Some remarks about Stopping Times

Chris Barnett

September 6, 2001

We assume at the outset that we work within a complete filtered Probability Space, (Ω, F_{T} , P) , with time set [0, T] and filtration (F_{t} ), for t ∈ [0, T]. The usual hypotheses hold.

Definition 1 A random variable, τ : Ω ⟶ [0, T] is a stopping time (of the filtration) if for each t ∈ [0, T] the event {τ ≤ t} ∈ F_{t}.

Although the idea of a stopping time stands on its’ own, it most commonly arises as the (random) time of some event. Typically, the first time that a process behaves in a certain way. Since the σ-field F_{t} is interpreted as the information available, that is, known, up to time t ∈ [0, T], then a stopping time is a random time whose occurence or not is known at time t ∈ [0, T]. One uses stopping times in conjunction with processes by “stopping” a process at the stopping time;

Definition 2 Let X = (X_{t}) be a stochastic process adapted to our filtration and τ a stopping time. We define the process X stopped at time τ to be the random variable

X_{τ}(ω) = X_{τ(ω)}(ω).

There is a related idea which goes by the dangerously confusing name, the stopped process, this object is a process

X^{τ} = (X_{τ∧t}).

So, X^{τ} is the process obtained by running along each path, t ⟶ X_{t}(ω), of X, until time τ occurs, at which point the value of the process freezes at X_{τ}(ω).

Theorem 1 Let τ : Ω ⟶ [0, T]. Then τ is a stopping time if and only if for t ∈ [0, T],

{τ < t} ∈ F_{t} .

Proof 1 Assume the events {τ < t} lie in F_{t} . Let n be a natural number. For t ∈ [0, T) and all large enough n,

\{τ ≤ t\} = ∩_{n}\{τ < t + \frac{1}{n}\}

and so

\{τ ≤ t\} ∈ ∩_{n}F_{t+ \frac{1}{n}} = F_{t}

because the filtration is increasing and right continuous. For t = T, we have

{τ ≤ T } = {τ < T } ∪ {τ = T }
= {τ < T } ∪ Ω \ {τ < T }
∈ F_{T} .

Conversely, for t = 0 the set {τ < 0} = ∅ ∈ F_{0}. For t > 0 we know τ(ω) < t if and only if there is a natural number, n(ω) such that τ(ω) ≤ t − \frac{1}{n(ω)}. It follows that for all large enough n,

\{τ(ω) < t\} = ∪_{n}\{τ ≤ t − \frac{1}{n}\}.

Of course \{τ ≤ t − \frac{1}{n}\} ∈ F_{t− \frac{1}{n}} which is in turn lies in F_{t} . Their union over all large enough n will lie in F_{t} because of the sequential closure properties of σ -fields.

Stopping times are a very useful tool for the analysis of stochastic processes. This analysis often involves dealing with combinations of stopping times and so it is natural to examine how stopping times may be combined. The following theorem is a non-exhaustive list, its proof is an exercise for the reader.

Theorem 2 Let σ and τ be stopping times. Then the following are stopping times too:

• σ ∧ τ = min{σ, τ}

• σ ∨ τ = max{σ, τ}

• σ + τ

• ασ, where α > 1 Also, each t ∈ [0, T] is a stopping time.

Just as there is a σ -field of events known up to the deterministic time t, so the same is true for a stopping time τ. This idea is captured in the following definition.

Definition 3 Let τ be a stopping time and denote by F_{τ} the collection of events

{E ∈ F_{T} : E ∩ {τ ≤ t} ∈ F_{t} , ∀t}.

We call F_{τ} the σ - algebra of events up to time τ. This will be shortened to “the time τ σ -field” or similar. As an exercise one may confirm that for t ∈ [0, T], the “time t σ -field” of the random time, t, is exactly F_{t} . To continue our development we will need

Theorem 3 Let ρ and τ be stopping times.

• ρ is F_{ρ} measurable.

• If ρ ≤ τ then F_{ρ} ⊆ F_{τ}.

