Cauchy density function for Brownian motion












0












$begingroup$


Let ${W(t):tgeq0}$ be a Brownian motion, and let ${mathcal{F}_{t},tgeq0}$ be its natural filtration. Let ${W_{2}(t):tgeq0}$ be a Brownian motion, independent of ${W(t):tgeq0}$. Denote, for $a>0$, $$tau_{a}=inf{tgeq0:W(t)=a}.$$ Using that the probability density of the first hitting time of $a>0$ for Brownian motion is given by $$f(t)=
left{
begin{array}{ll}
displaystylefrac{ae^{-a^2/2t}}{sqrt{2pi t^{3}}}&text{if }t>0\
0&text{otherwise}.
end{array}
right.
$$
Show that the probability denisty of $W_{2}(tau_{a})$ is given by the Cauchy density function $$f(y)=frac{a}{pi(a^{2}+y^{2})},quad yinmathbb{R}.$$



First, I do not fully understand the question. Is the Cauchy density function the probability density function of the second Brownian motion at time $tau_{a}$, when the first Brownian motion is stopped? The distribution of a Brownian motion should have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case.






brownian-motion martingales stopping-times







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$endgroup$

















    0












    $begingroup$


    Let ${W(t):tgeq0}$ be a Brownian motion, and let ${mathcal{F}_{t},tgeq0}$ be its natural filtration. Let ${W_{2}(t):tgeq0}$ be a Brownian motion, independent of ${W(t):tgeq0}$. Denote, for $a>0$, $$tau_{a}=inf{tgeq0:W(t)=a}.$$ Using that the probability density of the first hitting time of $a>0$ for Brownian motion is given by $$f(t)=
    left{
    begin{array}{ll}
    displaystylefrac{ae^{-a^2/2t}}{sqrt{2pi t^{3}}}&text{if }t>0\
    0&text{otherwise}.
    end{array}
    right.
    $$
    Show that the probability denisty of $W_{2}(tau_{a})$ is given by the Cauchy density function $$f(y)=frac{a}{pi(a^{2}+y^{2})},quad yinmathbb{R}.$$



    First, I do not fully understand the question. Is the Cauchy density function the probability density function of the second Brownian motion at time $tau_{a}$, when the first Brownian motion is stopped? The distribution of a Brownian motion should have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case.






    brownian-motion martingales stopping-times







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let ${W(t):tgeq0}$ be a Brownian motion, and let ${mathcal{F}_{t},tgeq0}$ be its natural filtration. Let ${W_{2}(t):tgeq0}$ be a Brownian motion, independent of ${W(t):tgeq0}$. Denote, for $a>0$, $$tau_{a}=inf{tgeq0:W(t)=a}.$$ Using that the probability density of the first hitting time of $a>0$ for Brownian motion is given by $$f(t)=
      left{
      begin{array}{ll}
      displaystylefrac{ae^{-a^2/2t}}{sqrt{2pi t^{3}}}&text{if }t>0\
      0&text{otherwise}.
      end{array}
      right.
      $$
      Show that the probability denisty of $W_{2}(tau_{a})$ is given by the Cauchy density function $$f(y)=frac{a}{pi(a^{2}+y^{2})},quad yinmathbb{R}.$$



      First, I do not fully understand the question. Is the Cauchy density function the probability density function of the second Brownian motion at time $tau_{a}$, when the first Brownian motion is stopped? The distribution of a Brownian motion should have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case.






