A note on a.s. finiteness of perpetual integral functionals of diffusions

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A note on a.s. finiteness of perpetual integral

functionals of diffusions

Paavo Salminen, Marc Yor

To cite this version:

Paavo Salminen, Marc Yor. A note on a.s. finiteness of perpetual integral functionals of diffusions.

2005. �hal-00013789�

ccsd-00013789, version 1 - 14 Nov 2005

fun tionals of diusions

Paavo Salminen

ÅboAkademi University,

Mathemati alDepartment,

Fänriksgatan 3 B,

FIN-20500Åbo,Finland,

email: phsalminabo.

Mar Yor

UniversitéPierre etMarieCurie,

Laboratoire de Probabilités

etModèles aléatoires,

4,Pla e Jussieu, Case 188

F-75252Paris Cedex05,Fran e

Abstra t

Inthis note, with thehelp of the boundary lassi ation of diu-sions, we derive a riterion of the onvergen e of perpetual integral

fun tionals of transient real-valued diusions.

In the parti ular ase of transient Bessel pro esses, we note that this riterion agrees with the one obtained via Jeulin's onvergen e

lemma.

Keywords: Brownian motion, random time hange, exit boundary,

lo al time,additive fun tional,sto hasti dierential equation.

AMS Classi ation: 60J65, 60J60.

1. Consider a diusion

Y

on an open interval

I = (l, r)

determined by the SDE

dY

t

= σ(Y

t

) dW

t

+ b(Y

t

) dt,

where

W

isastandardWienerpro essdenedina ompleteprobabilityspa e

(Ω, F , {F

t

}, P).

It is assumed that

σ

and

b

are ontinuous and

σ(x) > 0

for

all

x ∈ I

. We assumealsothat

Y

is transient and

lim

t→ζ

Y

t

= r

a.s.,

where

ζ

isthe life timeof

Y.

Hen e, if

ζ < ∞

then

ζ = H

r

(Y ) := inf{t : Y

t

= r}.

For the speed and the s ale measure of

Y

we use

m

Y

_{(dx) = 2 σ}

2

_(x)e

B

Y

_(x)

dx and S

Y

_{(dx) = e}

−

_B

Y

_(x)

dx,

(2) respe tively,where

B

Y

_{(x) = 2}

Z

x

b(z)

σ

2

_(z)

dz.

(3)

Let

f

be a positive and ontinuous fun tion dened on

I,

and onsider the perpetual integralfun tional

A

ζ

(f ) :=

Z

ζ

0

f (Y

s

) ds.

Weareinterestedinndingne essaryandsu ient onditionsfora.s.

nite-nessof

A

ζ

(f ).

When

Y

isaBrownianmotionwithdrift

µ > 0

su ha ondition

is that the fun tion

f

is integrable at

+∞

(see Engelbert and Senf [4℄ and Salminenand Yor[9℄). This onditionis derived in[9℄via Ray-Knight

theo-rems and the stationarity property of the lo altime pro esses (whi hmakes

Jeulin's lemma appli able). In this note a ondition (see Theorem 2) valid

forgeneral

Y

isdedu edbyexploitingthefa tthat

A

H

x

(f )

for

x < r

an,via

random time hange, be seen as the rst hitting time of a point for another

diusion.

2. The next proposition presents the key result onne ting perpetual

integral fun tionals to rst hitting times. The result is a generalization of

Proposition 2.1in[8℄ dis ussed in Propositions2.1and 2.3in[2℄.

Proposition 1. Let

Y

and

f

be as above, and assume that there exists a twotimes ontinuously dierentiable fun tion

g

su h that

f (x) = g

′

(x)σ(x)

2

,

x ∈ I.

(4) Set for

t > 0

A

t

:=

Z

t

0

f (Y

s

) ds.

(5)

and let

{a

t

: 0 ≤ t < A

ζ

}

denotethe inverse of

A,

that is,

a

t

:= min



s : A

s

> t

Then the pro ess

Z

given by

Z

t

:= g (Y

a

t

) ,

t ∈ [0, A

ζ

),

is a diusionsatisfying the SDE

dZ

t

= df

W

t

+ G(g

−

1

(Z

t

)) dt,

t ∈ [0, A

ζ

).

where

W

f

t

isa Brownianmotion and

G(x) =

1

f (x)



1

2

σ(x)

2

_g

′′

(x) + b(x) g

′

(x)



.

Moreover, for

l < x < y < r

A

H

y

(Y )

= inf{t : Z

t

= g(y)} =: H

g(y)

(Z) a.s.

(6)

with

Y

0

= x

and

Z

0

= g(x).

3. Toxideas,assumethatthefun tion

g

asintrodu edinProposition1 isin reasing. Wedene

g(r) := lim

x→r

g(x),

andusethesame onventionfor any in reasing fun tion dened on

(l, r).

The state spa e of the diusion

Z

is the interval

(g(l), g(r))

and a.s.

lim

t→ζ(Z)

Z

t

= g(r).

