Consider a probability space (Ω, F, P ) with a filtration {Ft }t≥0 satisfying
the usual conditions.
Recall that we have defined a d-dimensional Brownian motion (starting at
0) as a stochastic process {Bt }t≥0 such that
• P (B0 = 0) = 1
• Bt − Bs is independent of Fs for all t ≥ s
• Bt − Bs is Gaussian with mean 0 and variance (t − s)Id×d where Id×d is
the d-by-d identity matrix.
In this note we will prove the following:
Lemma 0.1. A stochastic process {Xt }t≥0 such that P (X0 = 0) = 1 is a
Brownian motion if and only if
h
i
2
1
E eiu·(Xt −Xs ) |Fs = e− 2 (t−s)|u| P − a.s.
where · denotes the inner product in Rd and | · | denotes Euclidian norm in Rd .
Proof. One implication is easy: assume {Bt }t≥0 is a Brownain motion. Since
Bt − Bs is independent of Fs , by assumption, we get that
h
i
h
E eiu·(Bt −Bs ) |Fs = E eiu·(Bt −Bs ) ]
1
2
= e− 2 (t−s)|u|
where we have used that Bt − Bs is Gaussian in the last equality.
Conversely, assume {Xt }t≥0 satisfies
h
i
2
1
E eiu·(Xt −Xs ) |Fs = e− 2 (t−s)|u| P − a.s.
Let us show that (Xt − Xs ) is independent of Fs : this is the same as showing
that
P ((Xt − Xs ∈ A) ∩ C) = P (Xt − Xs ∈ A) P (C)
for all Borel sets A ∈ B(Rd ) and sets C ∈ Fs .
To this end, for a fixed C ∈ Fs we define two measures on (Rd , B(Rd )):
µ(A) := P ((Xt − Xs ∈ A) ∩ C) and ν(A) := P (Xt − Xs ∈ A) P (C) .
Note that for a measurable and, say, bounded function f : Rd → R we have
Z
Z
f (x)dµ(x) = E[f (Xt − Xs )1C ] and
f (x)dν(x) = E[f (Xt − Xs )]P (C).
Rd
Rd
The above equalities are immediate for charateristic functions and follows for
step functions by linearity. The general case follows by taking limits of step
functions.
1
Then the characteristic function of µ satisfies, for u ∈ Rd ,
Z
h
i
h h
ii
µ̂(u) =
eiu·x dµ(x)
= E eiu·(Xt −Xs ) 1C = E E eiu·(Xt −Xs ) 1C |Fs
Rd
h h
i i
h 1
i
2
2
1
= E E eiu·(Xt −Xs ) |Fs 1C
= E e− 2 (t−s)|u| 1C = e− 2 (t−s)|u| E[1C ]
1
2
= e− 2 (t−s)|u| P (C)
where we have used the assumptions on {Xt }t≥0 and that C ∈ Fs , so that the
indicator function 1C is Fs -measurable.
The characteristic function of ν is readily seen to be
Z
i
h h
h
ii
ν̂(u) =
eiu·x dν(x) = E eiu·(Xt −Xs ) P (C) = E E eiu·(Xt −Xs ) |Fs P (C)
Rd
i
h 1
2
2
1
= e− 2 (t−s)|u| P (C)
= E e− 2 (t−s)|u| P (C)
so that µ̂ = ν̂. It follows that µ = ν which proves that Xt − Xs is independent
of Fs .
So see that Xt − Xs is Gaussian note that
i
h
i
h h
ii
h 1
2
2
1
E eiu·(Xt −Xs ) = E E eiu·(Xt −Xs ) |Fs = E e− 2 (t−s)|u| = e− 2 (t−s)|u|
so Xt − Xs is Gaussian with mean 0 and variance (t − s)Id×d .
It is an easy exercise to see that for t0 ≥ 0, the stochastic process B̃t :=
Bt+t0 − Bt0 is again a Brownian motion. However, using the above lemma we
can prove a stronger result:
Lemma 0.2. Let τ be a bounded stopping time. Define
B̃t := Bt+τ − Bτ .
Then {B̃t }t≥0 is a Brownian motion w.r.t. the filtration {Ft+τ }t≥0 . In particular it is independent of Fτ .
Proof. Notice that the process
1
Mt = exp{iu · Bt + |u|2 t}
2
is a martingale. From the Optional Stopping theorem applied to the stopping
times τ + t and τ + s,
E[Mτ +t |Fτ +s ] = Mτ +s
or
1
2
E[eiu·Bτ +t + 2 |u|
(τ +t)
1
|Fτ +s ] = eiu·Bτ +s + 2 |u|
2
(τ +s)
.
Rearranging the terms we get
1
2
E[eiu·(Bτ +t −Bτ +s ) |Fτ +s ] = e− 2 (t−s)|u| .
We conclude from the previous Lemma that B̃t = Bt+τ − Bτ is a Brownian
motion.
2