What is the quadratic variation of $e^{-tlambda}B_{frac{e^{2lambda t}-1}{2lambda}}$?












1












$begingroup$


Let us define $$X_t:=e^{-tlambda}B_{frac{e^{2lambda t}-1}{2lambda}},$$
where $B$ is a Brownian motion and $lambda>0$
How can I calculate quadratic variation $<X>_t$?



The first thing that comes to mind is to use Ito's formula and then calculate the quad. var. of the stochastic integral, as other parts are cont. differentiable and their quad. var is $0$. But the problem is, that I do not know how to apply it to this situation, as I know it only for functions with non-time changed BM's.



Is there a theorem that applies to this situation, should I consider replacing the changed time with a new variable?






brownian-motion stochastic-analysis quadratic-variation







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let us define $$X_t:=e^{-tlambda}B_{frac{e^{2lambda t}-1}{2lambda}},$$
    where $B$ is a Brownian motion and $lambda>0$
    How can I calculate quadratic variation $<X>_t$?



    The first thing that comes to mind is to use Ito's formula and then calculate the quad. var. of the stochastic integral, as other parts are cont. differentiable and their quad. var is $0$. But the problem is, that I do not know how to apply it to this situation, as I know it only for functions with non-time changed BM's.



    Is there a theorem that applies to this situation, should I consider replacing the changed time with a new variable?






    brownian-motion stochastic-analysis quadratic-variation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let us define $$X_t:=e^{-tlambda}B_{frac{e^{2lambda t}-1}{2lambda}},$$
      where $B$ is a Brownian motion and $lambda>0$
      How can I calculate quadratic variation $<X>_t$?



      The first thing that comes to mind is to use Ito's formula and then calculate the quad. var. of the stochastic integral, as other parts are cont. differentiable and their quad. var is $0$. But the problem is, that I do not know how to apply it to this situation, as I know it only for functions with non-time changed BM's.



      Is there a theorem that applies to this situation, should I consider replacing the changed time with a new variable?






      brownian-motion stochastic-analysis quadratic-variation







      share|cite|improve this question









      $endgroup$




      Let us define $$X_t:=e^{-tlambda}B_{frac{e^{2lambda t}-1}{2lambda}},$$
      where $B$ is a Brownian motion and $lambda>0$
      How can I calculate quadratic variation $<X>_t$?



      The first thing that comes to mind is to use Ito's formula and then calculate the quad. var. of the stochastic integral, as other parts are cont. differentiable and their quad. var is $0$. But the problem is, that I do not know how to apply it to this situation, as I know it only for functions with non-time changed BM's.



      Is there a theorem that applies to this situation, should I consider replacing the changed time with a new variable?






      brownian-motion stochastic-analysis quadratic-variation




      brownian-motion stochastic-analysis quadratic-variation






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 13 at 22:27









      RavonripRavonrip

      898




      898






















          1 Answer
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          $begingroup$

          Since $B_t$ is a standard Brownian motion, we always have
          $$
          left<Bright>_t=t
          $$
          for all forms of $t$, even if $t$ is written as a function of other variables. More precisely, if
          $$
          Y_t=B_{g(t)}
          $$
          for some strictly monotone function $g$ with $g(0)=0$, we have
          $$
          left<Yright>_t=g(t).
          $$
          Now, note that this $Y$ is a martingale, i.e.,
          $$
          mathbb{E}left(Y_t|Y_sright)=mathbb{E}left(B_{g(t)}|B_{g(s)}right)=B_{g(s)}=Y_s
          $$
          for all $tge sge 0$. Hence by martingale representation theorem, there must exist some predictable process $sigma_t$, such that
          $$
          {rm d}Y_t=sigma_t,{rm d}W_t,
          $$
          where, in case of any ambiguity, $W_t$ here denotes another standard Brownian motion. Note that by this formula, it is straightforward that
          $$
          {rm d}left<Yright>_t=sigma_t^2,{rm d}t.
          $$
          Hence, combined with the above $left<Yright>_t=g(t)$, it is obvious that
          $$
          sigma_t=sqrt{g'(t)},
          $$
          for which
          $$
          {rm d}Y_t=sqrt{g'(t)},{rm d}W_t.
          $$



