More natural example of measurable but not progressive process

Question

All examples of measurable but not progressive processes I have ever seen seemed to be based on the huge difference between $\mathcal{F}$ and $\mathcal{F}_\infty$. Here is what I mean.

Consider Example 19.5 from Counterexamples in Probability by Stoyanov: let $X_t ( \omega ) = 1_{t = \omega}$ be a process relative to $(\mathbb{R}_+, \mathcal{F}, \mathcal{F}_t)$, where $\mathcal{F} = \mathcal{B} ( \mathbb{R}_+ )$, whereas $\mathcal{F}_t$ does not depend on $t$ and is generated by points of $\mathbb{R}^d$ (that is, $A \in \mathcal{F}_t$ iff $A$ or $A^c$ is countable). Then it is easy to see that

1. $X^{-1} ( \{ 1 \} ) = \{ ( t, \omega ) \ \colon \ t = \omega \in \mathbb{R}_+ \}$ belongs to $\mathcal{B}(\mathbb{R}_+) \otimes \mathcal{F}$, which means that $X$ is measurable,
2. $X_t$ is $\mathcal{F}_t$-adapted,
3. $X^{-1} ( \{ 1 \} ) \cap [0, t] \times \Omega$ does not belong to $\mathcal{B}( [0, t] ) \otimes \mathcal{F}_t$, which means that $X$ is not progressive.

Essentially the same example one can find in Capacités et processus stochastiques by Claude Dellacherie (page 47): or a slight variation thereof on p.47 here (Example 1.38). In the latter $\mathcal{F}$ is the Lebesgue sigma-algebra on $[0, 1]$ and $\mathcal{F}_t = \mathcal{L}_0$ is generated by the null-sets.

Both examples seem somewhat artificial, because they only work because $\mathcal{F}$ is much bigger and most importantly has little to do with $\mathcal{F}_t$.

My question is: are there examples of measurable but not progressive processes with respect to $\mathcal{F} = \mathcal{F}_\infty$?

Answer 1

An example due to G. Letta in "Un exemple de processus mesurable adapté non progressif" [Séminaire de Probabilités XXII, 449–453, Lecture Notes in Math. 1321, 1988] is illuminating.

Take $\Omega$ to be the space of lower semi-continuous functions $\omega:[0,\infty)\to\Bbb R$, such that $\int_0^t|\omega(s)|\,ds<\infty$ for each $t>0$. Let $H_s(\omega):=\omega(s)$ be the coordinate process on $\Omega$, and $(\mathcal F_t)_{t\ge0}$ the filtration generated by $(H_t)_{t\ge 0}$. Let $\mathcal J$ denote the collection of compact intervals $J\subset[0,\infty)$ and for $J\in\mathcal J$ define $H_J(\omega):=\inf\{H_s(\omega): s\in J\}$. Finally, define $\mathcal G:=\sigma\{H_J: J\in\mathcal J\}$. Then for $c\in\Bbb R$, because $s\mapsto H_s(\omega)$ is lower semi-continuous, $$ \{(s,\omega)\in[0,\infty)\times\Omega: H_s(\omega)>c\}=\cup_{J\in\mathcal J_0}\left(J\times\{H_J>c\}\right), $$ where $\mathcal J_0$ is the subcollection of $\mathcal J$ comprising those intervals with rational endpoints. It follows that $H$ is $\mathcal B[0,\infty)\otimes\mathcal G$ measurable, and clearly $\mathcal F_t\subset\mathcal G$ for each $t>0$.

Consider now, for $t>0$ fixed, the integral $X_t:=\int_0^t H_s\,ds$. If $H$ were $(\mathcal F_t)$-progressive then $X_t$ would be $\mathcal F_t$ measurable. In this case there would be a countable set $S\subset(0,t]$ with $X_t$ measurable over $\sigma\{H_s:s\in S\}$. If $\omega$ and $\omega'$ were distinct elements of $\Omega$ with $\omega(s)=\omega'(s)$ for all $s\in S$, then we would clearly have $X_t(\omega)=X_t(\omega')$. The following choice of $\omega$ and $\omega'$ leads to a contradiction: $\omega(s)=1$ for all $s>0$; $\omega'$ the indicator of an open set $U\supset S$ of Lebesgue measure strictly less than $t$. Thus, $H$ is not $(\mathcal F_t)$-progressive.

