Skip to main content
2 of 6
Re-formatted question

Can one define a bounded noise process by conditioning standard Gaussian white noise on the assumption of boundedness?

Background of the question.

One of the problems that arises with Wiener-driven (and more general Lévy-driven) models of noisy systems is that, due to the extremely rapid decay of tails of infinitely divisible distributions, the theoretical long-term predictions of the model may be affected by the almost-sure occurrence of arbitrarily extreme freak events that have no physical meaning for the problems intended to be addressed by the model. One way to get round this is to work explicitly with the long-but-finite-time predictions of the model; but another way round this is to work with a bounded-noise model that approximates driving by a Wiener process.

One common bounded-noise approximation is dichotomous Markov noise ([1], see also [2]):

  • in place of the driving $\,\circ \,dW_t$ from a Wiener process $(W_t)_{t \in \mathbb{R}}$, we instead have the driving $\sqrt{\lambda}(-1)^{N_{\lambda t}} dt$ for some (large) $\lambda>0$, where $N_0$ is odd or even with equal probability and $(N_\tau-N_0)_{\tau \in \mathbb{R}}$ is a Poisson process of intensity $1$ independent of $N_0$.

However, I suspect it might sometimes be of interest to have a bounded-noise driving process whose regularity properties are "essentially the same" as those of the Wiener process. The question below formalises this.

[1] Van Den Broeck, C. On the relation between white shot noise, Gaussian white noise, and the dichotomic Markov process. J Stat Phys 31, 467–483 (1983). https://doi.org/10.1007/BF01019494
[2] https://math.stackexchange.com/questions/3643865/


THE QUESTION.

Some notations:

  • Given a closed connected set $G \subset \mathbb{R}$ with $0 \in G$, write $C_0(G,\mathbb{R})$ for the set of continuous functions $\omega \colon G \to \mathbb{R}$ with $\omega(0)=0$. We equip $C_0(G,\mathbb{R})$ with the topology of uniform convergence on bounded sets. Note that the Borel $\sigma$-algebra is given by $$ \mathcal{B}(C_0(G,\mathbb{R})) \ = \ \sigma(\omega \mapsto \omega(t) \, : \, t \in G) \ = \ \sigma(\omega \mapsto \omega(t) \, : \, t \in \mathbb{Q} \cap G). $$
  • For each closed connected $G \ni 0$, define the probability measure $\mathbb{P}^G$ on $C_0(G,\mathbb{R})$ to be the Wiener measure.

Now fix $\Delta>0$ and $M>0$, and for any closed connected $G \ni 0$, let $$ A_G \ := \ \{ \omega \in C_0(G,\mathbb{R}) \, : \, \text{for all $s,t \in G$, } |s-t| \leq \Delta \ \Longrightarrow \ |\omega(s)-\omega(t)| \leq M. \} $$ Note that $A_G$ is a closed subset of $C_0(G,\mathbb{R})$.

Is it the case that for each $\tau>0$ there is a probability measure $\tilde{\mathbb{P}}^{[0,\tau]}$ on $A_{[0,\tau]}$ such that for any sequences $s_n,t_n \!\uparrow\! \infty$ and any bounded continuous $g \colon A_{[0,\tau]} \to \mathbb{R}$, $$ \frac{1}{\mathbb{P}^{[-s_n,t_n]}(A_{[-s_n,t_n]})} \int_{A_{[-s_n,t_n]}} g(\omega|_{[0,\tau]}) \, \mathbb{P}^{[-s_n,t_n]}(d\omega) \, \to \, \int_{A_{[0,\tau]}} g \, d\tilde{\mathbb{P}}^{[0,\tau]} $$ as $n \to \infty$?

If so, is $\tilde{\mathbb{P}}^{[0,\tau]}$ absolutely continuous with respect to $\left.\mathbb{P}^{[0,\tau]}\right|_{\mathcal{B}(A_{[0,\tau]})}$?

Note that the absolute continuity in the second question implies in particular that $\tilde{\mathbb{P}}^{[0,\tau]}$-almost every sample path has the same local regularity properties (e.g. $\alpha$-Hölder-continuous for all $\alpha \in (0,\frac{1}{2})$ but not for $\alpha=\frac{1}{2}$) as the generic sample paths of a Wiener process on $[0,\tau]$.

Now if the answer to the first question is yes, then obviously the intervals $[0,\tau]$ can be generalised to any compact intervals $G \ni 0$. In this case, it is not hard to show that one can define a probability measure $\tilde{\mathbb{P}}$ on $A_\mathbb{R}$ whose natural projection onto $A_G$ coincides with $\tilde{\mathbb{P}}^G$ for any compact interval $G \ni 0$. (See immediately below for a proof.)


Proof of the last claim above.

First note that the family of probability measures $\tilde{\mathbb{P}}^G$ across compact intervals $G \ni 0$ is consistent: for compact intervals $G_2 \supset G_1 \ni 0$, to verify that $\tilde{\mathbb{P}}^{G_2}$ projects onto $\tilde{\mathbb{P}}^{G_1}$, it is sufficient to verify that $\tilde{\mathbb{P}}^{G_1}$ agrees with the projection of $\tilde{\mathbb{P}}^{G_2}$ for every bounded continuous functions $g_1 \colon A_{G_1} \to \mathbb{R}$; but for any such $g_1$, the map $g_2 \colon A_{G_2} \to \mathbb{R}$ sending $\omega \mapsto g_1(\omega|_{G_1})$ is a bounded continuous function and for any compact interval $G_3 \supset G_2$, for any $\omega \in A_{G_3}$ we have $g_2(\omega|_{G_2})=g_1(\omega|_{G_1})$. Having established this consistency enables us to apply the Kolmogorov extension theorem to obtain a probability measure $\tilde{\tilde{\mathbb{P}}}$ on $(\mathbb{R}^\mathbb{R},\mathcal{B}(\mathbb{R})^{\otimes \mathbb{R}})$ whose natural projection onto $\mathbb{R}^G$ agrees with $\tilde{\mathbb{P}}^G(\,\cdot\,\cap A_G)$ for each compact interval $G \ni 0$. For $\tilde{\tilde{\mathbb{P}}}$-almost every $\omega \colon \mathbb{R} \to \mathbb{R}$, we have that $\omega(0)=0$ and the restriction $\omega|_\mathbb{Q}$ is uniformly continuous on bounded sets, and hence there exists $\kappa(\omega) \in A_\mathbb{R}$ agreeing with $\omega$ on $\mathbb{Q}$. So take $\tilde{\mathbb{P}}$ to be the image measure of $\tilde{\tilde{\mathbb{P}}}$ under $\kappa(\cdot)$.


First thought on tackling the question. My strong intuition is that for each $E \in \mathcal{B}(A_{[0,\tau]})$, there exists a number $p(E) \in [0,1]$ such that for any sequences $s_n,t_n \!\uparrow\! \infty$, $$ \frac{1}{\mathbb{P}^{[-s_n,t_n]}(A_{[-s_n,t_n]})} \int_{A_{[-s_n,t_n]}} \mathbf{1}_E(\omega|_{[0,\tau]}) \, \mathbb{P}^{[-s_n,t_n]}(d\omega) \, \to \, p(E) $$ as $n \to \infty$. If so, the Vitali-Hahn-Saks theorem implies that $p(\cdot)$ is a probability measure, and so we can take $\tilde{\mathbb{P}}^{[0,\tau]}$ to be $p(\cdot)$ since this "strong convergence" implies the "weak convergence" being asked about in the question. Furthermore, this will then immediately yield a positive answer to the second question.