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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 ($dM_t^n = \sum_{t_i} (M_{t+}^n - M_{t-}^n) \delta(t-t_i)$).

The above explanation is entirely "formal" and may be useless computationally.

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.

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 $L2$ (??).

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 ($dM_t^n = \sum_{t_i} (M_{t+}^n - M_{t-}^n) \delta(t-t_i)$).

The above explanation is entirely "formal" and may be useless computationally.

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 $L2$ (??).

However, the "derivative" of the function may exist in the following sense.

For a given interval $[a,b]$, consider two cases: 1. $M_t$ is flat (i.e. no jump occurs in $[a,b]$) 2. At least a jump occurs in $[a,b]$.

For the first case, it can be easily shown that $h(t) - h(a) = \int_a^t M_t f(t)^{M_t-1} d f(t)$. The integral exists in the "classical" sense because $f(t)$ is increasing and right-continuous. (not sure; please verify)

For the second case, even if $f(t)$ is constant, $f(t)^{M_t}$ will be discontinuous where $M_t$ "jumps". The accumulative size of "jumps" up to $t$ is given by $\int_a^t f(t)^{M_t} dM_t / M_t$. Again the latter integral seems to exist because $M_t$ is increasing and right-continuous. (Possible problem where $M_t = 0$)

Thus it seems that, $h(t) - h(a) = \int_a^t M_t f(t)^{M_t-1} d f(t) + \int_a^t f(t)^{M_t} dM_t/M_t$, the convergence being almost surely (or almost everywhere). (The latter formula may be very incorrect; please verify)

The idea is to obtain the set of points where either $f(t)$ or $M_t$ jumps. Of course it may not work mathematically.

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 $L2$ (??).