I have the following quantity:
$$ g(t)=(f(t))^{M_{t}}, $$
where $M_{t}$ is a jump process which is neither Markovian nor Levy, and $f(t)$ is a positive, increasing but limited, right-continuous function.
How can I differentiate $g$ with respect to $t$? Is there a kind of generalized Ito's lemma?
Thanks in advance.
If you're talking about convergence under the $L2$-norm, I think the answer is probably no.
For example, take $M_t = 1$, and $g(t)$ as a step function. The "derivative" of $g(t)$ is a sum of delta functions, which are not in $L2$ (??).
I believe, for a "reasonable" jump process $M_t$ and a "reasonable" function $g(t)$ (e.g. differentiable function + countable/finite number of jumps), a derivative may be taken, though it may be hard to obtain a closed-form formula for that.
For example, consider the process $(M_t)^n$. If we "represent" the process $dM_t$ by $dM_t = \sum_{t_i} \Delta(t_i) \delta(t-t_i)$, where $\{ t_i \}$ is the set of times at which jumps occur (i.e. and which depend on the outcome $\omega$ of the probability space) and $\Delta(t_i) = M_{t+} - M_{t-}$, then the derivative of $M_t^n$ will be obvious ($dM_t^n = \sum_{t_i} \left[ (M_{t+})^n - (M_{t-})^n \right] \delta(t-t_i)$ ).
The above explanation is entirely "formal" and may be useless computationally.
