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(Philosophically, the following question is of a similar flavour to A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary, but more "advanced".)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t_1,\ldots,t_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1,\omega),\ldots,X(t_n,\omega) \right)$

is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map

$\hspace{5mm} \omega \mapsto \left( X(t_1+\tau,\omega),\ldots,X(t_n+\tau,\omega) \right)$.

Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t_1,\ldots,t_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map

$\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t_1+t,\omega),\ldots,X(t_n+t,\omega) \right) \in A )$

(where $\textrm{Leb}$ denotes the Lebesgue measure)?

How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?

Now I've asked the above question in quite a general form, but what I ultimately want is: if $(X_t)_{t \geq 0}$ is a measurable stationary stochastic process in continuous time, then is the discrete-time stochastic process $\left( \int_n^{n+1} g(X_t) \, dt \right)_{n \geq 0}$ necessarily also stationary (where $g$ is a bounded measurable function)??

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Yes.

I assume that $E$ is a Polish space, or a standard Borel space that can be turned into a Polish space by some auxillary topology.

Basically, you can view your measurable stochastic process as a random element in the space $L^0(\mathbb{R}_+, E)$ of equivalence classes of $E$-valued measurable functions. $L^0(\mathbb{R}_+, E)$ comes equipped with the Polish topology of convergence in measure on $[0,n]$ for all $n \in \mathbb{N}$ and its corresponding standard Borel $\sigma$-algebra. While doing that you lose some information (e.g. the value for some specific $t$), but you will be able to calculate your Lebesgue measures as measurable functions on $L^0$.

Now I claim that

The distribution of the $L^0$ version of $X$ can be recovered from the joint distribution of $(X_t, t \ge 0)$.

Note that this will immediately imply a positive answer to your question.

Here is a sketch of a proof of my claim. Consider the functions $L^0(\mathbb{R}_+, E) \to \mathbb{R}$ of the following form:

$$F_{f,T}: x \mapsto \intop_0^T f(x_t) dt,$$

where $f: E \to [0,1]$ is bounded and Borel, and $T \ge 0$. These functions generate the whole Borel $\sigma$-algebra on $L^0$, because the level sets $\{F_{f,T} < \mathrm{const}\}$ for continuous $f: E \to [0,1]$ generate the topology.

Now it remains to pin down the joint distribution of $F_{f,T}(X)$. This is done by observing that the moments can be calculated using the joint distribution of $(X_t)$ by Fubini's theorem:

$$\mathsf{E} \intop_0^T f_1(X_t) dt \dots \intop_0^T f_n(X_t) dt = \intop_{[0,T]^n} \mathsf{E} f_1(X_{t_1}) \dots f_n(X_{t_n}) dt_1 \dots dt_n$$

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  • $\begingroup$ Thank you for your reply. I do not understand why stationarity implies the existence of the measure-preserving transformation $T_s$ on $\Omega$ that you described. [If I have such a measure-preserving transformation then the answer to my question trivially becomes yes: my second $\mathbb{R}$-valued map is just the composition of my first one with $T_\tau$---and the two maps are certainly measurable since the measure of a section of a measurable set depends measurably on where the section is taken.] $\endgroup$ – Julian Newman Sep 23 '14 at 17:54
  • $\begingroup$ If you have in mind that one can "without loss of generality" replace $\Omega$ with the product space $E^{[0,\infty)}$ (equipped with the law $\rho$ of the stochastic process $\mathbf{X}$), then this would surely require that the map $(t,f)\mapsto f(t)$ from $[0,\infty)\times E^{[0,\infty)}$ to $E$ is measurable with respect to the $\mathrm{Leb}\otimes\rho$-completion of $\mathcal{B}([0,\infty))\otimes E^{\otimes [0,\infty)}$ (or something like that!). But is that true? Or have I missed the mark altogether? $\endgroup$ – Julian Newman Sep 23 '14 at 17:56
  • $\begingroup$ There is indeed such a map, under the assumptions that a) $X$ generates the whole σ-algebra on $\Omega$, b) $(\Omega,\mathcal{F},\mathsf{P})$ is standard (see the references in en.wikipedia.org/wiki/…). But we don't actually need it. Please see the updated answer... $\endgroup$ – Alexander Shamov Sep 23 '14 at 19:12
  • $\begingroup$ Thank you very much. I think this is a very nice argument. (I had noticed that the first moment of the r.v. in question can be determined from the finite-dimensional distributions, but hadn't managed to think beyond that observation.) $\endgroup$ – Julian Newman Sep 24 '14 at 0:13
  • $\begingroup$ Can I please cite this MathOverflow post (or if not, then just cite you) within a document that I am writing? (I do not currently intend to submit the document for publication, but only to post it on my website.) $\endgroup$ – Julian Newman Sep 24 '14 at 17:13

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