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 ℱ_{t}measurable, then X is said to be adaptedto the filtration
{ℱt}
. Denoteby ℱ_{t+}the intersection of all ℱ_{s}for s > t. The filtration isnormalif
ℱ_{0}contains all A ∈ℱ such that P
(A )
= 0
ℱ_{t} = ℱ_{t+}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])
×ℱ_{t}measurable.
Thus X is progressively measurable means
(s,ω) → X[0,t](s)X (s,ω)
is B
([0,t])
×ℱ_{t} 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, G_{t}.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 ℱ_{t}. Thus there is no harm in assuming astochastic process adapted to a filtration can be modified so the filtration is normal. Alsonote that ℱ_{t}defined this way will be complete so if A ∈ℱ_{t}has P
(A)
= 0 and if B ⊆ A,then B ∈ℱ_{t}also. 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 eachG_{s}for s > t. Hence B ∈G_{s}for each s > t and is therefore one of the sets inℱ_{t}.
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])
×ℱ_{t} 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 X_{E}is 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])
×ℱ_{t} measurable, the
ordinary product measure and so fixing any s ∈
[0,t]
, it follows the resulting function of
ω is ℱ_{t} measurable. In particular, this is true upon fixing s = t. Thus ω → X
(t,ω)
is ℱ_{t}
measurable and so X
(t)
is adapted.
A set E ⊆
[0,T]
× Ω is progressively measurable means that X_{E} is progressively
measurable. That is X_{E} restricted to
[0,t]
× Ω is ℬ
([0,t])
×ℱ_{t} measurable. In other
words, E is progressively measurable if
E ∩ ([0,t]× Ω ) ∈ ℬ([0,t])× ℱt.
If E_{i} is progressively measurable, does it follow that E ≡∪_{i=1}^{∞}E_{i} 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])
×ℱ_{t}. If E is progressively measurable, is
E^{C}?
∈ℬ([0,t])×ℱt ∈ℬ([0,t])×ℱt
C ◜-----◞◟-----◝ ◜--◞◟-◝
E ∩ ([0,t]× Ω)∪ (E ∩([0,t]× Ω)) = [0,t]× Ω
and so E^{C}∩
([0,t]× Ω)
∈ℬ
([0,t])
×ℱ_{t}. 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])
×ℱ_{r}. Then if b ≥ a and t ≥ r, then Qis also in ℬ
([0,b])
×ℱ_{t}.
Proof: Consider a measurable rectangle A×B where A ∈ℬ
([0,a])
and B ∈ℱ_{r}. Is it
true that A×B ∈ℬ
([0,b])
×ℱ_{t}? 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 ∈ℱ_{r} is in
ℬ
([0,b])
×ℱ_{t} and in fact it is a measurable rectangle in ℬ
([0,b])
×ℱ_{t}. Now let K denote
all these measurable rectangles A × B where A ∈ℬ
([0,a])
and B ∈ℱ_{r}. Let G (new
G) denote those sets Q of ℬ
([0,a])
×ℱ_{r} which are in ℬ
([0,b])
×ℱ_{t}. 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])
×ℱ_{t}. Therefore, this
one is also in ℬ
([0,b])
×ℱ_{t}. If Q_{i}∈G and the Q_{i} are disjoint, then ∪_{i}Q_{i} is also in
ℬ
([0,b])
×ℱ_{t} and so G is closed with respect to countable disjoint unions and
complements. Hence G⊇ σ
(K)
= ℬ
([0,a])
×ℱ_{r} 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)
∈ L^{1}
([0,T ],E )
for each ω. Here ℱ_{t}is 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])
×ℱ_{t} measurable? We know that Q ∩
[0,s]
× Ω is
ℬ
([0,s])
×ℱ_{s} measurable and hence ℬ
([0,t])
×ℱ_{t} measurable. When you integrate a
product measurable function, you do get one which is product measurable.
Therefore, this function must be ℬ
([0,t])
×ℱ_{t} measurable. This shows that
(t,ω)
→∫_{0}^{t}X_{Q}
(s,ω)
ds is progressively measurable. Here is a claim which was just
used.
Claim:If Q is ℬ
([0,t])
×ℱ_{t} measurable, then
(s,ω )
→∫_{0}^{s}X_{Q}dr is also
ℬ
([0,t])
×ℱ_{t} measurable.
Proof of claim:First consider A × B where A ∈ℬ
([0,t])
and B ∈ℱ_{t}.
Then
∫ s ∫ s ∫ s
0 XA ×Bdr = 0 XAXBdr = XB (ω ) 0 XA (s)dr
This is clearly ℬ
([0,t])
×ℱ_{t} measurable. It is the product of a continuous function of s
with the indicator function of a set in ℱ_{t}. 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])
×ℱ_{t} where K_{t} denotes the measurable rectangles
A × B where B ∈ℱ_{t} 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)
→∫_{0}^{s}X_{Q}
(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])
×ℱ_{t} 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 )
→∫_{0}^{t}Xds
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 ℱ_{t} be a filtration on
[0,T]
and extend the filtration to be
equal to ℱ_{0} and ℱ_{T} 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 ℱ_{T}×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])
×ℱ_{t−h} measurable. Hence
S + h ≡ Q + h ∩ Ω × (−∞,t] is ℬ
((− ∞,t])
×ℱ_{t−h} measurable. To see this, note
that if B × A ∈ℬ