Clearly,letting

y → r

in (6) it follows that

A

H

r

(Y )

= inf{t : Z

t

= g(r)} a.s.,

(7)

where both sides in(7) are eithernite or innite. Now we have

Theorem 2. For

Y, A, f

and

g

asabove itholdsthat

A

ζ

is a.s. niteif and only if for the diusion

Z

the boundary point

g(r)

is anexit boundary, i.e.,

Z

g(r)

S

Z

_(dα)

Z

α

m

Z

_{(dβ) < ∞,}

(8)

where the s ale

S

Z

and the speed

m

Z

of the diusion

Z

are given by

S

Z

_{(dα) = e}

−

B

Z

_(α)

dα and m

Z

_{(dβ) = 2 e}

B

Z

_(β)

dβ

with

B

Z

(β) = 2

Z

β

G ◦ g

−

₁

(z) dz.

The ondition (8) isequivalent with the ondition

Z

r

S

Y

_{(r) − S}

Y

_(v)

_{f (v) m}

Y

_{(dv) < ∞.}

its exit boundary with positiveprobability and anexit boundary annot be

unattainable(see[5℄or[1℄). This ombinedwith(7)andthe hara terization

of an exitboundary(see [1℄ No. II.6 p.14) provesthe rst laim. It remains

to showthat (8) and (9) are equivalent. We have

B

Z

_{(α) = 2}

Z

g

−1

_(α)

G(u) g

′

(u) du

= 2

Z

g

−1

(α)



1

2

g

′′

(u)

g

′

(u)

+

b(u)

σ

2

_(u)



du

= log(g

′

(g

−

₁

(α))) + B

Y

_(g

−

₁

(α)).

Consequently,

S

Z

_{(dα) = e}

−

B

Z

_(α)

dα =

1

g

′

_(g

−

₁

_(α))

exp −B

Y

_(g

−

1

(α))

dα

and

m

Z

_{(dα) = 2 e}

B

Z

_(α)

dα = 2 g

′

(g

−

1

(α)) exp B

Y

_(g

−

1

(α))

dα.

Substitutingrst

α = g(u)

intheouterintegralin(8)andafterthis

β = g(v)

in the inner integral yield

Z

g(r)

S

Z

(dα)

Z

α

m

Z

(dβ) = 2

Z

r

du e

−

B

Y

_(u)

Z

u

dv (g

′

(v))

2

e

B

Y

(u)

= 2

Z

r

dv (g

′

(v))

2

e

B

Y

(v)

Z

r

v

du e

−

B

Y

_(u)

by Fubini's theorem. Using the expressions given in (2) for the speed and

the s aleof

Y

and the relation(4) between

f

and

g

omplete the proof. 4. Itiseasytoderivea onditionthatthemeanof

A

ζ

(f )

isnite. Indeed,

E

x

(A

ζ

(f )) =

Z

∞

0

E

x

(f (Y

s

)) ds

=

Z

r

l

G

Y

0

(x, y) f (y) m

Y

(dy) < ∞,

(10) where

G

Y

0

is the Green kernel of

Y

w. r. t.

m

Y

_.

Under the assumption (1)

we may takefor

x ≥ y

(10).

5. Sin e the exit ondition (8) plays a ru ial rle in our approa h we

dis uss here shortly two proofs of this ondition, thus making the paper as

self- ontained aspossible.

Let

Y

be an arbitrary regular diusion livingon the interval

I

with the end points

l

and

r.

The s ale fun tion of

Y

is denoted by

S

and the speed measure by

m.

It isalsoassumed that the killingmeasure of

Y

isidenti ally zero. Re all the denition due to Feller

r is exit

⇔

Z

r

S(dα)

Z

α

m(dβ) < ∞.

(11)

Note that by Fubini's theorem

Z

r

S(dα)

Z

α

m(dβ) =

Z

r

m(dβ)(S(r) − S(β)),

and, hen e,

S(r) < ∞

if

r

is exit. Moreover, if

r

is exit then

H

r

< ∞

with positiveprobability.

5.1. We give nowsome details of the proof of (11) following losely

Kallen-berg [7℄ (see also Breiman [3℄). For

l < a < b < r

let

H

ab

:= inf{t : Y

t

=

a or b}.

Then for

a < x < b

E

x

(H

ab

) =

Z

b

a

b

G

Y

₀

(x, z) m(dz),

(12) where

G

b

Y

0

is the (symmetri ) Green kernel of

Y

killed when it exits

(a, b),

i.e.,

b

G

Y

0

(x, z) =

(S(b) − S(x))(S(y) − S(a))

S(b) − S(a)

x ≥ y.

If

r

is exit there exists

h > 0

su h that

P

x

(H

r

< h) > 0

for any xed

x ∈ (a, r).

Usingthis propertyit an bededu ed (see[7℄p. 377)that forany

a ∈ (l, r)

E

x

(H

ar

) < ∞,

whi h,from (12), isseen tobe equivalentwith (11).

5.2. Another proof of (11) an be found inIt and M Kean [5℄ p. 130. To

present also this briey re all rst the formula

E

x

(exp(−λ H

b

)) =

ψ

λ

(x)

ψ

λ

(b)

where

λ > 0

and

ψ

λ

is an in reasing solution of the generalized dierential equation

d

dm

d

dS

u = λu.

