2164 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE
From this all that remains is to write the above as
limk→∞
E
(mnk
∑j=1
ξ
(tnk
j
)Ank(
tnkj
)−
mnk−1
∑j=0
ξ
(tnk
j+1
)Ank(
tnkj
))
= limk→∞
(E (ξ (a)Ank (a))+E
(mnk−1
∑j=1
(ξ
(tnk
j
)−ξ
(tnk
j+1
))Ank(
tnkj
)))
Now using the fact ξ is a martingale, this last term equals 0. Here is why.
E(
ξ
(tnk
j+1
)Ank(
tnkj
))= E
(E(
ξ
(tnk
j+1
)Ank(
tnkj
)|Ft
nkj
))= E
(Ank(
tnkj
)E(
ξ
(tnk
j+1
)|Ft
nkj
)|Ft
nkj
)= E
(Ank(
tnkj
)ξ
(tnk
j
)|Ft
nkj
)The first term converges to E (ξ (a)A(a)) because this was how A(a) was obtained, as aweak limit in L1 (Ω) of Ank (a). Also by Lemma 63.7.7,
E (ξ (a)A(a)) = E(∫
(0,a]ξ (s)dA(s)
)From 63.7.37 this has now shown that
E (ξ (a)A(a)) = E(∫
(0,a]ξ− (s)dA(s)
)To get the desired result on (0, t], apply what was just shown to a “stopped martingale”,
ξt (s)≡
{ξ (s) if s≤ tξ (t) if s > t
E(∫
(0,t]ξ (s)dA(s)
)+(A(a)−A(t))E (ξ (t))
= E(∫
(0,a]ξ
t (s)dA(s))
From what was shown above,
= E(∫
(0,a]ξ
t− (s)dA(s)
)= E
(∫(0,t]
ξ− (s)dA(s)+∫(t,a]
ξ (t)sA(s))
= E(∫
(0,t]ξ− (s)dA(s)
)+(A(a)−A(t))E (ξ (t))