Sequential testing problems for some diffusion processes

für Angewandte Analysis und Sto hastik

imFors hungsverbund Berline.V.

Preprint ISSN 0946  8633

Sequential testing problems for some diusion

pro esses Pavel Gapeev

1 , 2

submitted: O tober27,2006

1

WeierstrassInstitute

forAppliedAnalysisandSto hasti s

Mohrenstrasse39

10117Berlin

Germany

e-mail: gapeevwias-berlin.de

2

RussianA ademyofS ien es

InstituteofControlS ien es

ProfsoyuznayaStr. 65 117997Mos ow Russia No. 1178 Berlin2006

W

I

A

S

2000 Mathemati sSubje tClassi ation. 60G40,62M20,34K10,62C10,62L15,60J60.

Keywordsandphrases. Sequentialtesting,diusionpro ess,optimalstopping,free-boundary

problem,smooth-t ondition,It'sformula.

This resear hwas supportedbyDeuts heFors hungsgemeins haftthroughtheSFB649

Weierstraÿ-Institutfür Angewandte Analysis und Sto hastik (WIAS) Mohrenstraÿe 39 10117 Berlin Germany Fax: +49 302044975 E-Mail: preprintwias-berlin.de

We study the Bayesian problem of sequential testing of two simple

hy-potheses about the lo al drift of an observed diusion pro ess. The optimal

stoppingtime isfound asthe rst timewhen theaposteriori probability

pro- essleavesthe region dened by two sto hasti boundaries depending on the

observation pro ess. It is shown that under some nontrivial relationships on

the oe ients of the observed diusion the problem admits a losed form

solution. The method of proof is based on embedding the initial problem

into a two-dimensional optimal stopping problem and solving the equivalent

free-boundary problembymeansof the smooth-t onditions.

1. Introdu tion

The problem of sequential testing of two simple hypotheses about the lo al drift

µ(x)

of an observed diusion pro ess seeks to determine as soon as possible and withminimalerrorprobabilitiesif thetrue drift oe ientis either

µ

0

(x)

or

µ

1

(x)

. This problem admits two dierent formulations (see Wald [20℄). In the Bayesian

formulationitisassumed that the drift oe ient

µ(x)

has ana priorigiven

distri-bution,andinthevariationalformulationnoprobabilitsi assumptionismadeabout

the unknown drift

µ(x)

. In this paper we onlystudy the Bayesian formulation.

Bymeans of the Bayesian approa h,Waldand Wolfowitz[21℄-[22℄ proved the

opti-malityofthesequentialprobabilityratiotest (SPRT) inthe variationalformulation

oftheproblemforsequen es ofi.i.d. observations. Dvoretzky,Kiefer andWolfowitz

[2℄pointed out that if the ( ontinuous time) likelihoodratio pro ess has stationary

independentin rements,thentheSPRTremainsoptimalinthevariationalproblem.

Mikhalevi h[12℄ and Shiryaev [18℄ (see also [19; Chapter IV℄) obtained an expli it

solution of the Bayesian problem for an observed Wiener pro ess by redu ing the

initial optimal stopping problem to a free-boundary problem for an ordinary

se -ond order operator. A omplete proof of the statement of [2℄ (under some mild

assumptions) was given by Irle and S hmitz [7℄. Peskir and Shiryaev [14℄ obtained

an expli it solution of the Bayesian problem of testing hypotheses about the

in-tensity of an observed Poisson pro ess by solving a free-boundary problem for a

dierential-dieren e operator. Sequential testing problems for a ompound

Pois-son pro ess having exponentially distributed jumps were expli itly solved in [4℄.

Re ently,Dayanik andSezer [1℄obtained asolutionofthe Bayesiansequential

test-ing problem for a general ompound Poisson pro ess. A nite horizon version of

al drift of an observed diusion pro ess in the Bayesian formulation. In this ase

the optimal Bayes stopping time is the rst time when the aposterioriprobability

pro ess leaves the region dened by two sto hasti boundaries depending on the

observation pro ess.

In the present paper we make an embedding of the initial Bayesian problem into

an extended optimal stopping problem for a two-dimensional (time-homogeneous

strong) Markov diusion pro ess ( onsisting of the a posteriori probability pro ess

and the observation pro ess). We show that the ontinuation region (for the a

posterioriprobabilitypro ess)isdeterminedbytwosto hasti boundariesdepending

onthe observation pro esswhere thebehaviorof the boundaries is hara terizedby

the signal/noise ratio pro ess. In order to nd analyti expressions for the value

fun tion and the stopping boundaries under some spe ial nontrivial relationships

on oe ients of the observed diusion, we formulate an equivalent free-boundary

problem. Byapplyingsmooth-t onditionsweshowthatthefree-boundaryproblem

admits an expli it solution and the boundaries are uniquely determined from a

oupledsystem oftrans endentalequations. Then weverify thatthe solutionofthe

free-boundary problem turns out to be a solution of the initial extended optimal

stoppingproblem.

