Supremum distribution of Bessel process of drifting Brownian motion

0.23 MB·English

by  Andrzej Pyć

SUPREMUM DISTRIBUTION OF BESSEL PROCESS OF DRIFTING
BROWNIAN MOTION
A. PYC´, G. SERAFIN AND T. Z˙AK
Abstract. Let (Bt(1),Bt(2),Bt(3)+µt) be a three-dimensional Brownian motion with drift µ,
startingattheorigin. ThenXt = (Bt(1),Bt(2),Bt(3)+µt) ,itsdistancefromthestartingpoint,
k k
isadiffusionwithmanyapplications. Weinvestigatethedistributionofthesupremumof(Xt),
give an infinite-series formula for its density and an exact estimate by elementary functions.
5
1
0
2 1. introduction

n
a In his famous paper ([9], see also [8], page 436) David Williams showed how one can decom-
J pose the paths of a transient one-dimensional diffusion at its maximum (or minimum). One of

3
the best-known examples of such decomposition is that of B(t)+µt, Brownian motion with a
1
positive drift µ, as a Brownian motion with a negative drift B(t) µt and a diffusion with a
−
] generator
R
P 1 d2 d
h. (1.1) ∆µ = 2dx2 +µcoth(µx)dx.
t
a Namely, forµ > 0letB(t) µtbeaBrownianmotionwithconstant drift µ, Z adiffusionwith
m − − t
generator (1.1), started at zero, and let γ be a random variable with exponential distribution
[
with parameter 1 . Assume that B , Z and γ, defined on the same probability space, are
2µ t t
1
independent. Define τ = inf t > 0 : B µt = γ . Then the following process
v { t − − }
0
B µt, 0 t τ,
0 X = t − ≤ ≤
2 t Z γ, t > τ,
t τ
3 (cid:26) − −
0 is B +µt, a Brownian motion with positive drift µ. Williams also showed that Z can be viewed
t t
.
1 as a Brownian motion with drift conditioned to hit infinity before zero.
0
5 Later on a process with generator (1.1), known as a hyperbolic Bessel proces ([7], p. 357) or
1 as a Bessel process of drifting Brownian motion and denoted BES(3,µ) ([6], [3]), appeared also
:
v in many papers as a drifting Brownian motion conditioned to not hit zero. Recently, it was
i
X proved in [1] that it can be obtained as a deterministic involution of Brownian motion with drift
r µ. If µ = 1 then (Z ) is a radial part of a hyperbolic Brownian motion in three-dimensional
a t
hyperbolic space.
The transition density function of (Z ) is well-known, compare [6] or [3] but to our best
t
knowledge, distribiution of different functionals of this process have not been investigated yet.
In this paper we investigate process (Z ) killed on exiting interval (0, r ) and give a formula
t 0
describing distribution of M = sup Z , the supremum of the process (Z ). Because the
t s t s t
formula is given as an infinite series,≤we give its exact estimate using elementary functions.
Moreover, our method of estimation applied to a function ss (v,t) used in a handbook by
y
Borodin and Salminen [2] give very precise estimate of this function (cf. Remark after Theorem
9).
2010 Mathematics Subject Classification. 60J60.
Key words and phrases. Drifting Brownian motion, Bessel process, supremum distribution.
1
2 A. PYC´, G. SERAFINAND T. Z˙AK
2. Transition density of a Bessel process of drifting Brownian motion,
killed on exiting interval (0, r )
0
Let Z(t) be a diffusion on (0, ) generated by operator (1.1) with µ > 0. The speed measure
∞
of this diffusion is equal to m(dy) = sinh2(µy)dy, hence with respect to m(dy), the transition
density of Z(t) has the following form (cf. [6]):
e µ2t/2 e (y x)2/(2t) e (y+x)2/(2t)
− − − −
−
(2.2) p(t;x,y) = .
√(cid:16)2πtsinh(µx)sinh(µy) (cid:17)
By (Zr0) we will denote the process killed on exiting (0, r ). For positive µ, the process starting
t 0
from x > 0 cannot reach zero and drifts to infinity so that almost all trajectories will be killed
at r . Even if (Z ) starts from zero, with probability one it will never visit zero again.