(14)

Letting

b → r

in(13) it isseen that

r is exit

⇔

lim

b→r

ψ

λ

(b) < ∞.

Let

ψ

+

λ

denote the (right) derivative of

ψ

λ

with respe t to

S.

Sin e

ψ

λ

is in reasing itholds that

ψ

+

λ

> 0.

Thefa t that

ψ

λ

solves(14) yieldsfor

z < r

ψ

_λ

+

(r) − ψ

+

_λ

(z) = λ

Z

r

z

ψ

λ

(a) m(da).

In parti ular,

ψ

+

λ

is in reasing and

ψ

+

λ

(r) > 0.

Hen e, assuming now that

ψ

λ

(r) < ∞

we obtain

S(r) < ∞,

and, further,

λ ψ

λ

(z)

Z

r

z

S(dα)

Z

α

z

m(dβ) ≤ λ

Z

r

z

S(dα)

Z

α

z

ψ

λ

(β)m(dβ)

=

Z

r

z

S(dα) ψ

_λ

+

(α) − ψ

_λ

+

(z)

= ψ

λ

(r) − ψ

λ

(z) − ψ

+

λ

(z) (S(r) − S(z)) < ∞,

whi hyields the ondition onthe right hand side of (11). Assume next that

the ondition onthe right hand side of (11) holds, and onsider for

z < β

0 ≤ (ψ

λ

(β))

−

₁

ψ

_λ

+

(β) − ψ

_λ

+

(z)

= (ψ

λ

(β))

−

₁

Z

β

z

ψ

λ

(α)m(dα).

Integrating over

β

gives

log(ψ

λ

(r)) − log(ψ

λ

(z)) − ψ

λ

+

(z)

Z

r

z

(ψ

λ

(β))

−

₁

S(dβ)

=

Z

r

z

S(dβ) (ψ

λ

(β))

−

₁

Z

β

z

ψ

λ

(α)m(dα)

≤

Z

r

z

S(dβ)

Z

β

z

m(dα) < ∞,

dimension parameter

δ > 2.

Let

R

denotethis pro ess. Itis wellknown that

lim

t→∞

R

t

= +∞

and that

R

solves the SDE

dR

t

= dW

t

+

δ − 1

2R

t

dt,

where

W

is a standard Brownian motion. Here the fun tion

B

R

( f. (3))

takesthe form

B

R

(v) = (δ − 1) log v,

and, onsequently,

Z

∞

dv (g

′

(v))

2

e

B

R

(v)

Z

∞

v

du e

−

B

R

_(u)

=

Z

∞

dv (g

′

(v))

2

v

δ−1

Z

∞

v

du u

−

_δ+1

=

Z

∞

dv (g

′

(v))

2

v

δ−1

1

δ − 2

v

−

_δ+2

leading to

Z

∞

0

f (R

t

) dt < ∞

⇔

Z

∞

u f (u) du < +∞.

Anotherwaytoderivethis onditionisvialo altimesandJeulin'slemma

[6℄. Indeed, bythe o upationtime formulaand Ray-Knighttheorem forthe

total lo altimes of

R

(see, e.g. [10℄Theorem 4.1p. 52) we have

Z

∞

0

f (R

s

) ds

(d)

=

Z

∞

0

f (a)

ρ

a

γ

γ a

γ−1

da

=

1

γ

Z

∞

0

a f (a)

ρ

a

γ

a

γ

da

where

δ = 2 + γ

and

ρ

is a squared 2-dimensional Bessel pro ess. Using the s aling property, it is seen that the distribution of the random variable

ρ

a

γ

/a

γ

doesnot depend on

a.

Hen e,weobtainbyJeulin'slemmathatif the fun tion

a 7→ a f (a), a > 0,

is lo allyintegrableon

[0, ∞)

then

Z

∞

0

f (R

s

) ds < ∞

⇔

Z

∞

Z

∞

0

g(W

s

(µ)

) ds < ∞

⇔

Z

∞

g(x) dx < ∞.

(16)

where

g

is any non-negative lo ally integrable fun tion and

W

(µ)

denotes a

Brownian motion with drift

µ > 0.

To see this, write

g(x) = f (e

x

₎

and use Lamperti's representation

exp(W

_s

(µ)

) = R

(µ)

A

(µ)

s

,

s ≥ 0,

where

A

(µ)

s

=

Z

s

0

du exp(2W

u

(µ)

)

and

R

(µ)

is aBesselpro ess withdimension

d = 2(1 + µ)

startingfrom1,we obtain ( f. [8℄Remark 3.3.(3))

Z

∞

0

f (exp(W

s

(µ)

)) ds =

Z

∞

0

R

(µ)

u

−

2

f (R

u

(µ)

) du a.s.,

and, in order toget (16) it nowonly remains touse the equivalen e(15).

We wish to underline the fa t that in Theorem 2 it is assumed that the

fun tion

f

is ontinuouswhereas the approa hvia Jeulin'slemma,whi hwe developed above,demands onlylo alintegrability.

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