On infinitesimal generator when the volatility of Brownian motion is given as a function of time












2












$begingroup$


Consider standard Brownian motion.
$dS_t=muspace d_t+sigmaspace dB_t$

For the process, the operator are given below.
$Af(x)=mu frac{df}{ds}+frac{sigma^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$



I would like to consider the case where σ is a function of time,for example $sigma(t)=cos(t)$



so,the brownian motion become,
$dS_t=muspace d_t+sigma(t)space dB_t$



In this case,Is it ok to think that the operator become the following?
$Af(x)=mu frac{df}{ds}+frac{sigma(t)^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$






stochastic-processes stochastic-calculus brownian-motion infinitesimals







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Consider standard Brownian motion.
    $dS_t=muspace d_t+sigmaspace dB_t$

    For the process, the operator are given below.
    $Af(x)=mu frac{df}{ds}+frac{sigma^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$



    I would like to consider the case where σ is a function of time,for example $sigma(t)=cos(t)$



    so,the brownian motion become,
    $dS_t=muspace d_t+sigma(t)space dB_t$



    In this case,Is it ok to think that the operator become the following?
    $Af(x)=mu frac{df}{ds}+frac{sigma(t)^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$






    stochastic-processes stochastic-calculus brownian-motion infinitesimals







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Consider standard Brownian motion.
      $dS_t=muspace d_t+sigmaspace dB_t$

      For the process, the operator are given below.
      $Af(x)=mu frac{df}{ds}+frac{sigma^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$



      I would like to consider the case where σ is a function of time,for example $sigma(t)=cos(t)$



      so,the brownian motion become,
      $dS_t=muspace d_t+sigma(t)space dB_t$



      In this case,Is it ok to think that the operator become the following?
      $Af(x)=mu frac{df}{ds}+frac{sigma(t)^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$






      stochastic-processes stochastic-calculus brownian-motion infinitesimals







      share|cite|improve this question











      $endgroup$




      Consider standard Brownian motion.
      $dS_t=muspace d_t+sigmaspace dB_t$

      For the process, the operator are given below.
      $Af(x)=mu frac{df}{ds}+frac{sigma^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$



      I would like to consider the case where σ is a function of time,for example $sigma(t)=cos(t)$



      so,the brownian motion become,
      $dS_t=muspace d_t+sigma(t)space dB_t$



      In this case,Is it ok to think that the operator become the following?
      $Af(x)=mu frac{df}{ds}+frac{sigma(t)^2}{2}frac{d^2f}{ds^2}+frac{df}{dt}$






      stochastic-processes stochastic-calculus brownian-motion infinitesimals




      stochastic-processes stochastic-calculus brownian-motion infinitesimals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 31 at 23:48







      Xminer

















      asked Jan 31 at 4:20









      XminerXminer

      1134




      1134






















          1 Answer
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          $begingroup$

          The answer to your question is yes (more or less). The short answer as to why is "because of Ito's lemma". A much more detailed explanation is given below, which I highly recommend taking the time to read and understand.



          Assumption. $sigma : [0, infty) rightarrow mathbb{R}$ is a continuous function of time.



          Let $f$ be any twice-continuously differentiable and compactly supported real-valued function.
          The infinitesimal generator at time $t$ (as applied to $f$) is defined as
          $$
          mathcal{A}_{t}f(x)=lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}
          $$
          where the process $X^{t,x}$ satisfies the SDE you wrote down:
          $$
          X_{s}=x+int_{t}^{s}mu dr+int_{t}^{s}sigma(r)dB_{r}qquadtext{for }s geq t.
          $$



          Use Ito's lemma on $f$ to get
          $$
          f(X_{t+h}^{t,x})-f(x)=int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds+int_{t}^{t+h}sigma(s)frac{partial f}{partial x}(X_{s}^{t,x})dB_{s}.
          $$
          Take expectations of both sides to get
          $$
          mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)=mathbb{E}left[int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})dsright]
          $$
          where we have used the fact that the expectation of the Ito integral is zero.
          Using the mean value theorem for integrals we get
          $$
          int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds=hleft(mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right)
          $$
          where $cequiv c(omega)$, which depends on the sample path $omega$, is a point between $t$ and $t+h$.
          Since $f$ is compactly supported, the dominated convergence theorem gives
          $$
          lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}=mathbb{E}left[lim_{hdownarrow0}left{ mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right} right].
          $$
          Using continuity and the facts that $sigma(c)rightarrow sigma(t)$ and $X_{c}rightarrow x$ as $h downarrow 0$ we can take the limit to conclude
          $$
          mathcal{A}_tf(x)=mufrac{partial f}{partial x}(x)+frac{(sigma(t))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(x).
          $$






