Almost Sure

23 November 09

Sigma Algebras at a Stopping Time

The previous post introduced the notion of a stopping time {\tau}. A stochastic process {X} can be sampled at such random times and, if the process is jointly measurable, {X_\tau} will be a measurable random variable. It is usual to study adapted processes, where {X_t} is measurable with respect to the sigma-algebra {\mathcal{F}_t} at that time. Then, it is natural to extend the notion of adapted processes to random times and ask the following. What is the sigma-algebra of observable events at the random time {\tau}, and is {X_\tau} measurable with respect to this? The idea is that if a set {A} is observable at time {\tau} then for any time {t}, its restriction to the set {\{\tau\le t\}} should be in {\mathcal{F}_t}. As always, we work with respect to a filtered probability space {(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})}. The sigma-algebra at the stopping time {\tau} is then,

\displaystyle  \mathcal{F}_\tau=\left\{A\in\mathcal{F}_\infty\colon A\cap\{\tau\le t\}\in\mathcal{F}_t{\rm\ for\ all\ }t\ge 0\right\}.

The restriction to sets in {\mathcal{F}_\infty} is to take account of the possibility that the stopping time can be infinite, and it ensures that {A=A\cap\{\tau\le\infty\}\in\mathcal{F}_\infty}. From this definition, a random variable {U} us {\mathcal{F}_\tau}-measurable if and only if {1_{\{\tau\le t\}}U} is {\mathcal{F}_t}-measurable for all times {t\in{\mathbb R}_+\cup\{\infty\}}.

Similarly, we can ask what is the set of events observable strictly before the stopping time. For any time {t}, then this sigma-algebra should include {\mathcal{F}_t} restricted to the event {\{t<\tau\}}. This suggests the following definition,

\displaystyle  \mathcal{F}_{\tau-}=\sigma\left(\left\{ A\cap\{t<\tau\}\colon t\ge 0,A\in\mathcal{F}_t \right\}\cup\mathcal{F}_0\right).

The notation {\sigma(\cdot)} denotes the sigma-algebra generated by a collection of sets, and in this definition the collection of elements of {\mathcal{F}_0} are included in the sigma-algebra so that we are consistent with the convention {\mathcal{F}_{0-}=\mathcal{F}_0} used in these notes.

With these definitions, the question of whether or not a process {X} is {\mathcal{F}_\tau}-measurable at a stopping time {\tau} can be answered. There is one minor issue here though; stopping times can be infinite whereas stochastic processes in these notes are defined on the time index set {{\mathbb R}_+}. We could just restrict to the set {\{\tau<\infty\}}, but it is handy to allow the processes to take values at infinity. So, for the moment we consider a processes {X_t} where the time index {t} runs over {\bar{\mathbb R}_+\equiv{\mathbb R}_+\cup\{\infty\}}, and say that {X} is a predictable, optional or progressive process if it satisfies the respective property restricted to times in {{\mathbb R}_+} and {X_\infty} is {\mathcal{F}_\infty}-measurable.

Lemma 1 Let {X} be a stochastic process and {\tau} be a stopping time.

  • If {X} is progressively measurable then {X_\tau} is {\mathcal{F}_\tau}-measurable.
  • If {X} is predictable then {X_\tau} is {\mathcal{F}_{\tau-}}-measurable.

Proof: If {X} is progressive then, as proven in the previous post, the stopped process {X^\tau} is also progressive and, hence, is adapted. It follows that {1_{\{\tau\le t\}}X_\tau=1_{\{\tau\le t\}}X^\tau_t} is {\mathcal{F}_t}-measurable which, from the definition above, implies that {1_{\{\tau<\infty\}}X_\tau} is {\mathcal{F}_{\tau}}-measurable.

Furthermore, {1_{\{\tau=\infty\}}X_\tau} is {\mathcal{F}_\infty}-measurable and is zero when restricted to the set {{\{\tau\le t\}}} for all {t\in{\mathbb R}_+}, so is also {\mathcal{F}_\tau}-measurable.

Now, consider a predictable process {X}. Write {\mathcal{\bar P}} for the predictable sigma-algebra on {\bar{\mathbb R}_+\times\Omega}. That is, the subsets of {S\subseteq\bar{\mathbb R}_+\times\Omega} which are predictable when restricted to {{\mathbb R}_+\times\Omega} and such that {\{\omega\in\Omega\colon(\infty,\omega)\in S\}} is {\mathcal{F}_\infty}-measurable. Then, {X} is {\mathcal{\bar P}}-measurable. By the functional monotone class theorem, it is enough to prove the result for processes of the form {X_t(\omega)=1_{\{(t,\omega)\in S\}}} for some pi-system of sets generating {\mathcal{\bar P}}.