0 t
Transition density pr0(t;x,y) of (Zr0), with respect to m(dy), is a solution of the following
t
Dirichlet problem:
∂ pr0(t;x,y) = ∆ pr0(t;x,y) t > 0, x (0,r ), y (0,r ),
∂t µ ∈ 0 ∈ 0
(2.3) pr0(t;x,r0) = 0 t > 0, x (0,r0),
∈
lim pr0(t;x,y) = δ (y) x (0,r ), y (0,r ).
t 0 x 0 0
→ ∈ ∈
Because of killing, pr0(t;x,r ) = 0 for all t > 0. Moreover, if µ > 0 then by (2.2) we
 0
have limsup p(t;x,y) < for all t, x > 0 and, because pr0(t;x,y) p(t;x,y), hence
limsup pry0→(t0;x,y) < . ∞ ≤
We wy→ill0use the separa∞tion variable technique used in mathematical physics. Suppose that
pr0(t;x,y) = Y(y)T(t).
Then, by (1.1), the first equation of the system (2.3) takes on the form
1
(2.4) Y(y)T (t) = Y (y)+µcoth(µy)Y (y) T(t).
′ ′′ ′
2
(cid:18) (cid:19)
We add 1(λ2 +µ2)T(t)Y(y) to both sides of (2.4) and get
2
1 1 1
Y(y)[T (t)+ (λ2 +µ2)T(t)] = T(t)[ Y (y)+µcoth(µy)Y (y)+ (λ2 +µ2)Y(y)].
′ ′′ ′
2 2 2
The solution of the differential equation T (t)+ 1(λ2 +µ2)T(t) = 0 is T(t) = c e−(λ2+µ2)t. If in
′ 2 1 2
the second equation
Y (y)+2µcoth(µy)Y (y)+(λ2 +µ2)Y(y) = 0
′′ ′
we substitute Y(y) = u(y)/sinh(µy), we get
1
(u (y)+λ2u(y)) = 0.
′′
sinhµy
By the above discussion we know that limsup Y(y) < and Y(y) = u(y)/sinh(µy), hence
lim u(y) = 0. This implies that with the by→ou0ndary co∞nditions u(0) = u(r ) = 0 equation
y 0 0
u (y→) + λ2u(y) = 0 is a special case of the general Sturm – Liouville problem [5]. It is well
′′
known (cf. [5], Theorem 4.1 and Exercise 4.2, page 337) that this Sturm – Liouville problem for
u(y) has a solution if and only if λ = nπ, n = 1,2,... and this solution (up to a multiplicative
r0
constant) is given by u (y) = sin(nπy/r ). Thus we may expand pr0(t;x,y) as
n 0
∞ a (x) nπy (n2π2/r2 +µ2)t
(2.5) pr0(t;x,y) = n sin exp − 0 .
sinh(µy) r 2
n=1(cid:20)(cid:18) (cid:18) 0 (cid:19)(cid:19) (cid:18) (cid:19)(cid:21)
X
SUPREMUM DISTRIBUTION OF BESSEL PROCESS OF DRIFTING BROWNIAN MOTION 3
sin(kπy)
To determine coefficient a (x) let usmultiply theequation(2.5) bya (x) r0 andintegrate
n k sinh(µy)
the product over (0,r ) with respect to the measure sinh2(µy)dy:
0
(2.6) r0pr0(t;x,y)a (x)sin(krπ0y) sinh2(µy)dy =
0 k sinh(µy)
= r0 ∞ an(x)aRk(x) sin nπy sin kπy exp −(n2π2/r02+µ2)t sinh2(µy)dy.
0 n=1 sinh2(µy) r0 r0 2
h(cid:16) (cid:16) (cid:17) (cid:16) (cid:17)(cid:17) (cid:16) (cid:17)i
Observe thRat P
r0 sin(nπy) sin(kπy) r
a (x)a (x)sinh2(µy) r0 r0 dr = a2(x) 0δn,
n k sinh(µy)sinh(µy) n 2 k
Z0
where δn is the Kronecker delta. Hence the right-hand side of (2.6) is equal to
k
r µ2t n2π2t
0a2(x)exp − .
2 n 2 − 2r2
(cid:18) 0 (cid:19)
(n2π2/r2+)t
Now, let t 0. We have exp 0 1 and, using the third condition from (2.3) for
→ − 2 →
the left-hand side of (2.6), we(cid:16)get (cid:17)
a (x)sin kπx
k r0 = r0a2(x),
sinh(µ(cid:16)x) (cid:17) 2 k
hence
2sin kπx
a (x) = r0 .
k r sin(cid:16)h(µx(cid:17))
0
To sum up, we have just proved the following theorem.
