Almost Sure

28 February 17

Pathwise Regularity of Optional and Predictable Processes

As I have mentioned before in these notes, when working with processes in continuous time, it is important to select a good modification. Typically, this means that we work with processes which are left or right continuous. However, in general, it can be difficult to show that the paths of a process satisfy such pathwise regularity. In this post I show that for optional and predictable processes, the section theorems introduced in the previous post can be used to considerably simplify the situation. Although they are interesting results in their own right, the main application in these notes will be to optional and predictable projection. Once the projections are defined, the results from this post will imply that they preserve certain continuity properties of the process paths.

Suppose, for example, that we have a continuous-time process X which we want to show to be right-continuous. It is certainly necessary that, for any sequence of times {t_n\in{\mathbb R}_+} decreasing to a limit {t}, {X_{t_n}} almost-surely tends to {X_t}. However, even if we can prove this for every possible decreasing sequence {t_n}, it does not follow that X is right-continuous. As a counterexample, if {\tau\colon\Omega\rightarrow{\mathbb R}} is any continuously distributed random time, then the process {X_t=1_{\{t\le \tau\}}} is not right-continuous. However, so long as the distribution of {\tau} has no atoms, X is almost-surely continuous at each fixed time t. It is remarkable, then, that if we generalise to look at sequences of stopping times, then convergence in probability along decreasing sequences of stopping times is enough to guarantee everywhere right-continuity of the process. At least, it is enough so long as we restrict consideration to optional processes.

As usual, we work with respect to a complete filtered probability space {(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge0},{\mathbb P})}. Two processes are considered to be the same if they are equal up to evanescence, and any pathwise property is said to hold if it holds up to evanescence. That is, a process is right-continuous if and only is it is everywhere right-continuous on a set of probability 1. All processes will be taken to be real-valued, and a process is said to have left (or right) limits if its left (or right) limits exist everywhere, up to evanescence, and are finite.

Theorem 1 Let X be an optional process. Then,

  1. X is right-continuous if and only if {X_{\tau_n}\rightarrow X_\tau} in probability, for each uniformly bounded sequence {\tau_n} of stopping times decreasing to a limit {\tau}.
  2. X has right limits if and only if {X_{\tau_n}} converges in probability, for each uniformly bounded decreasing sequence {\tau_n} of stopping times.
  3. X has left limits if and only if {X_{\tau_n}} converges in probability, for each uniformly bounded increasing sequence {\tau_n} of stopping times.

The `only if’ parts of these statements is immediate, since convergence everywhere trivially implies convergence in probability. The importance of this theorem is in the `if’ directions. That is, it gives sufficient conditions to guarantee that the sample paths satisfy the respective regularity properties.

Note that conditions for left-continuity are absent from the statements of Theorem 1. In fact, left-continuity does not follow from the corresponding property along sequences of stopping times. Consider, for example, a Poisson process, X. This is right-continuous but not left-continuous. However, its jumps occur at totally inaccessible times. This implies that, for any sequence {\tau_n} of stopping times increasing to a finite limit {\tau}, it is true that {X_{\tau_n}} converges almost surely to {X_\tau}. In light of such examples, it is even more remarkable that right-continuity and the existence of left and right limits can be determined by just looking at convergence in probability along monotonic sequences of stopping times. Theorem 1 will be proven below, using the optional section theorem.

For predictable processes, we can restrict attention to predictable stopping times. In this case, we obtain a condition for left-continuity as well as for right-continuity.

Theorem 2 Let X be a predictable process. Then,

  1. X is right-continuous if and only if {X_{\tau_n}\rightarrow X_\tau} in probability, for each uniformly bounded sequence {\tau_n} of predictable stopping times decreasing to a limit {\tau}.
  2. X is left-continuous if and only if {X_{\tau_n}\rightarrow X_\tau} in probability, for each uniformly bounded sequence {\tau_n} of predictable stopping times increasing to a limit {\tau}.
  3. X has right limits if and only if {X_{\tau_n}} converges in probability, for each uniformly bounded decreasing sequence {\tau_n} of predictable stopping times.
  4. X has left limits if and only if {X_{\tau_n}} converges in probability, for each uniformly bounded increasing sequence {\tau_n} of predictable stopping times.

