The estimates of the mean first exit time from
the ball for the α–stable Ornstein–Uhlenbeck
processes.
Tomasz Jakubowski ∗
Institute of Mathematics and Computer Science
Wroclaw University of Technology
Wybrzeże Wyspiańskiego 27, 50-370 Wroclaw, Poland
Université d’Angers, Département de Mathématiques
2 Boulevard Lavoisier 49045 Angers, France
e-mail: [email protected]
Abstract
We consider the α–stable Ornstein–Uhlenbeck process as a solution of the Langevin equation where the Brownian motion is replaced
by isotropic α–stable process. We give sharp estimates for the expectation of the first exit time from the center of a ball B(x, r) for all
x ∈ Rd and r > 0. We compare these results with the case of the
Ornstein–Uhlenbeck diffusion process.
keywords: α-stable process, Ornstein-Uhlenbeck process, exit time.
AMS Subject Classification: 60G52
1.
Introduction
Let us consider the Langevin stochastic differential equation
bt , λ > 0,
dXt = −λXt dt + dX
X0 = x.
∗
partially supported by KBN and MEN
1
bt is the d-dimensional Brownian motion and Xt is called
In classical case X
the Ornstein–Uhlenbeck diffusion process. Let τD = inf{t ≥ 0 : Xt ∈ D}
be the first exit time of the process Xt from the set D. The properties of
τD were intensively studied in literature, especially the Laplace transform,
density and expected value of τD for various sets D (e.g. [1], [8], [9], [16],
[18], [19]).
bt we obtain the
If we replace the Brownian motion by another process X
bt . Such extensions were previously
Ornstein–Uhlenbeck process related to X
studied in [6], [10], [13], [14], [11]. In this paper we are interested in the case
bt is the isotropic α-stable process (see details in Section 2). We will
where X
give the sharp estimates for Ex (τB(x,r) ) for all x ∈ Rd and r > 0. The main
results of this paper are stated in 3 theorems.
Let M be a constant sufficiently large, i.e. M is a constant depending
on d, α, λ determined by some conditions which appear in the proofs of the
Theorems 2 and 3.
Theorem 1. Let 0 < r < M . There exists a constant C1 depending on
d, α, λ, M such that for all x ∈ Rd it holds
r
r
1
α
x
α
r ∧
≤ E (τB(x,r) ) ≤ C1 r ∧
C1
|x|
|x|
Theorem 2. Let r ≥ M . There exists a constant C2 depending on d, α, λ, M
such that for all x ∈ Rd satisfying |x| ≥ r it holds
1
|x|
|x|
x
ln
≤ E (τB(x,r) ) ≤ C2 ln
C2
|x| − r + 1
|x| − r + 1
Theorem 3. Let r ≥ M . There exists a constant C3 depending on d, α, λ, M
such that for all x ∈ Rd satisfying |x| < r it holds
1
((r − |x|)α + ln r) ≤ Ex (τB(x,r) ) ≤ C3 ((r − |x|)α + ln r)
C3
The work is organized as follows. In Section 2 we give the notation,
definitions and basic facts used in this paper. Section 3 is devoted to the
2
case r < M and we prove the Theorem 1. In Section 4 we consider r ≥ M
and we prove the Theorems 2 and 3. The proofs of the theorems are long
and the methods are different for each considered case. Therefore the proofs
are presented in several lemmas concerning the particular cases.
2.
Preliminaries
bt , Px ) the isotropic, α-stable Lévy process with index of
We denote by (X
stability α ∈ (0, 2) with values in Rd and characteristic function
α
E0 eiξXt = e−t|ξ| ,
ξ ∈ Rd , t ≥ 0.
As usual, Ex denotes the expectation with respect to the distribution Px
bt , Px ) is a Markov process with
of the process starting from x ∈ Rd . (X
R
bt ∈ D) =
transition density pb(t, x, y), it means Px (X
pb(t, x, y)dy. The
D
function pb(t, x, y) possesses the following scaling property
pb(t, x, y) = ad pb(aα t, ax, ay) .
(2.1)
Let us consider the stochastic differential equation
bt ,
dXt = −λXt dt + dX
λ > 0,
(2.2)
where the stochastic differentials are in the sense of semimartingales (see
[15]). The equation (2.2) is equivalent to the integral equation Xt − X0 =
Rt
bt . The solution of the equation (2.2) is given in Proposition
−λ 0 Xs ds + X
4.
Proposition 4.
λ(s−t)
Xt = e
Z
t
bu ,
eλ(u−t) dX
Xs +
for 0 ≤ s ≤ t ,
s
where the integration is in Stieltjes sense, with respect to the trajectories of
bt . In particular
X
Z t
−λt
bu ,
Xt = e X0 +
eλ(u−t) dX
for 0 ≤ t .
(2.3)
0
3
Proof. We multiply both sides of (2.2) by eλt and obtain
bu .
eλu dXu + eλu λXu du = eλu dX
Consequently
bu .
d(eλu Xu ) = eλu dX
Integrating both sides from s to t we get
Z t
λt
λs
bu .
e Xt − e Xs =
eλu dX
s
which gives (2.3).
Integrating by parts the equation (2.3) we obtain
Z t
t
−λt
λ(s−t) b bs ds .
eλ(s−t) X
Xt − e X0 = e
Xs − λ
0
0
b0 = 0 we get
Since X
Corollary 5.
−λt
Xt = e
t
Z
bs ds .
eλ(s−t) X
bt − λ
X0 + X
(2.4)
0
The process Xt is called the Ornstein-Uhlenbeck α-stable process. We
denote by Xtx the process Xt starting from x. From Lemma 4 we get
Xtx − e−λt x = Xty − e−λt y ,
x, y ∈ Rd .
(2.5)
Xt is a Markov process (see [12], [17]) with the following transition density
p(t, x, y).
Proposition 6. The transition density of the process Xt is equal to
−αλt
p(t, x, y) = pb 1−eαλ , e−λt x, y .
(2.6)
Proof. We adapt the proof given for α = 1 in [6]. Let h(u) ≥ 0 be a continRt
bu . Let us note that
uously differentiable function and Z = s h(u)dX
L
Z = lim Zn = lim
n→∞
n→∞
n−1
X
k=0
Znk = lim
n→∞
4
n−1
X
k=0
bu − X
bu ,
h(uk ) X
k+1
k
where s = u0 < u1 < · · · < un = t is a partition of the interval [s, t]
b has
and convergence is in probability (see i.e. [17]). Since the process X
independent increments, the density of Zn is equal to the convolution of the
b has also stationary increments, so by (2.1),
densities of its summands. X
density of each summand equals
d 1
fZnk (y) =
pb uk − uk−1 , 0, h(uy k ) = pb ((uk − uk−1 )h(uk )α , 0, y) .
h(uk )
The law of Zn is equal to convolution of the laws of Znk . Hence the characteristic function of Zn is equal to
ϕZn (ξ) =
n
Y
k=1
ϕZnk (ξ) =
n
Y
e−(uk −uk−1 )h(uk )
α |ξ|α
= e−
Pn
k=1
−(uk −uk−1 )h(uk )α |ξ|α
k=1
Letting n → ∞ we obtain
α
Rt
α
ϕZ (ξ) = e−|ξ| s h(u) du .