Theorem 1. Transition density (with respect to the measure sinh2(µy)dy) of the Bessel process
of drifting Brownian motion, starting from the point x (0,r ) and killed at r , is given by the
0 0
∈
following formula
(2.7) pr0(t;x,y) = ∞ 2sin nrπ0x sin nrπ0y exp (n2π2/r02 +µ2)t .
 r sin(cid:16)h(µx(cid:17))sin(cid:16)h(µy(cid:17))  − 2 
n=1 0 (cid:18) (cid:19)
X
  
We will give another representation of the transition density, using the Poisson summation
formula (cf. 13.4 in [10]): for any function g absolutely integrable on ( , )
−∞ ∞
(2.8) ∞ g(n) = ∞ ∞ g(x)e 2πikxdx.
−
nX=−∞ kX=−∞Z−∞
First, observe that series (2.7) can be written as
pr0(t;x,y) = 1 ∞ 2sin nrπ0x sin nrπ0y exp (n2π2/r02 +µ2)t .
2  r sin(cid:16)h(µx(cid:17))sin(cid:16)h(µy(cid:17))  − 2 
n= 0 (cid:18) (cid:19)
X−∞
  
For the Poisson summation formula we need to compute the Fourier transform of any term
of the above series, taking n as a variable. Put z instead of n, then we have to evaluate the
following integral:
zπx zπy −z2π2t
∞ sin sin e 2r02 e−2πikzdz.
r r
Z (cid:18) 0 (cid:19) (cid:18) 0 (cid:19)
−∞
4 A. PYC´, G. SERAFINAND T. Z˙AK
In order to compute it, we use trigonometric identity e 2πikz = cos(2πkz) isin(2πkz) and
−
−
observe that the integral with sin(2πkz) is zero, hence we have to compute only
zπx zπy −z2π2t
∞ sin sin cos(2πkz)e 2r02 dz.
r r
Z (cid:18) 0 (cid:19) (cid:18) 0 (cid:19)
−∞
Now we use identity 2sinaxsinbx = cos((a b)x) cos((a+b)x) and a formula (3.898.2) from
− −
[4] to get:
zπx zπy −z2π2t
∞ sin sin e 2r02 e−2πikzdz =
r r
Z (cid:18) 0 (cid:19) (cid:18) 0 (cid:19)
−∞
= r0 e−(y−x2+t2kr0)2 e−(y+x2−t2kr0)2 +e−(y−x2−t2kr0)2 e−(y+x2+t2kr0)2 .
2√2πt − −
(cid:18) (cid:19)
The Poisson summation formula (2.8) gives us the following:
Theorem 2. Transition density (with respect to the measure sinh2(µy)dy) of the Bessel process
of drifting Brownian motion, starting from the point x (0,r ) and killed at r , is given by the
0 0
∈
following formula
−µ2t
pr0(t;x,y) = e 2 ∞ e−(y−x2+t2kr0)2 e−(y+x2+t2kr0)2 .
√2πtsinh(µx)sinh(µy) −
k= (cid:20) (cid:21)
X−∞
3. Exit time
Let us consider M = sup Z , the supremum of the Bessel process of drifting Brownian
t s t s
motion. The distribution of (≤M ) is closely related to the distribution of the time when the
t
process (Z ) exits the interval (0, r ). Recall that for µ > 0 process (Z ) exits (0, r ) at the
t 0 t 0
point r . For r > 0 let us define τ = inf s : Z > r . Distribution of (M ) and the survival
0 0 r0 { s 0} t
probability of the killed process (Zr0) are related by the following formula:
t
r0
Px(M < r ) = Px(τ > t) = pr0(t;x,y)sinh2(µy)dy.
t 0 r0
Z0
For fixed µ and r series (2.7) is uniformly convergent for x, y [0, r ], because there
0 0
∈
holds inequality sin(nπx/r0)sin(nπy/r0) c n2 and n2exp( n2π2/(2r2)) < . Thus we may
| sinh(µx)sinh(µy) | ≤ µ,r0 − 0 ∞
integrate the series term by term. Integrating n-th term, we get (compare [4], formula 2.671)
P
r0 nπy ( 1)n+1πr nsinh(µr )
0 0
sin sinh(µy)dy = − ,
r (n2π2 +µ2r2)
Z0 (cid:18) 0 (cid:19) 0
so that we may write the following:
Theorem 3. For t, r > 0 and x (0, r ) the following formula holds
0 0
∈
Px(M < r ) = Px(τ > t) =
t 0 r0
∞ ( 1)n+12πnsinh(µr )sin(nπx/r ) (n2π2/r2 +µ2)t
(3.9) − 0 0 exp 0 .
