Solution to stochastic differential equation $dY = YdW$, where $Y(0)=1$?












0












$begingroup$


How to find the solution to stochastic differential equation (SDE):



$dY = YdW$,



where $W(t)$ is Brownian motion and $Y(0)=1$? The solution to the corresponding ordinary differential equation is easy to find:



$Y=e^{W(t)}$



However the solution to the SDE, according to Ito's formula, is supposed to be:



$Y=e^{W(t)-t/2}$,



and I don't understand how to find this solution. I tried to compare Ito's formula (left-hand side) with the SDE (right-hand side):



$(y'b+frac{y''}{2})dt+y'dW = YdW$,



where $b$ is some function of $t$, but wasn't able to make progress. Also I tried to take the derivative of $Y$:



$Y'=(W'(t)-1/2)e^{W(t)-t/2}$,



but could't see how this relates to Ito's formula (e.g. there is no $y''$ term).






stochastic-calculus stochastic-integrals







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    How to find the solution to stochastic differential equation (SDE):



    $dY = YdW$,



    where $W(t)$ is Brownian motion and $Y(0)=1$? The solution to the corresponding ordinary differential equation is easy to find:



    $Y=e^{W(t)}$



    However the solution to the SDE, according to Ito's formula, is supposed to be:



    $Y=e^{W(t)-t/2}$,



    and I don't understand how to find this solution. I tried to compare Ito's formula (left-hand side) with the SDE (right-hand side):



    $(y'b+frac{y''}{2})dt+y'dW = YdW$,



    where $b$ is some function of $t$, but wasn't able to make progress. Also I tried to take the derivative of $Y$:



    $Y'=(W'(t)-1/2)e^{W(t)-t/2}$,



    but could't see how this relates to Ito's formula (e.g. there is no $y''$ term).






    stochastic-calculus stochastic-integrals







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      How to find the solution to stochastic differential equation (SDE):



      $dY = YdW$,



      where $W(t)$ is Brownian motion and $Y(0)=1$? The solution to the corresponding ordinary differential equation is easy to find:



      $Y=e^{W(t)}$



      However the solution to the SDE, according to Ito's formula, is supposed to be:



      $Y=e^{W(t)-t/2}$,



      and I don't understand how to find this solution. I tried to compare Ito's formula (left-hand side) with the SDE (right-hand side):



      $(y'b+frac{y''}{2})dt+y'dW = YdW$,



      where $b$ is some function of $t$, but wasn't able to make progress. Also I tried to take the derivative of $Y$:



      $Y'=(W'(t)-1/2)e^{W(t)-t/2}$,



      but could't see how this relates to Ito's formula (e.g. there is no $y''$ term).






      stochastic-calculus stochastic-integrals







      share|cite|improve this question









      $endgroup$




      How to find the solution to stochastic differential equation (SDE):



      $dY = YdW$,



      where $W(t)$ is Brownian motion and $Y(0)=1$? The solution to the corresponding ordinary differential equation is easy to find:



      $Y=e^{W(t)}$



      However the solution to the SDE, according to Ito's formula, is supposed to be:



      $Y=e^{W(t)-t/2}$,



      and I don't understand how to find this solution. I tried to compare Ito's formula (left-hand side) with the SDE (right-hand side):



      $(y'b+frac{y''}{2})dt+y'dW = YdW$,



      where $b$ is some function of $t$, but wasn't able to make progress. Also I tried to take the derivative of $Y$:



      $Y'=(W'(t)-1/2)e^{W(t)-t/2}$,



      but could't see how this relates to Ito's formula (e.g. there is no $y''$ term).






      stochastic-calculus stochastic-integrals




      stochastic-calculus stochastic-integrals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 26 at 21:10









      user117933user117933

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          1 Answer
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          $begingroup$

          If you know what the solution should look like then you can just apply Ito's formula to the thing you expect the solution to be and check it works. To apply Ito's formula you want to write $Y$ as a function of $W(t)$ and $t$. We have
          $Y_t = f(t,W(t))$ where $f(t,x) = e^{x - t/2}$. Then applying Ito's formula gives
          $$dY_t = frac{partial f}{partial t} dt + frac{partial f}{partial x} dW_t + frac12 frac{partial^2 f}{partial x^2} dt$$
          since $W(t)$ has quadratic variation $t$. Computing the partial derivatives gives
          begin{align}
          dY_t &= - frac12 f(t,W(t)) dt + f(t,W(t)) dW(t) + frac12 f(t,W(t))dt \& = f(t,W(t))dW(t) = Y_t dW(t)
          end{align}
          as desired.