2. Formulation and solution of the Bayesian problem

In the Bayesian formulationof the problem(see [19; Chapter IV, Se tion 2℄for the

ase of Wiener pro ess) it is assumed that we observe a traje tory of the diusion

pro ess

X = (X

t

)

t≥0

withdrift

µ

0

(x)+θ(µ

1

(x)−µ

0

(x))

wheretherandomparameter

θ

may be

1

or

0

with probability

π

or

1 − π

, respe tively.

2.1. Forapre iseprobabilisti formulation ofthe Bayesian problemitis onvenient

to assume that all our onsiderations take pla e on a probability spa e

(Ω, F , P

π

)

wherethe probability measure

P

π

has the following stru ture:

P

π

= πP

1

+ (1 − π)P

0

(2.1)

for any

π ∈ [0, 1]

. Let

θ

be a random variable taking two values

1

and

0

with

probabilities

P

π

[θ = 1] = π

and

P

π

[θ = 0] = 1 − π

, and let

W = (W

t

)

t≥0

be a standardWienerpro ess startedatzerounder

P

π

. Itisassumed that

θ

and

W

are independent.

It is further assumed that we observe a ontinuous pro ess

X = (X

t

)

t≥0

with the (open) state spa e

E ⊆ R

and solving the sto hasti dierential equation:

dX

t

= [µ

0

(X

t

) + θ(µ

1

(X

t

) − µ

0

(X

t

))] dt + σ(X

t

) dW

t

(X

0

= x)

(2.2) wherethe fun tions

µ

i

(x)

and

σ(x)

are Lips hitz ontinuouson

E

,i.e., thereexists a onstant

C > 0

su h that:

for all

x, x

′

_{∈ E}

and

i = 0, 1

. Thus, from [11; Theorem 4.6℄ it follows that under

xed

θ = i

equation (2.2) has a unique strong solution, and hen e,

P

π

[X ∈ · |θ =

i] = P

i

[X ∈ · ]

isthe distribution lawof a homogeneous diusion pro ess (starting at some xed point

x ∈ E

) with drift

µ

i

(x)

and diusion oe ient

σ

2

_(x)

for

i = 0, 1

. Wewillalsoassumethat either

µ

0

(x) < µ

1

(x)

or

µ

0

(x) > µ

1

(x)

holdsand

σ

2

_{(x) > 0}

for all

x ∈ E

. Let

π

and

1 − π

play the role of apriopi probabilities of

the statisti alhypotheses:

H

1

: θ = 1

and

H

0

: θ = 0

(2.4)

respe tively.

Beingbased upon the ontinuous observation of

X

our task is to test sequentially

the hypotheses

H

1

and

H

0

with a minimalloss. For this, we onsider a sequential de ision rule

(τ, d)

where

τ

is a stopping time of the observed pro ess

X

(i.e.,

a stopping time with respe t to the natural ltration

F

X

t

= σ{X

s

| 0 ≤ s ≤ t}

generatedby thepro ess

X

for

t ≥ 0

),and

d

isan

F

X

τ

-measurablefun tiontaking onvalues

0

and

1

. Afterstoppingthe observations attime

τ

, theterminalde ision

fun tion indi ates whi h hypothesis should be a epted a ording to the following

rule: if

d = 1

we a ept

H

1

, and if

d = 0

we a ept

H

0

. The problem onsists of omputingthe riskfun tion:

V (π) = inf

(τ,d)

E

π

[τ + aI(d = 0, θ = 1) + bI(d = 1, θ = 0)]

(2.5)

and nding the optimal de ision rule

(τ

∗

, d

∗

)

, alled the

π

-Bayes de ision rule, at whi h the inmum in

(2.5)

is attained. Here

E

π

[τ ]

is the average ost of the observations, and

aP

π

[d = 0, θ = 1] + bP

π

[d = 1, θ = 0]

is the average loss due to a wrongterminalde ision, where

a > 0

and

b > 0

are some given onstants.

