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THEORY PROBAB. APPL. c 2004 Society for Industrial and Applied MathematicsVol. 48, No. 2, pp. 288303 Translated from Russian Journal

MARTINGALES AND FIRST-PASSAGE TIMES FOR

ORNSTEINUHLENBECK PROCESSES WITH

A JUMP COMPONENT

A. NOVIKOV

(Translated by the author)

Dedicated to the Centennial of A. N. Kolmogorov

Abstract. Using martingale technique, we show that a distribution of the first-passage timeover a level for the OrnsteinUhlenbeck process with jumps is exponentially bounded. In the caseof absence of positive jumps, the Laplace transform for this passage time is found. Further, themaximal inequalities are also given when the marginal distribution is stable.

Key words. exponential martingales, first-passage times, OrnsteinUhlenbeck process, Laplacetransform, moment Walds identity, maximal inequalities, stable distribution

DOI. 10.1137/S0040585X97980403

1. Introduction. Let Xt, t 0, solve a linear stochastic equation

Xt = X0

t0

Xs ds + Yt, t 0,(1)

where {Yt, t 0} is a Levy process (that is, a process with independent homogeneousincrements; see, e.g., [1], [2]) and X0 is a nonrandom initial value.

We consider here only the stable case, that is, when > 0. In the literature,this process is cited as an important example of a different class of random processes:shot noise processes (see, e.g., [3]), filtered Poisson processes [4], generalized OrnsteinUhlenbeck (OU) processes (see [5], [6], and [7]), etc.

One of the problems for models of that sort is to determine a distribution, ormoments for the one-sided passage time

b = inf{t > 0 : Xt > b}, b > X 0.

Different approaches were used for studying this problem: integral equations (see,e.g., [1], [10], and [21]); martingale techniques ([7], [8], and [9]), etc. In this paper, wealso apply the martingale technique, namely a special parametric family of martingales(Theorem 1). We find the Laplace transform of b provided that the Levy process Ytpossesses negative jumps only (Theorem 2; under somewhat less general conditionsthis result is known from [7] and [8]; see also Remark 3).

When the process Yt has positive jumps, the Laplace transform of b, as well asits moments, are unknown, except for exponential distribution of positive jumps [10],or uniform distribution [22] of ones.

Received by the editors January 23, 2003. This work was supported by ARC Large GrantA0010474.

http://www.siam.org/journals/tvp/48-2/98040.htmlSteklov Mathematical Institute RAN, Gubkin St. 8, 119991 Moscow, Russia, and Department

of Mathematical Sciences, University of Technology, P.O. Box 123, Broadway, Sydney, NSW 2007,Australia ([email protected]).

288

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MARTINGALES AND FIRST-PASSAGE TIMES 289

In this paper, we prove that the distribution of b is exponentially bounded forall b under the assumption that the process Yt has diffusion component or positive

jumps (Theorem 3). This result is known from [9] under the additional assumptionof finiteness of the mathematical expectation of Yt. Moreover, with the help of the

moment Wald identity (section 4), we derive a lower bound for the expectation E(b).Notice that the moment Wald identity can also be used for deriving asymptotic ex-pansions of Eb as b (see [22]). In section 5, we use this moment identity forderiving the moment inequalities for supt(Xt) even for an arbitrary stopping time provided that Yt obeys one-sided stable distribution, i.e., in the absence of positive

jumps (Theorem 5). The proof of this result uses techniques from [11], where themoment inequalities for supt |Xt| are given for the Gaussian OU-process.

2. An exponential martingale family. We assume that the Levy process Ytand all other random objects are defined on a probability space ( , F, P) supplied bya filtration (including the assumption of right-continuity, etc.).

Recall that any Levy process

Yt = mt + Wt + Zt,(2)

where m and are constants, Wt is a standard Brownian motion, and Zt is a dis-continuous process with independent homogeneous increments and paths from theSkorokhod space.

