Applying Ito's formula to $D^x(t)=(partial/partial x)S(t)$












1












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Let $gamma:[0,T]times (0,infty)to mathbb{R}$ be continuous, bounded above, $(partial/partial s [gamma(t,s)])$ is continuous in $(s,t)$. Consider the following process $$dS_{gamma}^x(t)=S^x_{gamma}(t)[r(t)dt+gamma(t,S^x_{gamma}(t))dW(t)]$$ where $W(t)$ is a one dimensional Brownian motion and $r(t)$ is an adapted process and $S^x_{gamma}(0)=x$.




Now define the process $D^x(t)=(partial/partial x)S^x_{gamma}(t)$. And this apparently satisfies $$dD^x(t)=D^x(t)big[r(t)dt+frac{partial}{partial s}rho(t,S^x_{gamma}(t))dW(t)big]rightarrow(1)$$ where $rho(t,s):=sgamma(t,s). $ I am not quite sure how equation (1) can be derived since $gamma$ depends on $S^x_{gamma}$. Maybe the idea is to apply Ito's formula to $D^x(t)$ by defining a function $f(t,s):=(partial/partial x)s $ and then the ito rule gives us $df=f_tdt+f_sdS^x_{gamma}(t)+frac{1}{2}f_{ss}dS^x_{gamma}(t)dS^x_{gamma}(t)$. I am not sure if what I am doing is even slightly correct. Any help is appreciated!






stochastic-calculus brownian-motion finance







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$endgroup$












  • $begingroup$
    See Equation (2.16) in p. 261 and the proof around it: books.google.com/…
    $endgroup$
    – AddSup
    Jan 23 at 10:43










  • $begingroup$
    In case you're just talking about how to compute the volatility, you just need to differentiate it: $frac{partial}{partial x}(S^xgamma(S^x))=D^xgamma(S^x)+S^xleft(frac{partial}{partial s}gamma(S^x)right)D^x=D^xleft(gamma(S^x)+S^xfrac{partial}{partial s}gamma(S^x)right)$.
    $endgroup$
    – AddSup
    Jan 23 at 17:30












  • $begingroup$
    @AddSup I want to find $dD^x$
    $endgroup$
    – Heisenberg
    Jan 24 at 19:32










  • $begingroup$
    I'm saying, the volatility of $dD^x$ is calculated as above, and the drift is calculated, of course, in the same manner: $frac{partial}{partial x}(S^xr)=D^xr$.
    $endgroup$
    – AddSup
    Jan 25 at 4:29
















1












$begingroup$


Let $gamma:[0,T]times (0,infty)to mathbb{R}$ be continuous, bounded above, $(partial/partial s [gamma(t,s)])$ is continuous in $(s,t)$. Consider the following process $$dS_{gamma}^x(t)=S^x_{gamma}(t)[r(t)dt+gamma(t,S^x_{gamma}(t))dW(t)]$$ where $W(t)$ is a one dimensional Brownian motion and $r(t)$ is an adapted process and $S^x_{gamma}(0)=x$.




Now define the process $D^x(t)=(partial/partial x)S^x_{gamma}(t)$. And this apparently satisfies $$dD^x(t)=D^x(t)big[r(t)dt+frac{partial}{partial s}rho(t,S^x_{gamma}(t))dW(t)big]rightarrow(1)$$ where $rho(t,s):=sgamma(t,s). $ I am not quite sure how equation (1) can be derived since $gamma$ depends on $S^x_{gamma}$. Maybe the idea is to apply Ito's formula to $D^x(t)$ by defining a function $f(t,s):=(partial/partial x)s $ and then the ito rule gives us $df=f_tdt+f_sdS^x_{gamma}(t)+frac{1}{2}f_{ss}dS^x_{gamma}(t)dS^x_{gamma}(t)$. I am not sure if what I am doing is even slightly correct. Any help is appreciated!






stochastic-calculus brownian-motion finance







share|cite|improve this question











$endgroup$












  • $begingroup$
    See Equation (2.16) in p. 261 and the proof around it: books.google.com/…
    $endgroup$
    – AddSup
    Jan 23 at 10:43