2.2. By means of standard arguments (see [19; pages 166-167℄) one an redu e the

Bayesian problem(2.5) to the optimalstopping problem:

V (π) = inf

τ

E

π

[τ + g

a,b

(π

τ

)]

(2.6)

for the a posteriori probability pro ess

π

t

= P

π

[θ = 1|F

X

t

]

for

t ≥ 0

with

P

π

[π

0

=

π] = 1

. Here

g

a,b

(π) = aπ ∧ b(1 − π)

for

π ∈ [0, 1]

,andthe optimalde isionfun tion is given by

d

∗

= 1

if

π

τ

∗

≥ c

, and

d

∗

= 0

if

π

τ

∗

< c

, where here and in the sequel weset

c = b/(a + b)

.

2.3. Sin e for

i = 0, 1

ondition (2.3) is assumed to be satised, by means of

Girsanov theorem for diusion-type pro esses [11; Theorem 7.19℄ we get that the

loglikelihoodratio pro ess

Z = (Z

t

)

t≥0

dened aslogarithmof the Radon-Nikodym derivative:

Z

t

= log

d(P

1

|F

t

X

)

d(P

0

|F

t

X

)

(2.7) (here

P

i

|F

X

t

denotes the restri tion of

P

i

to

F

X

t

for

i = 0, 1

) takes the form:

Z

t

=

Z

t

0

µ

1

(X

s

) − µ

0

(X

s

)

σ

2

_(X

s

)

dX

s

−

1

2

Z

t

0

µ

2

1

(X

s

) − µ

2

0

(X

s

)

σ

2

_(X

s

)

ds

(2.8)

for all

t ≥ 0

. A ording to the arguments in [19; pages 180-181℄, the a posteriori

probabilitypro ess

(π

t

)

t≥0

an be expressed as:

π

t

=



π

1 − π

e

Z

t



_



1 +

π

1 − π

e

Z

t



(2.9)

and,by virtue of It's formula(see, e.g., [11; Theorem 4.4℄), it solves the equation:

dπ

t

=

µ

1

(X

t

) − µ

0

(X

t

)

σ(X

t

)

π

t

(1 − π

t

) dW

t

(π

0

= π)

(2.10)

where,bymeansofP.Lévy'stheorem[17;ChapterIV,Theorem3.6℄,theinnovation

pro ess

W = (W

t

)

t≥0

dened by:

W

t

=

Z

t

0

dX

s

σ(X

s

)

−

Z

t

0



µ

0

(X

s

)

σ(X

s

)

+ π

s

µ

1

(X

s

) − µ

0

(X

s

)

σ(X

s

)



ds

(2.11)

is a standard Wiener pro ess. Therefore, from (2.11) it follows that the pro ess

X = (X

t

)

t≥0

admitsthe representation:

dX

t

= [µ

0

(X

t

) + π

t

(µ

1

(X

t

) − µ

0

(X

t

))] dt + σ(X

t

) dW

t

(X

0

= x).

(2.12)

Letus suppose that the signal/noiseratio fun tion

r(x)

dened by:

r(x) =

µ

1

(x) − µ

0

(x)

σ(x)

(2.13)

isalsoLips hitz ontinuous,i.e. thereexists a onstant

C

′

_{> 0}

su h that ondition:

[r(x) − r(x

′

_)]

2

_{≤ C}

′

_{[x − x}

′

_]

2

(2.14)

holdsforall

x, x

′

_{∈ E}

,andthere are onstants

r

∗

and

r

∗

su hthatthe inequalities:

0 < r

∗

≤ r(x) ≤ r

∗

< ∞

(2.15)

are satised for all

x ∈ E

. Hen e, by means of Remark to [11; Theorem 4.6℄ (see

also[13; Theorem 5.2.1℄), we on lude that the pro ess

(π

t

, X

t

)

t≥0

turns out to be a unique strong solution of the (two-dimensional) sto hasti dierential equation

(2.10)+(2.12),and thus, byvirtue of[13; Theorem 7.2.4℄,itisa(time-homogeneous

strong) Markov pro ess with respe t to its natural ltration whi h obviously

oin-sideswith

(F

X

t

)

t≥0

. Therefore, theinmumin(2.6) istaken overallstoppingtimes of

(π

t

, X

t

)

t≥0

being a Markovian su ient statisti in the problem (see [19; Chap-ter II, Se tion15℄).

2.4. For the problem (2.6) let us onsider the following extended optimal stopping

problemfor the Markov pro ess

(π

t

, X

t

)

t≥0

:

V (π, x) = inf

τ

E

π,x

[τ + g

a,b

(π

τ

)]

where

P

π,x

is a measure of the diusion pro ess

(π

t

, X

t

)

t≥0

starting at the point

(π, x)

and solving the (two-dimensional) equation (2.10)+(2.12), and the inmum in (2.16) is taken over all stopping times

τ

of the pro ess

(π

t

, X

t

)

t≥0

su h that

E

π,x

[τ ] < ∞

for all

(π, x) ∈ [0, 1] × E

.

