Applying Itô's formula to logged Ornstein-Uhlenbeck process











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I have the following O-U process:
$$d log z_t =-nu log z_t d_t + barsigma dW_t tag{1}$$



and want to apply Itô's formula as:



$$dy=df(x)=bigg(mu(x)f'(x)+frac{1}{2}sigma^2(x)f''(x) bigg)dt + sigma (x)f'(x) tag{2}$$



Since the process is logged, I tried considering a non logged version as



$$d z_t =-nu z_t d_t + sigma dW_t tag{3}$$



and applying the transformation $Z_t=log z_t$. Thenby applying Itô's formula I obtain:



$$ begin{aligned} d Z_t &= bigg(-nu z_t frac{1}{z_t} + frac{1}{2}sigma^2 frac{-1}{z_t^2}bigg)dt+sigma frac{1}{z_t}dW\
&=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wend{aligned} tag{4}$$



Going back to logs:



$$ log z_t=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wtag{5}$$



However, I was supposed to obtain



$$mu(z)=bigg(-nu log(z)+frac{bar{sigma}^2}{2} bigg)z, text{ }sigma(z)=barsigma z tag{6}$$



which I do not see.



Any help is highly appreciated



Edit: The solution is given Did's hint:



Define a flexible guess for $d z_t$ as
$$dz_t=mu(z_t)dt+sigma(z_t)dW$$
I now define $y(t)=log(z_t)$ and apply Itô's formula to the above as
$$dlog z_t = bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z^2} bigg)dt + sigma (z_t)frac{1}{z_t}$$
Recalling that I have the definition of $d log z_t$ from (1), for the drift coefficient
$$begin{aligned}-nu z_tlog z_t dt&= bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z_t^2} bigg)dt\
Leftrightarrow mu(z_t)&=-nu z_tlog(z_t)+frac{sigma^2}{2 z_t}end{aligned}tag{*}$$

Solving for the diffusion coefficient yields
$$barsigma=sigma(z_t) frac{1}{z_t} Leftrightarrow sigma(z_t)=bar{sigma}z_t $$
such that (*) becomes
$$begin{aligned}mu(z_t)&=-nu z_t log z_t+frac{(bar{sigma}z_t)^2}{2 z_t}\
&=bigg(-nu log z_t +frac{bar{sigma}^2}{2}bigg)z_t
end{aligned}$$










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  • 1




    Why do you assume that one should have $$d z_t =-nu z_t d_t + sigma dW_t $$ for some constants $nu$ and $sigma$? You showed we do not... Try instead $$d z_t =b(z_t)d_t + a(z_t)dW_t $$ for some unknown functions $a$ and $b$.
    – Did
    Nov 17 at 13:00












  • you're probably right, but excuse my ignorance: How would I apply that? So for example: $$d log(z_t) = ( b(z_t)cdot(1/z_t)- (1/2)cdot(a(z_t)^2/(log(z_t)^2)dt + a(z) cdot(1/ log z_t)dW$$ and solve how?
    – user469216
    Nov 17 at 13:11








  • 1




    Identify this with $dlog z_t$ in your post and solve for $a$ and $b$ (what else?).
    – Did
    Nov 17 at 13:26












  • yes of course, I guess I stared myself blind at the final equation disregarding that I actually knew what d log(z_t) is!
    – user469216
    Nov 17 at 13:28















up vote
0
down vote

favorite












I have the following O-U process:
$$d log z_t =-nu log z_t d_t + barsigma dW_t tag{1}$$



and want to apply Itô's formula as:



$$dy=df(x)=bigg(mu(x)f'(x)+frac{1}{2}sigma^2(x)f''(x) bigg)dt + sigma (x)f'(x) tag{2}$$



Since the process is logged, I tried considering a non logged version as



$$d z_t =-nu z_t d_t + sigma dW_t tag{3}$$



and applying the transformation $Z_t=log z_t$. Thenby applying Itô's formula I obtain:



$$ begin{aligned} d Z_t &= bigg(-nu z_t frac{1}{z_t} + frac{1}{2}sigma^2 frac{-1}{z_t^2}bigg)dt+sigma frac{1}{z_t}dW\
&=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wend{aligned} tag{4}$$



Going back to logs:



$$ log z_t=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wtag{5}$$



However, I was supposed to obtain



$$mu(z)=bigg(-nu log(z)+frac{bar{sigma}^2}{2} bigg)z, text{ }sigma(z)=barsigma z tag{6}$$



which I do not see.