A (unique) solution of (1) has the following representation in terms of stochasticintegrals with respect to Wt and Zt:

Xt = X0et + et

t0

es dYs

=m

+

X0

m

et + et

t0

es dWs + et

t0

es dZs.(3)

It is well known that the jump component Zt of the Levy process can be represented

in terms of integrals with respect to a Poisson random measure p(dx, ds) (generatedby jumps of Yt), and a Levy canonical measure of jumps (dx):

Zt =

t0

xI{|x| < 1}

p(dx, ds) (dx) ds

+

t0

xI

|x| 1

p(dx, ds).(4)

Here I{ } is an indicator function and the Levy measure (dx) must satisfy thefollowing condition:

min(x2, 1)(dx) < .(5)

In what follows we define a class of martingales as a parametric family of Xtfor the special case when the process Yt obeys all exponential moments:

EeuYt = exp

t(u)

< for all u 0,(6)

where, as is well known [1], the cumulant function (u) has the following represen-tation:

(u) = m u +2

2u2 +

eux ux I{|x| < 1} 1

(dx).

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Additionally, we suppose that

E log(1 + Y1 ) < (7)

(henceforth, a+ = max(a, 0), a = (a)+). We shall see that this condition is

sufficient and necessary for finiteness of the following function (u), which will beused in a definition of martingales below:

(u) =1

u0

v1(v) dv =1

m u +

2

4u2 + I1(u) + I2(u)

, u 0,(8)

where

I1(u) =

u0

v1

(evx vxI{|x| < 1} 1)I{x > 1} (dx)

dv,

I2(u) =

u0

v1

(evx 1)I{x 1} (dx)

dv.

By conditions (5) and (6) the integral I1(u) is well defined and finite. It is convenient

to express the integrand in I2(u) as follows:u0

v1(evx 1) dv =

xu0

y1(ey 1) dy

=

10

y1(1 ey) dy

xu1

y1(1 ey) dy

= EulerGamma log(xu)

xu

y1ey dy,

where EulerGamma is the Euler constant (in notation of the package Mathemat-ica [23]):

EulerGamma= 1

0 y

1

(1 e

y

) dy

1 y

1

e

y

dy

(see also [12, formula 8.367.12]). Therefore,

I2(u) = D

log(u) +

xu

y1ey dy

I{x 1} (dx),(9)

where D =

[EulerGamma+ log(x)

I{x 1} (dx). Obviously, (7) providesthe existence of integral

log(x)I{x 1} (dx) so that, by (5), I2(u) is well

defined and finite if and only if (7) holds true.For definition of the above-mentioned parametric martingale family we introduce

also the following martingale function

H(, x) =

0

eux(u)u1 du, > 0.

The following simple estimate for an asymptotic limit of the function (u) willallow us to find simple conditions for finiteness of the function H(, x).

Lemma 1. Let conditions (6) and (7) hold. If

> 0, or ((0, )) > 0, or

|x| I{1 < x < 0} (dx) = ,(10)

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then

limu

(u)

u= .(11)

If

= 0, ((0, )) = 0,

|x|I{1 < x < 0} (dx) < ,(12)

then

limu

(u)

u=

1

m +

|x|I{1 < x < 0} (dx)

.(13)

Proof. From (9) it follows that

I2(u) = log(u)

(, 1]

+ O(1), u .(14)

By the inequality ez zI{|z| < 1} 1 z2I{z > 0}/2, we find that

I1(u) u2

4

x2I{x > 0} (dx),

and, hence,

(u) mu +u2

4

2 +

x2I{x > 0} (dx)

+ O(log(u)).

So, if > 0 or ((0, )) > 0, then by this lower bound we obtain (11).Now assume (12). Then

(u) = mu + u

0

ev x v x 1

v dv I{1 < x < 0} (dx) + I2(u).By virtue of (14), the integral I2(u) is of the order O(log(u)). Note that for x < 0and v > 0 the following inequalities hold:

0 evx vx 1

v x.

Taking into account the assumption

|x|I{1 < x < 0} (dx) < , the dominatedconvergence theorem, and the lHospital rule, we find

limu

1

u

u0

ev x v x 1

vdv I{1 < x < 0} (dx) =

|x|I{1 < x < 0} (dx).

