continuous local martingale brownian motion












2












$begingroup$


$B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
$X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



$f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
I would be thankful for any help.






probability-theory finance local-martingales







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$endgroup$

















    2












    $begingroup$


    $B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
    $X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



    $f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
    I would be thankful for any help.






    probability-theory finance local-martingales







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      $B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
      $X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



      $f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
      I would be thankful for any help.






      probability-theory finance local-martingales







      share|cite|improve this question











      $endgroup$




      $B$ is a one-dimensional Brownian motion and $X_t$ is defined as$\$
      $X_t:=f_{1-t}(B_t)$, $0le t<1$ and $0$, $1le t<infty$ where $f_s(x)=frac{1}{sqrt{2pi s}}e^{-frac{x^2}{2s}}$. I have to show that $X$ is a continuous local martingale.



      $f$ is continuous as a combination of continuous functions. I tried to use Ito's Formula, but i cannot see how it helps me. I also tried to show that it is a martingale by using stopping times.
      I would be thankful for any help.






      probability-theory finance local-martingales




      probability-theory finance local-martingales






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 17 '16 at 21:57







      Mathfreak

















      asked Apr 17 '16 at 12:56









      MathfreakMathfreak

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      9210






















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          $begingroup$

          Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
          $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
          and according to Ito's lemma:
          $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
          Computing the partial derivatives:
          begin{align}
          frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
          frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
          frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
          end{align}
          It follows that
          $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
          Hence, we obtain
          $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
          which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






          share|cite|improve this answer











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            $begingroup$

            Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
            $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
            and according to Ito's lemma:
            $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
            Computing the partial derivatives:
            begin{align}
            frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
            frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
            frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
            end{align}
            It follows that
            $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
            Hence, we obtain
            $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
            which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
              $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
              and according to Ito's lemma:
              $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
              Computing the partial derivatives:
              begin{align}
              frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
              frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
              frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
              end{align}
              It follows that
              $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
              Hence, we obtain
              $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
              which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
                $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
                and according to Ito's lemma:
                $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
                Computing the partial derivatives:
                begin{align}
                frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
                frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
                frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
                end{align}
                It follows that
                $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
                Hence, we obtain
                $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
                which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).






                share|cite|improve this answer











                $endgroup$



                Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have
                $$X_t = f_{1-t}(B_t) = frac{1}{sqrt{2pi (1-t)}} expleft{frac{-B_t^2}{2(1-t)}right }:=F(t,B_t),$$
                and according to Ito's lemma:
                $$dX_t = dF(t,B_t) = left(frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t^2}right)dt+ frac{partial F}{partial B_t}dW_t.$$
                Computing the partial derivatives:
                begin{align}
                frac{partial F}{partial t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{2sqrt{2pi(1-t)}}left(frac{1}{1-t}-frac{B_t^2}{(1-t)^2}right) \
                frac{partial F}{partial B_t} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-B_t}{1-t}right) \
                frac{partial^2 F}{partial B_t^2} &= frac{expleft{frac{-B_t^2}{2(1-t)}right}}{sqrt{2pi(1-t)}} left(frac{-1}{1-t}+frac{B_t^2}{(1-t)^2}right).
                end{align}
                It follows that
                $$frac{partial F}{partial t}+frac{1}{2}frac{partial^2 F}{partial B_t} = 0.$$
                Hence, we obtain
                $$dX_t = -frac{B_t}{1-t}X_tdW_t,$$
                which implies that $X_t$ is a (continuous) local martingale ($X_t$ is a stochastic integral w.r.t Brownian motion).







                share|cite|improve this answer














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                edited Dec 17 '18 at 22:24

























                answered Apr 17 '16 at 22:32









                SironSiron

                1,205614




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