63.7. DOOB MEYER DECOMPOSITION 2159
Theorem 63.7.12 Let {X (t)} be a submartingale of class DL. Then there exists a martin-gale, {M (t)} and an increasing submartingale, {A(t)} such that for each t,
X (t) = M (t)+A(t) .
If {A(t)} is chosen to be natural and A(0) = 0, then with this condition, {M (t)} and{A(t)} are unique.
Proof: First I will show uniqueness. Suppose then that
X (t) = M (t)+A(t) = M′ (t)+A′ (t)
where M,M′ and A,A′ satisfy the given conditions. Let t > 0 and consider s ∈ [0, t] . Then
A(s)−A′ (s) = M′ (s)−M (s)
Since A,A′ are natural, it follows that for ξ (t) a right continuous bounded martingale,
E(ξ (t)
(A(t)−A′ (t)
))= E
(∫(0,t]
ξ− (s)dA(s))−E
(∫(0,t]
ξ− (s)dA′ (s))
= E
(limn→∞
mn
∑k=1
ξ(tnk−1)(
A(tnk )−A
(tnk−1))−
mn
∑k=1
ξ(tnk−1)(
A′ (tnk )−A′
(tnk−1)))
where{
tnk
}mnk=0 is a sequence of partitions of [0, t] such that these are equally spaced points,
limn→∞ tnk+1−tn
k = 0, and{
tnk
}mnk=0 ⊆
{tn+1k
}mn+1k=0 . Then since A(t) and A′ (t) are increasing,
the absolute value of each sum is bounded above by an expression of the form
CA(t) or CA′ (t)
and so the dominated convergence theorem can be applied to get the above expression toequal
limn→∞
E
(mn
∑k=1
ξ(tnk−1)(
A(tnk )−A
(tnk−1))−
mn
∑k=1
ξ(tnk−1)(
A′ (tnk )−A′
(tnk−1)))
Now using X = A+M and X = A′+M′
= limn→∞
E
(mn
∑k=1
ξ(tnk−1)(
M (tnk )−M
(tnk−1))−
mn
∑k=1
ξ(tnk−1)(
M′ (tnk )−M′
(tnk−1)))
.
Both terms in the above equal 0. Here is why.
E(ξ(tnk−1)
M (tnk ))
= E(
E(
ξ(tnk−1)
M (tnk ) |Ftn
k−1
))= E
(ξ(tnk−1)
E(
M (tnk ) |Ftn
k−1
))= E
(ξ(tnk−1)
M(tnk−1))
.