  • $begingroup$
    In case you're just talking about how to compute the volatility, you just need to differentiate it: $frac{partial}{partial x}(S^xgamma(S^x))=D^xgamma(S^x)+S^xleft(frac{partial}{partial s}gamma(S^x)right)D^x=D^xleft(gamma(S^x)+S^xfrac{partial}{partial s}gamma(S^x)right)$.
    $endgroup$
    – AddSup
    Jan 23 at 17:30












  • $begingroup$
    @AddSup I want to find $dD^x$
    $endgroup$
    – Heisenberg
    Jan 24 at 19:32










  • $begingroup$
    I'm saying, the volatility of $dD^x$ is calculated as above, and the drift is calculated, of course, in the same manner: $frac{partial}{partial x}(S^xr)=D^xr$.
    $endgroup$
    – AddSup
    Jan 25 at 4:29














1












1








1





$begingroup$


Let $gamma:[0,T]times (0,infty)to mathbb{R}$ be continuous, bounded above, $(partial/partial s [gamma(t,s)])$ is continuous in $(s,t)$. Consider the following process $$dS_{gamma}^x(t)=S^x_{gamma}(t)[r(t)dt+gamma(t,S^x_{gamma}(t))dW(t)]$$ where $W(t)$ is a one dimensional Brownian motion and $r(t)$ is an adapted process and $S^x_{gamma}(0)=x$.




Now define the process $D^x(t)=(partial/partial x)S^x_{gamma}(t)$. And this apparently satisfies $$dD^x(t)=D^x(t)big[r(t)dt+frac{partial}{partial s}rho(t,S^x_{gamma}(t))dW(t)big]rightarrow(1)$$ where $rho(t,s):=sgamma(t,s). $ I am not quite sure how equation (1) can be derived since $gamma$ depends on $S^x_{gamma}$. Maybe the idea is to apply Ito's formula to $D^x(t)$ by defining a function $f(t,s):=(partial/partial x)s $ and then the ito rule gives us $df=f_tdt+f_sdS^x_{gamma}(t)+frac{1}{2}f_{ss}dS^x_{gamma}(t)dS^x_{gamma}(t)$. I am not sure if what I am doing is even slightly correct. Any help is appreciated!






stochastic-calculus brownian-motion finance







share|cite|improve this question











$endgroup$




Let $gamma:[0,T]times (0,infty)to mathbb{R}$ be continuous, bounded above, $(partial/partial s [gamma(t,s)])$ is continuous in $(s,t)$. Consider the following process $$dS_{gamma}^x(t)=S^x_{gamma}(t)[r(t)dt+gamma(t,S^x_{gamma}(t))dW(t)]$$ where $W(t)$ is a one dimensional Brownian motion and $r(t)$ is an adapted process and $S^x_{gamma}(0)=x$.




Now define the process $D^x(t)=(partial/partial x)S^x_{gamma}(t)$. And this apparently satisfies $$dD^x(t)=D^x(t)big[r(t)dt+frac{partial}{partial s}rho(t,S^x_{gamma}(t))dW(t)big]rightarrow(1)$$ where $rho(t,s):=sgamma(t,s). $ I am not quite sure how equation (1) can be derived since $gamma$ depends on $S^x_{gamma}$. Maybe the idea is to apply Ito's formula to $D^x(t)$ by defining a function $f(t,s):=(partial/partial x)s $ and then the ito rule gives us $df=f_tdt+f_sdS^x_{gamma}(t)+frac{1}{2}f_{ss}dS^x_{gamma}(t)dS^x_{gamma}(t)$. I am not sure if what I am doing is even slightly correct. Any help is appreciated!






stochastic-calculus brownian-motion finance




stochastic-calculus brownian-motion finance






share|cite|improve this question















share|cite|improve this question













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edited Jan 25 at 5:08







Heisenberg

















asked Jan 23 at 4:40









HeisenbergHeisenberg

1,3131741




1,3131741












  • $begingroup$
    See Equation (2.16) in p. 261 and the proof around it: books.google.com/…
    $endgroup$
    – AddSup
    Jan 23 at 10:43