The predictable sigma algebra is generated by the sets {S} of the following forms,

  1. {S=(t,\infty]\times A} for times {t\in{\mathbb R}_+} and {A\in\mathcal{F}_t}. If {X=1_S} then { X_\tau=1_{\{t<\tau\}\cap A} } which, by definition, is {\mathcal{F}_{\tau-}}-measurable.
  2. {S=\{0\}\times A} for {A\in\mathcal{F}_0}. If {X=1_S} then { X_\tau= 1_{\{\tau=0\}\cap A} } which is {\mathcal{F}_0}-measurable, and so is also {\mathcal{F}_{\tau-}}-measurable.

\Box

So, the `adaptedness’ of measurable processes extends to stopping times. In fact, it is possible to go further and use this as an alternative definition of these sigma-algebras.

Lemma 2 Let {U} be a random variable and {\tau} be a stopping time. Then,

  • {U} is {\mathcal{F}_\tau}-measurable if and only if {U=X_\tau} for some progressively measurable (resp. optional) process {X}.
  • {U} is {\mathcal{F}_{\tau-}}-measurable if and only if {U=X_{\tau}} for some predictable process {X}.

Proof: If {U} is {\mathcal{F}_\tau}-measurable, then the process {X_{t}=1_{\{t\ge\tau\}}U} is adapted and right-continuous. Therefore, it is optional (and hence, progressive) and clearly {U=X_\tau}.

For the second statement, consider the set {V} of random variables which can be expressed as {X_\tau} for a predictable process {X}. The functional monotone class theorem can be used to show that {V} contains all {\mathcal{F}_{\tau-}}-measurable random variables. First, {V} is clearly closed under taking linear combinations. Second, if {U_n\in V} is increasing to the limit {U} then there exists predictable processes {X^n} with {U_n=X^n_\tau}. Then, {U=\limsup_nX^n_\tau} is also in {V}.

Finally, it just needs to be shown that {1_S\in V} for all {S} in a pi-system generating {\mathcal{F}_{\tau-}}. By definition, the following sets generate {\mathcal{F}_{\tau-}}.

  • {S\in\mathcal{F}_0}. In this case, {1_S=X_\tau} with {X_t(\omega)=1_{\{\omega\in S\}}}.
  • {S=A\cap\{t<\tau\}} for {t\in{\mathbb R}_+} and {A\in\mathcal{F}_t}. Then, {1_S=X_\tau} with {X_s(\omega)=1_{\{s>t,\omega\in A\}}}.

In both these cases, {X} is left-continuous and adapted and, hence, is predictable. \Box

This result gives the main motivation for the definitions of {\mathcal{F}_\tau} and {\mathcal{F}_{\tau-}}. For the remainder of this post, I state and prove several simple results which are useful for general applications of stopping times.

Lemma 3 Any stopping time {\tau} is both {\mathcal{F}_{\tau}} and {\mathcal{F}_{\tau-}}-measurable.

Proof: The deterministic process {X_t\equiv t} is trivially adapted and both left and right-continuous, so it is predictable and optional. Consequently, by the previous lemma, {\tau=X_{\tau}} is {\mathcal{F}_\tau} and {\mathcal{F}_{\tau-}}-measurable. \Box

Next, the sigma-algebras are increasing in the sense that we would hope.

Lemma 4 For any stopping time {\tau},

\displaystyle  \mathcal{F}_{\tau-}\subseteq\mathcal{F}_\tau.

If {\sigma\le\tau} is any other stopping time then,

\displaystyle  \mathcal{F}_\sigma\subseteq\mathcal{F}_\tau,\ \mathcal{F}_{\sigma-}\subseteq\mathcal{F}_{\tau-}.

If, furthermore, {\sigma<\tau} whenever {\tau\not\in\{0,\infty\}} then {\mathcal{F}_\sigma\subseteq\mathcal{F}_{\tau-}}.

Proof: This proof makes use of Lemma 2. First, by the lemma, every {\mathcal{F}_{\tau-}}-measurable set {A} can be written in the form {1_A=X_\tau} for a predictable process {X}. However, as predictable processes are progressive, {A} will also be in {\mathcal{F}_\tau}.