• If E belongs to F_{ρ} then E ∩ {ρ ≤ τ} belongs to F_{τ}.

Proof 2 Let B be a Borel set in R. The set ρ^{−1}(B) will be in F_{ρ} if for each t ∈ [0, T],

ρ^{−1}(B) ∩ {ρ ≤ t} ∈ F_{t}.

But

ρ^{−1}(B) ∩ {ρ ≤ t} = {ρ ∈ B ∩ [0, t]}

and of course, B ∩ [0, t] is a Borel set in [0, t]. It is enough to show that the inverse image of a Borel set in [0, t] lies in F_{t}. In view of the definition of a stopping time, the inverse image of every sub-interval of [0, t] lies in F_{t}. Consider the collection of sets

R = {A ⊆ [0, t] : ρ^{−1}(A) ∈ F_{t}}.

One can verify that this collection of sets forms a σ -algebra on [0, t].

•

ρ^{−1}([0, t]) = {ρ ≤ t} ∈ F_{t}.

So [0, t] ∈ R • If A ∈ R then

ρ^{−1}([0, t] \ A) = {ρ ≤ t} \ ρ^{−1}(A) ∈ F_{t}.

Because F_{t} is a σ -field.

• If (A_{n}) is a sequence of sets in R then

ρ^{−1}(∪_{n}A_{n}) = ∪_{n}ρ^{−1}(A_{n}) ∈ F_{t}

because F_{t} is closed under countable unions.

So R is a σ -field on [0, t] which contains the intervals. So R contains the Borel σ -field on [0, t] and therefore the inverse image of any Borel set in [0, t] lies in F_{t}.

Suppose now that ρ ≤ τ. A set, E, in F_{ρ} satisfies

E ∩ {ρ ≤ t} ∈ F_{t}.

We note that {τ ≤ t} ⊆ {ρ ≤ t} therefore

E ∩ {τ ≤ t} = (E ∩ {ρ ≤ t}) ∩ {τ ≤ t} ∈ F_{t}

because both parts of the last intersection lie in F_{t}. So

F_{ρ} ⊆ F_{τ}.

For the last part note that

E ∩ {ρ ≤ τ} ∩ {τ ≤ t} = (E ∩ {ρ ≤ t}) ∩ {τ ≤ t} ∩ {ρ ∧ t ≤ τ ∧ t}

each of the three sets on the left are F_{t} measurable; the first because E is in F_{ρ}, the second because τ is a stopping time and the third because, by the first two parts, each of ρ ∧ t and τ ∧ t are F_{t} random variables.

As an example of a stopping time consider a Borel set, B, in R, and a measurable process, X. The first hitting time of B by X:

D_{B}(ω) = inf{t ∈ [0, T] : X_{t}(ω) ∈ B}

where inf ∅ = +∞ is a stopping time^{1}.

1The case D_{B}(ω) = ∞ is the event that B is never hit by X. This gives us a technical problem if T < ∞ because D_{B} takes values outside the time set. Don’t worry, one can fix it.

Theorem 4 Let X = (X_{t}) be a progressively measurable process and τ a stopping time. Then X_{τ} is an F_{τ} measurable random variable. Moreover, the process X^{τ} is progressively measurable.

Proof 3 To prove the first part we need to show that for every Borel set, B, in R and every t ∈ [0, T],

X_{τ}^{−1}(B) ∩ {τ ≤ t} ∈ F_{t}.