2.5. Nowletusdeterminethestru tureof theoptimalstoppingtimeintheproblem

(2.16).

(i) First, applying applying It-Tanaka-Meyer formula (see, e.g., [8; Chapter V,

(5.52)℄or[16; ChapterIV, Theorem 51℄)tothe fun tion

g

a,b

(π) = aπ ∧ b(1 − π)

,we get:

g

a,b

(π

t

) = g

a,b

(π) +

Z

t

0

(g

a,b

)

π

(π

s

) ds +

1

2

Z

t

0

∆

π

(g

a,b

)

π

(π

s

) dℓ

c

s

(π) + N

c

t

(2.17) where

R

t

0

∆

π

(g

a,b

)

π

(π

s

)dℓ

c

s

(π) = (−b − a)ℓ

c

t

(π)

, the pro ess

(ℓ

c

t

(π))

t≥0

is the lo al time of

(π

t

)

t≥0

atthe point

c

given by:

ℓ

c

t

(π) = lim

ε↓0

1

2ε

Z

t

0

I(c − ε < π

s

< c + ε) r

2

(X

s

)π

s

2

(1 − π

s

)

2

ds

(2.18)

asalimitinprobability,andforany

(F

X

t

)

t≥0

-stoppingtime

τ

satisfying

E

π,x

[τ ] < ∞

the pro ess

(N

c

τ ∧t

, F

t

X

, P

π,x

)

t≥0

dened by

N

c

τ ∧t

=

R

τ ∧t

0

(g

a,b

)

π

(π

s

)I(π

s

6= c)r(X

s

)π

s

(1 − π

s

)dW

s

isa ontinuous (uniformly integrable) martingale.

Let us x some

(π, x)

from the ontinuation region

C

and let

τ

∗

= τ

∗

(π, x)

de-notethe optimalstoppingtime inthe problem(2.16). ByapplyingDoob's optional

samplingtheorem (see, e.g., [9; Chapter I, Theorem 1.39℄or [17; Chapter II,

Theo-rem 3.1℄) and by using (2.17), itfollows that:

V (π, x) = E

π,x

[τ

∗

+ g

a,b

(π

τ

∗

)] = g

a,b

(π) + E

π,x



τ

∗

−

1

2

(a + b)ℓ

c

τ

∗

(π)



(2.19)

and hen e, by virtue of general optimal stopping theory for Markov pro esses (see

[19; Chapter III℄),we have:

V (π, x) − g

a,b

(π) = E

π,x



τ

∗

−

1

2

(a + b)ℓ

c

τ

∗

(π)



< 0.

(2.20)

Then taking any

π

′

su h that

π < π

′

_{≤ c}

or

c ≤ π

′

_{< π}

and using the expli it

expression (2.9), from(2.17)-(2.18) we obtain:

V (π

′

_{, x) − g}

a,b

(π

′

) ≤ E

π

′

_,x



τ

∗

−

1

2

(a + b)ℓ

c

τ

∗

(π

′

₎



≤ E

π,x



τ

∗

−

1

2

(a + b)ℓ

c

τ

∗

(π)



(2.21)

and thus, by means of (2.20), we see that

(π

′

_{, x) ∈ C}

. Therefore, a ording to the

generaloptimalstoppingtheory(see, e.g., [19℄ and [15℄),these arguments(together

[10℄) show that there exists a ouple of fun tions

(g

0

(x), g

1

(x))

,

x ∈ E

, su h that

0 ≤ g

0

(x) ≤ c ≤ g

1

(x) ≤ 1

, and the ontinuation region for the optimal stopping problem(2.16) isan open set of the form:

C = {(π, x) ∈ [0, 1] × E | π ∈ hg

0

(x), g

1

(x)i}

(2.22) and the stoppingregion is the losureof the set:

D = {(π, x) ∈ [0, 1] × E | π ∈ [0, g

0

(x)i ∪ hg

1

(x), 1]}.

