5Brownian motion

III Advanced Probability



5.3 Transience and recurrence
Theorem. Let (B
t
)
t0
be a Brownian motion in R
d
.
If
d
= 1, then (
B
t
)
t0
is point recurrent, i.e. for each
x, z R
, the set
{t 0 : B
t
= z} is unbounded P
x
-almost surely.
If
d
= 2, then (
B
t
)
t0
is neighbourhood recurrent, i.e. for each
x R
2
and
U R
2
open, the set
{t
0 :
B
t
U}
is unbounded
P
x
-almost surely.
However, the process does not visit points, i.e. for all x, z R
d
, we have
P
X
(B
t
= z for some t > 0) = 0.
If
d
3, then (
B
t
)
t0
is transient, i.e.
|B
t
|
as
t P
x
-almost
surely.
Proof.
This is trivial, since
inf
t0
B
t
=
−∞
and
sup
t0
B
t
=
almost surely,
and (B
t
)
t0
is continuous.
It is enough to prove for
x
= 0. Let 0
< ε < R <
and
ϕ C
2
b
(
R
2
) such
that
ϕ
(
x
) =
log |x|
for
ε |x| R
. It is an easy exercise to check that this
is harmonic inside the annulus. By the theorem we didn’t prove, we know
M
t
= ϕ(B
t
)
1
2
Z
t
0
ϕ(B
s
) ds
is a martingale. For
λ
0, let
S
λ
=
inf{t
0 :
|B
t
|
=
λ}
. If
ε |x| R
,
then
H
=
S
ε
S
R
is
P
X
-almost surely finite. Then
M
H
is a bounded
martingale. By optional stopping, we have
E
x
(log |B
H
|) = log |x|.
But the LHS is
log εP(S
ε
< S
R
) + log RP(S
R
< S
ε
).
So we find that
P
x
(S
ε
< S
R
) =
log R log |x|
log R log ε
. ()
Note that if we let
R
, then
S
R
almost surely. Using (
), this
implies
P
X
(
S
ε
<
) = 1, and this does not depend on
x
. So we are done.
To prove that (
B
t
)
t0
does not visit points, let
ε
0 in (
) and then
R for x 6= z = 0.
It is enough to consider the case
d
= 3. As before, let
ϕ C
2
b
(
R
3
) be such
that
ϕ(x) =
1
|x|
for ε x 2. Then ϕ(x) = 0 for ε x R. As before, we get
P
x
(S
ε
< S
R
) =
|x|
1
|R|
1
ε
1
R
1
.
As R , we have
P
x
(S
ε
< ) =
ε
x
.
Now let
A
n
= {B
t
n for all t B
T
n
3
}.
Then
P
0
(A
c
n
) =
1
n
2
.
So by Borel–Cantelli, we know only finitely of
A
c
n
occur almost surely. So
infinitely many of the A
n
hold. This guarantees our process .