          Thanks to this result, we may find Ito's formula for
          $$
          X_t=f(t),B_{g(t)}=f(t),Y_t
          $$
          with a general function $f$. We have
          $$
          {rm d}X_t=f'(t),Y_t,{rm d}t+f(t),{rm d}Y_t=f'(t),Y_t,{rm d}t+f(t)sqrt{g'(t)},{rm d}W_t,
          $$
          which indicates that
          $$
          {rm d}left<Xright>_t=f^2(t),g'(t).
          $$
          Therefore,
          $$
          left<Xright>_t=int_0^tf^2(s),g'(s),{rm d}s.
          $$



          Hope this explanation could be somewhat helpful for you.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            1 Answer
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            2












            $begingroup$

            Since $B_t$ is a standard Brownian motion, we always have
            $$
            left<Bright>_t=t
            $$
            for all forms of $t$, even if $t$ is written as a function of other variables. More precisely, if
            $$
            Y_t=B_{g(t)}
            $$
            for some strictly monotone function $g$ with $g(0)=0$, we have
            $$
            left<Yright>_t=g(t).
            $$
            Now, note that this $Y$ is a martingale, i.e.,
            $$
            mathbb{E}left(Y_t|Y_sright)=mathbb{E}left(B_{g(t)}|B_{g(s)}right)=B_{g(s)}=Y_s
            $$
            for all $tge sge 0$. Hence by martingale representation theorem, there must exist some predictable process $sigma_t$, such that
            $$
            {rm d}Y_t=sigma_t,{rm d}W_t,
            $$
            where, in case of any ambiguity, $W_t$ here denotes another standard Brownian motion. Note that by this formula, it is straightforward that
            $$
            {rm d}left<Yright>_t=sigma_t^2,{rm d}t.
            $$
            Hence, combined with the above $left<Yright>_t=g(t)$, it is obvious that
            $$
            sigma_t=sqrt{g'(t)},
            $$
            for which
            $$
            {rm d}Y_t=sqrt{g'(t)},{rm d}W_t.
            $$



            Thanks to this result, we may find Ito's formula for
            $$
            X_t=f(t),B_{g(t)}=f(t),Y_t
            $$
            with a general function $f$. We have
            $$
            {rm d}X_t=f'(t),Y_t,{rm d}t+f(t),{rm d}Y_t=f'(t),Y_t,{rm d}t+f(t)sqrt{g'(t)},{rm d}W_t,
            $$
            which indicates that
            $$
            {rm d}left<Xright>_t=f^2(t),g'(t).
            $$
            Therefore,
            $$
            left<Xright>_t=int_0^tf^2(s),g'(s),{rm d}s.
            $$



            Hope this explanation could be somewhat helpful for you.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Since $B_t$ is a standard Brownian motion, we always have
              $$
              left<Bright>_t=t
              $$
              for all forms of $t$, even if $t$ is written as a function of other variables. More precisely, if
              $$
              Y_t=B_{g(t)}
              $$
              for some strictly monotone function $g$ with $g(0)=0$, we have
              $$
              left<Yright>_t=g(t).
              $$
              Now, note that this $Y$ is a martingale, i.e.,
              $$
              mathbb{E}left(Y_t|Y_sright)=mathbb{E}left(B_{g(t)}|B_{g(s)}right)=B_{g(s)}=Y_s
              $$
              for all $tge sge 0$. Hence by martingale representation theorem, there must exist some predictable process $sigma_t$, such that
              $$
              {rm d}Y_t=sigma_t,{rm d}W_t,
              $$
              where, in case of any ambiguity, $W_t$ here denotes another standard Brownian motion. Note that by this formula, it is straightforward that
              $$
              {rm d}left<Yright>_t=sigma_t^2,{rm d}t.
              $$
              Hence, combined with the above $left<Yright>_t=g(t)$, it is obvious that
              $$
              sigma_t=sqrt{g'(t)},
              $$
              for which
              $$
              {rm d}Y_t=sqrt{g'(t)},{rm d}W_t.
              $$



              Thanks to this result, we may find Ito's formula for
              $$
              X_t=f(t),B_{g(t)}=f(t),Y_t
              $$
              with a general function $f$. We have
              $$
              {rm d}X_t=f'(t),Y_t,{rm d}t+f(t),{rm d}Y_t=f'(t),Y_t,{rm d}t+f(t)sqrt{g'(t)},{rm d}W_t,
              $$
              which indicates that
              $$
              {rm d}left<Xright>_t=f^2(t),g'(t).
              $$
              Therefore,
              $$
              left<Xright>_t=int_0^tf^2(s),g'(s),{rm d}s.
              $$