Thus, (13) holds.If

|x|I{1 < x < 0} (dx) = , the same arguments lead to the followingestimate with any > 0:

limu

(u)

u

1

m +

|x|I{1 < x < } (dx)

.

Letting here 0, we obtain (11). Lemma 1 is proved.

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Notice that by Lemma 1 the function H(, x) is finite for any real x if condi-tion (10) holds or if

= 0, ((0, )) = 0, m + |x|I{1 < x < 0} (dx) > x.(15)Remark 1. If

= 0, ((0, )) = 0, m +

|x|I{1 < x < 0} (dx) X0,(16)

then for b > X0 the stopping time b = .Indeed, the condition ((0, )) = 0 implies that the process Xt does not have

any positive jumps at all. So, since the continuous part of Xt is not smaller than Xtitself, by (3), (4), and (16) we have the following deterministic upper bound:

Xt m

+

X0

m

et + et

t0

es

|x| I{1 < x < 0} (dx)

ds

=1

m + |x|I{1 < x < 0} (dx)+

X0

m

1

|x|I{1 < x < 0} (dx)

e t X0.

Thus, under b > X0 we have supt>0 Xt < b and so b = .Theorem 1. Let conditions (6) and (7) hold. Further, assume (10) or (15) with

x = X0 hold. Then etH(, Xt), t 0

, > 0,

is the martingale.

Proof. Using standard tools of stochastic analysis (see, e.g., [1] or [13]) we obtainthat, under conditions (6) and (7), the process

t(u) = exp

uetXt

t0

(ues) ds

(17)

is a martingale. The fact that this process is a local martingale can be checked usingrepresentation (3) and the generalized Ito formula. The uniform integrability (underassumption (6)), and, hence, the validity of the martingale property, is a consequenceof the exponential identity

E exp

t0

f(s) dYs qt

= 1,

where

qt = mt0

f(s) ds + 22t0

f2(s) ds + t0

ef(s)x f(s) xI{|x| < 1} 1(dx) dsand f(s) is a bounded deterministic function.

Note that t0

(ues) ds =1

uetu

(v)

vdv = (uet) (u).

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Since E(t(u)) = t(0) = exp{uX0}, from the latter formula it follows that

E exp{uXt} = exp

uX0et + (u) (uet)

.(18)

Applying the Fubini theorem and then introducing a new variable z = uet

, weobtain

E(H(, Xt)) =

0

E(euXt(u)) u1 du

=

0

euX0et(uet) u1 du = etH(, X0) < .(19)

The finiteness of the function H(, X0) is due to Lemma 1 and condition (10) or (15)with x = X0.

By (17), for all s t we have

E

t(u) | Fs

= s(u) a.s.

Now, integrating both sides of the above equality with respect to Q(du) =e(u)u1 du ( > 0), with > 0, over the interval (0, ) we obtain

0

E(t(u) | Fs) e(u)u1 du =

0

s(u) e(u)u1 du

= es0

euesXs(ue

s)(ues)1 d(ues) = esH(, Xs).(20)

Hence, by the Fubini theorem, applied to the left-hand side of (20), we obtain therequired martingale property

E

etH(, Xt) | Fs

= esH(, Xs) a.s.

Theorem 1 is proved.Remark 2. The idea of constructing a special parametric martingale family is

not new. A similar method was used in the papers [14] and [15] for boundary crossingproblems related to a Brownian motion, and in the papers [7] and [16] for boundarycrossing problems related to a stable Levy process.

By the optional stopping theorem and Theorem 1 we have the following identity:For any stopping time and fixed t <

E

emin(,t) H(, Xmin(,t))

= H(, X0), > 0,(21)

we apply (21) to derive an explicit formula for the Laplace transform of b providedthat the process Yt does not have positive jumps. Before proving this formula we haveto introduce further notation. Set

H(, x) = 0

eux(u)u1 du, > 0,with

(u) = m u + 2u24

+ I1(u)

log(u) +

xu

ey

ydy

I{x 1} (dx).