  • $begingroup$
    In case you're just talking about how to compute the volatility, you just need to differentiate it: $frac{partial}{partial x}(S^xgamma(S^x))=D^xgamma(S^x)+S^xleft(frac{partial}{partial s}gamma(S^x)right)D^x=D^xleft(gamma(S^x)+S^xfrac{partial}{partial s}gamma(S^x)right)$.
    $endgroup$
    – AddSup
    Jan 23 at 17:30












  • $begingroup$
    @AddSup I want to find $dD^x$
    $endgroup$
    – Heisenberg
    Jan 24 at 19:32










  • $begingroup$
    I'm saying, the volatility of $dD^x$ is calculated as above, and the drift is calculated, of course, in the same manner: $frac{partial}{partial x}(S^xr)=D^xr$.
    $endgroup$
    – AddSup
    Jan 25 at 4:29


















  • $begingroup$
    See Equation (2.16) in p. 261 and the proof around it: books.google.com/…
    $endgroup$
    – AddSup
    Jan 23 at 10:43










  • $begingroup$
    In case you're just talking about how to compute the volatility, you just need to differentiate it: $frac{partial}{partial x}(S^xgamma(S^x))=D^xgamma(S^x)+S^xleft(frac{partial}{partial s}gamma(S^x)right)D^x=D^xleft(gamma(S^x)+S^xfrac{partial}{partial s}gamma(S^x)right)$.
    $endgroup$
    – AddSup
    Jan 23 at 17:30












  • $begingroup$
    @AddSup I want to find $dD^x$
    $endgroup$
    – Heisenberg
    Jan 24 at 19:32










  • $begingroup$
    I'm saying, the volatility of $dD^x$ is calculated as above, and the drift is calculated, of course, in the same manner: $frac{partial}{partial x}(S^xr)=D^xr$.
    $endgroup$
    – AddSup
    Jan 25 at 4:29
















$begingroup$
See Equation (2.16) in p. 261 and the proof around it: books.google.com/…
$endgroup$
– AddSup
Jan 23 at 10:43




$begingroup$
See Equation (2.16) in p. 261 and the proof around it: books.google.com/…
$endgroup$
– AddSup
Jan 23 at 10:43












$begingroup$
In case you're just talking about how to compute the volatility, you just need to differentiate it: $frac{partial}{partial x}(S^xgamma(S^x))=D^xgamma(S^x)+S^xleft(frac{partial}{partial s}gamma(S^x)right)D^x=D^xleft(gamma(S^x)+S^xfrac{partial}{partial s}gamma(S^x)right)$.
$endgroup$
– AddSup
Jan 23 at 17:30






$begingroup$
In case you're just talking about how to compute the volatility, you just need to differentiate it: $frac{partial}{partial x}(S^xgamma(S^x))=D^xgamma(S^x)+S^xleft(frac{partial}{partial s}gamma(S^x)right)D^x=D^xleft(gamma(S^x)+S^xfrac{partial}{partial s}gamma(S^x)right)$.
$endgroup$
– AddSup
Jan 23 at 17:30














$begingroup$
@AddSup I want to find $dD^x$
$endgroup$
– Heisenberg
Jan 24 at 19:32




$begingroup$
@AddSup I want to find $dD^x$
$endgroup$
– Heisenberg
Jan 24 at 19:32












$begingroup$
I'm saying, the volatility of $dD^x$ is calculated as above, and the drift is calculated, of course, in the same manner: $frac{partial}{partial x}(S^xr)=D^xr$.
$endgroup$
– AddSup
Jan 25 at 4:29




$begingroup$
I'm saying, the volatility of $dD^x$ is calculated as above, and the drift is calculated, of course, in the same manner: $frac{partial}{partial x}(S^xr)=D^xr$.
$endgroup$
– AddSup
Jan 25 at 4:29










1 Answer
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$begingroup$

For your Ito process $S_{gamma}^x(t)$, we focus on its time-differentiation as we write ${rm d}S_{gamma}^x(t)$. Since $x$ is an auxiliary parameter independent from $t$, we have that ${rm d}$ and $partial/partial x$ commute with each other, i.e.,
$$
frac{partial}{partial x}{rm d}S_{gamma}^x(t)={rm d}left(frac{partial}{partial x}S_{gamma}^x(t)right)={rm d}D^x(t).
$$
This is not a magic; it is just as if
$$
frac{partial}{partial x}left(frac{partial}{partial t}f(x,t)right)=frac{partial}{partial t}left(frac{partial}{partial x}f(x,t)right)
$$
for some differentiable function $f=f(x,t)$.



