Instead of having a sequence of σ algebras, one can consider an increasing collection of σ
algebras indexed by t ∈ ℝ. This is called a filtration.
Definition 61.3.1Let X be a stochastic process defined on an interval, I =
[0,T ]
or [0,∞). Suppose the probability space,
(Ω,ℱ, P)
has an increasing family of σalgebras,
{ℱt}
. This is called a filtration.If for arbitrary t ∈ I the random variableX
(t)
is ℱtmeasurable, then X is said to be adaptedto the filtration
{ℱt}
. Denoteby ℱt+the intersection of all ℱsfor s > t. The filtration isnormalif
ℱ0contains all A ∈ℱ such that P
(A )
= 0
ℱt = ℱt+for all t ∈ I.
X is called progressively measurableif for every t ∈ I, the mapping
(s,ω) ∈ [0,t]× Ω, (s,ω) → X (s,ω)
is B
([0,t])
×ℱtmeasurable.
Thus X is progressively measurable means
(s,ω) → X[0,t](s)X (s,ω)
is B
([0,t])
×ℱt measurable. As an example of a normal filtration, here is an
example.
Example 61.3.2For example, you could have a stochastic process, X
(t)
and you coulddefine
--------------
Gt ≡ σ (X (s) : s ≤ t),
the completion of the smallest σ algebra such that each X
(s)
is measurablefor all s ≤ t. This gives an example of a filtration to which X
(t)
is adaptedwhich satisfies 1. More generally, suppose X
(t)
is adapted to a filtration, Gt.Define
ℱt ≡ ∩s>tGs
Then
ℱt+ ≡ ∩s>tℱs = ∩s>t ∩r>s Gr = ∩s>tGs ≡ ℱt.
and each X
(t)
is measurable with respect to ℱt. Thus there is no harm in assuming astochastic process adapted to a filtration can be modified so the filtration is normal. Alsonote that ℱtdefined this way will be complete so if A ∈ℱthas P
(A)
= 0 and if B ⊆ A,then B ∈ℱtalso. This is because this relation between the sets and the probability ofA being zero, holds for this pair of sets when considered as elements of eachGsfor s > t. Hence B ∈Gsfor each s > t and is therefore one of the sets inℱt.
What is the description of a progressively measurable set?
PICT
It means that for Q progressively measurable, Q ∩
[0,t]
× Ω as shown in the above
picture is ℬ
([0,t])
×ℱt measurable. It is like saying a little more descriptively that the
function is progressively product measurable.
I shall generally assume the filtration is normal.
Observation 61.3.3If X is progressively measurable, then it is adapted.Furthermore the progressively measurable sets, those E ∩
[0,T ]
× Ω for which XEis progressively measurable form a σ algebra.
To see why this is, consider X progressively measurable and fix t. Then
(s,ω)
→ X
(s,ω)
for
(s,ω)
∈
[0,t]
× Ω is given to be ℬ
([0,t])
×ℱt measurable, the
ordinary product measure and so fixing any s ∈
[0,t]
, it follows the resulting function of
ω is ℱt measurable. In particular, this is true upon fixing s = t. Thus ω → X
(t,ω)
is ℱt
measurable and so X
(t)
is adapted.
A set E ⊆
[0,T]
× Ω is progressively measurable means that XE is progressively
measurable. That is XE restricted to
[0,t]
× Ω is ℬ
([0,t])
×ℱt measurable. In other
words, E is progressively measurable if
E ∩ ([0,t]× Ω ) ∈ ℬ([0,t])× ℱt.
If Ei is progressively measurable, does it follow that E ≡∪i=1∞Ei is also progressively
measurable? Yes.
E ∩ ([0,t]× Ω ) = ∪∞ Ei ∩ ([0,t]× Ω) ∈ ℬ ([0,t])× ℱt
i=1
because each set in the union is in ℬ
([0,t])
×ℱt. If E is progressively measurable, is
EC?
∈ℬ([0,t])×ℱt ∈ℬ([0,t])×ℱt
C ◜-----◞◟-----◝ ◜--◞◟-◝
E ∩ ([0,t]× Ω)∪ (E ∩([0,t]× Ω)) = [0,t]× Ω
and so EC∩
([0,t]× Ω)
∈ℬ
([0,t])
×ℱt. Thus the progressively measurable sets are a σ
algebra.
Another observation of interest is in the following lemma.
Lemma 61.3.4Suppose Q is in ℬ
([0,a])
×ℱr. Then if b ≥ a and t ≥ r, then Qis also in ℬ
([0,b])
×ℱt.
Proof: Consider a measurable rectangle A×B where A ∈ℬ
([0,a])
and B ∈ℱr. Is it
true that A×B ∈ℬ
([0,b])
×ℱt? This reduces to the question whether A ∈ℬ
([0,b])
. If A
is an interval, it is clear that A ∈ℬ
([0,b])
. Consider the π system of intervals and let G
denote those Borel sets A ∈ℬ
([0,a])
such that A ∈ℬ
([0,b])
. If A ∈G, then
[0,b]
∖A ∈ℬ
([0,b])
by assumption (the difference of Borel sets is surely Borel). However,
this set equals
([0,a]∖A )∪ (a,b]
and so
[0,b] = ([0,a]∖ A)∪ (a,b]∪ A
The set on the left is in ℬ
([0,b])
and the sets on the right are disjoint and two of them
are also in ℬ
([0,b])
. Therefore, the third,
([0,a]∖ A)
is in ℬ
([0,b])
. It is obvious that G is
closed with respect to countable disjoint unions. Therefore, by Lemma 10.12.3, Dynkin’s
lemma, G⊇ σ
(Intervals)
= ℬ
([0,a])
.