sinh(µx)(n2π2 +µ2r2) − 2
n=1(cid:20)(cid:18) 0 (cid:19) (cid:18) (cid:19)(cid:21)
X
If we differentiate the above series term by term with respect to t we get
πsinh(µr ) µ2t ∞ n2π2
0 exp ( 1)nnsin(nπx/r )exp t .
sinh(µx)r2 − 2 − 0 − 2r2
0 (cid:18) (cid:19)n=1 (cid:18) 0 (cid:19)
X
SUPREMUM DISTRIBUTION OF BESSEL PROCESS OF DRIFTING BROWNIAN MOTION 5
Fixing ε > 0 we can observe that the series of the derivatives is convergent uniformly for
t [ε, ) so that differentiation term by term is justified. Since Px(τ dt) = ∂ Px(τ > t),
∈ ∞ r0 ∈ −∂t r0
the exit time density is given by the following formula.
Theorem 4. For fixed r > 0, 0 < x < r and any t > 0
0 0
πsinh(µr ) µ2t ∞ n2π2
(3.10) Px(τ dt) = 0 exp ( 1)n+1nsin(nπx/r )exp t .
r0 ∈ sinh(µx)r2 − 2 − 0 − 2r2
0 (cid:18) (cid:19)n=1 (cid:18) 0 (cid:19)
X
Inasimilar way aswedidit forthedensity ofthekilled process, using thePoisson summation
formula we can obtain another representation of the exit time density. Note that the series in
formula (3.10) can be written down in the following form:
∞ n2π2 1 ∞ n2π2
( 1)n+1nsin(nπx/r )exp t = ( 1)n+1nsin(nπx/r )exp t .
− 0 − 2r2 2 − 0 − 2r2
n=1 (cid:18) 0 (cid:19) n= (cid:18) 0 (cid:19)
X X−∞
In order to use (2.8), we will compute the Fourier transform of its n-th term, taking z in place
of n. First we take only cosine of exp( 2πikz), next we integrate by parts and finally we use
−
([4], formula 3.896.4) to get
zπy z2π2t
∞
(3.11) zsin exp − exp( 2πikz)dz =
r 2r2 −
Z (cid:18) 0 (cid:19) (cid:18) 0 (cid:19)
−∞
r2 (y +2kr )2 (y 2kr )2
0 (y +2kr )exp − 0 +(y 2kr )exp − − 0 .
0 0
√2(πt)3/2 2t − 2t
(cid:20) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21)
Observe that
nπ(r x)
sin 0 − = ( 1)n+1sin(nπx/r ),
0
r −
(cid:18) 0 (cid:19)
hence putting y = r x in (3.11) and using Poisson formula, we get the second representation
0
−
of the exit time density.
Theorem 5. For fixed r > 0, 0 < x < r and any t > 0
0 0
Px(τ dt) = sinh(µr0)e−µ2t/2 ∞ (r x+2kr )exp −(r0 −x+2kr0)2 .
r0 ∈ sinh(µx)√2πt3/2 0 − 0 2t
k= (cid:18) (cid:19)
X−∞
4. Mean exit time
Now we want to compute Ex(τ ) — the mean exit time of (Z ) from (0, r ). We will use the
r0 t 0
formula
Ex(τ ) = ∞tPx(τ dt),
r0 r0 ∈
Z0
so that we need to compute the integral
sinh(µr0) ∞ e−µ2t/2 ∞ (r0 x+2kr0)2
(r x+2kr )exp − − dt.
0 0
√2πsinh(µx) √t − 2t
Z0 k= (cid:18) (cid:19)
X−∞
Integrating a single term we use formula (3.471(15)) from [4] and get
1 (r +2kr x)2 µ2t √2π
∞ 0 0
exp − − dt = exp( µ r +2kr x )dt,
0 0
√t 2t − 2 µ − | − |
Z0 (cid:18) (cid:19)
6 A. PYC´, G. SERAFINAND T. Z˙AK
which gives
sinh(µr ) ∞
(4.12) Ex(τ ) = 0 [(r x+2kr )exp( µ r x+2kr )] =
r0 µsinh(µx) 0 − 0 − | 0 − 0|
k=
X−∞
sinh(µr ) ∞
0
[(r x+2kr )exp(µ( r +x 2kr ))]+
0 0 0 0
µsinh(µx) − − −
k=1
X
sinh(µr ) ∞
0
[(r x 2kr )exp(µ(r x 2kr ))]+(r x)exp( µ(r x)) .
