4
$\begingroup$

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $X = (X_t)_{t \in \mathbb{R}^d}$ a classical stochastic process defined on $\Omega$. One says that a process $Y$ defined on $\Omega$ is a modification of $X$ if for all $t$, $\mathbb{P} (X_t=Y_t) = 1$. This concept enables to discuss the regularity of a stochastic process (ex: Kolmogorov-Chentsov theorem) while the probabilistic characterization of $X$ by its finite-dimensional marginals is not sufficient.

On the other hand, in the theory of generalized stochastic processes (ref: Gelfand & Vilenkin, Generalized functions, Vol. 4, Ch. III), a generalized stochastic process on $\Omega$ is defined as a (coherent) collection $(X_\varphi)_{\varphi \in \mathcal{D}}$ of random variables. $X$ is then a random element of $\mathcal{D}'$ and defines a probability measure on $\mathcal{D}'$ (image of $\mathbb{P}$ by $X$, noted $\mathbb{P}_X$).

There is not such a notion of a modification of a generalized stochastic process in the literature (according to my knowledge). I interpret this by thinking that this notion is useless for generalized stochastic processes.

My question is then: Is there a link (an equivalence?) between the concept of modifications of classical processes and their equality in the sense of generalized stochastic processes?

To precise my question, I have to recall that a classical process $X$ is also a generalized stochastic process assuming a reasonable condition (being Bochner integrable for instance) using the relation: $$\langle X,\varphi \rangle = \int X_t\varphi(t) \mathrm{d}t.$$

Moreover, the equality of generalized stochastic processes is as follow. One can say that a (deterministic) generalized function $u \in \mathcal{D}'$ is continuous if there exists a continuous function $f$ such that $\langle u,\varphi \rangle = \int f\varphi$ for all $\varphi \in \mathcal{D}$. In that case, $f = u$ in the sense of generalized functions. The equality of two generalized stochastic processes has to be understood in that sense: $X = Y$ if $\mathbb{P} \left( \{ \forall \varphi \in \mathcal{D}, \ \langle X,\varphi \rangle = \langle Y, \varphi\rangle \}\right) = 1$.

Thanks for attention.

$\endgroup$
  • 1
    $\begingroup$ It seems to me that the direct analogy of a modification to a generalized process would require a measure on $\mathcal{D}$ so you could talk about $\langle X,\varphi \rangle = \langle Y, \varphi \rangle$ for almost all $\varphi \in \mathcal{D}$. It is not clear to me that this interpretation would give anything particularly useful however. Or even be meaningful. $\endgroup$ – BSteinhurst Dec 17 '13 at 17:46

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.