R
t
α
Thus the density of Z is equal to pb s h(u) du, 0, y . Putting s = 0, h(u) =
Rt
−αλt
eλ(u−t) and X0 = x we obtain 0 h(u)α du = 1−eαλ . Applying this to (2.3)
we get (2.6).
To simplify the notation we will denote (t)p =
1−e−pt
p
for p > 0.
Definition 7. For an open set D ⊂ Rd , we define a Markov time
τD = inf{t ≥ 0 : Xt ∈ Dc },
the first exit time from D of the process Xt . In the same way we define
bt ∈ Dc }.
τbD = inf{t ≥ 0 : X
If it is not stated otherwise all constants depend only on d, α, λ. We
use the convention that the constants noted by capital letters don’t change
and the ones noted by small letters change from lemma to lemma. As usual
a ∧ b denotes the minimum of a, b, a ∨ b denotes the maximum of a, b and
aD = {az : z ∈ D} for all a > 0 and D ⊂ Rd .
We will need some facts concerning the isotropic α-stable processes.
5
Lemma 8 (Anderson). For all z ∈ Rd we have
bt ∈ B(x, r)) ≤ Px (X
bt ∈ B(x, r))
Pz (X
or equivalently
Z
Z
B(x,r)
pb(t, z, y)dy ≤
B(x,r)
pb(t, x, y)dy .
The next facts are taken from [3]. For convenience of the reader we give
the proof of the second and the third one.
Fact 1. There exists a constant K1 such that
t
t
−1
−d/α
−d/α
≤ pb(t, x, y) ≤ K1
.
K1
∧t
∧t
|x − y|d+α
|x − y|d+α
Fact 2. There exists a constant K2 < 1 such that for all r > 0 and x ∈ Rd
we have
1
Px (b
τB(x,r) > K2 rα ) ≥ .
2
L
bt =
Proof. We take K2 such that E0 (b
τB(0,1) > K2 ) ≥ 12 . By (2.1) we have rX
brα t with respect to P0 and we get
X
1
bt ∈ B(0, 1) : ∀t ≤ K2
≤ P0 τbB(0,1) > K2 = P0 X
2
0
b
= P Xtrα ∈ B(0, r) : ∀t ≤ K2 = P0 τbB(0,r) > K2 rα
= Px τbB(x,r) > K2 rα .
Fact 3. For t < rα there exists a constant K3 such that
bt 6∈ B(x, r)) ≥ K3 tr−α .
Px (X
Proof. Denote B = B(x, r). If t < rα we have |x − y|α > t for y ∈ B c . Hence
by Fact 1 and substituting x − y = r(x − z), we have
6
Z
Z
t
pb(t, x, y) dy ≥
K1−1
dy
|x − y|d+α
Bc
Bc
Z
trd
−1
= K1
dz
d+α |x − z|d+α
B(x,1)c r
Z
dz
−1
tr−α = K3 tr−α
= K1
d+α
|z|
c
B(0,1)
x
bt 6∈ B(x, r)) =
P (X
At the end of this section we give the explicit formula for Ez (b
τB(x,r) )
established in [7] (see also [4]).
Ez (b
τB(x,r) ) = Cd,α (r2 − |x − z|2 ))α/2 ,
(2.7)
where Cd,α is certain constant depending on d and α.
3.
Estimates for small r
In this section we give the proof of the Theorem 1. Hence, we will consider
the exit time from the ball of the radius r bounded by some constant M . We
will see, that for α ≥ 1 the results do not differ from the symmetric α-stable
case, while for α < 1 the asymptotics are different (see Section 5.). We will
start with the the following lemma.
Lemma 9. Let B = B(x, r) be a ball in Rd . Then there exists a constant C4
such that for all z ∈ B, r > 0 and t > 0 we have
−α
Pz (τB > t) ≤ e−C4 tr .
Proof. We will use the formula
Z
Z
z
t
p n , z, x1 dx1
p
P (τB > t) ≤ lim
n→∞
B
B
t
, x1 , x2
n
Denote β = αλ. Using (2.6) and Lemma 8 we get
7
(3.8)
Z
dx2 . . .
p
B
t
, xn−1 , xn
n
dxn .
Z
lim
pb nt β , e−λt/n z, x1 dx1 ·
n→∞ B
Z · pb nt β , e−λt/n x1 , x2 dx2 · . . . ·
ZB −λt/n
t
xn−1 , xn dxn .
· pb n β , e
B
Z n
t
pb n β , x, y dy
≤ lim
z
P (τB > t) ≤
n→∞
B
t
For large n, n β < rα , so we may apply Fact 3 to the expression above.
Putting K3 = C4 we get
Z n
n
z
t
P (τB > t) ≤ lim
pb n β , x, y dy
≤ lim 1 − C4 nt β r−α
n→∞
= e−C4
n→∞
B
tr−α
.
We give also an alternative proof of the Lemma 9:
Second proof of Lemma 9. Let us denote by ∆Xt = Xt − Xt− the jump of
the process Xt . Clearly if the process in time t makes a jump bigger then
2r then there exists s ≤ t such that Xs 6∈ B(x, r) for all starting point z.
Hence Pz (τB(x,r) > t) ≤ Pz (|Xs − Xs− | < 2r, ∀s≤t ). From Corollary 5 we see
bt , so Pz (|Xs − Xs− | <
that the jumps of the process Xt are the same as X
bs − X
bs− | < 2r, ∀s≤t ) The waiting time Z for the first jump
2r, ∀s≤t ) = Pz (|X
b bigger then 2r has the exponential distribution with parameter
of X
Z
p=
ν(y)dy = c̃2 r−α ,
|y|>2r
b Therefore
where ν(x) = c|x|−α−d is the density of the Lévy measure of X.
Z ∞
−α
z
P (τB(x,r) > t) ≤ P(Z > t) =
pe−sp ds = e−tp = e−c2 tr .
t
8
We easily get the corollary
Corollary 10. For all s ≤ t and z ∈ B(x, r) we have
−α
Pz (τB(x,r) > t) ≤ Pz (τB(x,r) > t − s)e−C4 sr ,
where C4 is the constant from Lemma 9.
Proof. By Markov property and Lemma 9 we obtain
Pz (τB(x,r) > t) = Ez 1{τB(x,r) >t−s} PXt−s (τB(x,r) > s)
−α
≤ Pz (τB(x,r) > t − s)e−C1 sr .
Lemma 11. The process Xt may be expressed in the following way
d
−λt b
(Xt )t≥0 = e X(eβt −1)/β
,
(3.9)
t≥0
where β = αλ.
Proof. It is enough to prove the equality for finite dimension distributions. So
we have to show that for any t1 < t2 < . . . < tn and Borel sets A1 , A2 , . . . , An
it holds
Px (Xt1 ∈ A1 , Xt2 ∈ A2 , . . . , Xtn ∈ An ) =
b(eβt1 −1)/β ∈ eλt1 A1 , X(eβt2 −1)/β ∈ eλt2 A2 , . . . , X(eβtn −1)/β ∈ eλtn An )
= Px (X
We will use the formula (2.6).