But

X_{τ}^{−1}(B) ∩ {τ ≤ t} = X_{τ∧t}^{−1}(B) ∩ {τ ≤ t}

and if we know that X^{τ} = (X_{τ∧t}) is progressively measurable then X_{τ∧t}^{−1}(B) will be an F_{τ∧t} set. This σ -field lies inside F_{t} so we would be finished. So it is enough to prove the second part of the theorem. By the previous result the stopping time τ ∧ t is an F_{t} measurable random variable. The map of [0, t] × Ω into itself defined, rather curiously, as

(s, ω) ⟶ (τ ∧ t(ω), ω)

is measurable when both its domain and range have the σ -field B_{t} ⊗ F_{t}. To see this you must resist the impulse to look for something difficult: If Y is any F_{t} measurable random variable and M is the map of [0, t] × Ω into itself defined by

(s, ω) ⟶ (Y (ω), ω)

then the inverse image under M of the rectangle, B × E, where B ∈ B_{t} and E ∈ F_{t} is the (rectangular) set [0, t] × (Y^{−1}(B) ∩ E). Of course, the collection of sets in [0, t] × Ω whose inverse image under M lies in B_{t} ⊗ F_{t} is a σ -field containing the generators of B_{t} ⊗ F_{t}, and therefore B_{t} ⊗ F_{t} itself. This shows that M is measurable when both its domain and range have this product σ -field. Returning to what we want to prove; we know that

(s, ω) ⟶ (τ(ω) ∧ t, ω)

is B_{t} ⊗ F_{t} measurable and we also know that the restriction of X to [0, t] × Ω is (B_{t} ⊗ F_{t}, B) measurable. The conclusion is that the composite map, (s, ω) ⟶ X(τ(ω) ∧ t, ω) is (B_{t} ⊗ F_{t}, B) measurable. So X^{τ} is a progressively measurable process.

To show that stopping times are of some use one can turn to the kind of process most often occuring in Mathematical Finance, namely martingales, and prove a very useful result.

Theorem 5 Let X = (X_{t}) be a process adapted to the filtration of our probability space. If there is a real constant, c, such that for every stopping time, τ, of the filtration we have

E(X_{τ}) = c.

Then the process X is a martingale. Conversely, should X be a martingale then {E(X_{τ}) : τ is a stopping time } is a singleton.

Proof 4 Let E ∈ F_{t} and define τ by the prescription that τ = t on the set E and τ = T on Ω \ E. Then

X_{τ} = X_{t}I_{E} + X_{T} I_{Ω\E}.

At the same time

X_{T} = X_{T} I_{E} + X_{T} I_{Ω\E}.

Now if E(X_{τ}) = E(X_{T}) then it follows that

E(X_{t}I_{E}) = E(X_{T} I_{E}).

This holds for every E ∈ F_{t} and it follows from the definition of conditional expectation that

M_{t}(X_{T}) = X_{t} P, a.s.

So X is a martingale. We will prove the converse at a later stage.

Let T < ∞ and τ a stopping time of the filtration. From τ and a partition, π, of [0, T], we can construct a simple stopping time τ_{π} ( “derived from τ and the partition π”) which is larger than τ. We simply define,

τ_{π} = 0I_{{τ=0}} + ∑_{i=1}^{n} t_{i}I_{{t_{i−1}<τ≤t_{i}}}.

Here π = {t_{i}} and 0 = t_{0} < t_{1} < … < t_{n} = T From its definition it is clear that τ_{π} ≥ τ. The point of all of this is that if we choose a sequence of, say, dyadic partitions of [0, T] each refining those that precede it then we construct a monotone sequence of simple stopping times which decend to τ^{2}. So our first conclusion is that any stopping time can be approximated from above by a sequence of (simple) stopping times. This prompts the question as to whether the same is true from below? Under some restrictions the answer is no and this leads to some important theory of stopping times. However, we are not going to follow this thread right now. We are going to use the observation above to prove a result about X_{τ} when X is a martingale.

Theorem 6 Let T < ∞, τ a stopping time of the filtration and X an L^{2} martingale. Let M_{τ} be the conditional expectation onto the σ -algebra F_{τ}. Then

X_{τ} = M_{τ}(X_{T}).

Lemma 1 Let π = {t_{i}} be a (dyadic) partition of [0, T] with t_{m} = T, τ_{π} the stopping time derived from τ and the partition π, and let P_{τ_{π}} be the operator defined for Y ∈ L^{2} by

P_{τ_{π}}(Y) = M_{0}(Y)I_{{τ=0}} + ∑_{π} M_{t_{i}}(Y)I_{{t_{i−1}<τ≤t_{i}}}.

