60.5. OPTIONAL STOPPING TIMES AND MARTINGALES 1965

Lemma 60.5.5 Let {X (k)}∞

k=0 be a submartingale adapted to the increasing sequence ofσ 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 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, consider

{X (k)−A(k)} .

Why is this a martingale?

E (X (k+1)−A(k+1) |Fk)

= E (X (k+1) |Fk)−A(k+1)

= E (X (k+1) |Fk)−k

∑j=0

E (X ( j+1)−X ( j) |F j)

= E (X (k+1) |Fk)−E (X (k+1)−X (k) |Fk)

−k−1

∑j=0

E (X ( j+1)−X ( j) |F j)

= X (k)−k−1

∑j=0

E (X ( j+1)−X ( j) |F j) = X (k)−A(k)