29.4. OPTIONAL SAMPLING AND STOPPING TIMES 799

Proposition 29.4.8 Let M be a martingale having values in some separable Banachspace. Let τ be a bounded stopping time and let σ be another stopping time. Then every-thing makes sense in the following formula and

M (σ ∧ τ)≡Mσ∧τ = E (M (τ) |Fσ ) a.e.

29.4.2 Optional Sampling Theorem for Sub-MartingalesWhat about the case where {X (n)} is a sub-martingale? Shouldn’t there be somethinglike the conclusion of Proposition 29.4.8? This requires a very interesting theorem whichinvolves the decomposition of a sub-martingale into a sum. Recall {X (k)}∞

k=1 is a sub-martingale if

E (X (k+1) |Fk)≥ X (k)

where the Fk are an increasing sequence of σ algebras in the usual way. The followingis the very interesting result about writing a sub-martingale as the sum of an increasingprocess and a martingale.

Lemma 29.4.9 Let {X (k)}∞

k=0 be a sub-martingale adapted to the increasing sequenceof σ algebras, {Fk} . Then there exists a unique increasing process {A(k)}∞

k=0 such thatA(0) = 0 and A(k+1) is Fk measurable for all k and a martingale, {M (k)}∞

k=0 such that

X (k) = A(k)+M (k) .

Furthermore, for τ a stopping time, A(τ) is Fτ measurable.

Proof: Define ∑−1k=0≡ 0. First consider the uniqueness assertion. Suppose A is a process

which does what it is supposed to do.

n−1

∑k=0

E (X (k+1)−X (k) |Fk) =n−1

∑k=0

E (A(k+1)−A(k) |Fk)

+n−1

∑k=0

E (M (k+1)−M (k) |Fk)

Then since {M (k)} is a martingale,

n−1

∑k=0

E (X (k+1)−X (k) |Fk) =n−1

∑k=0

A(k+1)−A(k) = A(n)

This shows uniqueness and gives a formula for A(n) assuming it exists. It is only a matterof verifying this does work. Define

A(n)≡n−1

∑k=0

E (X (k+1)−X (k) |Fk) , A(0) = 0.

Then A is increasing because from the definition,

A(n+1)−A(n) = E (X (n+1)−X (n) |Fn)≥ 0.

Also from the definition above, A(n) is Fn−1 measurable, so consider

{X (k)−A(k)} .