Now suppose that {A} is {\mathcal{F}_\sigma} (resp. {\mathcal{F}_{\sigma-}}) measurable. Then, there is a progressive (resp. predictable) process satisfying {X_\tau=1_A}. As the stopped process {X^\sigma} is also progressive (resp. predictable) it follows that {1_A=X^\sigma_\tau} is {\mathcal{F}_\tau} (resp. {\mathcal{F}_{\tau-}})-measurable.

Finally, suppose that {\sigma<\tau} whenever {\tau\not\in\{0,\infty\}} and that {A\in\mathcal{F}_\sigma}. Then

\displaystyle  X_t\equiv 1_{A\cap\{\sigma<t<\infty{\rm\ or\ }\sigma=0\}}+1_{A\cap\{t=\infty\}}

is a left-continuous and adapted at finite times, and {X_\infty} is {\mathcal{F}_\infty}-measurable. Hence, it is predictable process and {1_A=X_\tau} is {\mathcal{F}_{\tau-}}-measurable. \Box

The sigma-algebras satisfy the expected left and right-limits. In the following lemma, the first statement says that right-continuity of a filtration extends to arbitrary stopping times. The second says that {\mathcal{F}_{\tau-}} can indeed be interpreted as a left-limit. However, this statement does not say anything much for arbitrary stopping times, because it is not in general possible to strictly approximate them from the left in this way. If such a sequence {\tau_n} does indeed exist then the stopping time is called predictable.

Lemma 5 Let {\tau_n\rightarrow\tau} be stopping times. Then

  • If the filtration {\{\mathcal{F}_t\}} is right-continuous and {\tau_n\ge\tau} for each {n} then

    \displaystyle  \bigcap_n\mathcal{F}_{\tau_n}=\mathcal{F}_\tau.

  • If {\tau_n\le\tau}, with a strict inequality whenever {\tau\not\in\{0,\infty\}}, then

    \displaystyle  \sigma\left(\bigcup_n\mathcal{F}_{\tau_n}\right) =\sigma\left(\bigcup_n\mathcal{F}_{\tau_n-}\right)=\mathcal{F}_{\tau-}.

Proof: Starting with the first statement, we know that {\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}}. So, it just needs to be shown that any {A\in\cap_n\mathcal{F}_{\tau_n}} is in {\mathcal{F}_\tau}. Any such set satisfies

\displaystyle  A\cap\{\tau< t\} = \bigcup_n(A\cap\{\tau_n<t\})\in\mathcal{F}_t.

Then, by right-continuity of the filtration, for any {m\ge 1}

\displaystyle  A\cap\{\tau\le t\}=\bigcap_{n=m}^\infty(A\cap\{\tau<t+1/n\})\in\mathcal{F}_{t+}=\mathcal{F}_t

as required.

For the second statement, we know that {\mathcal{F}_{\tau_n-}\subseteq\mathcal{F}_{\tau_n}\subseteq\mathcal{F}_{\tau-}}, so it is only necessary to prove that there is a generating set for {\mathcal{F}_{\tau-}} lying in {\mathcal{G}\equiv\sigma(\cup_n\mathcal{F}_{\tau_n-})}. As {\mathcal{F}_0\subseteq\mathcal{G}} it is enough to consider sets of the form {S=A\cap\{t<\tau\}} for {A\in\mathcal{F}_t}. However

\displaystyle  A\cap\{t<\tau\}=\bigcup_n(A\cap\{t<\tau_n\})\in\mathcal{G}

as required. \Box

As should be the case, the definition of the sigma-algebras at a constant stopping time is consistent with the filtration.

Lemma 6 If {\tau\colon\Omega\rightarrow\bar{\mathbb R}_+} is equal to the constant value {t} then,

\displaystyle  \mathcal{F}_\tau=\mathcal{F}_t,\ \mathcal{F}_{\tau-}=\mathcal{F}_{t-}.

Proof: If {A\in\mathcal{F}_t} then for all times {s},

\displaystyle  A\cap\{\tau\le s\}=\begin{cases} A\in\mathcal{F}_t\subseteq\mathcal{F}_s,&\textrm{if }s\ge t,\\ \emptyset\in\mathcal{F}_s,&\textrm{if }s<t, \end{cases}

showing that {A\in\mathcal{F}_\tau}. Conversely, if {A\in\mathcal{F}_\tau} then {A=A\cap\{\tau\le t\}\in\mathcal{F}_t} as required.

