3The stochastic integral
III Stochastic Calculus and Applications
3.6 The L´evy characterization
A more major application of the stochastic integral is the following convenient
characterization of Brownian motion:
Theorem
(L´evy’s characterization of Brownian motion)
.
Let (
X
1
, . . . , X
d
) be
continuous local martingales. Suppose that
X
0
= 0 and that
hX
i
, X
j
i
t
=
δ
ij
t
for all
i, j
= 1
, . . . , d
and
t ≥
0. Then (
X
1
, . . . , X
d
) is a standard
d
-dimensional
Brownian motion.
This might seem like a rather artificial condition, but it turns out to be quite
useful in practice (though less so in this course). The point is that we know that
hH ·Mi
t
=
H
2
t
·hM i
t
, and in particular if we are integrating things with respect
to Brownian motions of some sort, we know
hB
t
i
t
=
t
, and so we are left with
some explicit, familiar integral to do.
Proof.
Let 0
≤ s < t
. It suffices to check that
X
t
−X
s
is independent of
F
s
and
X
t
− X
s
∼ N(0, (t − s)I).
Claim. E(e
iθ·(X
t
−X
s
)
| F
s
) = e
−
1
2
|θ|
2
(t−s)
for all θ ∈ R
d
and s < t.
This is sufficient, since the right-hand side is independent of
F
s
, hence so is
the left-hand side, and the Fourier transform characterizes the distribution.
To check this, for θ ∈ R
d
, we define
Y
t
= θ · X
t
=
d
X
i=1
θ
i
X
i
t
.
Then Y is a continuous local martingale, and we have
hY i
t
= hY, Y i
t
=
d
X
i,j=1
θ
j
θ
k
hX
j
, X
k
i
t
= |θ|
2
t.
by assumption. Let
Z
t
= e
iY
t
+
1
2
hY i
t
= e
iθ·X
t
+
1
2
|θ|
2
t
.
By Itˆo’s formula, with X = iY +
1
2
hY i
t
and f(x) = e
x
, we get
dZ
t
= Z
t
idY
t
−
1
2
dhY i
t
+
1
2
dhY i
t
= iZ
t
dY
t
.
So this implies
Z
is a continuous local martingale. Moreover, since
Z
is bounded
on bounded intervals of
t
, we know
Z
is in fact a martingale, and
Z
0
= 1. Then
by definition of a martingale, we have
E(Z
t
| F
s
) = Z
s
,
And unwrapping the definition of Z
t
shows that the result follows.
In general, the quadratic variation of a process doesn’t have to be linear in
t
.
It turns out if the quadratic variation increases to infinity, then the martingale
is still a Brownian motion up to reparametrization.
Theorem
(Dubins–Schwarz)
.
Let
M
be a continuous local martingale with
M
0
= 0 and hMi
∞
= ∞. Let
T
s
= inf{t ≥ 0 : hM i
t
> s},
the right-continuous inverse of
hMi
t
. Let
B
s
=
M
T
s
and
G
s
=
F
T
s
. Then
T
s
is a
(F
t
) stopping time, hM i
T
s
= s for all s ≥ 0, B is a (G
s
)-Brownian motion, and
M
t
= B
hMi
t
.
Proof.
Since
hMi
is continuous and adapted, and
hMi
∞
=
∞
, we know
T
s
is a
stopping time and T
s
< ∞ for all s ≥ 0.
Claim. (G
s
) is a filtration obeying the usual conditions, and G
∞
= F
∞
Indeed, if A ∈ G
s
and s < t, then
A ∩ {T
t
≤ u} = A ∩ {T
s
≤ u} ∩ {T
t
≤ u} ∈ F
u
,
using that
A ∩ {T
s
≤ u} ∈ F
u
since
A ∈ G
s
. Then right-continuity follows from
that of (F
t
) and the right-continuity of s 7→ T
s
.
Claim. B is adapted to (G
s
).
In general, if
X
is c´adl´ag and
T
is a stopping time, then
X
T
1
{T <∞}
∈ F
T
.
Apply this is with X = M , T = T
s
and F
T
= G
s
. Thus B
s
∈ G
s
.
Claim. B is continuous.
Here this is actually something to verify, because
s 7→ T
s
is only right contin-
uous, not necessarily continuous. Thus, we only know
B
s
is right continuous,
and we have to check it is left continuous.
Now B is left-continuous at s iff B
s
= B
s
−
, iff M
T
s
= M
T
s−
. Now we have
T
s−
= inf{t ≥ 0 : hM i
t
≥ s}.
If
T
s
=
T
s−
, then there is nothing to show. Thus, we may assume
T
s
> T
s−
.
Then we have
hMi
T
s
=
hMi
T
s−
. Since
hMi
t
is increasing, it means
hMi
T
s
is
constant in [T
s−
, T
s
]. We will later prove that
Lemma. M is constant on [a, b] iff hMi being constant on [a, b].
So we know that if T
s
> T
s−
, then M
T
s
= M
T
s
−
. So B is left continuous.
We then have to show that B is a martingale.
Claim. (M
2
− hMi)
T
s
is a uniformly integrable martingale.
To see this, observe that
hM
T
s
i
∞
=
hMi
T
s
=
s
, and so
M
T
s
is bounded. So
(M
2
− hMi)
T
s
is a uniformly integrable martingale.
We now apply the optional stopping theorem, which tells us
E(B
s
| G
r
) = E(M
T
s
∞
| G
s
) = M
T
t
= B
t
.
So B
t
is a martingale. Moreover,
E(B
2
s
− s | G
r
) = E((M
2
− hMi)
T
s
| F
T
r
) = M
2
T
r
− hMi
T
r
= B
2
r
− r.
So
B
2
t
− t
is a martingale, so by the characterizing property of the quadratic
variation,
hBi
t
=
t
. So by L´evy’s criterion, this is a Brownian motion in one
dimension.
The theorem is only true for martingales in one dimension. In two dimen-
sions, this need not be true, because the time change needed for the horizontal
and vertical may not agree. However, in the example sheet, we see that the
holomorphic image of a Brownian motion is still a Brownian motion up to a
time change.
Lemma. M is constant on [a, b] iff hMi being constant on [a, b].
Proof.
It is clear that if
M
is constant, then so is
hMi
. To prove the converse,
by continuity, it suffices to prove that for any fixed a < b,
{M
t
= M
a
for all t ∈ [a, b]} ⊇ {hMi
b
= hMi
a
} almost surely.
We set N
t
= M
t
− M
t
∧ a. Then hN i
t
= hMi
t
− hMi
t∧a
. Define
T
ε
= inf{t ≥ 0 : hN i
t
≥ ε}.
Then since N
2
− hNi is a local martingale, we know that
E(N
2
t∧T
ε
) = E(hNi
t∧T
ε
) ≤ ε.
Now observe that on the event
{hMi
b
=
hMi
a
}
, we have
hNi
b
= 0. So for
t ∈ [a, b], we have
E(1
{hMi
b
=hMi
a
}
N
2
t
) = E(1
{hMi
b
=hMi
a
N
2
t∧T
ε
) = E(hNi
t∧T
ε
) = 0.