2078 CHAPTER 62. STOCHASTIC PROCESSES

and also Xn = Mn +An.

Proof: Let A1 ≡ 0 and define

An ≡n

∑k=2

E (Xk−Xk−1|Fk−1) .

It follows An is Fn−1 measurable. Since {Xk} is a submartingale, An is increasing because

An+1−An = E (Xn+1−Xn|Fn)≥ 0 (62.6.25)

It is a submartingale because

E (An|Fn−1) = E

(n

∑k=2

E (Xk−Xk−1|Fk−1) |Fn−1

)

=n

∑k=2

E (Xk−Xk−1|Fk−1)≡ An ≥ An−1

Now let Mn be defined byXn = Mn +An.

Then from 62.6.25,

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.

Now here is a version of the optional sampling theorem for submartingales.