61.7.2 The Optional Sampling Theorem Continuous Case
Next I want a version of the Doob optional sampling theorem which applies to
martingales defined on [0,L],L ≤∞. First recall Theorem 60.1.2 part of which is stated
as the following lemma.
Lemma 61.7.8Let f ∈ L^{1}
(Ω;E, ℱ)
where E is a separable Banach space. Then if G isa σ algebraG⊆ℱ,
||E (f|G )|| ≤ E (||f|||G).
Here is a lemma which is the main idea for the proofs of the optional sampling
theorem for the continuous case.
Lemma 61.7.9Let X
(t)
be a right continuous nonnegative submartingale such that thefiltration
{ℱt}
is normal. Recall this includes
ℱt = ∩s>tℱs.
Also let τ be a stopping time with values in
[0,T ]
. Let P_{n} =
{tn}
k
_{k=1}^{mn+1}be a sequenceof partitions of
[0,T]
which have the property that
Pn ⊆ Pn+1, nli→m∞||Pn || = 0,
where
||P || ≡ sup{||tn− tn || : k = 1,2,⋅⋅⋅,m }
n k k+1 n
Then let
m∑n n
τn(ω) ≡ tk+1Xτ−1((tnk,tnk+1])(ω )
k=0
It follows that τ_{n}is a stopping time and also the functions
|X (τn)|
are uniformlyintegrable. Furthermore,
|X (τ)|
is integrable.
Proof: First of all, say t ∈ (t_{k}^{n},t_{k+1}^{n}]. If t < t_{k+1}^{n}, then
[τ ≤ t] = [τ ≤ tn] ∈ ℱ n ⊆ ℱ
n k tk t
and if t = t_{k+1}^{n}, then
[ ] [ ]
τn ≤ tnk+1 = τ ≤ tnk+1 ∈ ℱtnk+1 = ℱt
and so τ_{n} is a stopping time. It follows from Proposition 61.7.5 that X
By Lemma 61.7.10, τ_{n} is a stopping time and the functions
||M (τn)||
are uniformly
integrable. Also
||M (τ)||
is integrable. Similarly
||M (τn ∧ σn)||
are uniformly integrable
where σ_{n} is defined similarly to τ_{n}.
Consider the main claim now. Letting σ,τ be stopping times with τ bounded, it
follows that for σ_{n} and τ_{n} as above, it follows from Theorem 61.6.5
M (σ ∧ τ) = E (M (τ )|ℱ )
n n n σn
Thus, taking A ∈ℱ_{σ} and recalling σ ≤ σ_{n} so that by Proposition 61.7.5, ℱ_{σ}⊆ℱ_{σn},
∫ ∫ ∫
M (σn ∧τn)dP = E (M (τn)|ℱσn)dP = M (τn)dP.
A A A
Now passing to a limit as n →∞, the Vitali convergence theorem, Theorem 9.5.3 on
Page 701 and the right continuity of M implies one can pass to the limit in the above and
conclude
Then by Lemma 61.7.10τ_{n} is a stopping time, the functions
|X (τ )|
n
are uniformly
integrable, and
|X (τ)|
is also integrable. For σ_{n} defined similarly to τ_{n}, it also follows
|X (τ ∧ σ )|
n n
are uniformly integrable.
Let A ∈ℱ_{σ}. Since σ ≤ σ_{n}, it follows that ℱ_{σ}⊆ℱ_{σn}. By the discrete optional
sampling theorem for submartingales, Theorem 61.6.7,
X (σn ∧ τn) ≤ E (X (τn)|ℱσn)
and so
∫ ∫ ∫
X (σn ∧ τn) dP ≤ E (X (τn)|ℱσn)dP = X (τn)dP
A A A
and now taking lim_{n→∞} of both sides and using the Vitali convergence theorem along
with the right continuity of X, it follows
∫ ∫ ∫
X (σ∧ τ)dP ≤ X (τ)dP ≡ E (X (τ) |ℱ )dP
A A A σ
By Proposition 61.7.5, ℱ_{σ∧τ}⊆ℱ_{σ}, and so since A ∈ℱ_{σ} was arbitrary,
E (X (τ)|ℱ σ) ≥ X (σ∧ τ) a.e. ■
Note that a function defined on a countable ordered set such as the integers or equally
spaced points is right continuous.
Here is an interesting lemma.
Lemma 61.7.13Suppose E
(|Xn |)
< ∞ for all n,X_{n}is ℱ_{n}measurable, ℱ_{n+1}⊆ℱ_{n}forall n ∈ ℕ, and there exist X_{∞}ℱ_{∞}measurable such that ℱ_{∞}⊆ℱ_{n}for alln and X_{0}ℱ_{0}measurable such that ℱ_{0}⊇ℱ_{n}for all n such that for all n ∈