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If (7) holds, then, by (9) and (8),

H(, x) = exp{D1} H(, x),

where the constant D is defined above. Notice also that H(, x) is finite even if (7)fails.Theorem 2. Let ((0, )) = 0. If

> 0 or m +

|x| I{1 < x < 0} (dx) > b,(22)

then P{b < } = 1 and

Eeb =H(, X0)H(, b) , > 0.

Proof. First we assume that (7) holds and consider identity (21) with = b. ByLemma 1 and (22), |H(, x)| < , x b. Under the absence of positive jumps, onthe set {b < t} we have Xb = b and, on the set {b t}, we have Xt b for allt 0. Hence, H(, Xmin(b,t)) H(, b) < and, by the Fatou lemma, we can passto the limit as t under the expectation in the left-hand side of (21). As theresult, we get

E [I{b < }eb ] =

H(, X0)

H(, b), > 0.(23)

It is easy to verify, integrating by parts, that

lim0

H(, x) = 1.(24)

Passing to the limit as 0 in (23), by the Fatou lemma we get P{b < } = 1.Since

H(, X0)H(, b)

= H(, X0)H(, b) , > 0,then, by (23) under (7), the statement of Theorem 1 is valid under the imposedassumption (7). If (7) fails, we may consider the OU-process XNt , which solves (1)with Yt replaced by the truncated Levy process

YNt = mt + Wt + ZNt ,

where

ZNt =

t0

xI{1 < x < 0}

p(dx, ds) (dx) ds

+

t

0 xI{N x 1}p(dx,ds).Denote by Nb the corresponding crossing time of the level b and by

HN(, x) thecorresponding martingale function. Obviously, Nb b a.s. as N . Noticethat (7) holds for YNt and we have

E eNb =

HN(, X0)HN(, b) , > 0,

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where the function HN(, x) is defined above. Now it is easy to check thatlim

N

HN(, x) =

H(, x) for any x b.

Theorem 2 is proved.Remark 3. In the case when (dx) 0 and > 0 the process Xt is Gaussian

and, of course, the result of Theorem 2 for this case is well known (see, e.g., [17]).Note also that for this special case it is possible to derive an analytical inversionof the Laplace transform of b based on the representation for the function H(, x)in terms of the parabolic cylinder function D(x), which is well studied (see [12,formula 9.241.2]).

In the case when (, 0) > 0 and > 0, Theorem 2 is proved by Hadjiev (see[7, p. 85, Theorem 2]), where a slightly different parametric martingale family is used.The case when (, 0) > 0 and = 0 is also discussed in [7] under an additionalcondition (see Hypothesis G in [7]).

3. Exponential boundedness of b. In this section, we give sufficient condi-tions, weaker than in [9], for the exponential moments of b to be finite. The existenceof exponential moments for the first-passage times from interval (a, b):

a,b = inf{t > 0 : Xt > b or Xt < a}, b > X 0 > a,

is established too.Theorem 3. Let condition (10) or (15) with x = b hold. Assume also that

E(Y1 ) < for some (0, 1).(25)

Then there exists > 0 such that

E eb < .

Proof. Assuming (6), we shall use an analytical continuation of the martingalefamily etH(, Xt), > 0, to (, 0) with involved in (25).

For (, 0), set

H(, x) =

0

(eux(u) 1) u1 du for (, 0).(26)

By (10) and Lemma 1, |H(, x)| < if and only if10

|(u)| u1du < .

Owing to (5) and (8), for > 1 we have

10

|(u)| u1du C+ C10

|I2(u)| u1du

(hereafter C is a positive generic constant).The inequality

1 ez Cz, z > 0, (0, 1],(27)

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provides the following bound:10

|I2(u)| u1 du =

10

u0

(1 evx)

vdv I{x 1} (dx) u1 du

C

|x|I{x 1} (dx)1

0

u1+ du

in which the latter integral in the right-hand side is finite for any (, 0),whereas (25) is equivalent to

|x|I{x 1} (dx) < .