0 0 0 0 0 0
µsinh(µx) − − − − − − −
" #
k=1
X
But
∞ (r x)
(r x)(eµ(r0 x) +eµ(x r0)) exp( 2µkr ) = 0 − eµ(r0 x) +eµ(x r0)
0 − − − − 0 e2µr0 1 − −
k=1 −
X (cid:0) (cid:1)
and
∞ 2r e2µr0
2r (eµ(x r0) eµ(r0 x)) kexp( 2µkr ) = 0 (eµ(r0 x) eµ(x r0)).
0 − − − − 0 (e2µr0 1)2 − − −
k=1 −
X
If we put them into (4.12), after some algebraic manipulation the formula for Ex(τ ) can be
B
simplified a lot. Namely, we get the following:
Theorem 6. For any fixed r > 0 and any starting point x (0, r )
0 0
∈
1
Ex(τ ) = (r coth(µr ) xcoth(µx)).
r0 µ 0 0 −
5. Estimates
Except for the last one, all the above formulas are given as series so that they are not
convenient for computations or applications. In this section we give exact approximations by
elementary functions of the transition density of the killed process, of the killing time and of
the density of the distribution of the supremum of the process. Notation f g means that
≈
there exist two absolute constants c and c such that for all possible values of variables and
1 2
parameters there holds c f < g < c f.
1 2
For simplicity let us denote γr0(t;x) = Px(τ dt). Recall that by Theorems 4 and 5
r0 ∈
γr0(t,x) =
(5.13) = πsinshin(µhr(µ0)xe)−r02µ2t/2 ∞n=1(−1)n+1nsin(nπx/r0)exp −n22rπ022t
(5.14) = ssiinnhh((µµrx0))√e−2πµt23t//22 P∞n= (r0 −x+2nr0)exp −(r0(cid:16)−x+2t2nr0)2(cid:17)
−∞
(cid:16) (cid:17)
and, by Theorem 2, P
pr0(t;x,y) =
(5.15) = e−µ2t/2 ∞ exp (y−x+2nr0)2 exp (y+x+2nr0)2 .
sinh(µx)sinh(µy)√2πt n= − 2t − − 2t
−∞
h (cid:16) (cid:17) (cid:16) (cid:17)i
P
We will start with estimates in the particular case r = 1. We also make separate calculations
0
for t (0, 1] and for t [1, ).
∈ 4 ∈ 4 ∞
SUPREMUM DISTRIBUTION OF BESSEL PROCESS OF DRIFTING BROWNIAN MOTION 7
Theorem 7. Let r = 1. For 0 < t 1 and 0 < x < 1
0 ≤ 4
γ1(t,x)
0,25 4,02.
≤ √2π1t3/2x(t1+−xx)sisnihn(hµµx)e−µ2t/2−(1−x)2/(2t) ≤
Proof. First we consider the case 0 < x 1 and use (5.14). It is enough to estimate only the
≤ 2
part of γ1(t,x) consisting of a series, multiplied by e(1−x)2/(2t), that is the following quantity
x
I = e(1−x)2/(2t) ∞ (2k +1 x)e (2k+1 x)2/(2t).
− −
x −
k=
X−∞
Observe that we can group terms of the series: k = 0 with k = 1, k = 1 with k = 2 and so
− −
on. In this way we get
(5.16) I = ∞ e 2k(k+1 x)/t(2k +1−x)−(2k +1+x)e−2(2k+1)x/t.
− −
x
k=0
X
Now
(t+x)I = (t+x) ∞ e 2k(k+1 x)/t(2k +1−x)−(2k +1+x)e−2(2k+1)x/t =
− −
x
k=0
X
∞ t
e 2k(k+1 x)/t 2k +1 x t+ (2k +1)(1 e 2(2k+1)x/t) (2k +1+x+t)e 2(2k+1)x/t .