On the other hand, if we define the $\sigma$-field $\mathcal G_t$ as we did $\mathcal G$ except that the intervals $J$ must be contained in $[0,t]$, then $(\mathcal G_t)_{t\ge 0}$ is a filtration, $\mathcal F_t\subset\mathcal G_t$ for each $t\ge 0$, and $\vee_t\mathcal G_t=\mathcal G$. Arguing as before, using lower semi-continuity, it can be shown that the restriction of $H$ to $[0,t]\times\Omega$ is $\mathcal B[0,t]\otimes\mathcal G_t$ measurable, for each $t\ge 0$. That is, $H$ is $(\mathcal G_t)$-progressive.

As is the other examples of the phenomena, a "gap" between the filtration generated by $H$ and a $\sigma$-field $\mathcal G$ for which $H$ is $\mathcal B[0,\infty)\otimes\mathcal G$ measurable seems to be crucial.

Answer 2

$\newcommand\F{\mathcal F}\newcommand\B{\mathcal B}\newcommand\R{\mathbb R}\newcommand\om\omega\newcommand\Om\Omega$Understanding your question:

are there examples of measurable but not progressive processes with respect to $\mathcal{F} = \mathcal{F}_\infty$?

literally, such an example can be obtained by the following slight modification of the example in your post:

Let $X(t,\om):= 1(t=\om)$ for $(t,\om)\in\R_+\times\Om$, $\Om:=\R_+$, and $\F:=\B(\R_+)$. For each $t\in[0,1]$, let $\F_t$ be the $\sigma$-algebra generated by the singleton subsets of $\R_+$ (or by the singleton subsets of $[0,t]$); for each $t\in(1,\infty)$, let $\F_t:=\F$.

Then $\F_\infty=\F$, the function $X\colon\R_+\times\Om\to\R$ is $\B(\R_+)\otimes\F$-measurable, and $(X(t,\cdot))$ is $(\F_t)$-adapted but not progressive.

In general, it seems that the filtration $(\F_t)$ has to be discontinuous in some sense for such examples to exist.

Answer 3

It seems that I have found an example for my own question, although I have some minor doubts. The following explanation is somewhat handwavy.

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Let us take the same process $X_t ( \omega ) = 1_{t = \omega}$ on $\Omega = T = \mathbb{R}_+$ and let the "continuous" part of the filtration lag in time. Take some continuous function $\varphi ( t )$ satisfying

1. $\varphi(0) = 0$
2. $0 < \varphi ( t ) < t$ (strictly) for all $t > 0$
3. $\varphi ( t )$ monotonically increases to $\infty$

and define the filtration as follows: $$ \mathcal{F}_t = \mathcal{G}_{\varphi(t)} \vee \mathcal{H}_t, $$ where $\mathcal{G}_t = \mathcal{B} [0, t]$ and $\mathcal{H}_t$ is the natural filtration of $X_t$, that is, the sigma-algebra generated by singletons of $[0, t]$.

The function $\varepsilon ( t ) = t - \varphi ( t )$ can then be interpreted as the time lag of $\mathcal{G}_t$ at time $t$.

One possible interpretation would be that we receive the detailed continuous information about the process with some time lag due to imperfections of our measuring equipment, whereas crude single point information is available right away.

Then, if I am not mistaken, $X_t$ is $\mathcal{F}_t$-adapted and (jointly) measurable with respect to $\mathcal{F} = \mathcal{F}_\infty$.

It isn't progressive, because $\mathcal{F}_t$ lacks a tiny piece of continuous information.

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I thought of modifying this example a bit further. Take any process $X_t$ and consider two different filtrations:

1. its natural filtration $\mathcal{H}_t$, with respect to which I assume $X$ to be not progressive,
2. its enlargement $\mathcal{G}_t$, with which $X$ is progressive.

Then, take a function $\varphi ( t )$ as described above and define $$ \mathcal{F}_t = \mathcal{G}_{\varphi ( t )} \vee \mathcal{H}_t. $$

The definition means that the information which makes $X$ progressive arrives with a certain lag. Since it ultimately does, the process is measurable, but at a given $t$ we do not have enough information about it.

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I didn't think the example through thoroughly, but the idea makes sense to me. Please tell me if something's wrong.

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Continuing the handwavy intuitive description of the example, let's say the information about the process comes from two sensors: one is rough but quick (information is available right now), another is detailed but slow (lag in time). The first makes the process adapted. Using the information from the second, we'll ultimately obtain the full and detailed picture.

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P.S. My "lagging filtration" example seems to be in a sense a refutation of the following remark in Dellacherie & Meyer, Volume A, page 141: