4

Problem: I do not understand the step, of how we define the stochastic integral.

In my lecture notes we define firstly infite sum then finite. On the page 132 of Karazats, Shreve (BM and Stochastic Calculus) I see in opposite order.

By Karazats, Shreve, we have $I(X) = X\cdot M$ defined as: \begin{align} I_t(X) &\stackrel{\triangle}{=} \sum^{n-1}_{i=0} \xi_i(M_{t_{i+1}} - M_{t_{i}}) + \xi_n(M_{t} - M_{t_{n}})\\ & = \boxed{\sum^{\infty}_{i=0} \xi_i(M_{t \wedge t_{i+1}} - M_{t \wedge t_{i}})}, \space t \in [0,\infty)\\ \end{align} One more question related to a Problem: What is the logical consequence of what? Why do we rewrite through differences of a kind $(M_{t \wedge t_{i+1}} - M_{t \wedge t_{i}})$?

I have a guess about this: is the 2nd sum given to see that the product: $X\cdot M$ looks like a simple process definition?, i.e.: $$ X_t = \xi_0 1_{\{0\}}(t) + \boxed{\sum^{\infty}_{i=0}\xi_i 1_{(t_i,t_{i+1}]}(t)}, \space t \in [0,\infty)\\ $$ where $t_0=0, \{t_n\}^\infty_{n=0} \to \infty$ as $n\to \infty$ and $\{\xi_n\}^\infty_{n=0}$ - uniformly bounded sequence of r.v-s., s.t. $\xi_i$ is $ \mathcal{F}_{t_i}$ measurable, $M$ - sq.int., cts. martingale. In both cases we look at $X$ - simple process.

FD_bfa
  • 4,757
StochasticIntegrationStudent
  • 291

1 Answers1

2

The definition here is like the procedure of the Lebesgue integration. We first define the integral of the "simple functions". Then prove that the "simple functions" are dense in the space we want. So we can define the integral by approximation.

Chichipingguo
  • 31
  • Dear @Chichipingguo, therefore I assume it is like this:first, we have finite approximation, then we take a limit of $n$ and obtain the infinite sum, which turns in an integral, yes? – StochasticIntegrationStudent Mar 10 '17 at 18:33