(2.23)

(ii)Nowforgiven

(π, x) ∈ C

letustake

x

′

_{∈ E}

su hthat

x

′

_{< x}

if

x < c

or

x < x

′

if

x > c

. Thenusing thefa tsthat

(π

t

, X

t

)

t≥0

isatime-homogeneousMarkov pro ess and

τ

∗

= τ

∗

(π, x)

doesnot depend on

x

′

, from(2.17)-(2.18) weobtain:

V (π, x

′

) − g

a,b

(π) ≤ E

π,x

′



τ

∗

−

1

2

(a + b)ℓ

c

τ

∗

(π)



(2.24)

≤ E

π,x



τ

∗

−

1

2

(a + b)ℓ

c

τ

∗

(π)



= V (π, x) − g

a,b

(π)

andhen e,by meansof(2.20),wesee that

(π, x

′

_{) ∈ C}

. Therefore,wemay on lude

in (2.22)-(2.23) that the boundary

x 7→ g

0

(x)

is in reasing (de reasing) and the boundary

x 7→ g

1

(x)

is de reasing (in reasing) on

E

when the fun tion

r(x)

is in reasing (de reasing),respe tively.

(iii)Next,letusobserve thatthevaluefun tion

V (π, x)

from(2.16)andthe

bound-aries

(g

0

(x), g

1

(x))

from(2.22)-(2.23)alsodependon

r(x)

anddenotethem hereby

V

∗

(x, π)

and

V

∗

_{(π, x)}

and

(A

∗

, B

∗

)

and

(A

∗

_{, B}

∗

₎

when

r(x) = r

∗

and

r(x) = r

∗

for

all

x ∈ E

, respe tively. Usingthe fa t that

x 7→ V (π, x)

is an in reasing

(de reas-ing) fun tion when

r(x)

is in reasing (de reasing) on

E

, and

V (π, x) = g

a,b

(π)

for all

π ∈ [0, g

0

(x)] ∪ [g

1

(x), 1]

, we on lude that

0 < A

∗

_{≤ g}

0

(x) ≤ A

∗

< c < B

∗

≤

g

1

(x) ≤ B

∗

< 1

for all

x ∈ E

. Here we note that if

r

∗

= r

∗

then

A

∗

_{= g}

0

(x) = A

∗

and

B

∗

= g

1

(x) = B

∗

for all

x ∈ E

, where

0 < A

∗

_{< A}

∗

< c < B

∗

< B

∗

< 1

are uniquely determinedfrom the system (4.85) in [19; Chapter IV℄.

2.6. Summarizing the fa ts proved in Subse tion 2.5 above we may on lude that

the following optimalde ision rule isoptimal inthe extended problem(2.16):

τ

∗

= inf{t ≥ 0 | π

t

∈ hg

/

0

(X

t

), g

1

(X

t

)i}

(2.25)

d

∗

=



1,

if

π

τ

∗

= g

1

(X

τ

∗

)

0,

if

π

τ

∗

= g

0

(X

τ

∗

)

(2.26)

(whenever

E

π,x

[τ

∗

] < ∞

) where the two boundaries

(g

0

(x), g

1

(x))

,

x ∈ E

, satisfy the following properties:

g

0

(x) : E → [0, 1]

is ontinuous and in reasing (de reasing) (2.27)

g

1

(x) : E → [0, 1]

is ontinuous and de reasing (in reasing) (2.28)

A

∗

_{≤ g}

whenever

r(x)

is an in reasing (de reasing) fun tion on

E

, respe tively. Here

(A

∗

, B

∗

)

and

(A

∗

_{, B}

∗

₎

satisfying

0 < A

∗

_{< A}

∗

< c < B

∗

< B

∗

< 1

are the op-timalstoppingpointsfor the orrespondinginnitehorizonproblemwith

r(x) = r

∗

and

r(x) = r

∗

for all

x ∈ E

, respe tively, uniquely determined from the system of

trans endentalequations (4.85) in[19; Chapter IV℄.

2.7. Let us further assume that the state spa e of the pro ess

X = (X

t

)

t≥0

is

E = h−ζ, ∞i

forsome

ζ ∈ R

xed, andunder onditionsof Subse tions2.1and2.3 the relationship:

µ

i

(x) =

η

i

σ

2

(x)

x + ζ

(2.30)

holds for all

x ∈ E

and some onstants

η

i

∈ R

,

i = 0, 1

, su h that

η

0

6= η

1

and

η

0

+ η

1

= 1

. Let usdene the pro ess

Y = (Y

t

)

t≥0

by:

Y

t

= log

π

t

1 − π

t

−

1

η

log

x + ζ

X

t

+ ζ

(2.31)

with

η = 1/(η

1

− η

0

)

. From the stru ture of (2.31) it is easily seen that there is a one-to-one orresponden e between the pro esses

(π

t

, X

t

)

t≥0

and

(π

t

, Y

t

)

t≥0

, and thus, thelatteris alsoa(time-homogeneousstrong)Markov pro ess withrespe t to

itsnaturalltration,whi h oin ides with

(F

X

t

)

t≥0

. Derivingthe expression for

X

t

from(2.31) and substituting itinto (2.10),we obtain:

dπ

t

=

σ (x + ζ)e

−ηY

t

[π

t

/(1 − π

t

)]

η

− ζ

η(x + ζ)e

−ηY

t

[π

t

/(1 − π

t

)]

η

π

t

(1 − π

t

) dW

t

(π

0

= π).