− − − −
− − x − −
k=0 (cid:18) (cid:19)
X
Below we will estimate separately the term for k = 0 and terms for k 1. The term for k = 0 is
≥
equal to g(x,t) = 1 x t+ t(1 e 2x/t) (1+x+t)e 2x/t. We will show that for 0 < x 1/2
− − x − − − − ≤
and 0 < t < 1/4 there holds g(x,t) 2. Indeed, a derivative
≤
∂g(x,t) e 2x/t
= − 2x2(1+x) tx(x(e2x/t 1) 2) t2(e2x/t 1)
∂x tx2 − − − − −
is negative for x > 0 and t >(cid:0)0, because e2x/t 1 > 2x + (2x)2 and this ineq(cid:1)uality implies
− t 2t2
that in the formula for the derivative, the quantity in the bracket is negative. Hence g(x,t)
≤
lim g(x,t) = 2 2t and this is less than 2 for t > 0.
x 0
In→order to estim−ate terms for k 1, we use assumptions 0 < t 1, 0 < x 1 and inequality
≥ ≤ 4 ≤ 2
1 e x < x, valid for x > 0 , and get
−
−
∞ t
(t+x)I 2+ e 2k2/t 2k +1+ (2k +1)(1 e 2(2k+1)x/t)
− −
≤ x − ≤
k=1 (cid:18) (cid:19)
X
∞ 14 ∞
2+ e 8k2(2k +1)(4k+3) 2+21e 8 + 8k2e 8k2
− − −
≤ 8 ≤
k=1 k=2
X X
14 ∞
2+21e 8 + ne n 2,01.
− −
8 ≤
n=32
X
To get an estimate from below, we will use inequality 1 e x x , valid for x > 1. Using
− − ≥ x+1 −
formula (5.16), we get for 0 < t 1 and 0 < x 1:
≤ 4 ≤ 2
I = ∞ e 2k(k+1 x)/t−2x+(2k +1+x)(1−e−2(2k+1)x/t)
− −
x ≥
k=0
X
∞ 2(2k +1)(2k+1 x) 2t ∞ 2k +1 x t
e 2k(k+1 x)/t − − e 2k(k+1 x)/t − −
− − − −
t+2(2k +1)x ≥ t+x ≥
k=0 k=0
X X
8 A. PYC´, G. SERAFINAND T. Z˙AK
1 ∞ 1 1
e 2k(k+1 x)/t(2k + ) ,
− −
t+x 4 ≥ 4(t+x)
k=0
X
so that (t+x)I 1. Both above estimates imply the following
≥ 4
t+x ∞
0,25 e(1 x)2/(2t) (2k +1 x)e (2k+1 x)2/(2t) 2,01.
− − −
≤ x − ≤
k=
X−∞
If, for 0 < x 1, we divide the middle term of the above inequality by (1 x) [1, 1), we
≤ 2 − ∈ 2
must multiply its left and right-hand sides by, respectively, 1 and 2. This proves the theorem
in this case.
Now the proof for the case 1 x < 1 and 0 < t 1. Similarly as in the proof of the first
2 ≤ ≤ 4
case, we will examine the following quantity:
e(1−x)2/(2t) ∞ (2k +1 x)e (2k+1 x)2/(2t) = 1 ∞ (2k +1 x)e (2k2+2k(1 x))/t =
− − − −
1 x − 1 x −
− k= − k=
X−∞ X−∞
1 ∞ ∞ 1
2k e (2k2+2k(1 x))/t e (2k2 2k(1 x))/t + e (2k2+2k(1 x))/t = A+B.
− − − − − − −
1 x − 1 x
− Xk=1 (cid:16) (cid:17) kX=−∞ −
First we estimate the series denoted by B: for 0 < 1 x 1 and 0 < t 1
− ≤ 2 ≤ 4
∞ ∞
B 1 = e (2k2+2k(1 x))/t +e (2k2 2k(1 x))/t e 2k2/t +e (2k2 k)/t
− − − − − − − −
− ≤ ≤
Xk=1(cid:16) (cid:17) Xk=1(cid:16) (cid:17)
∞ 1 1 ∞
e 8k2 +e 4k2 + + 2e k < 0,02,
− − −
≤ e8 e4
Xk=1(cid:16) (cid:17) kX=16
hence 1 < B < 1,02.
Nowwehavetoestimate 11xA. ObservethatA = ∞k=12k e−(2k2+2k(1−x))/t −e−(2k2−2k(1−x))/t
−
is negative for 1 < x < 1 because all terms of the series ar(cid:16)e negative. We will estimate the (cid:17)
2 P
following positive quantity
A ∞ e−(2k2−2k(1−x))/t e−(2k2+2k(1−x))/t
− = 2k − =
1 x 1 x
!