The function H(, x), (, 0), defined in (26), can be considered as an an-alytical continuation of H(, x), > 0. From (26) (see also (24)) it easily followsthat

lim0

H(, x) = 1.(28)

Just repeating the proof of Theorem 1 with H(, x), (, 0), we may prove thatthe process

et0

(euXt euX0) u1e(u) du + H(, X0) et

is a martingale. Then, by the optional stopping theorem for martingales for anyt < , we have

Eemin(b,t)

0

(euXmin(b,t) euX0) u1e(u) du + H(, X0)

= H(, X0).

(29)

First consider the case ((0, )) = 0. Then condition (6) holds obviously. SinceXmin(b,t) b and < 0, we have

Eemin(b,t)

0

(eub euX0) u1e(u) du + H(, X0) H(, X0).(30)

Further, whereas

0 0.8.

On account of (28), there is 1 (0, 0) such that 0.9 < 1H(1, X0) < 1.1 So, (30)provides

Ee1min(b,t) 1H(1, X0)

10

(eub euX0) u11e(u) du + 1H(1, X0) 0, then, choosing a positive constant A with ((0, A]) > 0, we in-troduce the OU-process XAt , generated by the Levy process Y

At with positive jumps

truncated by A, and the level crossing time Ab and notice that Ab b. Ap-

plying identity (29) with b replaced by Ab and properly defined functions A(u)

and HA(, X0), and taking into account XAmin(A

b,t) b + A and < 0, we get the

bound

Ee min(Ab ,t)

0

(eu(A+b) euX0) u1eA(u) du + HA(, X0)

HA(, X0).

The last part of the proof is similar to that for the case ((0, )) = 0. Theorem 3 isproved.

Corollary 1. Let > 0 or ((, )) > 0. Assume

E|Y1| < for some > 0.

Then there exists > 0 such that

Eea,b < .

Proof. Denote a = inf{t > 0 : Xt < a} and note that a,b = min(b, a). ByTheorem 3, applied to b and a, the desired result holds.

4. The moment Wald identity. The theorem below generalizes Theorem 2of [8].

Theorem 4. Denote T = inf{t 0 : Xt f(t)}, X0 < f(0), where f(t) is acontinuous deterministic function such that supt0 f(t) = M < . Let conditions (6)

and (25) hold. Further, assume condition (10) or (15) with x = M holds.Then

ET = E

0

(euXT euX0) u1e(u) du < .(31)

Proof. First we shall show that under (10) or (15) with x = M the process0

(euXt euX0) u1e(u) du t, t 0

(32)

is a martingale. Indeed, since {etH(, Xt), t 0} is the martingale for > 0, wehave

etE

H(, Xt) H(, X0)

| Fs

+ (et 1) H(, X0)

= es

H(, Xs) H(, X0)

+ (es 1) H(, X0) a.s.(33)

Under the conditions of Theorem 4

lim0

H(, z) H(, X0)

=0

(euz euX0) u1e(u) du

for any z, if (10) holds, or for any z < M, if (15) holds with x = M. Further, dueto (24), we have

lim0

(et 1) H(, X0) = t.

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Applying now the dominated convergence theorem we can interchange the symbolsof the limit as 0 and the conditional expectation in the left side of (33). Thus,passing to the limit as 0 in both parts of (33) we obtain the martingale prop-erty (32).

By the optional stopping theorem for martingales we find that

E min(, t) = E

0

(euXmin(,t) euX0) u1e(u) du(34)

is valid for any stopping time and fixed t < .To complete the proof, it remains only to verify that for computation of lim t

in the right-hand side of (34) with = T can be reduced to computation under theexpectation symbol. We verify that as follows. By Theorem 3,

ET < .(35)

Further, since (34), with = T, is equivalent to

E min(T, t) = EI{T t}0

(euXT euX0) u1e(u) du + et,

where et = E I{T > t}0 (e

uXt euX0) u1e(u) du. Assume for a moment that

limt

et = 0.(36)

Then, by the dominated convergence theorem, (31) holds true.For a verification of (36), we notice that Xt M on the set {T > t}. Therefore,

applying (27), we find that

|et| = EI{T > t, Xt X0}

0

(euXt euX0) u1e(u) du

EI{T > t, Xt < X0}0

euX0(1 eu(XtX0)) u1e(u) du

P{T > t, Xt X0}

0

(euM euX0) u1e(u) du

+ E

I{T > t, Xt < X0} C

0

euX0u1+e(u) du

(Xt X0)

C1P{T > t} + C2 E

I{T > t}|Xt X0|

,(37)

where

C1 =

0

(euM euX0) u1e(u) du, C 2 = C

0

euX0u1+e(u) du.

