The purpose of this section is to consider a special optional sampling theorem for
martingales which is superior to the one presented earlier. I have presented a different
treatment of the fundamental properties of stopping times also. See Kallenberg [?] for
more.
Definition 59.4.1Let
(Ω,ℱ,P )
be a probability space and let
{ℱn }
_{n=1}^{∞}be anincreasing sequence of σ algebras each contained in ℱ. A stopping timeis a measurablefunction, τ which maps Ω to ℕ,
τ−1(A) ∈ ℱ for all A ∈ P (ℕ ),
such that for all n ∈ ℕ,
[τ ≤ n] ∈ ℱ .
n
Note this is equivalent to saying
[τ = n] ∈ ℱn
because
[τ = n ] = [τ ≤ n]∖[τ ≤ n − 1].
For τ a stopping time define ℱ_{τ}as follows.
ℱ ≡ {A ∈ ℱ : A ∩ [τ ≤ n] ∈ ℱ for all n ∈ ℕ }
τ n
These sets in ℱ_{τ}are referred to as “prior” to τ.
First note that for τ a stopping time, ℱ_{τ} is a σ algebra. This is in the next
proposition.
Proposition 59.4.2For τ a stopping time, ℱ_{τ}is a σ algebra and if Y
(k)
is ℱ_{k}measurable for all k, then
ω → Y (τ (ω))
is ℱ_{τ}measurable.
Proof:Let A_{n}∈ℱ_{τ}. I need to show ∪_{n}A_{n}∈ℱ_{τ}. In other words, I need to show
that
∪nAn ∩[τ ≤ k] ∈ ℱk
The left side equals
∪n(An ∩ [τ ≤ k])
which is a countable union of sets of ℱ_{k} and so ℱ_{τ} is closed with respect to countable
unions. Next suppose A ∈ℱ_{τ}.
E (M (τ)|ℱ σ) = E (M (τ)|ℱi) = E(E (M (l)|ℱτ)|ℱi) = E (E (M (l)|ℱj)|ℱi)
If j ≤ i, this reduces to
E (M (l)|ℱ ) = M (j) = M (σ ∧τ).
j
If j > i, this reduces to
E (M (l)|ℱi) = M (i) = M (σ ∧τ)
and since this exhausts all possibilities for values of σ and τ, it follows
E (M (τ)|ℱσ) = M (σ ∧ τ) a.e.
This proves the theorem.
This is a really amazing theorem. Note it says M
(σ ∧τ)
= E
(M (τ)|ℱσ)
. This would
not be so surprising if it had said
M (σ∧ τ) = E (M (τ)|ℱσ∧τ).
What about submartingales? Recall
{X (k)}
_{k=1}^{∞} is a submartingale if
E (X (k+ 1)|ℱk) ≥ X (k)
where the ℱ_{k} are an increasing sequence of σ algebras in the usual way. The following is
a very interesting result.
Lemma 59.4.5Let
{X (k)}
_{k=0}^{∞}be a submartingale adapted to the increasingsequence of σ algebras,
{ℱk}
. Then there exists a unique increasing process
{A (k)}
_{k=0}^{∞}such that A
(0)
= 0 and A
(k + 1)
is ℱ_{k}measurable for all k and a martingale,
{M (k)}
_{k=0}^{∞}such that
X (k) = A(k)+ M (k).
Furthermore, for τ a stopping time, A
(τ)
is ℱ_{τ}measurable.
Proof:Define ∑_{k=0}^{−1}≠0. First consider the uniqueness assertion. Suppose A is a
process which does what is supposed to do.
n− 1 n− 1
∑ E (X (k +1) − X (k)|ℱ ) = ∑ E (A (k+ 1)− A (k)|ℱ )
k=0 k k=0 k
n−1
+∑ E(M (k + 1)− M (k)|ℱ )
k=0 k
Then since
{M (k)}
is a martingale,
n∑−1 n∑−1
E (X (k + 1)− X (k)|ℱk) = A (k + 1)− A (k) = A(n)
k=0 k=0
This shows uniqueness and gives a formula for A
(n)
assuming it exists. It is only a
matter of verifying this does work. Define
n−1
A (n) ≡ ∑ E (X (k + 1)− X (k)|ℱ ) , A (0) = 0.
k=0 k
Then A is increasing because from the definition,
A (n+ 1)− A (n) = E (X (n+ 1)− X (n)|ℱ ) ≥ 0.
n
Also from the definition above, A
(n)
is ℱ_{n−1} measurable, consider
{X (k)− A (k)} .
Why is this a martingale?
E(X (k+ 1)− A (k+ 1)|ℱk)
= E(X (k+ 1)|ℱk)− A (k+ 1)
k
= E(X (k+ 1)|ℱk)− ∑ E (X (j + 1)− X (j)|ℱj)
j=0
= E(X (k+ 1)|ℱ )− E (X (k+ 1)− X (k)|ℱ )
k k
k∑−1
− E (X (j + 1)− X (j)|ℱj )
j=0
k−∑ 1
= X (k) − E(X (j + 1)− X (j)|ℱj ) = X (k)− A (k)
j=0
Note the nonnegative integers could be replaced with any finite set or ordered
countable set of numbers with no change in the conclusions of this lemma or the above
optional sampling theorem.
Next consider the case of a submartingale.
Theorem 59.4.6Let
{X (k)}
be a submartingale with respect to the increasing sequenceof σ algebras,
{ℱk }
and let σ,τ be two stopping times such that τ is bounded. Then X
(τ)
defined as
ω → X (τ (ω ))
is integrable and
X (σ∧ τ) ≤ E (X (τ)|ℱ ).
σ
Proof:The claim about X
(τ)
being integrable is the same as in Theorem 61.6.5. If
τ ≤ l,