Topics In Analysis

61.3 Filtrations

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 61.3.1 Let X be a stochastic process defined on an interval, I =

or [0,∞). Suppose the probability space,

has an increasing family of σ algebras,

. This is called a filtration. If for arbitrary t ∈ I the random variable X

is ℱ_t measurable, then X is said to be adapted to the filtration

. Denote by ℱ_t+ the intersection of all ℱ_s for s > t. The filtration is normal if

ℱ₀ contains all A ∈ℱ such that P
(A )
= 0
ℱ_t = ℱ_t+ for all t ∈ I.

X is called progressively measurable if for every t ∈ I, the mapping

(s,ω) \in [0,t]\times Ω, (s,ω) \to X (s,ω)

is B

×ℱ_t measurable.

Thus X is progressively measurable means

(s,ω) \to X[0,t](s)X (s,ω)

is B

×ℱ_t measurable. As an example of a normal filtration, here is an example.

Example 61.3.2 For example, you could have a stochastic process, X

and you could define

-------------- Gt \equiv σ (X (s) : s \leq t),

the completion of the smallest σ algebra such that each X

is measurable for all s ≤ t. This gives an example of a filtration to which X

is adapted which satisfies 1. More generally, suppose X

is adapted to a filtration, G_t. Define

ℱt \equiv \caps>tGs

Then

ℱt+ \equiv \caps>tℱs = \caps>t \capr>s Gr = \caps>tGs \equiv ℱt.

and each X

is measurable with respect to ℱ_t. Thus there is no harm in assuming a stochastic process adapted to a filtration can be modified so the filtration is normal. Also note that ℱ_t defined this way will be complete so if A ∈ℱ_t has P

= 0 and if B ⊆ A, then B ∈ℱ_t also. This is because this relation between the sets and the probability of A being zero, holds for this pair of sets when considered as elements of each G_s for s > t. Hence B ∈G_s for each s > t and is therefore one of the sets in ℱ_t.

What is the description of a progressively measurable set?

It means that for Q progressively measurable, Q ∩

× Ω as shown in the above picture is ℬ

×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 61.3.3 If X is progressively measurable, then it is adapted. Furthermore the progressively measurable sets, those E ∩

× 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

→ X

for

∈

× Ω is given to be ℬ

×ℱ_t measurable, the ordinary product measure and so fixing any s ∈

, it follows the resulting function of ω is ℱ_t measurable. In particular, this is true upon fixing s = t. Thus ω → X

is ℱ_t measurable and so X

is adapted.

A set E ⊆

× Ω is progressively measurable means that X_E is progressively measurable. That is X_E restricted to

× Ω is ℬ

×ℱ_t measurable. In other words, E is progressively measurable if

E \cap ([0,t]\times Ω ) \in ℬ([0,t])\times ℱt.

If E_i is progressively measurable, does it follow that E ≡∪_i=1^∞E_i is also progressively measurable? Yes.

E \cap ([0,t]\times Ω ) = \cup\infty Ei \cap ([0,t]\times Ω) \in ℬ ([0,t])\times ℱt i=1

because each set in the union is in ℬ

×ℱ_t. If E is progressively measurable, is E^C?

\inℬ([0,t])\timesℱt \inℬ([0,t])\timesℱt C ◜-----◞◟-----◝ ◜--◞◟-◝ E \cap ([0,t]\times Ω)\cup (E \cap([0,t]\times Ω)) = [0,t]\times Ω

and so E^C ∩

∈ℬ

×ℱ_t. Thus the progressively measurable sets are a σ algebra.

Another observation of interest is in the following lemma.

Lemma 61.3.4 Suppose Q is in ℬ

×ℱ_r. Then if b ≥ a and t ≥ r, then Q is also in ℬ

×ℱ_t.