Px (Xt1 ∈ A1 , Xt2 ∈ A2 , . . . , Xtn ∈ An ) =
Z
Z
−λt1
pb((t1 )β , e
x, x1 )dx1
pb((t2 − t1 )β , e−λ(t2 −t1 ) x1 , x2 )dx2 . . .
A1
A2
Z
pb((tn − tn−1 )β , e−λ(tn −tn−1 ) xn−1 , xn )dxn = I .
An
9
Using the formulas pb(t, x, y) = ad pb(aα t, ax, ay), eβt (t − s)β = eβt (t)β − eβs (s)β
and change of variables xk = e−λtk yk we obtain
Z
Z
dλt1 dλt2
dλtn
βt1
λt1
e e
...e
pb(e (t1 )β , x, e x1 )
pb(eβt2 (t2 − t1 )β , eλt1 x1 , eλt2 x2 )
A1
A2
Z
...
pb(eβtn (tn − tn−1 )β , eλtn−1 xn−1 , eλtn xn )dx1 dx2 dxn =
An
Z
βt1
eλt1 A1
pb e
Z
(t1 )β , x, y1 dy1
Z
x
beβt1 (t ) ∈
P (X
1 β
eλt2 A2
pb eβt2 (t2 )β − eβt1 (t1 )β , y1 , y2 dy2 . . .
pb eβtn (tn )β − eβtn−1 (tn−1 )β , yn−1 , yn dyn =
eλtn An
beβt2 (t )
eλt1 A1 , X
2 β
beβtn (t ) ∈ eλtn An )
∈ eλt2 A2 , . . . , X
n β
and since eβtk (tk )β = (eβtk − 1)/β we obtain (3.9).
We will need a simple technical lemma
Lemma 12.
1. For t <
r
4|x|λ
we have e−λt x ∈ B(x, r/2).
2. If |x| ≥ 2r then for t ≥
a constant.
2r
λ|x|
we have e−λt x 6∈ B(x, κr), where κ > 1 is
Proof.
1. We have to prove that |x − e−λt x| < r/2. Since (1 − e−t )/t < 1 we have
|x − e−λt x| = λt|x|
1 − e−λt
r
r
≤ λ|x|
< .
λt
4|x|λ
2
2. Let κ = 2(1 − e−1 ). We need to show |e−λt x − x| ≥ κr. Since the
−t
2r
function 1−et is decreasing and |x|
< 1 we have
1 − e−λt
2r
|x|
≥
1 − e−2r/|x|
2r
|x|
Hence 1 − e−λt |x| > κr.
10
≥ 1 − e−1 =
κ
.
2
For the rest of the paper we will denote B = B(x, r).
1
r
2 α
< 2λ
Lemma 13. If t < K
r
∧
then
2λ
4λ|x|
1
Px (τB > t) ≥ .
2
(3.10)
Proof. By Lemma 11
αλs
e
−
1
λs
b
∈ e B, ∀s≤t
P (τB > t) = P X
αλ
b
= Px X(s)
∈ (αλs + 1)1/α B, ∀s≤ eαλt −1 .
x
x
αλ
We will show that for s <
eαλt −1
αλ
it holds
B(x, r/2) ⊂ (αλs + 1)1/α B .
(3.11)
If |x| ≤ r then (3.11) obviously holds. Suppose now that |x| > r. By
Lemma 12, e−λt x ∈ B(x, r/2), so e−λt |x| > |x| − r/2 and we get eλt <
|x|
< |x|−r/2
. Therefore
|x|−r/2
|x|−r
1/α
eαλt − 1
+1
> (|x|−r)(αλs+1)1/α .
|x|−r/2 > (|x|−r)e = (|x|−r) α
α
λt
Hence |x|(αλs + 1)1/α − |x| < r(αλs + 1)1/α − r/2 and it follows that |(αλs +
1)1/α x − x| < r(αλs + 1)1/α − r/2. So we obtain (3.11). Now we have
αλt
e
−
1
x
x
x
b
.
P (τB > t) ≥ P X(s)
∈ B(x, r/2), ∀s< eαλt −1 = P τbB(x,r/2) >
αλ
αλ
Hence
Px (τB > t) ≥ Px τbB(x,r/2) >
Since tλ ≤
1
2
.
≤ 2tλ and by Fact 2
τbB(x,r/2) > 2tλ ≥ Px τbB(x,r/2) > K2 rα ≥ 1/2.
we have
Px (τB > t) ≥ Px
eαλt −1
αλ
eαλt −1
αλ
11
From the last Lemma we immediately obtain the lower bound estimates
in Theorem 1, when M = 1.
Corollary 14. If r < 1 then there exists a constant C5 such that
C5 α r r ∧ |x| .
E (τB ) ≥
λ
r
we may use the Lemma 13,
∧ 81 . For t < 2 Cλ5 rα ∧ |x|
x
Proof. Let C5 =
so we obtain
K2
4
2
Z
x
E (τB ) ≥
C5 α r
r ∧
λ
|x|
0
C5
P (τB > t) dt ≥
λ
x
r
r ∧
|x|
α
.
Now we will deal with upper bound estimates. We start with a simple
lemma which gives the upper bound for all x and r with constant independent
of λ.
Lemma 15. Let C4 be the constant from Lemma 9. Then
Ex (τB ) ≤
1 α
r .
C4
(3.12)
Proof. By Lemma 9
x
Z
∞
Z
x
∞
P (τB > t) dt ≤
E (τB ) =
0
0
e−C4 tr
−α
dt =
rα
.
C4
r
So it suffices to prove that Ex (τB ) ≤ c |x|
for some constant c. At first we
will prove it for |x| ≥ 2r.
Lemma 16. If |x| ≥ 2r then there exists a constant C6 such that
Ex (τB ) ≤ C6
12
r
.
|x|
(3.13)
2r
Proof. Let t0 = λ|x|
. At the beginning we estimate Px (τB ≥ t0 ). For y ∈ B
we have |e−λt0 x − x| ≤ |e−λt0 x − y| + |x − y| < |e−λt0 x − y| + r. By Lemma
12 we know that |e−λt0 x − x| ≥ κr. Hence |e−λt0 x − y| > |e−λt0 x − x| − r ≥
|e−λt0 x − x|. Using (2.6) and Fact 1 we
|e−λt0 x − x| − κ1 |e−λt0 x − x| = κ−1
κ
)αλ
have p(t0 , x, y) ≤ K1 |e−λt(t00x−y|
d+α . Since t0 ≤ 1/λ we get
Z
Z
(t0 )αλ
x
x
P (τB ≥ t0 ) ≤ P (Xt0 ∈ B) =
p(t0 , x, y) dy ≤ K1
dy
−λt
0 x − y|d+α
B
B |e
Z
1
(t0 )αλ
dy
≤ K1
−λt
d+α
d+α
0
(1 − e
)
B ((κ − 1)|x|/κ)
rd
r1−α
≤ c1 d+α−1 d+α = c1
,
|x|
t0
|x|
for some constant c1 . By Corollary 10 we obtain
Z ∞
Z t0
x
x
P (τB ≥ t) dt +
Px (τB ≥ t) dt
E (τB ) =
t0
0
Z ∞
−α
≤ t0 +
Px (τB ≥ t − (t − t0 ))e−C4 (t−t0 )r dt
t0
Z ∞
−α
x
= t0 + P (τB ≥ t0 )
e−C4 tr dt
0
1−α α
2r
r
2
c1
r
r
≤
+ c1
=
+
.
