Is there a Stochastic Time derivative

Question

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 $t$.

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.

2. 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.