Therefore, act $partial/partial x$ on both sides of
$$
{rm d}S_{gamma}^x(t)=S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right),
$$
and we obtain
begin{align}
{rm d}D^x(t)&=frac{partial}{partial x}{rm d}S_{gamma}^x(t)\
&=frac{partial}{partial x}left(S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)right)\
&=frac{partial}{partial x}S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
&+S_{gamma}^x(t)frac{partial}{partial x}left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
&=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
&+S_{gamma}^x(t),frac{partial}{partial x}gamma(t,S_{gamma}^x(t)),{rm d}W(t)\
&=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
&+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))frac{partial}{partial x}S_{gamma}^x(t),{rm d}W(t)\
&=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
&+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t)),D^x(t),{rm d}W(t)\
&=D^x(t)left(r(t),{rm d}t+left(gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))right){rm d}W(t)right).
end{align}



Note that
begin{align}
&gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))\
&=left(gamma(t,s)+sfrac{partialgamma}{partial s}(t,s)right)bigg|_{s=S_{gamma}^x(t)}\
&=frac{partialrho}{partial s}(t,s)bigg|_{s=S_{gamma}^x(t)}\
&=frac{partialrho}{partial s}(t,S_{gamma}^x(t)).
end{align}
Hence, we eventually obtain
$$
{rm d}D^x(t)=D^x(t)left(r(t),{rm d}t+frac{partialrho}{partial s}(t,S_{gamma}^x(t)),{rm d}W(t)right),
$$
as is desired.



In the above derivation, we merely use the commutativity between ${rm d}$ and $partial/partial x$; Ito's formula is not necessary here.






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    1 Answer
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    active

    oldest

    votes









    1





    +100







    $begingroup$

    For your Ito process $S_{gamma}^x(t)$, we focus on its time-differentiation as we write ${rm d}S_{gamma}^x(t)$. Since $x$ is an auxiliary parameter independent from $t$, we have that ${rm d}$ and $partial/partial x$ commute with each other, i.e.,
    $$
    frac{partial}{partial x}{rm d}S_{gamma}^x(t)={rm d}left(frac{partial}{partial x}S_{gamma}^x(t)right)={rm d}D^x(t).
    $$
    This is not a magic; it is just as if
    $$
    frac{partial}{partial x}left(frac{partial}{partial t}f(x,t)right)=frac{partial}{partial t}left(frac{partial}{partial x}f(x,t)right)
    $$
    for some differentiable function $f=f(x,t)$.



    Therefore, act $partial/partial x$ on both sides of
    $$
    {rm d}S_{gamma}^x(t)=S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right),
    $$
    and we obtain
    begin{align}
    {rm d}D^x(t)&=frac{partial}{partial x}{rm d}S_{gamma}^x(t)\
    &=frac{partial}{partial x}left(S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)right)\
    &=frac{partial}{partial x}S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
    &+S_{gamma}^x(t)frac{partial}{partial x}left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
    &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
    &+S_{gamma}^x(t),frac{partial}{partial x}gamma(t,S_{gamma}^x(t)),{rm d}W(t)\
    &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
    &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))frac{partial}{partial x}S_{gamma}^x(t),{rm d}W(t)\
    &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
    &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t)),D^x(t),{rm d}W(t)\
    &=D^x(t)left(r(t),{rm d}t+left(gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))right){rm d}W(t)right).
    end{align}



    Note that
    begin{align}
    &gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))\
    &=left(gamma(t,s)+sfrac{partialgamma}{partial s}(t,s)right)bigg|_{s=S_{gamma}^x(t)}\
    &=frac{partialrho}{partial s}(t,s)bigg|_{s=S_{gamma}^x(t)}\
    &=frac{partialrho}{partial s}(t,S_{gamma}^x(t)).
    end{align}
    Hence, we eventually obtain
    $$
    {rm d}D^x(t)=D^x(t)left(r(t),{rm d}t+frac{partialrho}{partial s}(t,S_{gamma}^x(t)),{rm d}W(t)right),
    $$
    as is desired.



