61.6.1 Stopping Times And Their Properties
The optional sampling theorem is very useful in the study of martingales and
submartingales as will be shown.
First it is necessary to define the notion of a stopping time.
Definition 61.6.1 Let
be a probability space and let
n=1∞ be an
increasing sequence of σ algebras each contained in ℱ, called a discrete filtration. A
stopping time is a measurable function, τ which maps
Ω to ℕ,
such that for all n ∈ ℕ,
Note this is equivalent to saying
For τ a stopping time define ℱτ as follows.
These sets in ℱτ are referred to as “prior” to τ.
The most important example of a stopping time is the first hitting time.
Example 61.6.2 The first hitting time of an adapted process X
of a Borel set G is a
stopping time. This is defined as
To see this, note that
Proposition 61.6.3 For τ a stopping time, ℱτ is a σ algebra and if Y
ℱk measurable for all k,Y
having values in a separable Banach space E,
is ℱτ measurable.
Proof: Let An ∈ℱτ. I need to show ∪nAn ∈ℱτ. In other words, I need to show
The left side equals
which is a countable union of sets of ℱk and so ℱτ is closed with respect to countable
unions. Next suppose A ∈ℱτ.
and Ω ∩
Therefore, so is AC ∩
It remains to verify the last claim. Let B be an open set in E. Is
The following lemma contains the fundamental properties of stopping times for
Lemma 61.6.4 In the situation of Definition 61.6.1, let σ,τ be two stopping times.
- τ is ℱτ measurable
- ℱσ ∩
⊆ℱσ∧τ = ℱσ ∩ℱτ
- ℱτ = ℱk on
for all k. That is if A ∈ℱk, then A ∩
∈ℱτ and if
A ∈ℱτ, then A ∩
∈ℱk. In other words, the two σ algebras
are equal. Letting G denote this σ algebra, if g is either ℱτ or ℱk measurable then
its restriction to
is G measurable. Also if A ∈ℱτ, and Y ∈ L1
Proof: Consider the first claim.
and so τ
is the greatest integer less than or equal to
note that σ ∧ τ
is a stopping time because
Next consider the second claim. Let A ∈ℱσ. I want to show
In other words, I want to show
for all k. However, the set on the left equals
Now let A ∈ℱσ∧τ. I want to show it is in both ℱτ and ℱσ. To show it is in ℱτ
I need to show that
for all k. However,
and so this is in ℱk
. Thus A ∩
being the finite union of sets which are.
Similarly A ∩
for all k
and so A ∈ℱτ ∩ℱσ
Next let A ∈ℱτ ∩ℱσ. Then is it in ℱσ∧τ? Is A∩
? Of course this is
since both σ,τ are stopping times. This proves part 2.).
Now consider part 3.). Note that
is in both
. First consider the
claim it is in ℱτ
which is in ℱl. If l ≥ k, it reduces to
so it is in ℱτ
I need to show
means all sets of the form
. Let A ∈ℱτ.
Therefore, there exists B ∈ℱk such that B = A ∩
which shows ℱτ ∩
. Now let
because in case j < k, the set on the left is ∅ and if j ≥ k it reduces to A ∩
. Thus A ∩
Therefore, the two σ algebras of subsets of
are equal. Thus for A in either ℱτ or ℱk, A∩
is a set of both
if A ∈ℱk,
then from the above, there exists B ∈ℱτ
with similar reasoning holding if A ∈ℱτ. In other words, if g is ℱτ or ℱk measurable,
then the restriction of g to
is measurable with respect to
be an arbitrary random variable in L1
It follows, since
is in both
Since this holds for an arbitrary set in ℱτ ∩
The assertion that
and that a function
which is ℱτ
measurable when restricted to
is the main result in the above lemma and this fact leads to the amazing Doob optional
sampling theorem below. Also note that if Y
is any process defined on the positive
, then by definition, Y
on the set
constant on this set.