λ|x|
|x| C4
λ C4 |x|
Proof of Theorem 1. Let r< M . The
case M ≤ 1was already proved hence
r
2
α
, hence by Corollary 14
let M > 1. Since r ∧ |x| ≤ M (r/M )α ∧ r/M
|x|
Ex (τB ) ≥ Ex (τB(x,r/M ) ) ≥
C5 (r/M )α ∧
λ
r/M
|x|
≥
C5 α
r ∧
λM 2
r
|x|
.
The Lemmas 15 and 16 yield upper bound estimates for |x| ≥ 2r. If |x| < 2r
r
then |x|
≥ 12 ≥ rα /(2M α ) and Lemma 15 gives
r
x
α
α
α
E (τB ) ≤ C4 r ≤ 2M C4 r ∧
.
|x|
Taking C1 = 2M α C4 ∨
λM 2
C5
∨ C6 we obtain Theorem 1.
13
4.
Estimates for large r
In this section we prove the Theorems 2 and 3. In the proof of theorem 2 we
consider two cases r ≤ |x| < r + 1 and |x| ≥ r + 1. The first one is considered
in Lemmas 19 and 20, the second one in Lemmas 21 and 22. Similarly the
proof of Theorem 3 is given in Lemma 23 (the case |x| ≥ r/2) and in Lemmas
24, 25 (the case r/2 < |x| < r). At the beginning of this section we will deal
with unbounded domains like cones. It will help us to obtain the estimates
from above for the exit time from balls.
We will frequently use the formula following from Lemma 11
b eαλs −1 ∈ eλs B, ∀s≤t )
P(τB > t) = P(Xs ∈ B, ∀s≤t ) = P(X
αλ
1/α
bs ∈ (1 + αλs) B, ∀ eαλt −1
(4.14)
= P X
s≤
αλ
We see that if we replace B by a infinite cone
q
2
Γ = {z = (z1 , . . . , zd ) ∈ Rd : z12 + . . . + zd−1
η < zd },
η ∈ R,
(4.15)
having the property Γ = aΓ for all a > 0, we may write (4.14) in the following
way
x
P (τΓ > t) = P
x
bs ∈ Γ, ∀
X
s≤
eαλt −1
α
=P
x
eαλt − 1
τbΓ >
α
.
In [2] the authors consider the moments of τbΓ , the first exit time from an
infinite cone. One of the results is the estimate of Px (b
τΓ > t) (lemma 4.2)
K4−1 M (x)t−β/α ≤ Px (b
τΓ > t) ≤ K4 M (x)t−β/α ,
t > |x|α ,
(4.16)
where β ∈ (0, α) is a constant depending on η, M (x) is the Martin kernel of
Γ at infinity (see [2]) and K4 is a constant depending on α, Γ. We know that
if η = 0 (Γ is a half-space) then β = α/2. The function M is normalized to
one in the point x0 = (0, . . . , 0, 1), e.g. M (x0 ) = 1 (see [2]). Using Lemma
14
11 and (4.14) we have
∞
αλt
e
−
1
bs ∈ Γ, ∀s ≤
P X
dt
E (τΓ ) =
αλ
0
Z ∞
bs ∈ Γ, ∀s ≤ u) du
=
Px (X
1 + αλu
Z0 ∞
du
Px (b
τΓ > u)
=
1 + αλu
0
x
Z
x
(4.17)
Since Px (b
τΓ > t) ≤ K4−1 M (x)t−β/α for t > |x|α we obtain that for all x ∈ Γ
it holds Ex (τΓ ) < ∞. From (4.17) we easily get
1 x
E (ln(1 + αλb
τΓ )) .
(4.18)
αλ
Using (4.16) and (4.17) we may estimate Ex (τΓ ) for all x ∈ Γ, when η = 0,
e.g. Γ is a half-space.
Ex (τΓ ) =
Lemma 17. Let Γ be a half-space {z ∈ Rd : zd > 0} with η = 0 and let x ∈ Γ
be such that δΓ (x) = |x| (where δΓ (x) = dist(x, ∂Γ)). There exists a constant
C7 such that
C7−1 ln(1 + |x|α/2 ) ≤ Ex (τΓ ) ≤ C7 ln(1 + |x|α/2 ) .
(4.19)
bs ∈
Proof. Using the scaling property (2.1) we have Px (b
τΓ > u) = Px (X
x/|x| b
x/|x|
−α
Γ, ∀s≤u ) = P
(Xs|x|−α ∈ Γ, ∀s≤u ) = P
(b
τΓ > u|x| ). Hence substituting in (4.17) u|x|−α = t we get
Z ∞
Z ∞
|x|α dt
du
x/|x|
x/|x|
−α
=
P
(b
τΓ > t)
P
(b
τΓ > u|x| )
I =
1 + αλu
1 + αλ|x|α t
0
0
We note that x/|x| = x0 . Since β = α/2 and M (x/|x|) = 1, from (4.16) we
get for t > 1
K4−1 t−1/2 ≤ Px/|x| (b
τΓ > t) ≤ K4 t−1/2 .
R ∞ x/|x|
dt
< ∞.
Hence we may denote c1 = 1 P
(b
τΓ > t) αλt
For |x| > 1 we have
Z 1
Z ∞
|x|α dt
dt
x
+
Px0 (b
τΓ > t)
E (τΓ ) ≤
α
αλt
1
0 1 + αλ|x| t
1
≤
ln(1 + αλ|x|α ) + c1 ≤ c2 ln(1 + |x|α/2 ) ,
αλ
15
for some constant c2 . On the other hand there exists a constant c3 such that
Z 1
|x|α dt
x
τΓ > 1)
Px0 (b
E (τΓ ) ≥
1 + αλ|x|α t
0
1
τΓ > 1)
ln(1 + αλ|x|α ) ≥ c3 ln(1 + |x|α/2 ) .
= Px0 (b
αλ
So we get (4.19) for |x| ≥ 1. In fact in this case, Ex (τΓ ) ≈ ln(1 + |x|), but we
want to have common estimates for all x. If 0 < |x| < 1 we have
Z 1
Z ∞
|x|α dt
|x|α dt
x0
x0
x
τΓ > t)
(b
τ
>
t)
P (b
P
E (τΓ ) =
+
Γ
1 + αλ|x|α t
1 + αλ|x|α t
0
1
Z 1
Z ∞
|x|α dt
≤
|x|α dt + K4
t−1/2
1 + α|x|α λt
0
1
Z ∞
K4 ds
|x|α/2
= |x|α +
(1 + αλs)s1/2
|x|α
Z ∞
K4 ds
≤ |x|α + |x|α/2
≤ c4 |x|α/2 ≤ c5 ln(1 + |x|α/2 ) .