    In the above derivation, we merely use the commutativity between ${rm d}$ and $partial/partial x$; Ito's formula is not necessary here.






    share|cite|improve this answer









    $endgroup$


















      1





      +100







      $begingroup$

      For your Ito process $S_{gamma}^x(t)$, we focus on its time-differentiation as we write ${rm d}S_{gamma}^x(t)$. Since $x$ is an auxiliary parameter independent from $t$, we have that ${rm d}$ and $partial/partial x$ commute with each other, i.e.,
      $$
      frac{partial}{partial x}{rm d}S_{gamma}^x(t)={rm d}left(frac{partial}{partial x}S_{gamma}^x(t)right)={rm d}D^x(t).
      $$
      This is not a magic; it is just as if
      $$
      frac{partial}{partial x}left(frac{partial}{partial t}f(x,t)right)=frac{partial}{partial t}left(frac{partial}{partial x}f(x,t)right)
      $$
      for some differentiable function $f=f(x,t)$.



      Therefore, act $partial/partial x$ on both sides of
      $$
      {rm d}S_{gamma}^x(t)=S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right),
      $$
      and we obtain
      begin{align}
      {rm d}D^x(t)&=frac{partial}{partial x}{rm d}S_{gamma}^x(t)\
      &=frac{partial}{partial x}left(S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)right)\
      &=frac{partial}{partial x}S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
      &+S_{gamma}^x(t)frac{partial}{partial x}left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
      &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
      &+S_{gamma}^x(t),frac{partial}{partial x}gamma(t,S_{gamma}^x(t)),{rm d}W(t)\
      &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
      &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))frac{partial}{partial x}S_{gamma}^x(t),{rm d}W(t)\
      &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
      &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t)),D^x(t),{rm d}W(t)\
      &=D^x(t)left(r(t),{rm d}t+left(gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))right){rm d}W(t)right).
      end{align}



      Note that
      begin{align}
      &gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))\
      &=left(gamma(t,s)+sfrac{partialgamma}{partial s}(t,s)right)bigg|_{s=S_{gamma}^x(t)}\
      &=frac{partialrho}{partial s}(t,s)bigg|_{s=S_{gamma}^x(t)}\
      &=frac{partialrho}{partial s}(t,S_{gamma}^x(t)).
      end{align}
      Hence, we eventually obtain
      $$
      {rm d}D^x(t)=D^x(t)left(r(t),{rm d}t+frac{partialrho}{partial s}(t,S_{gamma}^x(t)),{rm d}W(t)right),
      $$
      as is desired.



      In the above derivation, we merely use the commutativity between ${rm d}$ and $partial/partial x$; Ito's formula is not necessary here.






      share|cite|improve this answer









      $endgroup$
















        1





        +100







        1





        +100



        1




        +100



        $begingroup$

        For your Ito process $S_{gamma}^x(t)$, we focus on its time-differentiation as we write ${rm d}S_{gamma}^x(t)$. Since $x$ is an auxiliary parameter independent from $t$, we have that ${rm d}$ and $partial/partial x$ commute with each other, i.e.,
        $$
        frac{partial}{partial x}{rm d}S_{gamma}^x(t)={rm d}left(frac{partial}{partial x}S_{gamma}^x(t)right)={rm d}D^x(t).
        $$
        This is not a magic; it is just as if
        $$
        frac{partial}{partial x}left(frac{partial}{partial t}f(x,t)right)=frac{partial}{partial t}left(frac{partial}{partial x}f(x,t)right)
        $$
        for some differentiable function $f=f(x,t)$.