          If you didn't know the right solution beforehand, one way you might proceed is as follows. Notice that right hand side of Ito's formula is the same as you get in the usual calculus except for the second order Ito correction term. This means you might hope that you can start with a solution to the corresponding ODE and guess the right change to cancel out the second order correction. Here this method works.



          We have
          $$de^{W(t)} = e^{W(t)} dW(t) + frac12 e^{W(t)} dt$$
          Now a natural guess is to assume that you can separate the variable and write $Y(t) = f(t) e^{W(t)}$ for some $f$. By the product rule for Ito calculus (or an application of Ito's formula)
          $$d(f(t)e^{W(t)}) = e^{W(t)}f'(t) dt + f(t)e^{W(t)} dW(t) + frac12 f(t) e^{W(t)} dt$$
          and so $Y(t) = f(t) e^{W(t)}$ is the desired solution if and only if $f'(t) = -frac12 f(t)$ which is if and only if $f(t) = e^{-t/2}$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            If you know what the solution should look like then you can just apply Ito's formula to the thing you expect the solution to be and check it works. To apply Ito's formula you want to write $Y$ as a function of $W(t)$ and $t$. We have
            $Y_t = f(t,W(t))$ where $f(t,x) = e^{x - t/2}$. Then applying Ito's formula gives
            $$dY_t = frac{partial f}{partial t} dt + frac{partial f}{partial x} dW_t + frac12 frac{partial^2 f}{partial x^2} dt$$
            since $W(t)$ has quadratic variation $t$. Computing the partial derivatives gives
            begin{align}
            dY_t &= - frac12 f(t,W(t)) dt + f(t,W(t)) dW(t) + frac12 f(t,W(t))dt \& = f(t,W(t))dW(t) = Y_t dW(t)
            end{align}
            as desired.




            If you didn't know the right solution beforehand, one way you might proceed is as follows. Notice that right hand side of Ito's formula is the same as you get in the usual calculus except for the second order Ito correction term. This means you might hope that you can start with a solution to the corresponding ODE and guess the right change to cancel out the second order correction. Here this method works.



            We have
            $$de^{W(t)} = e^{W(t)} dW(t) + frac12 e^{W(t)} dt$$
            Now a natural guess is to assume that you can separate the variable and write $Y(t) = f(t) e^{W(t)}$ for some $f$. By the product rule for Ito calculus (or an application of Ito's formula)
            $$d(f(t)e^{W(t)}) = e^{W(t)}f'(t) dt + f(t)e^{W(t)} dW(t) + frac12 f(t) e^{W(t)} dt$$
            and so $Y(t) = f(t) e^{W(t)}$ is the desired solution if and only if $f'(t) = -frac12 f(t)$ which is if and only if $f(t) = e^{-t/2}$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              If you know what the solution should look like then you can just apply Ito's formula to the thing you expect the solution to be and check it works. To apply Ito's formula you want to write $Y$ as a function of $W(t)$ and $t$. We have
              $Y_t = f(t,W(t))$ where $f(t,x) = e^{x - t/2}$. Then applying Ito's formula gives
              $$dY_t = frac{partial f}{partial t} dt + frac{partial f}{partial x} dW_t + frac12 frac{partial^2 f}{partial x^2} dt$$
              since $W(t)$ has quadratic variation $t$. Computing the partial derivatives gives
              begin{align}
              dY_t &= - frac12 f(t,W(t)) dt + f(t,W(t)) dW(t) + frac12 f(t,W(t))dt \& = f(t,W(t))dW(t) = Y_t dW(t)
              end{align}
              as desired.




              If you didn't know the right solution beforehand, one way you might proceed is as follows. Notice that right hand side of Ito's formula is the same as you get in the usual calculus except for the second order Ito correction term. This means you might hope that you can start with a solution to the corresponding ODE and guess the right change to cancel out the second order correction. Here this method works.



