2074 CHAPTER 62. STOCHASTIC PROCESSES

Lemma 62.6.4 In the situation of Definition 62.6.1, let σ ,τ be two stopping times. Then

1. τ is Fτ measurable

2. Fσ ∩ [σ ≤ τ]⊆Fσ∧τ = Fσ ∩Fτ

3. Fτ =Fk on [τ = k] for all k. That is if A∈Fk, then A∩ [τ = k]∈Fτ and if A∈Fτ ,then A∩ [τ = k] ∈Fk. In other words, the two σ algebras

[τ = k]∩Fτ , [τ = k]∩Fk

are equal. Letting G denote this σ algebra, if g is either Fτ or Fk measurable thenits restriction to [τ = k] is G measurable. Also if A ∈Fτ , and Y ∈ L1 (Ω;E) ,∫

A∩[τ=k]E (Y |Fτ)dP =

∫A∩[τ=k]

E (Y |Fk)dP

andE (Y |Fτ) = E (Y |Fk) a.e.

on [τ = k].

Proof: Consider the first claim. [τ ≤ l]∩ [τ ≤ m] = [τ ≤ ⌊l⌋∧m] ∈F[l]∧m ⊆Fm andso τ is Fτ measurable. Here ⌊l⌋ is the greatest integer less than or equal to l. Next notethat σ ∧ τ is a stopping time because

[σ ∧ τ ≤ k] = [σ ≤ k]∪ [τ ≤ k] ∈Fk

Next consider the second claim. Let A ∈Fσ . I want to show

A∩ [σ ≤ τ] ∈Fτ∧σ (62.6.23)

In other words, I want to show

A∩ [σ ≤ τ]∩ [τ ∧σ ≤ k] ∈Fk (62.6.24)

for all k. However, the set on the left equals

A∩ [σ ≤ τ]∩ [σ ≤ k]

= ∪kj=1A∩ [σ = j]∩ [τ ≥ j]∩ [σ ≤ k]

= ∪kj=1A∩ [σ = j]∩ [τ ≤ j−1]C ∩ [σ ≤ k] ∈Fk

Now let A ∈Fσ∧τ . I want to show it is in both Fτ and Fσ . To show it is in Fτ I needto show that

A∩ [τ ≤ k] ∈Fk

for all k. However,A∩ [τ = k] = ∪∞

i=1A∩ [σ = i]∩ [τ = k]