Any help is highly appreciated



Edit: The solution is given Did's hint:



Define a flexible guess for $d z_t$ as
$$dz_t=mu(z_t)dt+sigma(z_t)dW$$
I now define $y(t)=log(z_t)$ and apply Itô's formula to the above as
$$dlog z_t = bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z^2} bigg)dt + sigma (z_t)frac{1}{z_t}$$
Recalling that I have the definition of $d log z_t$ from (1), for the drift coefficient
$$begin{aligned}-nu z_tlog z_t dt&= bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z_t^2} bigg)dt\
Leftrightarrow mu(z_t)&=-nu z_tlog(z_t)+frac{sigma^2}{2 z_t}end{aligned}tag{*}$$

Solving for the diffusion coefficient yields
$$barsigma=sigma(z_t) frac{1}{z_t} Leftrightarrow sigma(z_t)=bar{sigma}z_t $$
such that (*) becomes
$$begin{aligned}mu(z_t)&=-nu z_t log z_t+frac{(bar{sigma}z_t)^2}{2 z_t}\
&=bigg(-nu log z_t +frac{bar{sigma}^2}{2}bigg)z_t
end{aligned}$$










share|cite|improve this question




















  • 1




    Why do you assume that one should have $$d z_t =-nu z_t d_t + sigma dW_t $$ for some constants $nu$ and $sigma$? You showed we do not... Try instead $$d z_t =b(z_t)d_t + a(z_t)dW_t $$ for some unknown functions $a$ and $b$.
    – Did
    Nov 17 at 13:00












  • you're probably right, but excuse my ignorance: How would I apply that? So for example: $$d log(z_t) = ( b(z_t)cdot(1/z_t)- (1/2)cdot(a(z_t)^2/(log(z_t)^2)dt + a(z) cdot(1/ log z_t)dW$$ and solve how?
    – user469216
    Nov 17 at 13:11








  • 1




    Identify this with $dlog z_t$ in your post and solve for $a$ and $b$ (what else?).
    – Did
    Nov 17 at 13:26












  • yes of course, I guess I stared myself blind at the final equation disregarding that I actually knew what d log(z_t) is!
    – user469216
    Nov 17 at 13:28













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the following O-U process:
$$d log z_t =-nu log z_t d_t + barsigma dW_t tag{1}$$



and want to apply Itô's formula as:



$$dy=df(x)=bigg(mu(x)f'(x)+frac{1}{2}sigma^2(x)f''(x) bigg)dt + sigma (x)f'(x) tag{2}$$



Since the process is logged, I tried considering a non logged version as



$$d z_t =-nu z_t d_t + sigma dW_t tag{3}$$



and applying the transformation $Z_t=log z_t$. Thenby applying Itô's formula I obtain:



$$ begin{aligned} d Z_t &= bigg(-nu z_t frac{1}{z_t} + frac{1}{2}sigma^2 frac{-1}{z_t^2}bigg)dt+sigma frac{1}{z_t}dW\
&=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wend{aligned} tag{4}$$



Going back to logs:



$$ log z_t=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wtag{5}$$



However, I was supposed to obtain



$$mu(z)=bigg(-nu log(z)+frac{bar{sigma}^2}{2} bigg)z, text{ }sigma(z)=barsigma z tag{6}$$



which I do not see.



Any help is highly appreciated



Edit: The solution is given Did's hint:



Define a flexible guess for $d z_t$ as
$$dz_t=mu(z_t)dt+sigma(z_t)dW$$
I now define $y(t)=log(z_t)$ and apply Itô's formula to the above as
$$dlog z_t = bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z^2} bigg)dt + sigma (z_t)frac{1}{z_t}$$
Recalling that I have the definition of $d log z_t$ from (1), for the drift coefficient
$$begin{aligned}-nu z_tlog z_t dt&= bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z_t^2} bigg)dt\
Leftrightarrow mu(z_t)&=-nu z_tlog(z_t)+frac{sigma^2}{2 z_t}end{aligned}tag{*}$$

Solving for the diffusion coefficient yields
$$barsigma=sigma(z_t) frac{1}{z_t} Leftrightarrow sigma(z_t)=bar{sigma}z_t $$
such that (*) becomes
$$begin{aligned}mu(z_t)&=-nu z_t log z_t+frac{(bar{sigma}z_t)^2}{2 z_t}\
&=bigg(-nu log z_t +frac{bar{sigma}^2}{2}bigg)z_t
end{aligned}$$










share|cite|improve this question















I have the following O-U process:
$$d log z_t =-nu log z_t d_t + barsigma dW_t tag{1}$$



and want to apply Itô's formula as:



$$dy=df(x)=bigg(mu(x)f'(x)+frac{1}{2}sigma^2(x)f''(x) bigg)dt + sigma (x)f'(x) tag{2}$$



Since the process is logged, I tried considering a non logged version as



$$d z_t =-nu z_t d_t + sigma dW_t tag{3}$$



and applying the transformation $Z_t=log z_t$. Thenby applying Itô's formula I obtain:



$$ begin{aligned} d Z_t &= bigg(-nu z_t frac{1}{z_t} + frac{1}{2}sigma^2 frac{-1}{z_t^2}bigg)dt+sigma frac{1}{z_t}dW\
&=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wend{aligned} tag{4}$$



Going back to logs:



$$ log z_t=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wtag{5}$$



However, I was supposed to obtain



$$mu(z)=bigg(-nu log(z)+frac{bar{sigma}^2}{2} bigg)z, text{ }sigma(z)=barsigma z tag{6}$$



which I do not see.



