2152 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE

Lemma 63.7.3 Let a stochastic process, {An} be natural. Then for every martingale,{Mn} ,

E (MnAn) = E

(n−1

∑j=1

M j(A j+1−A j

))

Proof: Start with the right side.

E

(n−1

∑j=1

M j(A j+1−A j

))= E

(n

∑j=2

M j−1A j−n−1

∑j=1

M jA j

)

= E

(n−1

∑j=2

A j(M j−1−M j

))+E (Mn−1An)

Then the first term equals zero because since A j is F j−1 measurable,∫Ω

A jM j−1dP−∫

Ω

A jM j =∫

Ω

A jE(M j|F j−1

)dP−

∫Ω

A jM jdP

=∫

Ω

E(A jM j|F j−1

)dP−

∫Ω

A jM jdP

=∫

Ω

A jM jdP−∫

Ω

A jM jdP = 0.

The last term equals∫Ω

Mn−1AndP =∫

Ω

E (Mn|Fn−1)AndP

=∫

Ω

E (MnAn|Fn−1)dP = E (MnAn) .

This proves the lemma.

Definition 63.7.4 Let A be an increasing function defined on R. By Theorem 4.3.4 on Page50 there exists a positive linear functional, L defined on Cc (R) given by

L f ≡∫ b

af dA where spt( f )⊆ [a,b]

where the integral is just the Riemann Stieltjes integral. Then by the Riesz representationtheorem, Theorem 12.3.2 on Page 288, there exists a unique Radon measure, µ whichextends this functional, as described in the Riesz representation theorem. Then for B ameasurable set, I will write either ∫

Bf dµ or

∫B

f dA

to denote the Lebesgue integral, ∫XB f dµ.