              Hope this explanation could be somewhat helpful for you.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Since $B_t$ is a standard Brownian motion, we always have
                $$
                left<Bright>_t=t
                $$
                for all forms of $t$, even if $t$ is written as a function of other variables. More precisely, if
                $$
                Y_t=B_{g(t)}
                $$
                for some strictly monotone function $g$ with $g(0)=0$, we have
                $$
                left<Yright>_t=g(t).
                $$
                Now, note that this $Y$ is a martingale, i.e.,
                $$
                mathbb{E}left(Y_t|Y_sright)=mathbb{E}left(B_{g(t)}|B_{g(s)}right)=B_{g(s)}=Y_s
                $$
                for all $tge sge 0$. Hence by martingale representation theorem, there must exist some predictable process $sigma_t$, such that
                $$
                {rm d}Y_t=sigma_t,{rm d}W_t,
                $$
                where, in case of any ambiguity, $W_t$ here denotes another standard Brownian motion. Note that by this formula, it is straightforward that
                $$
                {rm d}left<Yright>_t=sigma_t^2,{rm d}t.
                $$
                Hence, combined with the above $left<Yright>_t=g(t)$, it is obvious that
                $$
                sigma_t=sqrt{g'(t)},
                $$
                for which
                $$
                {rm d}Y_t=sqrt{g'(t)},{rm d}W_t.
                $$



                Thanks to this result, we may find Ito's formula for
                $$
                X_t=f(t),B_{g(t)}=f(t),Y_t
                $$
                with a general function $f$. We have
                $$
                {rm d}X_t=f'(t),Y_t,{rm d}t+f(t),{rm d}Y_t=f'(t),Y_t,{rm d}t+f(t)sqrt{g'(t)},{rm d}W_t,
                $$
                which indicates that
                $$
                {rm d}left<Xright>_t=f^2(t),g'(t).
                $$
                Therefore,
                $$
                left<Xright>_t=int_0^tf^2(s),g'(s),{rm d}s.
                $$



                Hope this explanation could be somewhat helpful for you.






                share|cite|improve this answer











                $endgroup$



                Since $B_t$ is a standard Brownian motion, we always have
                $$
                left<Bright>_t=t
                $$
                for all forms of $t$, even if $t$ is written as a function of other variables. More precisely, if
                $$
                Y_t=B_{g(t)}
                $$
                for some strictly monotone function $g$ with $g(0)=0$, we have
                $$
                left<Yright>_t=g(t).
                $$
                Now, note that this $Y$ is a martingale, i.e.,
                $$
                mathbb{E}left(Y_t|Y_sright)=mathbb{E}left(B_{g(t)}|B_{g(s)}right)=B_{g(s)}=Y_s
                $$
                for all $tge sge 0$. Hence by martingale representation theorem, there must exist some predictable process $sigma_t$, such that
                $$
                {rm d}Y_t=sigma_t,{rm d}W_t,
                $$
                where, in case of any ambiguity, $W_t$ here denotes another standard Brownian motion. Note that by this formula, it is straightforward that
                $$
                {rm d}left<Yright>_t=sigma_t^2,{rm d}t.
                $$
                Hence, combined with the above $left<Yright>_t=g(t)$, it is obvious that
                $$
                sigma_t=sqrt{g'(t)},
                $$
                for which
                $$
                {rm d}Y_t=sqrt{g'(t)},{rm d}W_t.
                $$



                Thanks to this result, we may find Ito's formula for
                $$
                X_t=f(t),B_{g(t)}=f(t),Y_t
                $$
                with a general function $f$. We have
                $$
                {rm d}X_t=f'(t),Y_t,{rm d}t+f(t),{rm d}Y_t=f'(t),Y_t,{rm d}t+f(t)sqrt{g'(t)},{rm d}W_t,
                $$
                which indicates that
                $$
                {rm d}left<Xright>_t=f^2(t),g'(t).
                $$
                Therefore,
                $$
                left<Xright>_t=int_0^tf^2(s),g'(s),{rm d}s.
                $$



                Hope this explanation could be somewhat helpful for you.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 14 at 0:56

























                answered Jan 14 at 0:41









                hypernovahypernova

                4,834414




                4,834414






























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