Again, the proof is given below, and relies on the predictable section theorem. (more…)

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29 November 16

The Section Theorems

Consider a probability space {(\Omega,\mathcal{F},{\mathbb P})} and a subset S of {{\mathbb R}_+\times\Omega}. The projection {\pi_\Omega(S)} is the set of {\omega\in\Omega} such that there exists a {t\in{\mathbb R}_+} with {(t,\omega)\in S}. We can ask whether there exists a map

\displaystyle  \tau\colon\pi_\Omega(S)\rightarrow{\mathbb R}_+

such that {(\tau(\omega),\omega)\in S}. From the definition of the projection, values of {\tau(\omega)} satisfying this exist for each individual {\omega}. By invoking the axiom of choice, then, we see that functions {\tau} with the required property do exist. However, to be of use for probability theory, it is important that {\tau} should be measurable. Whether or not there are measurable functions with the required properties is a much more difficult problem, and is answered affirmatively by the measurable selection theorem. For the question to have any hope of having a positive answer, we require S to be measurable, so that it lies in the product sigma-algebra {\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}}, with {\mathcal{B}({\mathbb R}_+)} denoting the Borel sigma-algebra on {{\mathbb R}_+}. Also, less obviously, the underlying probability space should be complete. Throughout this post, {(\Omega,\mathcal{F},{\mathbb P})} will be assumed to be a complete probability space.

It is convenient to extend {\tau} to the whole of {\Omega} by setting {\tau(\omega)=\infty} for {\omega} outside of {\pi_\Omega(S)}. Then, {\tau} is a map to the extended nonnegative reals {\bar{\mathbb R}_+={\mathbb R}_+\cup\{\infty\}} for which {\tau(\omega) < \infty} precisely when {\omega} is in {\pi_\Omega(S)}. Next, the graph of {\tau}, denoted by {[\tau]}, is defined to be the set of {(t,\omega)\in{\mathbb R}_+\times\Omega} with {t=\tau(\omega)}. The property that {(\tau(\omega),\omega)\in S} whenever {\tau(\omega) < \infty} is expressed succinctly by the inclusion {[\tau]\subseteq S}. With this notation, the measurable selection theorem is as follows.

Theorem 1 (Measurable Selection) For any {S\in\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}}, there exists a measurable {\tau\colon\Omega\rightarrow\bar{\mathbb R}_+} such that {[\tau]\subseteq S} and

\displaystyle  \left\{\tau < \infty\right\}=\pi_\Omega(S). (1)

As noted above, if it wasn’t for the measurability requirement then this theorem would just be a simple application of the axiom of choice. Requiring {\tau} to be measurable, on the other hand, makes the theorem much more difficult to prove. For instance, it would not hold if the underlying probability space was not required to be complete. Note also that, stated as above, measurable selection implies that the projection of S is equal to a measurable set {\{\tau < \infty\}}, so the measurable projection theorem is an immediate corollary. I will leave the proof of Theorem 1 for a later post, together with the proofs of the section theorems stated below.

A closely related problem is the following. Given a measurable space {(X,\mathcal{E})} and a measurable function, {f\colon X\rightarrow\Omega}, does there exist a measurable right-inverse on the image of {f}? This is asking for a measurable function, {g}, from {f(X)} to {X} such that {f(g(\omega))=\omega}. In the case where {(X,\mathcal{E})} is the Borel space {({\mathbb R}_+,\mathcal{B}({\mathbb R}_+))}, Theorem 1 says that it does exist. If S is the graph {\{(t,f(t))\colon t\in{\mathbb R}_+\}} then {\tau} will be the required right-inverse. In fact, as all uncountable Polish spaces are Borel-isomorphic to each other and, hence, to {{\mathbb R}_+}, this result applies whenever {(X,\mathcal{E})} is a Polish space together with its Borel sigma-algebra. (more…)

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