      brownian-motion martingales stopping-times







      share|cite|improve this question











      $endgroup$




      Let ${W(t):tgeq0}$ be a Brownian motion, and let ${mathcal{F}_{t},tgeq0}$ be its natural filtration. Let ${W_{2}(t):tgeq0}$ be a Brownian motion, independent of ${W(t):tgeq0}$. Denote, for $a>0$, $$tau_{a}=inf{tgeq0:W(t)=a}.$$ Using that the probability density of the first hitting time of $a>0$ for Brownian motion is given by $$f(t)=
      left{
      begin{array}{ll}
      displaystylefrac{ae^{-a^2/2t}}{sqrt{2pi t^{3}}}&text{if }t>0\
      0&text{otherwise}.
      end{array}
      right.
      $$
      Show that the probability denisty of $W_{2}(tau_{a})$ is given by the Cauchy density function $$f(y)=frac{a}{pi(a^{2}+y^{2})},quad yinmathbb{R}.$$



      First, I do not fully understand the question. Is the Cauchy density function the probability density function of the second Brownian motion at time $tau_{a}$, when the first Brownian motion is stopped? The distribution of a Brownian motion should have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case.






      brownian-motion martingales stopping-times




      brownian-motion martingales stopping-times






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 22 at 9:03









      AddSup

      4851317




      4851317










      asked Jan 21 at 16:30









      rs4rs35rs4rs35

      237




      237






















          1 Answer
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          $begingroup$

          "Is the Cauchy density function the probability function of the second Brownian motion at time $tau_a$, when the first Brownian motion is stopped?" Yes.



          "The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $a$ increases, so will $tau_a$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $a$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.



          And the solution to the problem:
          begin{align}
          f_{W_2(tau_a)}(y)dy
          &=P[W_2(tau_a)in(y,y+dy)]\
          &=int_{t=0}^infty P[W_2(tau_a)in(y,y+dy)|tau_a=t]f_{tau_a}(t)dt\
          &=int_{t=0}^infty frac{1}{sqrt{2pi t}}e^{-frac{y^2}{2t}}dyfrac{1}{sqrt{2pi t^3}} ae^{-frac{a^2}{2t}}dt\
          &= left(int_{0}^infty frac{a}{2pi t^2}e^{-frac{y^2+a^2}{2t}}dtright)dy,quadtext{let }s=frac{1}{t},\
          &= left(int_infty^0frac{as^2}{2pi}e^{-frac{y^2+a^2}{2}s}left(-frac{1}{s^2}right)ds
          right)dy\
          &=left(
          int_0^inftyfrac{a}{2pi}e^{-frac{y^2+a^2}{2}s}ds
          right)dy\
          & = frac{a}{pi(a^2+y^2)}dy.
          end{align}






          share|cite|improve this answer









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            1 Answer
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            1 Answer
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            active

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            active

            oldest

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            1












            $begingroup$

            "Is the Cauchy density function the probability function of the second Brownian motion at time $tau_a$, when the first Brownian motion is stopped?" Yes.



            "The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $a$ increases, so will $tau_a$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $a$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.



            And the solution to the problem:
            begin{align}
            f_{W_2(tau_a)}(y)dy
            &=P[W_2(tau_a)in(y,y+dy)]\
            &=int_{t=0}^infty P[W_2(tau_a)in(y,y+dy)|tau_a=t]f_{tau_a}(t)dt\
            &=int_{t=0}^infty frac{1}{sqrt{2pi t}}e^{-frac{y^2}{2t}}dyfrac{1}{sqrt{2pi t^3}} ae^{-frac{a^2}{2t}}dt\
            &= left(int_{0}^infty frac{a}{2pi t^2}e^{-frac{y^2+a^2}{2t}}dtright)dy,quadtext{let }s=frac{1}{t},\
            &= left(int_infty^0frac{as^2}{2pi}e^{-frac{y^2+a^2}{2}s}left(-frac{1}{s^2}right)ds
            right)dy\
            &=left(
            int_0^inftyfrac{a}{2pi}e^{-frac{y^2+a^2}{2}s}ds
            right)dy\
            & = frac{a}{pi(a^2+y^2)}dy.
            end{align}






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              "Is the Cauchy density function the probability function of the second Brownian motion at time $tau_a$, when the first Brownian motion is stopped?" Yes.



              "The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $a$ increases, so will $tau_a$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $a$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.



              And the solution to the problem:
              begin{align}
              f_{W_2(tau_a)}(y)dy
              &=P[W_2(tau_a)in(y,y+dy)]\
              &=int_{t=0}^infty P[W_2(tau_a)in(y,y+dy)|tau_a=t]f_{tau_a}(t)dt\
              &=int_{t=0}^infty frac{1}{sqrt{2pi t}}e^{-frac{y^2}{2t}}dyfrac{1}{sqrt{2pi t^3}} ae^{-frac{a^2}{2t}}dt\
              &= left(int_{0}^infty frac{a}{2pi t^2}e^{-frac{y^2+a^2}{2t}}dtright)dy,quadtext{let }s=frac{1}{t},\
              &= left(int_infty^0frac{as^2}{2pi}e^{-frac{y^2+a^2}{2}s}left(-frac{1}{s^2}right)ds
              right)dy\
              &=left(
              int_0^inftyfrac{a}{2pi}e^{-frac{y^2+a^2}{2}s}ds
              right)dy\
              & = frac{a}{pi(a^2+y^2)}dy.
              end{align}






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                "Is the Cauchy density function the probability function of the second Brownian motion at time $tau_a$, when the first Brownian motion is stopped?" Yes.



                "The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $a$ increases, so will $tau_a$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $a$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.



                And the solution to the problem:
                begin{align}
                f_{W_2(tau_a)}(y)dy
                &=P[W_2(tau_a)in(y,y+dy)]\
                &=int_{t=0}^infty P[W_2(tau_a)in(y,y+dy)|tau_a=t]f_{tau_a}(t)dt\
                &=int_{t=0}^infty frac{1}{sqrt{2pi t}}e^{-frac{y^2}{2t}}dyfrac{1}{sqrt{2pi t^3}} ae^{-frac{a^2}{2t}}dt\
                &= left(int_{0}^infty frac{a}{2pi t^2}e^{-frac{y^2+a^2}{2t}}dtright)dy,quadtext{let }s=frac{1}{t},\
                &= left(int_infty^0frac{as^2}{2pi}e^{-frac{y^2+a^2}{2}s}left(-frac{1}{s^2}right)ds
                right)dy\
                &=left(
                int_0^inftyfrac{a}{2pi}e^{-frac{y^2+a^2}{2}s}ds
                right)dy\
                & = frac{a}{pi(a^2+y^2)}dy.
                end{align}






                share|cite|improve this answer









                $endgroup$



                "Is the Cauchy density function the probability function of the second Brownian motion at time $tau_a$, when the first Brownian motion is stopped?" Yes.



                "The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $a$ increases, so will $tau_a$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $a$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.



                And the solution to the problem:
                begin{align}
                f_{W_2(tau_a)}(y)dy
                &=P[W_2(tau_a)in(y,y+dy)]\
                &=int_{t=0}^infty P[W_2(tau_a)in(y,y+dy)|tau_a=t]f_{tau_a}(t)dt\
                &=int_{t=0}^infty frac{1}{sqrt{2pi t}}e^{-frac{y^2}{2t}}dyfrac{1}{sqrt{2pi t^3}} ae^{-frac{a^2}{2t}}dt\
                &= left(int_{0}^infty frac{a}{2pi t^2}e^{-frac{y^2+a^2}{2t}}dtright)dy,quadtext{let }s=frac{1}{t},\
                &= left(int_infty^0frac{as^2}{2pi}e^{-frac{y^2+a^2}{2}s}left(-frac{1}{s^2}right)ds
                right)dy\
                &=left(
                int_0^inftyfrac{a}{2pi}e^{-frac{y^2+a^2}{2}s}ds
                right)dy\
                & = frac{a}{pi(a^2+y^2)}dy.
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 22 at 8:55









                AddSupAddSup

                4851317




                4851317






























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