− k=1 −
X
∞ 2ke 2k2/t e2k(1−x)/t −e−2k(1−x)/t = 4 ∞ 2k2e 2k2/tsinh(2k(1−x)/t).
− −
1 x t 2k(1 x)/t
k=1 (cid:18) − (cid:19) k=1 −
X X
For fixed t > 0 and k = 1,2,3,... function g(x) = sinh(2k(1−x)/t) is decreasing for 1 x < 1
2k(1 x)/t 2 ≤
−
hence its maximal value is attained for x = 1 and is equal to sinh(k/t) = ek/t e−k/t. Thus
−
2 k/t 2k/t
0 < −A = 4 ∞ 2k2e 2k2/tek/t −e−k/t 8 ∞ ke (2k2 k)/t 8 ∞ ke k2/t
− − − −
1 x t 2k/t ≤ ≤ ≤
− k=1 k=1 k=1
X X X
∞ 4 ∞
2 4k2e 4k2 2 + ne n 0,15.
− −
≤ e4 ≤
!
k=1 n=16
X X
Finally, using the estimates 0,15 < A < 0 and 1 < B < 1,02, we get the desired result: for
− 1 x
0 < t 1 and 1 x < 1 −
≤ 2 2 ≤
1
0,85 < A+B < 1,02,
1 x
−
SUPREMUM DISTRIBUTION OF BESSEL PROCESS OF DRIFTING BROWNIAN MOTION 9
which, for 1 x < 1, implies the following
2 ≤
γ1(t,x)
0,85 1,02.
≤ 1 (1 x) sinhµ e µ2t/2 (1 x)2/(2t) ≤
√2πt3/2 − sinh(µx) − − −
If we want to write factor x(1−x) instead of (1 x) in the denominator, we need to multiply the
t+x −
right-hand side constant by 3/2, because for 0 < t 1 and 1 x < 1 we have 2 x < 1.
≤ 4 2 ≤ 3 ≤ t+x
This gives the estimate for t 1 with constants 0,85 and 1,53. Finally, taking into account
≥ 4
estimates for 0 < x 1 and for 1 x < 1 we get the desired estimate, which ends the proof
≤ 2 2 ≤ (cid:3)
of the theorem.
Now the estimate for t 1.
≥ 4
Theorem 8. For t 1 and 0 < x < 1
≥ 4
γ1(t,x)
0,8 1,2.
≤ πsin(πx) sinh(µ) e (µ2+π2)t/2 ≤
sinh(µx) −
Proof. For t 1 we use formula (5.13) and the following inequality: for 0 < x < 1 there holds
≥ 4
|sin(kπx)| ≤ ksin(πx). The first term of the series ∞k=1(−1)k+1ksin(kπx)e−k2π2t/2 is much
larger than the sum of the absolute values of all the rest:
P
∞ ∞
( 1)k+1ksin(kπx)e k2π2t/2 sin(πx)e π2t/2 k2e (k2 1)π2t/2
− − − −
| − | ≤ ≤
k=2 k=2
X X
sin(πx)e π2t/2 ∞ ne (n 1)π2/8 e−π2/4(4eπ2/8 −3) sin(πx)e π2t/2 0,2sin(πx)e π2t/2.
− − − − −
≤ (eπ2/8 1)2 ≤
n=4 −
X
This means that
∞
0,8sin(πx)e π2t/2 ( 1)k+1ksin(kπx)e k2π2t/2 1,2sin(πx)e π2t/2,
− − −
≤ − ≤
k=1
X
hence for t 1/4 and 0 < x < 1
≥
sinhµ sinhµ
0,8πsin(πx) e µ2t/2 π2t/2 γ1(t,x) 1,2πsin(πx) e µ2t/2 π2t/2,
− − − −
sinh(µx) ≤ ≤ sinh(µx)
(cid:3)
which ends the proof.
Now, because for all x (0,1) there holds π2 < πsin(πx) < 4π, the above inequality implies
∈ x(1 x)
the following: −
γ1(t,x)
(5.17) 0,8π2 4,8π.