Then P_{τ_{π}} is exactly the conditional expectation onto the σ -field F_{τ_{π}}. Moreover

∥ P_{τ_{π}}(Y) ∥_{2}≤∥ Y ∥_{2} .

Proof 5 We prove this by showing that P_{τ_{π}} is an orthogonal projection operator which is both smaller and larger than the conditional expectation operator M_{τ_{π}}. First of all, because the sets {τ = 0}, {t_{0} < τ ≤ t_{1}} …, partition Ω and the conditional expectations are themselves (orthogonal) projections and τ is a stopping time, then it follows (write it out!) that P_{τ_{π}} is an orthogonal projection. Now consider the sets {τ_{π} ≤ t} for t ∈ [0, T]. We have

{τ_{π} ≤ t} = {τ = 0} if 0 ≤ t < t_{1}
= {τ = 0} ∪ {0 < τ ≤ t_{1}} if t_{1} ≤ t < t_{2}
·
·
= {τ = 0} ∪ ∪_{i=1}^{k}{t_{i−1} < τ ≤ t_{i}} if t_{k} ≤ t < t_{k+1}, k < m
= Ω if t = t_{m} = T

2As an exercise you might write out this construction explicitly and satisfy yourself that what I say is true

A set, F, in F_{τ_{π}} must satisfy

E ∩ {τ_{π} ≤ t} ∈ F_{t}, t ∈ [0, T].

Reading off what this means for 0 ≤ t < t_{1} we see that

F ∩ {τ = 0} ∈ F_{0}.

For t_{1} ≤ t < t_{2} we see that

F ∩ ({τ = 0} ∪ {0 < τ ≤ t_{1}}) ∈ F_{t_{1}}.

Taking into account the first relationship, this shows that

F ∩ {0 < τ ≤ t_{1}} ∈ F_{t_{1}}.

This process can be continued, formally by induction, for us to conclude that

F ∩ {t_{i−1} < τ ≤ t_{i}} ∈ F_{t_{i}}

for all i. As a consequence

P_{τ_{π}}(I_{F}) = M_{0}(I_{F})I_{{τ=0}} + ∑_{π} M_{t_{i}}(I_{F})I_{{t_{i−1}<τ≤t_{i}}}
= M_{0}(I_{F} I_{{τ=0}}) + ∑_{π} M_{t_{i}}(I_{F} I_{{t_{i−1}<τ≤t_{i}}})
= M_{0}(I_{F ∩{τ=0}}) + ∑_{π} M_{t_{i}}(I_{F ∩{t_{i−1}<τ≤t_{i}}})
= I_{F ∩{τ=0}} + ∑_{π} I_{F ∩{t_{i−1}<τ≤t_{i}}}
= I_{F}

This is enough to show that P_{τ_{π}} ≥ M_{τ_{π}}. On the other hand, given Y in L^{2} we make the observation the P_{τ_{π}}(Y) is simply the process (M_{t}(Y)) stopped at the random time τ_{π}. Calling this process Y = (Y_{t}) also^{3}, (of course the random variable Y = Y_{T}) this just states that

P_{τ_{π}}(Y_{T}) = Y_{τ_{π}}.

3The abuse of notation has a mathematical point; the L^{2} martingales and L^{2} itself are in one to one correspondence

We already know that Y_{τ_{π}} is F_{τ_{π}} measurable. So

M_{τ_{π}}(P_{τ_{π}}(Y)) = M_{τ_{π}}(Y_{τ_{π}}) = Y_{τ_{π}}.

