Homework Assignment #7
1: Let M1 (t), Ft , P and M2 (t), Ft , P be a pair of continuous local martingales. If σ {M1 (t) :
t ≥ 0} is independent of σ {M1 (t) : t ≥ 0} , show that hM1 , M2 i(t) = 0 for all t ≥ 0.
2: In hw #4, you derived a version of Burkholder’s inequality for Brownian stochastic integrals,
and it should be clear that the reasoning there extends to general continuous local martingales.
Nonetheless, it is interesting to give another derivation using Kunita and Watanabe’s verion of
Itô’s formula. The following outline is based on ideas introduced by A. Garsia, who proved that,
for each p ∈ (0, ∞), there exist 0 < cp ≤ Cp < ∞ such that, for any M ∈ Mloc (P; R) with
M (0) = 0 and any stopping time ζ,
1
p1
p1
(1)
cp EP hhM ii(ζ) 2 p ≤ EP kM kp[0,ζ) p ≤ Cp EP hhM ii(ζ) 2 p ,
an inequality that was first proved by R. Gundy and D. Burkholder. Follow these to a prove (1)
for p ∈ [2, ∞).
(i) The first step is to show that it suffices to treat the case in which both M and hhM ii are
uniformly bounded and ζ is equal to some constant T .
q p+1
p
(ii) Let p ∈ [2, ∞) be given, and set Cp = (p−1)
p−1 . Prove that the right hand side of (1)
holds with this choice of Cp .
p
Hint: Begin by making the reductions in (i). Given > 0, set F (x) = x2 + 2 2 , and apply
Doob’s inequality plus Itô’s formula to see that
EP kM kp[0,T ] ≤ (p0 )p EP F M (T )
"Z
#
T
(p0 )p p(p − 1) P
=
E
F00 M (τ ) hhM ii(dτ )
2
0
h
i
0 p
p −1
(p ) p(p − 1) P
≤
E kM k2[0,T ] + 2 2 hhM ii(T ) .
2
Now let & 0, and apply Hölder’s inequality.
p
1
(iii) Assume that M and hhM ii are bounded, set ξ(t) = hhM ii(t) 4 − 2 , and take M 0 = IξM .
After noting that
Z T
p
p
2
0
hhM ii(T ) =
hhM ii 2 −1 (τ ) hhM ii(dτ ) = hhM ii(T ) 2 ,
p
0
p
conclude that EP hhM ii(T ) 2 = p2 EP M 0 (T )2 .
(iv) Continuing part (iii), apply Itô’s formula to see that
Z T
p
p
1
1
−
0
M (T )hhM ii(T ) 4 2 = M (T ) +
M (τ ) dhhM ii(τ ) 4 − 2 ,
0
and conclude that kM 0 k[0,T ] ≤ 2kM k[0,T ] hhM ii(T )
1
to get the left hand side of (1) with cp = (2p)− 2 .
p
1
4−2
. Now combine this with the result in (iii)