Therefore, such a measurable rectangle A × B where A ∈ℬ
([0,a])
and B ∈ℱr is in
ℬ
([0,b])
×ℱt and in fact it is a measurable rectangle in ℬ
([0,b])
×ℱt. Now let K denote
all these measurable rectangles A × B where A ∈ℬ
([0,a])
and B ∈ℱr. Let G (new
G) denote those sets Q of ℬ
([0,a])
×ℱr which are in ℬ
([0,b])
×ℱt. Then if
Q ∈G,
Q ∪ ([0,a]× Ω ∖Q )∪(a,b]× Ω = [a,b]× Ω
Then the sets are disjoint and all but
[0,a]
× Ω ∖Q are in ℬ
([0,b])
×ℱt. Therefore, this
one is also in ℬ
([0,b])
×ℱt. If Qi∈G and the Qi are disjoint, then ∪iQi is also in
ℬ
([0,b])
×ℱt and so G is closed with respect to countable disjoint unions and
complements. Hence G⊇ σ
(K)
= ℬ
([0,a])
×ℱr which shows
ℬ ([0,a]) × ℱ ⊆ ℬ ([0,b])× ℱ ■
r t
A significant observation is the following which states that the integral of a
progressively measurable function is progressively measurable.
Proposition 61.3.5Suppose X :
[0,T ]
× Ω → E where E is a separable Banach space.Also suppose that X
(⋅,ω)
∈ L1
([0,T ],E )
for each ω. Here ℱtis a filtration and withrespect to this filtration, X is progressively measurable. Then
∫
t
(t,ω) → 0 X (s,ω )ds
is also progressively measurable.
Proof:Suppose Q ∈
[0,T ]
× Ω is progressively measurable. This means for each
t,
Q ∩ [0,t]× Ω ∈ ℬ ([0,t])× ℱt
What about
∫
(s,ω) ∈ [0,t]× Ω, (s,ω ) → sX dr?
0 Q
Is that function on the right ℬ
([0,t])
×ℱt measurable? We know that Q ∩
[0,s]
× Ω is
ℬ
([0,s])
×ℱs measurable and hence ℬ
([0,t])
×ℱt measurable. When you integrate a
product measurable function, you do get one which is product measurable.
Therefore, this function must be ℬ
([0,t])
×ℱt measurable. This shows that
(t,ω)
→∫0tXQ
(s,ω)
ds is progressively measurable. Here is a claim which was just
used.
Claim:If Q is ℬ
([0,t])
×ℱt measurable, then
(s,ω )
→∫0sXQdr is also
ℬ
([0,t])
×ℱt measurable.
Proof of claim:First consider A × B where A ∈ℬ
([0,t])
and B ∈ℱt.
Then
∫ s ∫ s ∫ s
0 XA ×Bdr = 0 XAXBdr = XB (ω ) 0 XA (s)dr
This is clearly ℬ
([0,t])
×ℱt measurable. It is the product of a continuous function of s
with the indicator function of a set in ℱt. Now let
{ ∫ }
G ≡ Q ∈ ℬ([0,t])× ℱ : (s,ω) → sX (r,ω) dr is ℬ ([0,t])× ℱ measurable
t 0 Q t
Then it was just shown that G contains the measurable rectangles. It is also clear
that G is closed with respect to countable disjoint unions and complements.
Therefore, G⊇ σ
(Kt )
= ℬ
([0,t])
×ℱt where Kt denotes the measurable rectangles
A × B where B ∈ℱt and A ∈ℬ
([0,t])
= ℬ
([0,T])
∩
[0,t]
. This proves the
claim.
Thus if Q is progressively measurable, it follows that
(s,ω)
→∫0sXQ
(r,ω)
dr ≡ f
(s,ω)
is progressively measurable because for
(s,ω )
∈
[0,t]
× Ω,
(s,ω)
→ f
(s,ω)
is
ℬ
([0,t])
×ℱt measurable. This is what was to be proved in this special case.
Now consider the conclusion of the proposition. By considering the positive and
negative parts of ϕ
(X )
for ϕ ∈ E′, and using Pettis theorem, it suffices to consider the
case where X ≥ 0. Then there exists an increasing sequence of progressively
measurable simple functions
{Xn}
converging pointwise to X. From what was just
shown,
∫ t
(t,ω) → Xnds
0
is progressively measurable. Hence, by the monotone convergence theorem,
(t,ω )
→∫0tXds
is also progressively measurable. ■
What else can you do to something which is progressively measurable and obtain
something which is progressively measurable? It turns out that shifts in time can preserve
progressive measurability. Let ℱt be a filtration on
[0,T]
and extend the filtration to be
equal to ℱ0 and ℱT for t < 0 and t > T, respectively. Recall the following definition of
progressively measurable sets.
Definition 61.3.6Denote by P those sets Q in ℱT×ℬ
([0,T])
such that fort ∈
[− ∞,T ]
Ω × (− ∞, t]∩ Q ∈ ℱt × ℬ ((− ∞, t]).
Lemma 61.3.7Define Q + h as
Q + h ≡ {(t+ h,ω) : (t,ω ) ∈ Q} .
Then if Q ∈P, it follows that Q + h ∈P.
Proof:This is most easily seen through the use of the following diagram. In this
diagram, Q is in P so it is progressively measurable.
PICT
By definition, S in the picture is ℬ
((− ∞,t− h])
×ℱt−h measurable. Hence
S + h ≡ Q + h ∩ Ω × (−∞,t] is ℬ
((− ∞,t])
×ℱt−h measurable. To see this, note
that if B × A ∈ℬ