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 $t$.
  • 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, 2016 at 20:26

1 Answer 1


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.
  • $\begingroup$ Is there any relation to the Malliavin derivative of either of these ideas? $\endgroup$
    – AIM_BLB
    Aug 5, 2017 at 15:43
  • $\begingroup$ What do you mean Batman? You mean its defined to make the adjoint relationship hold? $\endgroup$
    – AIM_BLB
    Aug 7, 2017 at 1:29

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