          share|cite|improve this answer











          $endgroup$














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            $begingroup$

            The answer to your question is yes (more or less). The short answer as to why is "because of Ito's lemma". A much more detailed explanation is given below, which I highly recommend taking the time to read and understand.



            Assumption. $sigma : [0, infty) rightarrow mathbb{R}$ is a continuous function of time.



            Let $f$ be any twice-continuously differentiable and compactly supported real-valued function.
            The infinitesimal generator at time $t$ (as applied to $f$) is defined as
            $$
            mathcal{A}_{t}f(x)=lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}
            $$
            where the process $X^{t,x}$ satisfies the SDE you wrote down:
            $$
            X_{s}=x+int_{t}^{s}mu dr+int_{t}^{s}sigma(r)dB_{r}qquadtext{for }s geq t.
            $$



            Use Ito's lemma on $f$ to get
            $$
            f(X_{t+h}^{t,x})-f(x)=int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds+int_{t}^{t+h}sigma(s)frac{partial f}{partial x}(X_{s}^{t,x})dB_{s}.
            $$
            Take expectations of both sides to get
            $$
            mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)=mathbb{E}left[int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})dsright]
            $$
            where we have used the fact that the expectation of the Ito integral is zero.
            Using the mean value theorem for integrals we get
            $$
            int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds=hleft(mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right)
            $$
            where $cequiv c(omega)$, which depends on the sample path $omega$, is a point between $t$ and $t+h$.
            Since $f$ is compactly supported, the dominated convergence theorem gives
            $$
            lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}=mathbb{E}left[lim_{hdownarrow0}left{ mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right} right].
            $$
            Using continuity and the facts that $sigma(c)rightarrow sigma(t)$ and $X_{c}rightarrow x$ as $h downarrow 0$ we can take the limit to conclude
            $$
            mathcal{A}_tf(x)=mufrac{partial f}{partial x}(x)+frac{(sigma(t))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(x).
            $$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              The answer to your question is yes (more or less). The short answer as to why is "because of Ito's lemma". A much more detailed explanation is given below, which I highly recommend taking the time to read and understand.



              Assumption. $sigma : [0, infty) rightarrow mathbb{R}$ is a continuous function of time.



              Let $f$ be any twice-continuously differentiable and compactly supported real-valued function.
              The infinitesimal generator at time $t$ (as applied to $f$) is defined as
              $$
              mathcal{A}_{t}f(x)=lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}
              $$
              where the process $X^{t,x}$ satisfies the SDE you wrote down:
              $$
              X_{s}=x+int_{t}^{s}mu dr+int_{t}^{s}sigma(r)dB_{r}qquadtext{for }s geq t.
              $$



              Use Ito's lemma on $f$ to get
              $$
              f(X_{t+h}^{t,x})-f(x)=int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds+int_{t}^{t+h}sigma(s)frac{partial f}{partial x}(X_{s}^{t,x})dB_{s}.
              $$
              Take expectations of both sides to get
              $$
              mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)=mathbb{E}left[int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})dsright]
              $$
              where we have used the fact that the expectation of the Ito integral is zero.
              Using the mean value theorem for integrals we get
              $$
              int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds=hleft(mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right)
              $$
              where $cequiv c(omega)$, which depends on the sample path $omega$, is a point between $t$ and $t+h$.
              Since $f$ is compactly supported, the dominated convergence theorem gives
              $$
              lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}=mathbb{E}left[lim_{hdownarrow0}left{ mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right} right].
              $$
              Using continuity and the facts that $sigma(c)rightarrow sigma(t)$ and $X_{c}rightarrow x$ as $h downarrow 0$ we can take the limit to conclude
              $$
              mathcal{A}_tf(x)=mufrac{partial f}{partial x}(x)+frac{(sigma(t))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(x).
              $$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                The answer to your question is yes (more or less). The short answer as to why is "because of Ito's lemma". A much more detailed explanation is given below, which I highly recommend taking the time to read and understand.



