63.7. DOOB MEYER DECOMPOSITION 2151

It is a submartingale because

E (An|Fn−1) = An ≥ An−1.

Now let Mn be defined byXn = Mn +An.

Then from 63.7.27,

E (Mn+1|Fn) = E (Xn+1|Fn)−E (An+1|Fn)

= E (Xn+1|Fn)−E (An+1−An|Fn)−An

= E (Xn+1|Fn)−E (E (Xn+1−Xn|Fn) |Fn)−An

= E (Xn+1|Fn)−E (Xn+1−Xn|Fn)−An

= E (Xn|Fn)−An

= Xn−An ≡Mn

This proves the existence part.It remains to verify uniqueness. Suppose then that

Xn = Mn +An = M′n +A′n

where {An} and {A′n} both satisfy the conditions of the theorem and {Mn} and {M′n} areboth martingales. Then

Mn−M′n = A′n−An

and so, since A′n−An is Fn−1 measurable and {Mn−M′n} is a martingale,

Mn−1−M′n−1 = E(Mn−M′n|Fn−1

)= E

(A′n−An|Fn−1

)= A′n−An = Mn−M′n.

Continuing this way shows Mn−M′n is a constant. However, since A′1−A1 = 0 = M1−M′1,it follows Mn = M′n and this proves uniqueness. This proves the theorem.

Definition 63.7.2 A stochastic process, {An} which satisfies the conditions of Theorem63.7.1,

An (ω)≤ An+1 (ω)

and

An is Fn−1 adapted for all n≥ 1

where F0 ≡F1 is said to be natural.The Doob Meyer theorem needs to be extended to continuous submartingales and this

will require another description of what it means for a stochastic process to be natural. Toget an idea of what this condition should be, here is a lemma.