Lemma 1.
Consider a positive defnite symmetric matrix
A
.
(
Det A
)
−
1
2
=
c
n
i
S
n

1
ds
[
< s,As >
]
n
2
Proof:
(
Det A
)
−
1
2
= (2
π
)
n
2
i
R
n
e
−
<x,Ax>
2
dx
= (2
π
)
n
2
i
S
n

1
e
−
r
2
<s,As>
2
r
n
−
1
dsdr
=
c
n
i
S
n

1
ds
[
]
n
2
It is thereFore su±cient to estimate For each
k
,
sup
x
:
b
x
b
=1
E
[
< x,Ax >
−
k
]
≤
C
k
to yield an estimate oF the Form
E
[(
Det A
)
−
k
]
≤
B
k
Weak formulation. With out estimates.
Let us note that the Malliavin covariance
A
(
t
) has the Form
A
(
t
) =
i
t
0
B
(
s,t, ω
)
a
(
x
(
s
))
B
∗
(
)
ds
with
B
(
) being the Jacobian oF the map
R
n
→
R
n
that maps the initial point
x
=
x
(
s,ω
) in
R
n
to
x
(
t
)
∈
R
n
. The problem with
A
(
t
) is that
B
(
) is NOT progressively
measurable. Moreover intrinsically
A
(
t
) is a quadratic Form on the cotangent space at
x
(
t, ω
). We can instead look at
B
(0
, t, ω
)
−
1
A
(
t
)
B
(0
, t, ω
)
−
1
∗
. Since
Det B
(0
, t,ω
) and its
inverse will have moments oF all orders, we can try to show that
C
(
t
) =
B
(0
, t, ω
)
−
1
A
(
t
)
B
(0
, t, ω
)
−
1
∗
has a nice inverse.
C
(
t
) has a better representation,
B
(0
, s,ω
) being progressively measur
able.
C
(
t
) =
i
t
0
B
(0
)
−
1
a
(
x
(
s
))
B
(0
)
−
1
∗
ds
Suppose
C
(1)
x
= 0 For some
x
=
{
x
i
}
. Then denoting by
h
i,j
(
) = (
B
(0
)
−
1
)
i,j
we
have
s
i,j
h
i,j
(
)
x
i
σ
j,k
(
x
(
s
))
≡
0
1
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k
= 1
,
2
, . . .n
. If we think of
σ
j,k
(
x
) as vector Felds
X
k
(
x
), then
s
i
x
i
q
i,k
(
s,ω
)
≡
0
for
k
= 1
,
2
, . . .n,
where
s
i
x
i
q
i,k
(
) =
< x,h
(
)
X
k
(
x
(
s
))
>
Now
ξ
(
s
) =
(
)
X
k
(
x
(
s
))
>
is a semimartingale and we can compute the Stratonovich di±erential
d
◦
(
h
(
)
X
k
(
x
(
s
)) = [
d
◦
h
(
)]
X
k
(
x
(
s
)) +
h
(
)[
d
◦
X
k
(
x
(
s
))]
=
s
r
e
k,r
(
)
◦
dβ
r
(
s
) +
e
k,
0
(
)
ds
One can compute easily
e
k,r
(
s
) =
h
(
s
)[
−
X
r
X
k
+
X
k
X
r
](
x
(
s
)) =
h
(
s
)[
X
k
, X
r
](
x
(
s
))
and
e
k,
0
(
) =
h
(
s
)[
−
X
0
X
k
+
X
k
X
0
](
x
(
s
)) =
h
(
s
)[
X
k
, X
0
](
x
(
s
))
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 Fall '06
 Varadhan
 Trigraph, Hyperbolic function, λ T

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