(2.32)

Dierentiatingby It's formulathe expression (2.31) and using the representations

(2.10)and(2.12) aswellasthe assumption(2.30) with

η

0

6= η

1

and

η

0

+ η

1

= 1

,we get

dY

t

= 0

and thus:

Y

t

= log

π

1 − π

(2.33)

forall

t ≥ 0

.

2.8. By means of standard arguments it is shown that under the assumptions of

Subse tion2.7 the optimalstopping problems (2.6) and (2.16) are equivalentto:

e

V (π, y) = inf

τ

E

π

[τ + g

a,b

(π

τ

)]

(2.34)

wherethe inmumis taken over allstoppingtimes

τ

ofthe pro ess

(π

t

, Y

t

)

t≥0

su h that

E

π

[τ ] < ∞

forall

(π, y) ∈ [0, 1] × R

and

y = log[π/(1 − π)]

for ea h

π ∈ h0, 1i

and

x ∈ E = h−ζ, ∞i

xed. It also follows that there exists a ouple of fun tions

(h

0

(y), h

1

(y))

,

y ∈ R

,su hthat the ontinuationregion

C

from(2.22)isequivalent to:

e

C = {(π, y) ∈ [0, 1] × R | π ∈ hh

0

(y), h

1

(y)i}

(2.35) and the set

D

from (2.23)is equivalentto:

e

forea h

y ∈ R

and

x ∈ E

xed.

2.9. If the assumption (2.30) with

η

0

6= η

1

and

η

0

+ η

1

= 1

holds, then by means of standard arguments it is shown that the innitesimaloperator

L

e

of the pro ess

(π

t

, Y

t

)

t≥0

from(2.32)-(2.33) a ts ona fun tion

F ∈ C

2,0

_{(h0, 1i × R)}

like:

(e

LF )(π, y) =

r

2

_{(x; π, y)}

2

π

2

_{(1 − π)}

2

∂

2

F

∂π

2

(π, y)

(2.37) with

r(x; π, y) =

σ ((x + ζ)e

−ηy

_{[π/(1 − π)]}

η

_{− ζ)}

η(x + ζ)e

−ηy

_{[π/(1 − π)]}

η

(2.38)

forall

(π, y) ∈ h0, 1i ∈ R

and ea h

x ∈ E = h−ζ, ∞i

xed.

Now let us use the results of general theory of optimal stopping problems for

on-tinuous time Markov pro esses (see, e.g., [6℄, [19; Chapter III, Se tion 8℄ and [15℄)

toformulate the orresponding free-boundary problem for the unknown value

fun -tion

(π, y) 7→ e

V (π, y)

from (2.16)(with

g

a,b

(π) = aπ ∧ b(1 − π)

) and the ouple of boundaries

(h

0

(y), h

1

(y))

,

y ∈ R

, from(2.35)-(2.36):

(e

L e

V )(π, y) = −1

for

(π, y) ∈ e

C

(2.39)

e

V (π, y)





_π=h

0

(y)+

= ah

0

(y),

V (π, y)

e





π=h

1

(y)−

= b(1 − h

1

(y))

(2.40)

∂ e

V

∂π

(π, y)





π=h

0

(y)+

= a,

∂ e

V

∂π

(π, y)





π=h

1

(y)−

= −b

(2.41)

e

V (π, y) = g

a,b

(π)

for

(π, y) ∈ e

D

(2.42)

e

V (π, y) ≤ g

a,b

(π)

for

(π, y) ∈ e

C

(2.43)

where

C

e

and

D

e

are given by (2.35) and (2.36), and the instantaneous stopping

onditions(2.40)andthe smooth-t onditions(2.41)areassumed tobesatisedfor

all

y ∈ R

and ea h

x ∈ E

xed.

Note that by Dynkin's superharmoni hara terization of the value fun tion (see

[3℄ and [19℄) it follows that

V (π, y)

e

from (2.34) is the largest fun tion satisfying

(2.39)-(2.40)and (2.42)-(2.43)for ea h

y ∈ R

and

x ∈ E

xed.