This states that P_{τ_{π}} ≤ M_{τ_{π}}. So the two are equal. For the last part note that the square of P_{τ_{π}}(Y) is

P_{τ_{π}}(Y)^{2} = M_{0}(Y)^{2}I_{{τ=0}} + ∑_{π} M_{t_{i}}(Y)^{2}I_{{t_{i−1}<τ≤t_{i}}}
≤ M_{0}(Y^{2})I_{{τ=0}} + ∑_{π} M_{t_{i}}(Y^{2})I_{{t_{i−1}<τ≤t_{i}}}
= M_{0}(Y^{2}I_{{τ=0}}) + ∑_{π} M_{t_{i}}(Y^{2}I_{{t_{i−1}<τ≤t_{i}}})

using the Schwartz inequality for conditional expectations, that the sets involved partition Ω and that τ is a stopping time. Taking the expectation, E, of the last term, noting that each conditional expectation preserves E, gives the result.

We turn to the proof of the main theorem.

Proof 6 Let E ∈ F_{τ} and let (τ_{n}) be a sequence of simple times, constructed as above, decending to τ. Now it is known that X is a right continuous process^{4} and therefore, by the dominated convergence theorem,

lim_{n⟶∞} X_{τ_{n}} ⟶ X_{τ},

the limit occurs in L^{2} norm and P a.s.. This shows also that X_{τ} is an element of L^{2} with

∥ X_{τ} ∥_{2}≤ lim inf_{n} ∥ X_{τ_{n}} ∥_{2}≤∥ X_{T} ∥_{2} .

Let π be the partition of [0, T] that defines the simple time τ_{n}. Then

X_{τ_{n}} = X_{0}I_{{τ=0}} + ∑_{π} X_{t_{i}}I_{{t_{i−1}<τ≤t_{i}}}
= M_{0}(X_{T})I_{{τ=0}} + ∑_{π} M_{t_{i}}(X_{T})I_{{t_{i−1}<τ≤t_{i}}}
= M_{τ_{n}}(X_{T})

4Well, there is another part of this course that proves this!

Now let E ∈ F_{τ} then

< M_{τ}(X_{T}), I_{E} > = < X_{T} , I_{E} >
= < M_{τ_{n}}(X_{T}), I_{E} > because F_{τ} ⊆ F_{τ_{n}}
= < X_{τ_{n}}, I_{E} >
⟶ < X_{τ}, I_{E} > .

Since both X_{τ} and M_{τ}(X_{T}) are F_{τ} measurable this shows that they are identical in L^{2}.

Corollary 1 If X is a martingale then {E(X_{τ}) : τ is a stopping time} is a singleton.

Proof 7 For any time τ,

E(X_{τ}) = E(M_{τ}(X_{T})) = E(X_{T}).

We have already observed that any stopping time is the limit of a decreasing sequence of simple stopping times. As we remarked at the time the same is not true the other way round. We will discuss this feature of the theory now, but first of all it will be useful to think about the filtration of σ -fields. Our filtration satisfies the usual conditions, that is, it is right continuous meaning, for t ∈ [0, T)

∩_{s>t}F_{s} = F_{t}.

You might think this kind of thing is automatic; after all taking the intersection in this fashion is just throwing away everything in the filtration which is not in F_{t} ....isn’t it?

Example 1 Let Ω = [0, 1] and we take [0, 1] as our time set also. For times t ∈ [0, \frac{1}{2}] let F_{t} = \{Ω∅\}. For times t ∈ (\frac{1}{2}, 1] we set F_{t} = B([0, 1]), the Borel sets of [0, 1]. So this is our filtration. Now

∩_{s> \frac{1}{2}} F_{s} = B([0, 1])

while F_{\frac{1}{2}} = \{Ω, ∅\}. So this filtration is not right continuous because it has a “big” jump in the available information immediately after t = \frac{1}{2}.

So, in a filtration of σ -fields there can be “jumps” in the available information. Indeed, for example, the σ -fields can change from being finite to infinite very rapidly. As an exercise you might like to modify the example so that the filtration is right continuous and has exactly this property. When there is a big change of the sort we have described it comes as no surprise that “the past” is of little use in predicting the “future”. Put a little more formally, predictions using the past information can be very inaccurate. However, there may be times, processes and events which can still be predicted by the past despite that, in general, not all of these can be.