Proof: Consider a measurable rectangle A×B where A ∈ℬ

and B ∈ℱ_r. Is it true that A×B ∈ℬ

×ℱ_t? This reduces to the question whether A ∈ℬ

. If A is an interval, it is clear that A ∈ℬ

. Consider the π system of intervals and let G denote those Borel sets A ∈ℬ

such that A ∈ℬ

. If A ∈G, then

∖A ∈ℬ

by assumption (the difference of Borel sets is surely Borel). However, this set equals

([0,a]∖A )\cup (a,b]

and so

[0,b] = ([0,a]∖ A)\cup (a,b]\cup A

The set on the left is in ℬ

and the sets on the right are disjoint and two of them are also in ℬ

. Therefore, the third,

is in 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

= ℬ

Therefore, such a measurable rectangle A × B where A ∈ℬ

and B ∈ℱ_r is in ℬ

×ℱ_t and in fact it is a measurable rectangle in ℬ

×ℱ_t. Now let K denote all these measurable rectangles A × B where A ∈ℬ

and B ∈ℱ_r. Let G (new G) denote those sets Q of ℬ

×ℱ_r which are in ℬ

×ℱ_t. Then if Q ∈G,

Q \cup ([0,a]\times Ω ∖Q )\cup(a,b]\times Ω = [a,b]\times Ω

Then the sets are disjoint and all but

× Ω ∖Q are in ℬ

×ℱ_t. Therefore, this one is also in ℬ

×ℱ_t. If Q_i ∈G and the Q_i are disjoint, then ∪_iQ_i is also in ℬ

×ℱ_t and so G is closed with respect to countable disjoint unions and complements. Hence G⊇ σ

= ℬ

×ℱ_r which shows

ℬ ([0,a]) \times ℱ \subseteq ℬ ([0,b])\times ℱ ■ r t

A significant observation is the following which states that the integral of a progressively measurable function is progressively measurable.

Proposition 61.3.5 Suppose X :

× Ω → E where E is a separable Banach space. Also suppose that X

∈ L¹

for each ω. Here ℱ_t is a filtration and with respect to this filtration, X is progressively measurable. Then

\int t (t,ω) \to 0 X (s,ω )ds

is also progressively measurable.

Proof: Suppose Q ∈

× Ω is progressively measurable. This means for each t,

Q \cap [0,t]\times Ω \in ℬ ([0,t])\times ℱt

What about

\int (s,ω) \in [0,t]\times Ω, (s,ω ) \to sX dr? 0 Q

Is that function on the right ℬ

×ℱ_t measurable? We know that Q ∩

× Ω is ℬ

×ℱ_s measurable and hence ℬ

×ℱ_t measurable. When you integrate a product measurable function, you do get one which is product measurable. Therefore, this function must be ℬ

×ℱ_t measurable. This shows that

→∫ ₀^tX_Q

ds is progressively measurable. Here is a claim which was just used.

Claim: If Q is ℬ

×ℱ_t measurable, then

→∫ ₀^sX_Qdr is also ℬ

×ℱ_t measurable.

Proof of claim: First consider A × B where A ∈ℬ

and B ∈ℱ_t. Then

\int s \int s \int s 0 XA \timesBdr = 0 XAXBdr = XB (ω ) 0 XA (s)dr

This is clearly ℬ

×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

{ \int } G \equiv Q \in ℬ([0,t])\times ℱ : (s,ω) \to sX (r,ω) dr is ℬ ([0,t])\times 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

= ℬ

×ℱ_t where K_t denotes the measurable rectangles A × B where B ∈ℱ_t and A ∈ℬ

= ℬ

∩

. This proves the claim.

Thus if Q is progressively measurable, it follows that

→∫ ₀^sX_Q

dr ≡ f

is progressively measurable because for

∈

× Ω,

→ f

is ℬ

×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

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

converging pointwise to X. From what was just shown,

\int t (t,ω) \to Xnds 0

is progressively measurable. Hence, by the monotone convergence theorem,

→∫ ₀^tXds 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

and extend the filtration to be equal to ℱ₀ and ℱ_T for t < 0 and t > T, respectively. Recall the following definition of progressively measurable sets.

Definition 61.3.6 Denote by P those sets Q in ℱ_T ×ℬ

such that for t ∈

Ω \times (- \infty, t]\cap Q \in ℱt \times ℬ ((- \infty, t]).

Lemma 61.3.7 Define Q + h as

Q + h \equiv {(t+ h,ω) : (t,ω ) \in 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.

By definition, S in the picture is ℬ

×ℱ_t−h measurable. Hence S + h ≡ Q + h ∩ Ω × (−∞,t] is ℬ

×ℱ_t−h measurable. To see this, note that if B × A ∈ℬ