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In Steven Shreve's book "Stochastic Calculus for Finance 2", page 126, a simple process $\Delta(t)$ is a stochastic process such that there is a partition of time $0 < t_1 < ... < t_n \leq T$, such that $\Delta(t)$ is a constant for $t_i \leq t < t_{i+1}$. However it is not clear whether this partition of time depends on $\omega\in \Omega$. Does anyone know the answer? Thanks.

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  • $\begingroup$ Page.......... ? $\endgroup$ – user16651 Jul 24 '16 at 18:16
  • $\begingroup$ @BehrouzMaleki 126 $\endgroup$ – Resorter Jul 24 '16 at 18:27
  • $\begingroup$ To give an usefull comment, the integration of process with random partition (think compound poisson process) are dealt in levy calculus with a specific integral. It is an advanced topic. So i wild guess would be that no, the partition is constant for a simple process. $\endgroup$ – lcrmorin Jul 25 '16 at 15:12
  • $\begingroup$ Hey I am new to this and have completed cal 1 and 2 with As. What level of math do I need to complete before I can start learning stochastic calc? $\endgroup$ – user3138766 Aug 22 '16 at 4:21
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I try to be precise. There is no dependence in $\omega$ for the $t_i$ sequence.

simple process

Let $t_0<t_1<\dots<t_n<t_{n+1}$ be a increasing sequence of real-numbers then $f$ is said to be a simple process on a filtered space $(\Omega,\mathcal{F})$ if:

$$f(t,\omega) = \sum_{i\geq 0}\mathbb{1}_{t\in[t_i,t_{i+1})}\xi_i(\omega)\text{ where }\xi_i\text{ is }\mathcal{F}_{t_i}\text{-measurable}$$

Here is an example where you can find this definition and how it is useful to build stochastic integrals: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Olson.pdf

simple function

if you talk about simple functions, then you will have:

$$f(t,\omega)=\sum_{k\geq 0}a_k \mathbb{1}_{A_k}$$ where $A_k$ is a borel set of $(\mathbb{R}_+\times \Omega,\text{Bor}(\mathbb{R}_+)\times\mathcal{F})$ but then it is not anymore a process with the notion of adaptability, it is just a random variable.

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  • $\begingroup$ So that would be a no as $t_i,t_{i+1}$ is not wrote as $t_i(\omega),t_{i+1}(\omega)$ $\endgroup$ – lcrmorin Jul 25 '16 at 13:27

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