Applying Ito's formula to $D^x(t)=(partial/partial x)S(t)$
$begingroup$
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
asked Jan 23 at 4:40
HeisenbergHeisenberg
1,3131741
$endgroup$
add a comment |
$begingroup$
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
asked Jan 23 at 4:40
HeisenbergHeisenberg
1,3131741
$endgroup$
$begingroup$
See Equation (2.16) in p. 261 and the proof around it: books.google.com/…
$endgroup$
– AddSup
Jan 23 at 10:43
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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
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@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
add a comment |
$begingroup$
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
asked Jan 23 at 4:40
HeisenbergHeisenberg
1,3131741
$endgroup$
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
asked Jan 23 at 4:40
HeisenbergHeisenberg
1,3131741
asked Jan 23 at 4:40
HeisenbergHeisenberg
1,3131741
edited Jan 25 at 5:08
Heisenberg
asked Jan 23 at 4:40
HeisenbergHeisenberg
1,3131741
asked Jan 23 at 4:40
HeisenbergHeisenberg
1,3131741
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
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.
answered Jan 25 at 18:31
hypernovahypernova
4,834414
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add a comment |
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1 Answer
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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.
answered Jan 25 at 18:31
hypernovahypernova
4,834414
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 25 at 18:31
hypernovahypernova
4,834414
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 25 at 18:31
hypernovahypernova
4,834414
$endgroup$
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.
answered Jan 25 at 18:31
hypernovahypernova
4,834414
answered Jan 25 at 18:31
hypernovahypernova
4,834414
answered Jan 25 at 18:31
hypernovahypernova
4,834414
answered Jan 25 at 18:31
hypernovahypernova
4,834414
4,834414
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$begingroup$
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