≤ x(1 x) sinhµ e µ2t/2 π2t/2 ≤
− sinh(µx) − −
Observe, that the above inequality differs from that given in Theorem 7 for the case 0 < x 1:
≤ 2
it does not contain factor 1/(√2πt3/2) in the denominator and instead of a factor e (1 x)2/(2t)
− −
it has a factor e π2t/2. If we want to have one estimate for all t > 0 and 0 < x < 1, we must
−
add such factors. Multiplying the estimating function from Theorem 7 by e π2t/2 (e π2/8, 1),
− −
∈
we must multiply the constant on the left-hand side by e π2/8 0,29. This operation changes
−
≈
constant 1 from Theorem 7 to e−π2/8 0,0728 > 0,07.
4 4 ≈
Ontheotherhand, theestimatingfunctionfromthedenominatorin(5.17)mustbemultiplied
by
(1+t)5/2
e (1 x)2/(2t).
− −
√2π(t+x)t3/2
10 A. PYC´, G. SERAFINAND T. Z˙AK
But for t 1 and 0 < x < 1 the above function is greater than 1/√2π 0,3989... > 0,39
≥ 4 ≈
and less than 25√5e−1/8 4,92... < 5 hence constant 4,8π from (5.17) must be changed to
4√2π ≈
5 4,8π = 24π and constant 0,8π2 from (5.17) must be changed to 0,8π2/√2π > 3,1499 > π.
·
In this way theorems 7 and 8 together imply the following:
Corollary 1. For all t > 0 and 0 < x < 1 the following inequality holds
γ1(t,x)
0,07 24π < 75,4.
≤ x(1−x) sinh(µ) (1+t)5/2 exp µ2+π2t (1−x)2 ≤
t+x sinh(µx)√2πt3/2 − 2 − 2t
(cid:16) (cid:17)
It is easy to notice that by (5.14) the following scaling property holds:
sinh(µx/r ) µ 1 t x
γr0(t,x) = 0 exp t 1 γ1 , .
sinh(µ)sinh(µx)r2 −2 − r2 r2 r
0 (cid:18) (cid:18) 0(cid:19)(cid:19) (cid:18) 0 0(cid:19)
This, together with the above Corollary proves the following.
Theorem 9. For µ, r , t > 0 and 0 < x < r we have
0 0
γr0(t,x)
(5.18) 0,07 75,4.
≤ x(r0−x)sinh(µr0) (r02+t)5/2 exp (r0µ)2+π2t (r0−x)2 ≤
t+r0x sinh(µx) √2πr04t3/2 − 2r02 − 2t
(cid:16) (cid:17)
Remark. Many formulasinthe book [2] are givenin the languageofa specialfunction ss (v,t) =
y
√2π1y3/2 ∞k= (t−v+2kt)e−(t−v+2kt)2/(2y), v ≤ t (see [2], page 641). Observe that sst(x,r0) is
−∞
precisely the series in our γr0(t,x) and was estimated above. Using our method it is possible
P
to give an estimate of ss (x,r ) for all possible values of variables (like we did it in Theorem 9
t 0
for γr0(t,x)) but estimates for different sets of t and x, given in Theorems 7 and 8, are much
more exact. For instance the proof of Theorem 7 gives the following: for r = 1, 0 < t 1 and
0 ≤ 4
0 < x 1 there holds
≤ 2
ss (x,1)
t
0,25 2,01.
≤ x e (1 x)2/(2t) ≤
√2πt3(t+x) − −
Now we estimate the density of the transition probability of the killed process.
Theorem 10. For fixed r > 0, all x,y (0, r ) and t > 0 there holds
0 0
∈
(r2 +t)5/2 xy (r x)(r y)
pr0(t;x,y) 0 1 1 0 − 0 −
≈ r5sinh(µx)sinh(µy)√t ∧ t ∧ t
0 (cid:18) (cid:19)
(cid:16) (cid:17)
(r µ)2 +π2 (x y)2
0
exp t − .
× − 2r2 − 2t
(cid:18) 0 (cid:19)
The constants in the above estimate can be taken c = 0,0029 and c = 2413.
1 2
Proof. Observe that
sinh(µx/r )sinh(µy/r ) µ2t 1 t x y
(5.19) pr0(t;x,y) = 0 0 exp 1 p1 ; , ,
sinh(µx)sinh(µy)r − 2 − r2 r2 r r
0 (cid:18) (cid:18) 0(cid:19)(cid:19) (cid:18) 0 0 0(cid:19)
thus it is sufficient to consider only the case when r = 1. Define the following function
0
1 ∞ (w +2k)2
λ(w) = (w +2k)exp , w R.
√2πt3/2 − 2t ∈
k= (cid:18) (cid:19)
X−∞