2150 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE

To save notation, let

Qn (t)≡mn−1

∑k=0

∣∣M (t ∧ tnk+1)−M (t ∧ tn

k )∣∣2H

Let η ,ε > 0 be given. Then there exists k large enough that P([τk = ∞])> 1−η/2. Thisis because the sets [τk = ∞] increase to Ω other than a set of measure zero. Then,[∣∣Qτk

n − [M]τk (t)∣∣> ε

]∩ [τk = ∞] = [|Qn− [M] (t)|> ε]∩ [τk = ∞]

Thus

P([|Qn− [M] (t)|> ε]) ≤ P([|Qn− [M] (t)|> ε]∩ [τk = ∞])

+P([τk < ∞])

≤ P([∣∣Qτk

n − [M]τk (t)∣∣> ε

])+η/2

From the first part, the convergence in probability of Qτkn (t) to [M]τk (t) follows from the

convergence in L2 (Ω) and so if n is large enough, the right side of the above inequality isless than η/2+η/2 = η . Since η was arbitrary, this proves convergence in probability.

63.7 Doob Meyer DecompositionThis section is on the Doob Meyer decomposition which is a way of starting with a sub-martingale and writing it as the sum of a martingale and an increasing adapted stochasticprocess of a certain form. This is more general than what was done above with the sub-martingales ||M (t)||2 for M (t) ∈M 2

T (H) where M is a continuous martingale. There aretwo forms for this theorem, one for discrete martingales and one for martingales defined onan interval of the real line which is much harder. According to [74], this material is foundin [78] however, I am following [74] for the continuous version of this theorem.

Theorem 63.7.1 Let {Xn} be a submartingale. Then there exists a unique stochastic pro-cess, {An} and martingale, {Mn} such that

1. An (ω)≤ An+1 (ω) , A1 (ω) = 0,

2. An is Fn−1 adapted for all n≥ 1 where F0 ≡F1.

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 (63.7.27)

2150 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALETo save notation, letmn—1On(t)= Y, |M(tAdh) —M(Ant)|;,k=0Let n,€ > 0 be given. Then there exists k large enough that P([t,; = ]) > 1— 1/2. Thisis because the sets [tT = °c] increase to Q other than a set of measure zero. Then,[|Ou’ — [M]* (t)| > €] A [te = 2] = []Qn — [M] (2)| > €] [te = IThusP((lQn—[M](t)| >€]) < P([]Qn—[M](1)| > €] 9 [te = J)+P ([T < o9])<P ([|Qi* — [M]" (t)| > €]) +n/2From the first part, the convergence in probability of Q;* (t) to [M]* (t) follows from theconvergence in L? (Q) and so if n is large enough, the right side of the above inequality isless than 1 /2+1/2= 1. Since 7 was arbitrary, this proves convergence in probability. J63.7 Doob Meyer DecompositionThis section is on the Doob Meyer decomposition which is a way of starting with a sub-martingale and writing it as the sum of a martingale and an increasing adapted stochasticprocess of a certain form. This is more general than what was done above with the sub-martingales ||M (t)||° for M(t) € .@? (H) where M is a continuous martingale. There aretwo forms for this theorem, one for discrete martingales and one for martingales defined onan interval of the real line which is much harder. According to [74], this material is foundin [78] however, I am following [74] for the continuous version of this theorem.Theorem 63.7.1 Let {X,,} be a submartingale. Then there exists a unique stochastic pro-cess, {A,} and martingale, {M,,} such that1. An(@) <An+1(@), Ar (@) =0,2. Ay is Fy_1 adapted for alln > 1 where Fo = F\.and also X, = M,+An-Proof: Let A; = 0 and definenAn =) E (Xe —Xk-1|Fi-1)-k=2It follows A, is ,_; measurable. Since {X;} is a submartingale, A, is increasing becauseAnti —An = E (Xn41 —Xn|-Fn) = 0 (63.7.27)