Set

Yt = Wt + t0

xI{|x| < 1}

p(dx, ds) (dx) ds

,

Yt = t0

xI{|x| 1}p(dx,ds).

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Since by (3) and (4),

Xt X0 = (1m X0)(1 e

t) + ett0

es d

Ys + e

t

t0

es d

Ys.

with the help of inequality |a + b + c| |a| + |b| + |c|, (0, 1), we find that

|Xt X0| |1m X0|

+ |Xt| + t0

|x|I{|x| 1}p(dx,ds),(38)

where Xt is an OU-process with X0 = 0, generated by the square-integrable martin-gale Yt.

Below, we shall show that for any stopping time and 2

E supt

|Xt| C,E/2.(39)By the property of the stochastic integral (see, e.g., [1] or [13]), for any stopping

time we have

E

0

|x|I{|x| 1}p(dx, ds) =

|x|I{|x| 1} (dx) E,

where, by (6) and (25), |x|I{|x| 1} (dx) < .(40)

Combining (35), (37), (38), (40), and (39) (inequality (39) will be proved in whatfollows), we conclude that E suptT |Xt x| < and in turn (36) holds.

To verify (39), we apply the Ito formula to

X2t = e

2tM2t , with

Mt = t0

es dYs,and find that

X2t = t0

e2s(2Ms dMs) +

t0

e2s d

[Ms, Ms] Ms, Ms

+

t0

M2s de2s + t

2 +

x2I{|x| < 1} (dx)

.

Here, the first and second integral terms are martingales, and the third one is negative.For any bounded stopping time , these facts provide

E X2 E()2 + x2I{|x| < 1} (dx).So, for 2, (39) is provided by Lenglarts domination principle (see, e.g., [18,p. 156]). Theorem 4 is proved.

Remark 4. Since Xb b, Theorem 4 provides

Eb

0

(eub euX0) u1e(u) du.

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Note that Theorems 1 and 4 involve condition (6), which does not hold for ex-ponentially distributed, as well as others of such type, positive jumps. However, thetruncation technique implementation (for large positive jumps) allows obtaining alower bound in this case as well. So, instead of (6) we assume

K = sup

u 0 : EeuYt = exp

t(u)

<

<

and define the function

(u) =1

u0

v1(v) dv, u < K.

Then, repeating the steps of the proof of Theorem 4, first for the case with truncatedjumps and then passing to the limit as the parameter of truncation increases to infinity,we obtain the lower bound

Eb K

0

(eub euX0) u1e(u) du.

Remark 5. Identity (31) might also be used for creating corresponding boundsfor two-sided stopping times a,b. If, for example, Yt is the process with a symmetricdistribution, X0 = 0, and (6) holds, then (34) holds for Xt and (Xt) as well, thatis, for any stopping time

E min(, t) = E

0

(euXmin(,t) 1) u1e(u) du.

Hence,

E min(, t) = E

0

(cosh(uXmin(,t)) 1) u1e(u) du.(41)

Note that the similar identity is used in [11] for the derivation of maximal inequalitiesfor the Gaussian OU-process. Since the Gaussian OU-process is continuous, from(41), as t , it follows that

Eb,b =

0

cosh(ub) 1

u1e(u) du < , (u) =

2u2

4.

5. Maximal inequalities for stable OU-processes. We consider now a spec-tral negative stable process Yt (see [1] or [19]) with

EeuY1 = exp{1u}, u 0, 1 < 2.(42)

This process Yt, and in turn Xt, does not have positive jumps. Moreover,

(u) = u/(2).

By (18),

E exp{uXt} = exp

uX0e

t +(1 et) u

2

.