This shows that {\mathcal{F}_{\tau}=\mathcal{F}_t}. The equality {\mathcal{F}_{\tau-}=\mathcal{F}_{t-}} follows by taking left limits and applying the previous lemma. \Box

Given two stopping times {\sigma,\tau} it follows from the definitions that {\{\sigma\le\tau\}\in\mathcal{F}_\tau} and {\{\sigma<\tau\}\in\mathcal{F}_{\tau-}}. So, {\{\sigma=\tau\}} is in {\mathcal{F}_\tau} and, by symmetry, is also in {\mathcal{F}_\sigma}. Furthermore, these two sigma-algebras coincide when restricted to this set.

Lemma 7 If {\sigma,\tau} are stopping times then

\displaystyle  \mathcal{F}_\sigma\vert_{\{\sigma=\tau\}}=\mathcal{F}_{\tau}\vert_{\{\sigma=\tau\}}.

Proof: If {A\subseteq\{\sigma=\tau\}} is {\mathcal{F}_\sigma} measurable then {A\cap\{\tau\le t\}=A\cap\{\sigma\le t\}} is in {\mathcal{F}_t} and {A\in\mathcal{F}_\tau}. The reverse inclusion follows by exchanging {\sigma} and {\tau}. \Box

Given a stopping time taking values in a countable set of times, the following result is often useful to show that a set is in the sigma algebra by checking it at each of the fixed times.

Lemma 8 Let {\tau,(\tau_n)_{n=1,2,\ldots}} be stopping times such that {\tau(\omega)\in\{\tau_1(\omega),\tau_2(\omega),\ldots\}} for all {\omega\in\Omega}.

A set {A\subseteq\Omega} is in {\mathcal{F}_\tau} if and only if {A\cap\{\tau_n=\tau\}\in\mathcal{F}_{\tau_n}} for each {n}.

Proof: By the previous lemma, if {A\in\mathcal{F}_\tau} then {A\cap\{\tau=\tau_n\}\in\mathcal{F}_{\tau_n}}. Conversely,

\displaystyle  A\cap\{\tau\le t\}=\bigcup_n((A\cap\{\tau=\tau_n\})\cap\{\tau_n\le t\})\in\mathcal{F}_t

as required. \Box

Finally, the following result is used to construct new stopping times out of old ones. If we wait until a time {\tau} occurs, and then decide to either use that time or not based on an {\mathcal{F}_\tau}-measurable event, the result is again a stopping time.

Lemma 9 Let {\tau} be a stopping time and {A\in\mathcal{F}_\tau}. Then,

\displaystyle  \tau_A(\omega)\equiv\begin{cases} \tau(\omega),&\textrm{if }\omega\in A,\\ \infty,&\textrm{otherwise} \end{cases}

is also a stopping time.

Proof: This follows from the following

\displaystyle  \left\{\tau_A\le t\right\}=A\cap\left\{\tau\le t\right\}\in\mathcal{F}_t

for all {t\in{\mathbb R}_+}. \Box

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6 Comments »

  1. Dear Almost sure,
    In lemma 7, do you want to show
    ... = \mathcal{F}_\tau |_{\{\sigma = \tau\}}
    Thanks

    Comment by kenneth — 30 April 10 @ 4:02 PM | Reply

  2. Dear Georges,
    Excuse me if I am wrong, but in the proof of Lemma 2, last but one line, shouldn’t it be S=A\cap\{t <\tau \} ?

    Comment by josh — 3 October 12 @ 9:40 AM | Reply

    • You’re right. I fixed it, thanks.

      Comment by George Lowther — 3 October 12 @ 11:32 AM | Reply

      • Thanks for your answer. Did you also by any chance see my question on the Stochastic Integral ?

        Comment by josh — 8 October 12 @ 9:40 AM | Reply

  3. Hello George,
    I recently meet the problem on consistency of probability measures. Given a sequence of probability measures Q_n, how is it possible to check that they are consistent? The definition seems impractical. What I want is a method that we can actually apply facing concrete examples. Also, where shall I start if I want to find counter-examples?
    Thank you very much!

    Comment by Zhenyu (Rocky) Cui — 1 November 12 @ 7:55 PM | Reply


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