Topics In Analysis

59.3.1 Stopping Times And Their Properties

First it is necessary to define the notion of a stopping time.

The following lemma is fundamental to understand.

Proof: Let A ∈ℱ_S. Then this means

A ∩ [S ≤ n] ∈ ℱ for all n. n

Then

A ∩ [T ≤ n ] = ∪n (A∩ [S ≤ i])∩[T ≤ n] (59.3.7) i=1

(59.3.7)

Suppose ω is in the set on the left. Then if T

< n, it is clearly in the set on the right. If T = n, then ω ∈ for some i ≤ n and it is also in . Thus the set on the left is contained in the set on the right. Next suppose ω is in the set on the right. Then ω ∈ and it only remains to verify ω ∈ A. However, ω ∈ A∩ for some i and so ω ∈ A also.

Now from 59.3.7 it follows A ∩

∈ℱ_n because

A ∩ [S ≤ i] ∈ ℱ ⊆ ℱ i n

and

∈ℱ_n because T is a stopping time. Since n is arbitrary, this shows ℱ_S ⊆ℱ_T.

It remains to verify ℱ_T is a σ algebra. Suppose

is a sequence of sets in ℱ_T. Then I need to show that ∩ ∈ℱ_j for all j.

∞ ∞ ∪i=1Ai ∩ [T ≤ j] = ∪ i=1 (Ai ∩[T ≤ j])

Now each

is in ℱ_j and so the countable union of these sets is also in ℱ_j. Next suppose A ∈ℱ_T. I need to verify A^C ∩ ∈ℱ_j for all j. However, ∈ℱ_j and Ω ∈ℱ_j so Ω ∈ℱ_T. Thus

( ) Ω ∩ [T ≤ j] = (A ∩ [T ≤ j]) ∪ AC ∩ [T ≤ j]

and so

( ) AC ∩ [T ≤ j] = Ω ∩ [T ≤ j]∖(A ∩ [T ≤ j]) ∈ ℱj.

This proves the lemma.

Proof: I assume the X_n have values in some topological space and each is measurable because the inverse image of an open set is in ℱ_n. I need to show X_T⁻¹

∩ ∈ℱ_n for all n whenever U is open.

X− 1(U ) = ∪ ∞ X −1(U) ∩[T = i]. T i=1 i

It follows X_T⁻¹

∈ℱ. Furthermore,

and so X_T is ℱ_T is measurable as claimed. This proves the lemma.

Proof: I will prove the case where

is a submartingale and note the other case will only involve replacing ≥ with =. First recall that from Lemma 59.3.2 ℱ_S ⊆ℱ_T. Also let m be an upper bound for T. Then it follows from this that

m ∫ E (|X |) = ∑ |X |dP < ∞ T i=1 [T=i] i

with a similar formula holding for E

. Thus it makes sense to speak of E.

I need to show that if B ∈ℱ_S, so that B ∩

∈ℱ_n for all n, then

∫ ∫ XTdP ≥ XSdP. (59.3.8) B B

(59.3.8)

It suffices to do this for B of the special form

B = A∩ [S = i]

because if this is done, then the result follows from summing over all possible values of S. Note that if B = A ∩

, then X_T = X_S = X_m and there is nothing to prove in 59.3.8 so it can be assumed i ≤ m − 1. Then let B be of this form.

∫ ∑m ∫ A∩[S=i]XTdP = A∩[S=i]∩[T=j]XTdP j=i

m∑−1∫ ∫ = XT dP + XmdP j=i A∩[S=i]∩[T=j] A∩[S=i]∩[T≥m]

And so

∫ m∑−1∫ ∫ XT dP = XT dP + XmdP (59.3.9) A∩[S=i] j=i A ∩[S=i]∩[T=j] A∩[S=i]∩[T≥m ]

(59.3.9)

provided m − 1 ≥ i because is a submartingale and

A ∩ [S = i]∩ [T ≤ m − 1]C ∈ ℱm −1

Now combine the top term of the sum with the term on the right to obtain

m∑−2∫ ∫ = XTdP + Xm −1dP j=i A ∩[S=i]∩[T=j] A∩[S=i]∩[T≥m −1]

which is exactly the same form as 59.3.9 except m is replaced with m − 1. Now repeat this process till you get the following inequality

∫ i+1 ∫ ∫ X dP ≥ ∑ X dP + X dP A ∩[S=i] T j=i A∩[S=i]∩[T=j] T A∩[S=i]∩[T ≥i+2] i+2

The right hand side equals

In the case where is a martingale, you replace every occurance of ≥ in the above argument with =. This proves the lemma.

This lemma is called the optional sampling theorem. Another version of this theorem is the case where you have an increasing sequence of stopping times,

_n=1^∞. Thus if is a sequence of random variables each ℱ_n measurable, the sequence is also a sequence of random variables such that X_Tn is measurable with respect to ℱ_Tn where ℱ_Tn is an increasing sequence of σ fields. In the case where X_n is a submartingale (martingale) it is reasonable to ask whether is also a submartingale (martingale). The optional sampling theorem says this is often the case.

Proof: This follows from Lemma 59.3.4

To see this is a stopping time,

[T ≤ l] = ∪l X −1(A ) ∈ ℱ . i=1 i l