1/2
(1
+
αλs)s
0
Similarly we show
x
∞
|x|α dt
1 + αλ|x|α t
1
Z ∞
Z ∞
K4−1 ds
K4−1 ds
α/2
α/2
≥
|x|
≥
|x|
(1 + αλs)s1/2
(1 + αλs)s1/2
1
|x|α
Z
E (τΓ ) ≥
Px0 (b
τΓ > t)
≥ c6 |x|α/2 ≥ c6 ln(1 + |x|α/2 ) .
−1
Taking C7 = (c2 + c5 ) ∨ (c−1
3 + c6 ) we get (4.19).
Using the same method as for the mean first exit time, we may express
the Laplace transform of τΓ as Ex (f (b
τΓ )), for some function f .
Remark 18. Let Γ be a cone defined as in (4.15). For all x ∈ Γ we have
− s
x −sτΓ
x
αλ
E (e
) = αλ E
1 + αλb
τΓ
.
(4.20)
16
Proof. Integrating by parts we obtain
Z ∞
x −sτΓ
se−su Px (τΓ > u) du
E (e
) = 1−
Z0 ∞
eαλu − 1
−su x
= 1−
se P τbΓ >
du
αλ
0
Z ∞
− s
dt
= 1−
s 1 + αλt αλ Px (b
τΓ > t)
1 + αλt
Z0 ∞
s
x
= 1−
τΓ > t) dt
s +1 P (b
0
1 + αλt αλ
Denoting by fτbxΓ (t) the density of τbΓ with respect to Px and again integrating
by parts we get
Z ∞
αλ
x
x −sτΓ
E (e
) = 1 − αλ +
s fτbΓ (t) dt
0
1 + αλt αλ
− s
x
= 1 − αλ + αλ E
1 + αλb
τΓ αλ .
Let us note that the process Xt is invariant by rotation around 0. It means
that if T : Rd → Rd denotes a rotation around 0 then p(t, x, y) = p(t, T x, T y).
Indeed p(t, x, y) = pb((t)αλ , e−λt x, y) = pb((t)αλ , e−λt T x, T y) = p(t, T x, T y).
For given x, by Tx we denote the rotation around 0 such that T x = x̃ =
(0, 0, . . . , |x|). Then for any set D and point x we have
Px (Xt ∈ D) = PTx x (Xt ∈ Tx D) ,
(4.21)
Ex (τD ) = ETx x (τTx D ) .
(4.22)
and consequently
Now we will pass to the proof of Theorem 2. At the beginning we will
deal with the case r = |x| > M , where M is certain constant.
Lemma 19. Let |x| = r > M , then there exists a constant C3 such that
C8−1 ln r ≤ Ex (τB ) ≤ C8 ln r
17
(4.23)
αλt
1
Proof. Let t0 = αλ
ln(1 + αλK2 rα ). Then e αλ0 −1 = K2 rα . Since |x| = r it
holds B(x, r) ⊂ cB(x, r) for any constant c ≥ 1. By (4.14) and Fact 2 we get
x
x
1/α
b
α
P (τB > t0 ) = P Xs ∈ (αλs + 1) B, ∀s≤K2 r
1
bs ∈ B, ∀s≤K2 rα = Px (b
≥ Px X
τB > K 2 r α ) ≥ .
2
2/α
1
We take M such that r > M > αλK
, so αλK2 rα > αλK2 M α/2 rα/2 ≥
2
rα/2 and we get
x
Z
E (τB ) ≥
0
t0
ln r
1
1
Px (τB > t) dt ≥ t0 ≥
ln rα/2 =
,
2
2αλ
4λ
which gives estimates from below. To obtain the estimates from above we
use the Lemma 17. Let Γ be as in (4.15) with η = 0. Then by (4.22)
Ex (τB ) = ETx x (τTx B ) ≤ ETx x (τΓ ) ≤ C7 ln(1 + rα/2 ) ≤ αC7 ln r,
because we may choose M > 2 which gives 1 + rα/2 ≤ rα . So taking C8 =
αC7 ∨ 4λ we obtain (4.23).
Now we will generalize a little the results of Lemma 19.
Lemma 20. Let r > M > 2. Then there exists a constant C9 such that for
all r ≤ |x| ≤ r + 1 we have
C9−1 ln r ≤ Ex (τB ) ≤ C9 ln r
(4.24)
Proof. For estimates from above it suffices to notice that B(x, r) ⊂ B(x, |x|).
Hence Ex (τB ) ≤ C8 ln |x| ≤ C8 ln(r2 ) ≤ 2C8 ln r. To obtain the estimates
from below we start with the case |x| = r + 1. We notice
that B(x1 , r1 ) ⊂
1
B(x2 , r2 ) if and only if |x1 − x2 | ≤ r2 − r1 . Let t0 = λ ln |x|−r/2
. Then for
|x|−r
t ≤ t0 it holds eλt ≤ |x|−r/2
, so |x − eλt x| ≤ eλt r − 2r . Hence B x, 2r ⊂ eλt B.
|x|−r
For t < t0 we have
λs
x
x
b eαλs −1 ∈ e B, ∀s<t
P (τB > t) = P X
αλ
x
b
≥ P X eαλs −1 ∈ B(x, r/2), ∀s<t
αλ
eαλt − 1
x
≥ P τbB(x,r/2) >
.
αλ
18
Let c1 = (4λ)−1 , then
|x| = r + 1 we obtain
Z t0
Px (τB > t) dt
(r/2+1)α −1
αλ
≥ c1 (r/2)α for r > M ≥ 4. Hence putting
0
Z
≥
0
≥
≥
≥
Px τbB(x,r/2)
eαλt − 1
>
αλ
Z
dt =
eαλt0 −1
αλ
0
(r/2+1)α −1
αλ
Px (b
τB(x,r/2) > t)
dt
αλt + 1
α
c1 (r/2)
Px (b
τB(x,r/2) > t)
Px (b
τB(x,r/2) > t)
dt ≥
dt
αλt + 1
αλt + 1
0
0
Z (c1 ∧K2 )(r/2)α x
P (b
τB(x,r/2) > t)
dt
αλt + 1
0
1 x
P τbB(x,r/2) > K2 (r/2)α ln (αλ(c1 ∧ K2 )(r/2)α + 1)
αλ
1
ln (αλ(c1 ∧ K2 )(r/2)α ) .
2αλ
Z
=
t0
Z
For M large enough and some constant c2 (for example c2 = α/2 and M ≥
4
we have ln (αλ(c1 ∧ K2 )(r/2)α ) ≥ c2 ln r. Now if r < |x| < r+1
(αλ(c1 ∧K2 ))2/α
c2
then Ex (τB ) ≥ Ex (τB(x,|x|−1) ) ≥ 2αλ
ln(|x| − 1) ≥ c3 ln r for some constant
c3 .
