0
$\begingroup$

So, $X_1(s)$ is a jump process, $X_2(s)$ is another jump process, $X_2^c(s)$ is the continuous part of $X_2(s)$.

And $\int_0^tX_1(s-)dX_2^c(s) = \int_0^tX_1(s)dX_2^c(s)$, is it because the integration on the jumps of the integrand $X_1$ is just zero?

Thank you!

$\endgroup$

1 Answer 1

0
$\begingroup$

First of all, note that the stochastic integral

$$\int_0^t X_1(s) \, dX_2^c(s)$$

is well-defined since $(X_2^c(t))_{t \geq 0}$ is continuous. Thus, by Itô's isometry

$$\mathbb{E} \left( \int_0^t (X_1(s-)-X_1(s)) \, dX_2^c(s) \right)^2 = \mathbb{E} \left(\int_0^t (X_1(s-)-X_1(s))^2 \, d\langle X_2^c \rangle_s \right) = 0$$

since $\{X_1(s) \neq X_1(s-)\}$ has measure zero (assuming that $(X_1(t))_{t \geq 0}$ is a cadlag process). Obviously, this implies

$$\int_0^t X_1(s-) \, dX_2^c(s) = \int_0^t X_1(s) \, dX_2^c(s)$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .