2k}leq 0}.$$

                3. Combining Step 1+2 gives $$mathbb{P}(X_n = 2k) = mathbb{P}(U_1 > 0,ldots,U_{2k}>0) mathbb{P}(V_1 leq 0,ldots, V_{2n-2k} leq 0).$$ Use the facts which you mentioned about $u_{2n}$ to prove that the right-hand side equals $frac{1}{2} u_{2k} u_{2n-2k}$.



                Hints for $mathbb{P}(X_n = 2k+1)$: Set $$T_j := sum_{i=2}^{j+1} epsilon_i$$ and denote by $Y_n$ the position of the first maximum of $(T_j)_{j leq 2n}$.




                1. Show that ${X_n =2k+1} = {Y_n = 2k}$.

                2. Deduce from the first part of this problem that $$mathbb{P}(Y_n=2k) = frac{1}{2} u_{2k} u_{2n-k}.$$

                3. Conclude.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 10:50

























                answered Jan 6 at 18:33









                sazsaz

                79.7k860124




                79.7k860124






























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