7
$\begingroup$

The Setup

Suppose I have a stochastic process $f(t,Z_t)$ where $Z_t$ solve the $d$-dimensional SDE $$ dZ_t = \mu(t,Z_t)dt + \sigma(t,Z_t)dW_t $$ and $f$ is a smooth function.


My Question

Is there a notion of time-derivative "$d_t$" of the process $f(t,Z_t)$ which satisfies:

  1. Some sort of chain rule like $$ \partial_t f(t,Z_t) * d_t(Z_t), $$ where $\partial_t$ is the usual derivative wrt $t$.
  2. If $Z_t$ is deterministic (ie: $\sigma(t,Z_t)=0$) and $\mu(t,z)$ is $C^1$ in $t$ then $$d_t=\partial_t,$$ ie: $d_t$ reduces to the usual derivative when $f(t,Z_t)$ is a smooth function of $t$.
$\endgroup$
  • 2
    $\begingroup$ Both properties seem to be in contradiction with existing notions of time derivatives. When Z_t is deterministic, its total time derivative would satisfy $\frac{d}{dt}f(t,Z_t)=\partial_t f(t,Z_t)+\partial_z f(t,Z_t) \dot Z_t$, where $\dot Z_t$ is the time derivative of $Z_t$. There is a straightforward generalization of that: the stochastic derivative $d$. It satisfies the Ito formula $df(t,Z_t)=\partial_t f(t,Z_t)dt+\partial_z f(t,Z_t) dZ_t+\frac{1}{2}\partial^2_z f(t,Z_t) dZ_t^2$. $\endgroup$ – S.Surace Dec 17 '16 at 20:26
2
$\begingroup$

The short answer is 'No'. You are conflating two issues here.

  1. For two processes $X,Y$ one may ask whether $dX/dY$ makes sense. For this to have a chance to work one needs $X$ and $Y$ to be of finite variation (on compact time sets). Then total variation of $X$ defines a measure on each path and one can ask whether variation of $X$ is absolutely continuous with respect to variation of $Y$ and take $dX/dY$ to be the Radon-Nikodym derivative if $X\ll Y$ (appropriately allowing for the fact that $dX$ and $dY$ are signed measures).

In this sense $d[Z,Z]/dt$ is meaningful and equals $\sigma^2(t,Z)$, while $dW/dt$ is not well defined.

  1. Now consider $f$ as a smooth function. When one takes the object $Y:=f(X)$ and one writes $f'(X)$ one does not mean $dY/dX$ in the sense of item 1. Instead one means a deterministic derivative of the deterministic function $f$. In your case $\partial f/\partial t$ is a deterministic partial derivative of a deterministic function of two variables. I doubt this partial derivative can be defined pathwise in any meaningful sense.
$\endgroup$
  • $\begingroup$ Is there any relation to the Malliavin derivative of either of these ideas? $\endgroup$ – AIM_BLB Aug 5 '17 at 15:43
  • $\begingroup$ What do you mean Batman? You mean its defined to make the adjoint relationship hold? $\endgroup$ – AIM_BLB Aug 7 '17 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.