0
$\begingroup$

Usually the stochastic integral is defined for processes indexed over $[0,\infty).$

I wonder about the standard way to define the integral for processes indexed over $[0,T].$ That is, for a continuous local martingale $M = (M_t)_{t \in [0,T]}$ and a progressive process (sufficiently integrable) $H = (H_t)_{t \in [0,T]} $ I want to define $$ \int_0^\cdot H_s dM_s, \quad t \in [0,T]. $$

I guess one could do two things: 1. Do the proof of the existence of the stochastic integral all over with processes defined on $[0,T]$ 2. Reduce it to the general case. That is, define $$ \tilde{M}_t = \begin{cases} M_t, t \in [0,T] \\ M_T, t \in [T, \infty) \end{cases} $$ Then define $\tilde{H}$ in the same way. And then define $$ \int_0^t H_s dM_s = \int_0^t \tilde{H}_s d \tilde{M}_s, \quad t \in [0,T]. $$

Do both approaches work?

$\endgroup$
  • 1
    $\begingroup$ Yeah, should work fine. $\endgroup$ – saz Nov 14 '18 at 18:48
  • $\begingroup$ Thanks. Regarding the second option: I would also like to get an existence and uniqueness theorem like in the $[0,\infty)$ case. What is the definition of a local martingale for $[0,T]$ ? $\endgroup$ – White Nov 14 '18 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.