1956 CHAPTER 60. CONDITIONAL, MARTINGALES

Similarly, M (σ ∧ τ) is integrable. Now let A ∈Fσ . Then using Lemma 60.3.2 as needed,∫A

M (σ ∧ τ) =n

∑i=1

∫A∩[τ=i]

M (σ ∧ i) =n

∑i=1

∞

∑j=1

∫A∩[τ=i]∩[σ= j]

M ( j∧ i)

=n

∑i=1

∞

∑j=1

∫A∩[τ=i]∩[σ= j]

E (M (i) |F j)

There are two cases here. If j ≤ i it is the martingale definition. If j > i the third termhas integrand equal to M (i) which is F j measurable so the formula is still valid. Then theabove equals

n

∑i=1

∞

∑j=1

∫A∩[τ=i]∩[σ= j]

E (M (i) |Fσ ) =∞

∑j=1

n

∑i=1

∫A∩[τ=i]∩[σ= j]

E (M (i) |Fσ )

=∞

∑j=1

∫A∩[σ= j]

E (M (τ) |Fσ ) =∫

AE (M (τ) |Fσ )

Since A is an arbitrary element of Fσ , this shows the optional sampling theorem thatM (σ ∧ τ) = E (M (τ) |Fσ ) .

Proposition 60.3.3 Let M be a martingale having values in some separable Banach space.Let τ be a bounded stopping time and let σ be another stopping time. Then everythingmakes sense in the following formula and

M (σ ∧ τ) = E (M (τ) |Fσ ) a.e.

60.4 Stopping TimesThe following lemma is fundamental to understand.

Lemma 60.4.1 In the situation of Definition 60.3.1, if S≤ T for two stopping times, S andT, then FS ⊆FT . Also FT is a σ algebra.

Proof: Let A ∈FS. Then this means

A∩ [S≤ n] ∈Fn for all n.

Then I claim thatA∩ [T ≤ n] = ∪n

i=1 (A∩ [S≤ i])∩ [T ≤ n] (60.4.7)

Suppose ω is in the set on the left. Then if T (ω) < n, it is clearly in the set on the right.If T (ω) = n, then ω ∈ [S≤ i] for some i ≤ n and it is also in [T ≤ n] . Thus the set on theleft is contained in the set on the right. Next suppose ω is in the set on the right. Thenω ∈ [T ≤ n] and it only remains to verify ω ∈ A. However, ω ∈ A∩ [S≤ i] for some i andso ω ∈ A also.

Now from 60.4.7 it follows A∩ [T ≤ n] ∈Fn because

A∩ [S≤ i] ∈Fi ⊆Fn