        Therefore, act $partial/partial x$ on both sides of
        $$
        {rm d}S_{gamma}^x(t)=S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right),
        $$
        and we obtain
        begin{align}
        {rm d}D^x(t)&=frac{partial}{partial x}{rm d}S_{gamma}^x(t)\
        &=frac{partial}{partial x}left(S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)right)\
        &=frac{partial}{partial x}S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t)frac{partial}{partial x}left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t),frac{partial}{partial x}gamma(t,S_{gamma}^x(t)),{rm d}W(t)\
        &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))frac{partial}{partial x}S_{gamma}^x(t),{rm d}W(t)\
        &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t)),D^x(t),{rm d}W(t)\
        &=D^x(t)left(r(t),{rm d}t+left(gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))right){rm d}W(t)right).
        end{align}



        Note that
        begin{align}
        &gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))\
        &=left(gamma(t,s)+sfrac{partialgamma}{partial s}(t,s)right)bigg|_{s=S_{gamma}^x(t)}\
        &=frac{partialrho}{partial s}(t,s)bigg|_{s=S_{gamma}^x(t)}\
        &=frac{partialrho}{partial s}(t,S_{gamma}^x(t)).
        end{align}
        Hence, we eventually obtain
        $$
        {rm d}D^x(t)=D^x(t)left(r(t),{rm d}t+frac{partialrho}{partial s}(t,S_{gamma}^x(t)),{rm d}W(t)right),
        $$
        as is desired.



        In the above derivation, we merely use the commutativity between ${rm d}$ and $partial/partial x$; Ito's formula is not necessary here.






        share|cite|improve this answer









        $endgroup$



        For your Ito process $S_{gamma}^x(t)$, we focus on its time-differentiation as we write ${rm d}S_{gamma}^x(t)$. Since $x$ is an auxiliary parameter independent from $t$, we have that ${rm d}$ and $partial/partial x$ commute with each other, i.e.,
        $$
        frac{partial}{partial x}{rm d}S_{gamma}^x(t)={rm d}left(frac{partial}{partial x}S_{gamma}^x(t)right)={rm d}D^x(t).
        $$
        This is not a magic; it is just as if
        $$
        frac{partial}{partial x}left(frac{partial}{partial t}f(x,t)right)=frac{partial}{partial t}left(frac{partial}{partial x}f(x,t)right)
        $$
        for some differentiable function $f=f(x,t)$.



        Therefore, act $partial/partial x$ on both sides of
        $$
        {rm d}S_{gamma}^x(t)=S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right),
        $$
        and we obtain
        begin{align}
        {rm d}D^x(t)&=frac{partial}{partial x}{rm d}S_{gamma}^x(t)\
        &=frac{partial}{partial x}left(S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)right)\
        &=frac{partial}{partial x}S_{gamma}^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t)frac{partial}{partial x}left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t),frac{partial}{partial x}gamma(t,S_{gamma}^x(t)),{rm d}W(t)\
        &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))frac{partial}{partial x}S_{gamma}^x(t),{rm d}W(t)\
        &=D^x(t)left(r(t),{rm d}t+gamma(t,S_{gamma}^x(t)),{rm d}W(t)right)\
        &+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t)),D^x(t),{rm d}W(t)\
        &=D^x(t)left(r(t),{rm d}t+left(gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))right){rm d}W(t)right).
        end{align}



        Note that
        begin{align}
        &gamma(t,S_{gamma}^x(t))+S_{gamma}^x(t),frac{partialgamma}{partial s}(t,S_{gamma}^x(t))\
        &=left(gamma(t,s)+sfrac{partialgamma}{partial s}(t,s)right)bigg|_{s=S_{gamma}^x(t)}\
        &=frac{partialrho}{partial s}(t,s)bigg|_{s=S_{gamma}^x(t)}\
        &=frac{partialrho}{partial s}(t,S_{gamma}^x(t)).
        end{align}
        Hence, we eventually obtain
        $$
        {rm d}D^x(t)=D^x(t)left(r(t),{rm d}t+frac{partialrho}{partial s}(t,S_{gamma}^x(t)),{rm d}W(t)right),
        $$
        as is desired.



        In the above derivation, we merely use the commutativity between ${rm d}$ and $partial/partial x$; Ito's formula is not necessary here.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 18:31









        hypernovahypernova

        4,834414




        4,834414






























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