2.10. Integrating the equation (2.39) with some

h

1

(y) ∈ hc, 1i

xed for any given

y ∈ R

and using the boundary onditions (2.40)-(2.41), weobtain:

e

V (π, y; h

1

(y)) = b(1 − h

1

(y)) −

Z

h

1

(y)

π

Z

h

1

(y)

u

2

r

2

_{(x; v, y)v}

2

_{(1 − v)}

2

dvdu

(2.44) with

r(x; π, y)

given by (2.38) for all

π ∈ h0, h

1

(y)]

and ea h

x ∈ E = h−ζ, ∞i

xed.

From (2.44) it is easily seen that for any

y ∈ R

given and xed the fun tion

π 7→

e

V (π, y; h

1

(y))

is on aveon

h0, 1i

,andhen e

V (h

e

′

1

(y), y; h

′′

1

(y)) < e

V (h

′

1

(y), y; h

′

1

(y))

for

0 < h

′

1

(y) < h

′′

1

(y) < 1

. Thismeansthatfordierent

h

′

1

(y)

and

h

′′

π 7→ e

V (π, y; h

′

1

(y))

and

π 7→ e

V (π, y; h

′′

1

(y))

have no points of interse tion on the whole interval

π ∈ h0, h

′

1

(y)]

. From (2.44) it also follows that

V (π, y; h

e

1

(y)) →

−∞

as

π ↓ 0

for all

h

1

(y) ∈ [c, 1i

and

V (π, y; 1−) < 0

e

for all

π ∈ h0, 1i

and

e

V (1−, y; 1−) = 0

. Inthis ase,for some

eh

1

(y) ∈ hc, 1i

the urve

π 7→ e

V (π, y; e

h

1

(y))

interse ts the line

π 7→ aπ

atsome point

h

0

(y) ∈ h0, ci

. Sin e for dierent

h

′

1

(y) ∈

hc, 1i

the urves

π 7→ e

V (π, y; h

′

1

(y))

do not interse t ea h other on the intervals

h0, h

′

1

(y)i

, we may on lude that there exists a unique point

h

1

(y)

obtained by moving the point

h

′

1

(y)

from

eh

1

(y)

and su h that in some point

h

0

(y) ∈ h0, ci

the boundary onditions (2.40)-(2.41) hold. It thus follows that the boundaries

(h

0

(y), h

1

(y))

are uniquely determined fromthe system:

b + a =

Z

h

1

(y)

h

0

(y)

2

r

2

_{(x; u, y)u}

2

_{(1 − u)}

2

du

(2.45)

b(1 − h

1

(y)) = ah

0

(y) −

Z

h

1

(y)

h

0

(y)

Z

h

1

(y)

u

2

r

2

_{(x; v, y)v}

2

_{(1 − v)}

2

dvdu

(2.46) forea h

y ∈ R

and

x ∈ E = h−ζ, ∞i

xed.

2.11. Makinguse ofthe fa tsproved aboveweare nowready toformulatethe main

resultof the paper.

Theorem 2.1. Suppose that onditions (2.3) and (2.14)-(2.15) hold for all

x ∈

E = h−ζ, ∞i

andsome

ζ ∈ R

xed, andassumption (2.30)issatisedwith

η

0

6= η

1

and

η

0

+ η

1

= 1

. Then in the Bayesian problem (2.6)+(2.16)+(2.34)of testing two simplehypotheses (2.4) for the pro ess (2.2) the value fun tion has the expression:

V (π) = V (π, x) = e

V (π, y) =

(

e

V (π, y; h

1

(y)),

if

π ∈ hh

0

(y), h

1

(y)i

g

a,b

(π),

if

π ∈ [0, h

0

(y)] ∪ [h

1

(y), 1]

(2.47)

and the optimal

π

-Bayes de ision rule is expli itly given by:

τ

∗

= inf{t ≥ 0 | π

t

∈ hh

/

0

(y), h

1

(y)i}

(2.48)

d

∗

=

(

1,

if

π

τ

∗

= h

1

(y)

0,

if

π

τ

∗

= h

0

(y)

(2.49)

wherethetwo boundaries

(h

0

(y), h

1

(y))

are hara terizedas aunique solutionofthe oupledsystem of equations (2.45)-(2.46)for

y = log[π/(1 − π)]

andea h

π ∈ h0, 1i

and

x ∈ E

xed.