R t λ(s−t)
b
bs ds + |(1 −
X
By (2.4) if X0 = x and |Xt − x| ≥ r then X
t−λ 0 e
R t λ(s−t)
b
−λt
b
Xs ds ≥ r − (1 − e−λt )|x|. From
e )x| ≥ r and we get Xt − λ 0 e
R
bs ds ≥ (r −
bt | ≥ (r − (1 − e−λt )|x|)/2 or λ t eλ(s−t) X
this we deduce that |X
0
R
−λt
bs | t λeλ(s−t) ds ≥
(1 − e )|x|)/2. In the second case it follows that sups≤t |X
0
R t λ(s−t)
−λt
r/2 − (1 − e )(|x|/2). Since 0 λe
ds < 1 we obtain that if |Xt − x| ≥ r
then there exists s ≤ t such that
bs | ≥
|X
r
|x|
− (1 − e−λt ) .
2
2
(4.25)
bs | < r − (1 − e−λt ) |x| then for all s ≤ t
Therefore if for all s ≤ t it holds |X
2
2
we have |Xt − x| < r. Hence if 2r − (1 − e−λt ) |x|
>
0
then
2
!
Px (τB(x,r) > t) ≥ P0
τb
„
«
|x|
r
B 0, −(1−e−λt )
2
2
19
>t
.
(4.26)
Using the last inequality we may obtain lower-bound estimates of Ex (τB ) in
certain cases.
Lemma 21. If |x| ≥ r + 1 then
x
E (τB(x,r) ) ≥ C10 ln
Proof. Let t0 =
1
λ
ln
|x|
|x|−r
|x|
|x| − r
.
. Then r = (1 − e−λt0 )|x| and we get
r
|x|
|x|
|x|
− (1 − e−λt0 /2 )
= (1 − e−λt0 )
− (1 − e−λt0 /2 )
> 0.
2
2
2
2
Hence we obtain by (4.26) and (2.1)
!
Px (τB > t0 /2) ≥ P0
„
“ √
” |x| «
r
B 0, − 1− (|x|−r)/|x|
2
2
τb
> t0 /2
−α !
r
|x|
t0
= P0 τbB(0,1) >
− 1 − (|x| − r)/|x|
2
2
2
p
−α
t0 .
= P0 τbB(0,1) > 2α−1 r + |x|(|x| − r) − |x|
The function
p
p
|x|2 − |x|r −|x| is increasing as a function of |x| on (r, ∞),
so
f (x) = r +
p
|x|2
−α 1 |x| − |x|r − |x|
ln
λ
|x| − r
is a decreasing function. Therefore since |x| ≥ r + 1 we get
Px (τB(x,r) > t0 /2) ≥
−α p
|x|
2α−1
0
ln |x|−r
≥ P τbB(0,1) > λ r + |x|(|x| − r) − |x|
−α
√
2α−1
0
≥ P τbB(0,1) > λ r + r + 1 − (r + 1)
ln(r + 1)
α−1 ln(r+1)
= P0 τbB(0,1) > λ2 √r+1−1
.
α
(
)
20
2α−1 ln(r+1)
√
α
λ( r+1−1)
Since
is decreasing for r ≥ e2/α we get for r ≥ M ≥ e2/α
Px (τB(x,r) > t0 /2) ≥ P0
τbB(0,1)
2α−1 ln(M + 1)
α
√
>
λ M +1−1
!
= c1 .
Hence
x
Z
E (τB ) ≥
t0 /2
x
P (τB(x,r)
0
t0
> t) dt ≥ Px (τB(x,r) >
2
t0
)
2
c1
ln
≥
2λ
|x|
|x| − r
.
Let S = {z = (z1 , . . . , zd ) ∈ Rd : |x| − r < zd < |x| + r} then by (4.22)
Ex (τB(x,r) ) = ETx x (τB(Tx x,r) ) ≤ ETx x (τS ) = E|x| (τ(|x|−r,|x|+r) ) ,
(4.27)
where the last expectation is taken with respect to one-dimensional process
Xt .
Lemma 22. There exists a constant C11 such that for |x| > r + 1 it holds
|x|
x
E (τB(x,r) ) ≤ C11 ln
.
(4.28)
|x| − r
Proof. According to (4.27) it suffices to prove (4.28) only for
d =1. We will
|x|
for some
prove more: for x > r + 1 we have Ex (τ(x−r,∞) ) ≤ C11 ln |x|−r
constant C11 . We assume that x > 0.
x
Let t0 = λ1 ln x−r
. From (2.5) we get Xtx −e−λt x = Xt0 . Hence if Xtx > x−r
and t ≥ t0 we have Xt0 > x−r−xe−λt and in consequence Xt0 > x−r−x x−r
=
x
0. Therefore
Px (Xs > x − r, ∀t0 <s<t ) ≤ P0 (Xs > 0, ∀t0 <s<t ) .
(4.29)
Now using (4.29) we may estimate
Z t0
Z ∞
x
x
Px (τ(x−r,∞) > t) dt
E (τ(x−r,∞) ) =
P (τ(x−r,∞) > t) dt +
0
t0
Z ∞
≤ t0 +
Px (Xs > x − r, ∀t0 ≤s≤t ) dt
Zt0∞
≤ t0 +
P0 (Xs > 0, ∀t0 ≤s≤t ) dt = t0 + I .
t0
21
It suffices to estimate the last integral I. We will use the Markov property,
Fubini-Tonelli theorem and Lemma 17
Z ∞
Z ∞
0
P (Xs > 0, ∀t0 ≤s≤t ) dt =
E0 PXt0 (Xs > 0, ∀s≤t−t0 ) 1{Xt0 >0} dt
t0
t0
Z ∞
Xt0
0
τ(0,∞) > t − t0 dt
P
= E 1{Xt0 >0}
t0
= E0 1{Xt0 >0} EXt0 τ(0,∞)
α/2
0
≤ C7 E 1{Xt0 >0} ln 1 + Xt0
Z ∞
ln 1 + z α/2 p(t0 , 0, z) dz .
= C7
0
Denote β = αλ. From Fact 1 we know that p(t0 , 0, z) ≤ K1
Since ln(1 + x) ≤ x we obtain
1/α
(t0 )β
Z
I ≤ K1 C7
z
α/2
−1/α
(t0 )β
Z
= K1 C7
1
(α/2+1)/α
−1/α
(t0 )β
(t0 )β
1 + α/2
(t0 )β
z 1+α
−1/α
∧ (t0 )β
.
∞
(t0 )β
dz
1/α
z 1+α
(t0 )β
∞
2
−α/2 − K1 C7 (t0 )β z
1/α
α
(t0 )
dz + K1 C7
0
z α/2
β
2
2
4α + 4
1/2
−1/2
1/2
= K1 C7
(t0 )β + K1 C7 (t0 )β (t0 )β
= K1 C7
(t0 )β .
2+α
α
(2 + α)α
Hence
1/2
1/2
Ex (τ(x−r,∞) ) ≤ t0 + c1 (t0 )β ≤ t0 + c1 t0
1/2
If |x| < 2r then t0
≤ t0 /(ln 2)1/2 and we obtain
Ex (τB ) ≤ E|x| (τ(|x|−r,∞) ) ≤ (1 + c1 /(ln 2)1/2 )t0 .
If |x| > 2r by Lemma 16 we get
r
r
|x|
≤ 2C6 ln 1 +
≤ 2C6 ln
.