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If = 2, X0 = 0, the process Xt is Gaussian. Then by [11] the following remarkableinequality is valid: For any stopping time

C1Elog (1 + ) Emaxt |Xt| C2Elog (1 + ),(43)where C1

13 , C2 3.3795.

We prove here an analogue of (43).Theorem 5. Let(42) hold and Xt solve (1) with X0 0. Then for any stopping

time and all p > 0

cpE

log(1 + )p(11/)

E

supt

Xt

p ap + CpE

log(1 + )

p(11/),(44)

where positive constants ap, cp, and Cp do not depend on .

For the proof of inequality (43), Graversen and Peskir [11] apply Walds momentidentity with the formula for E being similar to (41). Here, we apply identity (34),which is the one-sided analogue of the above-mentioned formula from [11]. We alsouse the following simple consequence of Lenglarts domination principle.

Lemma 2. Let Qt be a nonnegative, right continuous process and let At be anincreasing continuous process, A0 = 0. Assume that for all bounded stopping times

EQ EA.(45)

Then for all p > 0 and for all bounded stopping times there exist constants cp andCp such that

E log1 + supt Qtp

cp + CpE([log(1 + A)]p).(46)Proof. By Lenglarts principle, for any increasing continuous function H(x) with

H(0) = 0, (45) provides

E

supt

H(Qt)

E

H(A),(47)where

H(x) = x

x

1

sdH(s) + 2H(x).

Set H(x) = (log (1 + x))p, x 0. By lHospitals rule, limx0 H(x) = 0 andlimx

x

H(x)

x

1

sdH(s) = 0.

Hence, limx[ H(x)/H(x)] = 2 and there are constants cp and Cp such that H(x) cp + CpH(x). Therefore, (46) is implied by (47) with H(x) = (log (1 + x))

p.

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302 A. NOVIKOV

Proof of Theorem 5. Denote Xt = supst Xs. In the absence of positive jumps

for Xt, the process Xt is increasing and continuous. Then, (34) provides the following

inequality as valid for any bounded stopping times :

E E0

(euX

euX0) u1eu

/(

2

) du = EG(X) G(X0),where

G(y) =

0

(euy 1) u1eu/(2) du.

Hence, (45) is valid for Qt = t and At = G(Xt )G(x) the continuous increasing

process, A0 = 0. By Lemma 2,

E

log(1 + )p(11/)

cp + CpE

log

1 + G(X) G(X0)p(11/)

.

Thus, the lower bound in (44) will be held, if the inequality

log(1 + G(y)) C1 + C2y/(1), y > 0,

is valid. The latter bound readily follows from the well-known asymptotic relation

G(y) = exp

Cy/(1)

1 + o(1)

, y ,(48)

(see, e.g., [20, Chap. 3, Exercise 7.3]). The boundedness requirement for stoppingtime is easily removed by applying the localization technique.

The upper bound (44) is derived with the help of (34) which, jointly with an

obvious equality ex = ex+

1 + ex

, for any bounded stopping time gives

E

G(X+ ) G(X0) + E + E

0

(1 euX

) u1eu/(2) du.

Since

E

0

(1 euX

) u1eu/(2) du E(X )

0

eu/(2) du

and EX |X0|/+ CE (see, also the proof of Theorem 4), we find the followingestimate:

E

G(X+ ) c + CE.

Thus, (45) is valid with At = c + Ct and Qt = G(X+t ), where Qt is a nonnegative

right-continuous process. By Lemma 2,

E log 1 + G(X)

p(11/)

cp + CpE log(1 + )p(11/)

and it remains only to notice that (48) provides the following bound:

C+ log(1 + G(y)) Cy/(1), y > 0.

Theorem 5 is proved.Remark 6. For = 2 and X0 = 0, the application of (44) to maxtXt and

maxt(Xt) leads to (43) without specification of the constants cp and Cp.

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Acknowledgments. The author is thankful to R. Elliot, B. Ergashev, R. Liptser,E. Shinjikashvili, and A. N. Shiryaev for useful comments.

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