5
$\begingroup$

Can anyone suggest how to prove a right continuous stochastic process is measurable?

Thanks Indrajit

1 Answers 1

5

Let $(X_t)_{t\geq 0}$ be a right-continuous stochastic process. For every $n\in\mathbb{N}$ we define a process $(X_t^n)_{t\geq 0}$ given by $ X_t^n(\omega)=X_{i/2^n}(\omega)\quad \text{if } \tfrac{i-1}{2^n}\leq t<\tfrac{i}{2^n}\; \text{and } i\geq 1. $

Show that for each $n\in\mathbb{N}$ the process $(X_t^n)_{t\geq 0}$ is indeed a measurable process.

Convince yourself that the right-continuity ensures that $X_t(\omega)=\lim_{n\to\infty} X_t^n(\omega)$ for every $(t,\omega)\in [0,\infty)\times \Omega$. Conclude.


To show that $(X_t^n)_{t\geq 0}$ is a measurable process for every $n\in\mathbb{N}$ we note that $ \{(t,\omega)\mid X_t^n(\omega)\in B\}=\bigcup_{i=1}^\infty \left[ \tfrac{i-1}{2^n},\tfrac{i}{2^n}\right[\times \{X_{i/2^n}\in B\} $ for every $B\in\mathcal{B}(\mathbb{R})$.

  • 0
    @StefanHansen: well... that's embarrassingly obvious now :) Thanks.2014-11-04