              We have
              $$de^{W(t)} = e^{W(t)} dW(t) + frac12 e^{W(t)} dt$$
              Now a natural guess is to assume that you can separate the variable and write $Y(t) = f(t) e^{W(t)}$ for some $f$. By the product rule for Ito calculus (or an application of Ito's formula)
              $$d(f(t)e^{W(t)}) = e^{W(t)}f'(t) dt + f(t)e^{W(t)} dW(t) + frac12 f(t) e^{W(t)} dt$$
              and so $Y(t) = f(t) e^{W(t)}$ is the desired solution if and only if $f'(t) = -frac12 f(t)$ which is if and only if $f(t) = e^{-t/2}$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                If you know what the solution should look like then you can just apply Ito's formula to the thing you expect the solution to be and check it works. To apply Ito's formula you want to write $Y$ as a function of $W(t)$ and $t$. We have
                $Y_t = f(t,W(t))$ where $f(t,x) = e^{x - t/2}$. Then applying Ito's formula gives
                $$dY_t = frac{partial f}{partial t} dt + frac{partial f}{partial x} dW_t + frac12 frac{partial^2 f}{partial x^2} dt$$
                since $W(t)$ has quadratic variation $t$. Computing the partial derivatives gives
                begin{align}
                dY_t &= - frac12 f(t,W(t)) dt + f(t,W(t)) dW(t) + frac12 f(t,W(t))dt \& = f(t,W(t))dW(t) = Y_t dW(t)
                end{align}
                as desired.




                If you didn't know the right solution beforehand, one way you might proceed is as follows. Notice that right hand side of Ito's formula is the same as you get in the usual calculus except for the second order Ito correction term. This means you might hope that you can start with a solution to the corresponding ODE and guess the right change to cancel out the second order correction. Here this method works.



                We have
                $$de^{W(t)} = e^{W(t)} dW(t) + frac12 e^{W(t)} dt$$
                Now a natural guess is to assume that you can separate the variable and write $Y(t) = f(t) e^{W(t)}$ for some $f$. By the product rule for Ito calculus (or an application of Ito's formula)
                $$d(f(t)e^{W(t)}) = e^{W(t)}f'(t) dt + f(t)e^{W(t)} dW(t) + frac12 f(t) e^{W(t)} dt$$
                and so $Y(t) = f(t) e^{W(t)}$ is the desired solution if and only if $f'(t) = -frac12 f(t)$ which is if and only if $f(t) = e^{-t/2}$






                share|cite|improve this answer











                $endgroup$



                If you know what the solution should look like then you can just apply Ito's formula to the thing you expect the solution to be and check it works. To apply Ito's formula you want to write $Y$ as a function of $W(t)$ and $t$. We have
                $Y_t = f(t,W(t))$ where $f(t,x) = e^{x - t/2}$. Then applying Ito's formula gives
                $$dY_t = frac{partial f}{partial t} dt + frac{partial f}{partial x} dW_t + frac12 frac{partial^2 f}{partial x^2} dt$$
                since $W(t)$ has quadratic variation $t$. Computing the partial derivatives gives
                begin{align}
                dY_t &= - frac12 f(t,W(t)) dt + f(t,W(t)) dW(t) + frac12 f(t,W(t))dt \& = f(t,W(t))dW(t) = Y_t dW(t)
                end{align}
                as desired.




                If you didn't know the right solution beforehand, one way you might proceed is as follows. Notice that right hand side of Ito's formula is the same as you get in the usual calculus except for the second order Ito correction term. This means you might hope that you can start with a solution to the corresponding ODE and guess the right change to cancel out the second order correction. Here this method works.



                We have
                $$de^{W(t)} = e^{W(t)} dW(t) + frac12 e^{W(t)} dt$$
                Now a natural guess is to assume that you can separate the variable and write $Y(t) = f(t) e^{W(t)}$ for some $f$. By the product rule for Ito calculus (or an application of Ito's formula)
                $$d(f(t)e^{W(t)}) = e^{W(t)}f'(t) dt + f(t)e^{W(t)} dW(t) + frac12 f(t) e^{W(t)} dt$$
                and so $Y(t) = f(t) e^{W(t)}$ is the desired solution if and only if $f'(t) = -frac12 f(t)$ which is if and only if $f(t) = e^{-t/2}$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 26 at 21:28

























                answered Jan 26 at 21:17









                Rhys SteeleRhys Steele

                7,2901930




                7,2901930






























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