                Assumption. $sigma : [0, infty) rightarrow mathbb{R}$ is a continuous function of time.



                Let $f$ be any twice-continuously differentiable and compactly supported real-valued function.
                The infinitesimal generator at time $t$ (as applied to $f$) is defined as
                $$
                mathcal{A}_{t}f(x)=lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}
                $$
                where the process $X^{t,x}$ satisfies the SDE you wrote down:
                $$
                X_{s}=x+int_{t}^{s}mu dr+int_{t}^{s}sigma(r)dB_{r}qquadtext{for }s geq t.
                $$



                Use Ito's lemma on $f$ to get
                $$
                f(X_{t+h}^{t,x})-f(x)=int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds+int_{t}^{t+h}sigma(s)frac{partial f}{partial x}(X_{s}^{t,x})dB_{s}.
                $$
                Take expectations of both sides to get
                $$
                mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)=mathbb{E}left[int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})dsright]
                $$
                where we have used the fact that the expectation of the Ito integral is zero.
                Using the mean value theorem for integrals we get
                $$
                int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds=hleft(mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right)
                $$
                where $cequiv c(omega)$, which depends on the sample path $omega$, is a point between $t$ and $t+h$.
                Since $f$ is compactly supported, the dominated convergence theorem gives
                $$
                lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}=mathbb{E}left[lim_{hdownarrow0}left{ mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right} right].
                $$
                Using continuity and the facts that $sigma(c)rightarrow sigma(t)$ and $X_{c}rightarrow x$ as $h downarrow 0$ we can take the limit to conclude
                $$
                mathcal{A}_tf(x)=mufrac{partial f}{partial x}(x)+frac{(sigma(t))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(x).
                $$






                share|cite|improve this answer











                $endgroup$



                The answer to your question is yes (more or less). The short answer as to why is "because of Ito's lemma". A much more detailed explanation is given below, which I highly recommend taking the time to read and understand.



                Assumption. $sigma : [0, infty) rightarrow mathbb{R}$ is a continuous function of time.



                Let $f$ be any twice-continuously differentiable and compactly supported real-valued function.
                The infinitesimal generator at time $t$ (as applied to $f$) is defined as
                $$
                mathcal{A}_{t}f(x)=lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}
                $$
                where the process $X^{t,x}$ satisfies the SDE you wrote down:
                $$
                X_{s}=x+int_{t}^{s}mu dr+int_{t}^{s}sigma(r)dB_{r}qquadtext{for }s geq t.
                $$



                Use Ito's lemma on $f$ to get
                $$
                f(X_{t+h}^{t,x})-f(x)=int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds+int_{t}^{t+h}sigma(s)frac{partial f}{partial x}(X_{s}^{t,x})dB_{s}.
                $$
                Take expectations of both sides to get
                $$
                mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)=mathbb{E}left[int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})dsright]
                $$
                where we have used the fact that the expectation of the Ito integral is zero.
                Using the mean value theorem for integrals we get
                $$
                int_{t}^{t+h}mufrac{partial f}{partial x}(X_{s}^{t,x})+frac{(sigma(s))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{s}^{t,x})ds=hleft(mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right)
                $$
                where $cequiv c(omega)$, which depends on the sample path $omega$, is a point between $t$ and $t+h$.
                Since $f$ is compactly supported, the dominated convergence theorem gives
                $$
                lim_{hdownarrow0}frac{mathbb{E}left[f(X_{t+h}^{t,x})right]-f(x)}{h}=mathbb{E}left[lim_{hdownarrow0}left{ mufrac{partial f}{partial x}(X_{c}^{t,x})+frac{(sigma(c))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(X_{c}^{t,x})right} right].
                $$
                Using continuity and the facts that $sigma(c)rightarrow sigma(t)$ and $X_{c}rightarrow x$ as $h downarrow 0$ we can take the limit to conclude
                $$
                mathcal{A}_tf(x)=mufrac{partial f}{partial x}(x)+frac{(sigma(t))^{2}}{2}frac{partial^{2}f}{partial x^{2}}(x).
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 31 at 15:53

























                answered Jan 31 at 5:29









                parsiadparsiad

                18.6k32453




                18.6k32453






























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