A stochastic process is defined as a family of random variables $\{X_t: \Omega \to S, t \in T\}$ , where $\Omega$ is a probability space, $T$ is a set, and $S$ is a measurable space. Equivalently, a stochastic process can be defined as a mapping $X: \Omega \to S^T$ such that it is measurable with respect to the sigma algebra on the probability space $\Omega$ and the product sigma algebra on $S^T$.

Now let me try to define a different kind of "stochastic processes", by replacing the sigma algebra on $S^T$ with a different one. Suppose $A \subset S^T$ which is not necessarily measurable wrt the product sigma algebra on $S^T$. Now introduce the smallest sigma algebra $\mathcal M$ on $A$ such that $\forall t \in T$, the mapping $t: A\to S$ defined as $t(a): =a(t), \forall a \in A$ is measurable. Now define a "stochastic process" to be a measurable mapping $Y:\Omega \to A$ wrt the sigma algebras on $\Omega$ and on $A$. An example of such a "stochastic process" is a random measure, where $T$ is a sigma algebra, $S$ is $[0, \infty]$ and $A$ is a set of measures defined on $T$.

My questions:

  1. When viewing a "stochastic process" $Y:\Omega \to A$ as a mapping $Y:\Omega \to S^T$, is $Y$ not necessarily a stochastic process? If possible, what conditions can make $Y$ a stochastic process?

    For example, can a random measure $Y:\Omega \to A$ be viewed as a family of random variables $\{Y_C: \Omega \to [0, \infty], \forall C \in \mathcal M\}$?

  2. Is a stochastic process $X: \Omega \to S^T$ necessarily a "stochastic process"? In other words, on $S^T$, is the product sigma algebra the same as

    the smallest sigma algebra on $S^T$ such that $\forall t \in T$, the mapping $t: S^T\to S$ defined as $t(f): =f(t), \forall f \in S^T$ is measurable?

    Added: Yes, I think it is. Because product sigma algebra is defined to make projections measurable.

Thanks and regards!


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.