E (τB ) ≤ C6
|x|
|x|
|x| − r
x
The Theorem 2 follows now from Lemmas 20, 21 and 22. Now we pass to
the proof of the Theorem 3. We will start with the case when |x| ≤ 2r , so it
suffices to show that Ex (τB(x,r) ) ≥ crα for some constant c.
22
Lemma 23. If |x| ≤
r
2
then
Ex (τB(x,r) ) ≥ E0 (b
τB(0,r/4) ).
Proof. Since |x| < r/2 we have 2r ≥ (1 − e−λt )|x| which is equivalent to
r
− (1 − e−λt ) |x|
≥ 4r . Now by (4.26) we get
2
2
Px (τB(x,r) > t) ≥ P0 (b
τB (0, r −(1−e−λt ) |x| ) > t) ≥ P0 (b
τB(0,r/4) > t)
2
2
By integrating from 0 to infinity we get the assertion of the lemma.
Consequently by (2.7) and Lemma 15 we get for |x| < r/2,
Cd,α α
1 α
r .
r ≤ Ex (τB(x,r) ) ≤
α
4
C4
(4.30)
Lemma 24. Let M/2 < r/2 < |x| < r. There exists a constant C12 such
that
Ex (τB(x,r) ) ≤ C12 (ln r + (r − |x|)α ) .
(4.31)
Proof. According to (4.27) it suffices to show (4.31) only for d = 1. We
may assume that x > 0. Denote B = B(x, r) = (x − r, x + r) and I =
supz∈B Ez (τB ). By strong Markov property (we stop the process on exiting
(0, x + r)) and Lemma 17 we have
Xτ(0,x+r)
z
z
(τB )
I = sup E (τ(0,x+r) ) + E 1{Xτ(0,x+r) ∈(x−r,0]} E
z∈B
≤ c1 ln(1 + (x + r)α/2 ) +
sup Ey (τB ) sup Pz (Xτ(0,x+r) ∈ (x − r, 0))
y∈(x−r,0]
≤ c2 ln r +
z∈B
y
sup E (τB ) .
y∈(x−r,0]
Now we stop the process on exiting (x − r, r − x) and use the Lemma 15.
Xτ(x−r,r−x)
y
y
(τB )
E (τ(x−r,r−x) ) + E E
I ≤ c2 ln r + sup
y∈(x−r,0]
≤ c2 ln r +
sup Ey (τ(y−2(r−x), y+2(r−x)) )
y∈(x−r,0]
+
Xτ(x−r,r−x)
sup Ey E
(τB )
y∈(x−r,0]
≤ c2 ln r + c3 (r − x)α +
Xτ(x−r,r−x)
sup Ey E
y∈(x−r,0]
23
(τB ) .
Let us note that
sup
Xτ(x−r,r−x)
Ey E
(τB )
y∈(x−r,0]
=
sup Ey 1{Xτ(x−r,r−x) ∈[r−x,r+x)} E
Xτ(x−r,r−x)
(τB )
y∈(x−r,0]
≤
sup Py Xτ(x−r,r−x) ∈ [r − x, r + x)
y∈(x−r,0]
≤
Ez (τB )
z∈[r−x,r+x)
y
sup P Xτ(x−r,r−x)
∈ [r − x, ∞) sup Ez (τB )
z∈B
y∈(x−r,0]
It suffices to show that
sup
sup Py Xτ(x−r,r−x) ∈ [r − x, ∞) ≤ 21 . From (2.5)
y∈(x−r,0]
we know that
Xty = e−λt y + Xt0 .
So if y < 0 then Xty ≤ Xt0 . Denote λyD = inf{t ≥ 0 : Xty ∈ D} the first hitting
time of a domain D by the process Xty . Hence if y < 0 then
λy(−∞,x−r] ≤ λ0(−∞,x−r]
and
λy[r−x,∞) ≥ λ0[r−x,∞) .
So, by symmetry of the process Xt0 , we obtain for y ∈ (x − r, 0]
Py Xτ(x−r,r−x) ∈ [r − x, ∞) = P λy(−∞,x−r] > λy[r−x,∞)
1
≤ P λ0(−∞,x−r] > λ0[r−x,∞) = .
2
X
Hence sup Ey E τ(x−r,r−x) (τB ) ≤ 21 sup Ez (τB ) and we get
y∈(x−r,0)
z∈B
sup Ez (τB ) ≤ 2c2 ln r + 2c3 (r − x)α ≤ 2(c2 + c3 )(ln r + (r − x)α ) .
z∈B
Since supz∈B Ez (τB ) ≥ Ex (τB ) we obtain the assertion of the Lemma with
C12 = 2(c2 + c3 ).
To obtain estimates from below we use the same method as in Lemma
21.
Lemma 25. There exists a constant C13 such that for r/2 < |x| < r it holds
Ex (τB(x,r) ) ≥ C13 (ln r + (r − |x|)α ) .
24
(4.32)
Proof. Let t0 = ((r − |x|)α + ln r)/λ. By (4.26) it holds
!
Px (τB > t0 /4) ≥ P0
= P
τb
«
„
|x|
r
B 0, −(1−e−λt0 /4 )
2
2
0
−α
!
|x|
r
− 1 − e−λt0 /4
2
2
|x|
r
α
− 1 − e−((r−|x|) +ln r)/4
2
2
τbB(0,1) >
= P0
> t0 /4
τbB(0,1) >
t0
4
−α
t0
4
!
If (r − |x|)α ≥ ln r then we have
α α
r − |x| |x| −((r−|x|)α +ln r)/4
r − |x|
(r − |x|)α + ln r
≥
≥ c1
+
e
,
2
2
2
4λ
for c1 = λ/2α−1 . If (r − |x|)α < ln r then
α
α
1
r − |x| |x| −((r−|x|)α +ln r)/4
|x|
−(ln r+ln r)/4 α
≥ α |x|e
+
e
≥
2
2
2
2r1/2
ln r
rα/2
≥
c
≥
2
4α
2λ
(r − |x|)α + ln r
≥ c2
,
4λ
for some constant c2 depending on α and λ. Hence
−α
r
t0
1
−((r−|x|)α +ln r)/4 |x|
− 1−e
≤
,
2
2
4
c1 ∧ c2
and we get
x
P (τB > t0 /4) ≥ P
0
τbB(0,1)
1
>
c1 ∧ c2
= c3 .
Therefore
x
Z
E (τB(x,r) ) ≥
t0 /4
Px (τB > t) dt ≥
0
c3
((r − |x|)α + ln r) .
4λ
Now the Theorem 3 follows from (4.30) and Lemmas 24 and 25.
25
5.
Comparison – with isotropic stable process
and OU diffusion process
At the end we would like to compare the results with the case of the Ornstein–
Uhlenbeck diffusion process denoted by Yt . First we will give some well
known results concerning the first exit time of Yt and Brownian motion Bt .
We denote by σD and σ
bD the first exit time from D of Yt and Bt respectively.
Since the infinitesimal generator acting on the Ex (b
σD ) as the function of x
x
gives −1 for all x ∈ D and E (b
σD ) = 0 for x 6∈ D, we get the simple formula
Ez (b
σB(x,r) ) = (r2 − |x − z|2 )/d ,
z ∈ B(x, r) .
