define stochastic integral that does not start from 0

Question

This problem occurred when I was trying to solve Exercise 3.2.30 of Karatzas & Shreve's book, the exercise stated:

For $M\in\mathcal{M}^{c,loc},X\in\mathcal{P}^*$, and $Z$ an $\mathcal{F}_s$-measurable random variable, show that $\int_{s}^{t}ZX_u\;dM_u=Z\int_{s}^{t}X_u\;dM_u$.

I "solved" this exercise by calculating the quadratic variation on both sides, which are both $\int_{s}^{t}Z^2X_u^2\;d\langle M\rangle _u$, notationally.

Then I realized that the integral $\int_{s}^{t}ZX_u\;dM_u$ may not have a proper definition. I would interpret $\int_{s}^{t}ZX_u\;dM_u$ as the difference of two integrals, namely $\int_{0}^{t}ZX_u\;dM_u-\int_{0}^{s}ZX_u\;dM_u$. This interpretation worked well in many cases. But with $Z$ being an $\mathcal{F}_s$-measurable random variable (not $\mathcal{F}_0$-measurable), the process $\{ZX_t\}_{t\ge 0}$ does not satisfy the measurability conditions in the definition of stochastic integrals and therefore $\int_{0}^{t}ZX_u\;dM_u$ is not well-defined.

How to properly define the term $\int_{s}^{t}ZX_u\;dM_u$ then?

Answer 1

You can define

$$ \int_{s}^{t} X_u \, \mathrm{d}M_u = \int_{0}^{t} X_u \mathbf{1}_{[s,\infty)}(u) \, \mathrm{d}M_u. $$

Now if $Z$ is $\mathcal{F}_s$-measurable, then $ u \mapsto ZX_u\mathbf{1}_{[s,\infty)}(u) $ is also adapted to the filtration $(\mathcal{F}_u)_{u\geq 0}$, hence the integral can be defined.

Answer 2

If you really need the integral to start at $0$ (and mind, I don't know why this matters), then you can just consider $$\int_0^{t-s} Z X_{u+s}\,dM_{u+s}$$

These processes are measurable with respect to $\mathcal{G}_u = \mathcal{F}_{u+s}$ and $Z \in \mathcal{G}_0$.