I am quite new with Stochastic Calculus, and I was wondering if it is possible to (formally, i.e. mathematically rigorously) if it's possible to get the differential of the stochastic process that I will define below. So I have the initial assumptions of a probability space $(\Omega, \mathbb{P}, \mathcal{F} )$ together with a filtration $\left\{ \mathcal{F}_{t} \right\}$ and a Brownian motion $W$:
1 - We have a time interval $[0,T]$ and discrete set of time points that belong to this interval: $0=t_{0} < t_{1} < \cdots <t_{n}$.
2 - For these discrete time points we have random variables $X_{t_{1}}, \ldots X_{t_{n}}$. At this point, I assume that the r.v. $X$ have a common distribution (for example, normal with same mean and variance).
3 - We have a stochastic proces $r(t)$ whose s.d.e. is given by $dr(t)=\mu(t)dt + \sigma(t) dW(t)$.
Having these, I define the process:
$$ C(t,\omega) = \sum_{t:t_{i}\leq t} X_{t_{i}}(\omega) e^{\int_{t_{i}}^{ \min \left\{t_{i+1},t \right\} } r(s)ds} $$
Is it possible to (formally) get the differential of this jump-process? Basically, what is $dC(t)$?
Thanks
That should not be a big problem, as at each time it seems only one of the terms in the sum changes with time. So when we are at $t_i < t < t_{i+1}$ you will get $$ \mathrm dC(t) = X_{t_i}\cdot\mathrm d\mathrm{exp}\left(\int_{t_i}^t r(s)\mathrm ds\right) $$ which you can easily get from Ito's formula. You only need to work around $\mathrm dC(t_i)$, but your process does actually seem to be continuous, so I think nothing special will happen at those points.
Formally speaking, I think you need to assume that $X$ is a sequence of variables somehow adapted to the filtration you work with, and likely to assume its independence from $W$.