Any help is highly appreciated



Edit: The solution is given Did's hint:



Define a flexible guess for $d z_t$ as
$$dz_t=mu(z_t)dt+sigma(z_t)dW$$
I now define $y(t)=log(z_t)$ and apply Itô's formula to the above as
$$dlog z_t = bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z^2} bigg)dt + sigma (z_t)frac{1}{z_t}$$
Recalling that I have the definition of $d log z_t$ from (1), for the drift coefficient
$$begin{aligned}-nu z_tlog z_t dt&= bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z_t^2} bigg)dt\
Leftrightarrow mu(z_t)&=-nu z_tlog(z_t)+frac{sigma^2}{2 z_t}end{aligned}tag{*}$$

Solving for the diffusion coefficient yields
$$barsigma=sigma(z_t) frac{1}{z_t} Leftrightarrow sigma(z_t)=bar{sigma}z_t $$
such that (*) becomes
$$begin{aligned}mu(z_t)&=-nu z_t log z_t+frac{(bar{sigma}z_t)^2}{2 z_t}\
&=bigg(-nu log z_t +frac{bar{sigma}^2}{2}bigg)z_t
end{aligned}$$







stochastic-processes stochastic-calculus stochastic-analysis stochastic-pde






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edited Nov 17 at 16:14

























asked Nov 17 at 12:34









user469216

567




567








  • 1




    Why do you assume that one should have $$d z_t =-nu z_t d_t + sigma dW_t $$ for some constants $nu$ and $sigma$? You showed we do not... Try instead $$d z_t =b(z_t)d_t + a(z_t)dW_t $$ for some unknown functions $a$ and $b$.
    – Did
    Nov 17 at 13:00












  • you're probably right, but excuse my ignorance: How would I apply that? So for example: $$d log(z_t) = ( b(z_t)cdot(1/z_t)- (1/2)cdot(a(z_t)^2/(log(z_t)^2)dt + a(z) cdot(1/ log z_t)dW$$ and solve how?
    – user469216
    Nov 17 at 13:11








  • 1




    Identify this with $dlog z_t$ in your post and solve for $a$ and $b$ (what else?).
    – Did
    Nov 17 at 13:26












  • yes of course, I guess I stared myself blind at the final equation disregarding that I actually knew what d log(z_t) is!
    – user469216
    Nov 17 at 13:28














  • 1




    Why do you assume that one should have $$d z_t =-nu z_t d_t + sigma dW_t $$ for some constants $nu$ and $sigma$? You showed we do not... Try instead $$d z_t =b(z_t)d_t + a(z_t)dW_t $$ for some unknown functions $a$ and $b$.
    – Did
    Nov 17 at 13:00












  • you're probably right, but excuse my ignorance: How would I apply that? So for example: $$d log(z_t) = ( b(z_t)cdot(1/z_t)- (1/2)cdot(a(z_t)^2/(log(z_t)^2)dt + a(z) cdot(1/ log z_t)dW$$ and solve how?
    – user469216
    Nov 17 at 13:11








  • 1




    Identify this with $dlog z_t$ in your post and solve for $a$ and $b$ (what else?).
    – Did
    Nov 17 at 13:26












  • yes of course, I guess I stared myself blind at the final equation disregarding that I actually knew what d log(z_t) is!
    – user469216
    Nov 17 at 13:28








1




1




Why do you assume that one should have $$d z_t =-nu z_t d_t + sigma dW_t $$ for some constants $nu$ and $sigma$? You showed we do not... Try instead $$d z_t =b(z_t)d_t + a(z_t)dW_t $$ for some unknown functions $a$ and $b$.
– Did
Nov 17 at 13:00






Why do you assume that one should have $$d z_t =-nu z_t d_t + sigma dW_t $$ for some constants $nu$ and $sigma$? You showed we do not... Try instead $$d z_t =b(z_t)d_t + a(z_t)dW_t $$ for some unknown functions $a$ and $b$.
– Did
Nov 17 at 13:00














you're probably right, but excuse my ignorance: How would I apply that? So for example: $$d log(z_t) = ( b(z_t)cdot(1/z_t)- (1/2)cdot(a(z_t)^2/(log(z_t)^2)dt + a(z) cdot(1/ log z_t)dW$$ and solve how?
– user469216
Nov 17 at 13:11






you're probably right, but excuse my ignorance: How would I apply that? So for example: $$d log(z_t) = ( b(z_t)cdot(1/z_t)- (1/2)cdot(a(z_t)^2/(log(z_t)^2)dt + a(z) cdot(1/ log z_t)dW$$ and solve how?
– user469216
Nov 17 at 13:11






1




1




Identify this with $dlog z_t$ in your post and solve for $a$ and $b$ (what else?).
– Did
Nov 17 at 13:26






Identify this with $dlog z_t$ in your post and solve for $a$ and $b$ (what else?).
– Did
Nov 17 at 13:26














yes of course, I guess I stared myself blind at the final equation disregarding that I actually knew what d log(z_t) is!
– user469216
Nov 17 at 13:28




yes of course, I guess I stared myself blind at the final equation disregarding that I actually knew what d log(z_t) is!
– user469216
Nov 17 at 13:28















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