(5.33)
We may also solve equation LEx (σB(0,r) ) = −1 where L = 21 ∆ − λx · ∇x to
obtain the results for Ornstein–Uhlenbeck diffusion process [9]
1
d
d
Ex (σB(0,r) ) = (r2 2 F2 (1, 1; + 1, 2; λr2 ) − |x|2 2 F2 (1, 1; + 1, 2; λ|x|2 )) .
d
2
2
(5.34)
This result was also obtained in [8] by concerning the radial Ornstein–
Uhlenbeck diffusion process. We may also deduce the asymptotic behaviour
of Ex (σB(0,r) ) when x = 0 and r > M (see [9])
λr2
−1 e
C14 d
r
2
eλr
≤ E (σB(0,r) ) ≤ C14 d .
(5.35)
r
On the other hand from [5] we know that the Green function of the operator
L = 12 ∆ − λx · ∇x is comparable with the Green function of Brownian motion
for all bounded Lipschitz domains.
R Since the expectation of the first exit
time from the set D is equal to D GD (x, y)dy, we have got the result that
for r, |x| < M it holds
0
−1 z
C15
E (σ̂B(x,r) ) ≤ Ez (σB(x,r) ) ≤ C15 Ez (σ̂B(x,r) ) ,
(5.36)
where the constant C15 depends on d, λ and M . Hence from (5.33) and (5.36)
we obtain that for r, |x| < M it holds
1
C15 d
r2 ≤ Ex (σB(x,r) ) ≤
C15 2
r .
d
(5.37)
We would like to compare the results of this work with the α-stable case.
First we recall the formula (2.7)
Ex (b
τB(0,r) ) = Cd,α (r2 − |x|2 )α/2 .
26
We note that this formula holds also for α = 2, then the α-stable symmetric process becomes the Brownian motion. Let r, |x| < M then from the
Theorem 1 we have
1
r
r
α
x
α
r ∧
≤ E (τB(x,r) ) ≤ C1 r ∧
C1
|x|
|x|
r
If α ≥ 1 then rα ≤ M α |x|
and we get an analog of (5.37)
1
rα ≤ Ex (τB(x,r) ) ≤ C1 rα .
C1 M α
(5.38)
For α < 1 there is no more analogy with diffusion case. We may take a
(1−α)/2
sequence of balls B(xn , rn ) such that xn = rn
and rn → 0 when n → ∞.
(α+1)/2
rn
α
. Hence Exn (τB(xn ,rn ) )/rnα → 0
Then for rn < 1 we have rn ∧ |xn | = rn
while n → ∞ and (5.38) does not hold. This indicates that in the local study
bt differ substantially when α < 1. For 1 ≤ α ≤ 2 they
of Ex (τB ), Xt and X
have similar behaviour.
On the other hand for large r (r > M ), from Theorem 3 we have
1 α
r ≤ E0 (τB(0,r) ) ≤ C3 rα ,
C3
for all α < 2, which gives the same asymptotics as for E0 (b
τB(0,r) ) but completely different one then (5.35). Heuristically it may be explained in the
following way. The influence of the drift on the mean first exit time from
B(0, r) is much smaller then the influence of the jumps of Xt (see the second
proof of the Lemma 9). Hence, although the process Xt is attracted to 0
by the drift, it ”jumps out” from B(0, r) in the time rα . Since Ornstein–
Uhlenbeck diffusion has no jumps it has to reach the boundary overcoming
the drift and this is why Yt needs more time for exiting B(0, r).
Acknowledgments
I would like to thank P. Graczyk, K. Bogdan and L. Alili for help in preparation of this paper.
27
References
[1] L. Alili, P. Patie, and J. L. Pedersen. Representations of the first hitting
time density of an Ornstein-Uhlenbeck process. Stoch. Models, 21(4):967–
980, 2005.
[2] Rodrigo Bañuelos and Krzysztof Bogdan. Symmetric stable processes in
cones. Potential Anal., 21(3):263–288, 2004.
[3] Krzysztof Bogdan, Andrzej Stós, and Pawel Sztonyk. Harnack inequality
for stable processes on d-sets. Studia Math., 158(2):163–198, 2003.
[4] Krzysztof Burdzy and Tadeusz Kulczycki. Stable processes have thorns.
Ann. Probab., 31(1):170–194, 2003.
[5] M. Cranston and Z. Zhao. Conditional transformation of drift formula
and potential theory for 12 ∆ + b(·) · ∇. Comm. Math. Phys., 112(4):613–
625, 1987.
[6] Piotr Garbaczewski and Robert Olkiewicz. Ornstein-Uhlenbeck-Cauchy
process. J. Math. Phys., 41(10):6843–6860, 2000.
[7] R. K. Getoor. First passage times for symmetric stable processes in space.
Trans. Amer. Math. Soc., 101:75–90, 1961.
[8] Anja Göing-Jaeschke and Marc Yor. A survey and some generalizations
of Bessel processes. Bernoulli, 9(2):313–349, 2003.
[9] P. Graczyk and T. Jakubowski. Exit times and Poisson kernels of the
Ornstein-Uhlenbeck process. Preprint, 2005.
[10] M. Grigoriu and G. Samorodnitsky. Tails of solutions of certain nonlinear
stochastic differential equations driven by heavy tailed Lévy motions.
Stochastic Process. Appl., 105(1):69–97, 2003.
[11] Peter Imkeller and Ilya Pavlyukevich. First exit times of solutions of nonlinear stochastic differential equations driven by symmetric levy processes
with alpha-stable components. Preprint, 2004.
[12] Aleksander Janicki and Aleksander Weron. Simulation and chaotic behavior of α-stable stochastic processes, volume 178 of Monographs and
28
Textbooks in Pure and Applied Mathematics. Marcel Dekker Inc., New
York, 1994.
[13] A. A. Novikov. Martingales and first-exit times for the OrnsteinUhlenbeck process with jumps.
Teor. Veroyatnost. i Primenen.,
48(2):340–358, 2003.
[14] Piere Patie. On a martingale associated to generalized OrnsteinUhlenbeck processes and an application to finance. Stoch. Proc. App.,
15(4):593–607, 2005.
[15] Philip E. Protter. Stochastic integration and differential equations, volume 21 of Applications of Mathematics (New York). Springer-Verlag,
Berlin, second edition, 2004. Stochastic Modelling and Applied Probability.
[16] Luigi M. Ricciardi and Shunsuke Sato. First-passage-time density and
moments of the Ornstein-Uhlenbeck process. J. Appl. Probab., 25(1):43–
57, 1988.
[17] Gennady Samorodnitsky and Murad S. Taqqu. Stable non-Gaussian
random processes. Stochastic Modeling. Chapman & Hall, New York,
1994. Stochastic models with infinite variance.
[18] Shunsuke Sato. Evaluation of the first-passage time probability to
a square root boundary for the Wiener process. J. Appl. Probability,
14(4):850–856, 1977.
[19] Marlin U. Thomas. Some mean first-passage time approximations for the
Ornstein-Uhlenbeck process. J. Appl. Probability, 12(3):600–604, 1975.
29