Proof. Itremainstoshowthatthefun tion(2.47) oin ideswiththevaluefun tion

(2.34)andthat thestoppingtime

τ

∗

from(2.48)withtheboundaries

(h

0

(y), h

1

(y))

,

y ∈ R

, spe ied above is optimal. Let us denote by

V (π, y)

e

the right-hand side of the expression (2.47). It follows by onstru tionfrom the previous se tion that the

e

V (π

t

, y)

, we obtain:

e

V (π

t

, y) = e

V (π, y) +

Z

t

0

(e

L e

V )(π

s

, y)I(π

s

6= h

0

(y), π

s

6= h

1

(y)) ds + f

M

t

(2.50)

wherethe pro ess

( f

M

t

)

t≥0

dened by:

f

M

t

=

Z

t

0

∂ e

V

∂π

(π

s

, y)I(π

s

6= h

0

(y), π

s

6= h

1

(y)) r(X

t

)π

s

(1 − π

s

) dW

s

(2.51) isa ontinuous lo almartingale under

P

π

with respe t to

(F

X

t

)

t≥0

.

By using the arguments above it an be veried that

(e

L e

V )(π, y) ≥ −1

for all

(π, y) ∈ h0, 1i × R

su h that

π 6= h

0

(y)

and

π 6= h

1

(y)

. Moreover, by means of standard arguments and using the onstru tion of

V (π, y)

e

above it an be he ked

that the property (2.43) also holds that together with (2.39)-(2.40)+(2.42) yields

e

V (π, y) ≤ g

a,b

(π)

for all

(π, y) ∈ [0, 1] × R

. Observe that the time spent by the pro ess

π

at the boundaries

(h

0

(y), h

1

(y))

,

y ∈ R

, is of Lebesgue measure zero, that allows to extend

(e

L e

V )(π, y)

arbitrarily to

π = h

0

(y)

and to

π = h

1

(y)

and thus to ignore the indi ators in (2.50)-(2.51). Hen e, from the expressions (2.50)

and the stru ture of the stopping time in(2.48) itfollows that the inequalities:

τ + g

a,b

(π

τ

) ≥ τ + e

V (π

τ

, y) ≥ e

V (π, y) + f

M

τ

(2.52) hold for any stopping times

τ

of the pro ess

(π

t

)

t≥0

started at

π ∈ [0, 1]

and for ea h

y ∈ R

.

Let

(τ

n

)

n∈N

beanarbitrary lo alizingsequen es of stoppingtimes forthe pro esses

( f

M

t

)

t≥0

. Taking in (2.52) the expe tation with respe t to the measure

P

π

, by means of the optionalsampling theorem (see, e.g., [9; Chapter I, Theorem 1.39℄ or

[17; Chapter II, Theorem 3.1℄), we get:

E

π

[τ ∧ τ

n

+ g

a,b

(π

τ ∧τ

n

)] ≥ E

π

h

τ

∧ τ

n

+ e

V

(π

τ ∧τ

n

, y)

i

≥ e

V

(π, y) + E

π

 f

M

τ ∧τ

n



= e

V

(π, y)

(2.53)

for all

(π, y) ∈ [0, 1] × R

. Hen e, letting

n

gotoinnityand using Fatou's lemma,

forany stoppingtimes

τ

su h that

E

π

[τ ] < ∞

we obtainthat the inequalities:

E

π

[τ + g

a,b

(π

τ

)] ≥ E

π

h

τ + e

V (π

τ

, y)

i

≥ e

V (π, y)

(2.54)

are satisedfor all

(π, y) ∈ [0, 1] × R

.

Byvirtue of the fa t that the fun tion

V (π, y)

e

together with the boundaries

h

0

(y)

and

h

1

(y)

satisfy the system (2.39)-(2.43),by the stru ture of the stoppingtime

τ

∗

in(2.48) and the expressions (2.50)it follows that the equalities:

τ

∗

∧ τ

n

+ g

a,b

(π

τ

∗

∧τ

n

) = τ

∗

∧ τ

n

+ e

V (π

τ

∗

∧τ

n

, y) = e

V (π, y) + f

M

τ

∗

∧τ

n

(2.55) hold for all

(π, y) ∈ [0, 1] × R

and any lo alizing sequen e

(τ

n

)

n∈N

of

( f

M

t

)

t≥0

. Notethat, by means of standard arguments and using the stru ture of the pro ess

(2.32)andofthestoppingtime(2.48),wehave

E

π

[τ

∗

] < ∞

forall

π ∈ [0, 1]

. Hen e, letting

n

gotoinnityandusing onditions(2.39)-(2.40),we anapplytheLebesgue

bounded onvergen e theorem for (2.55) toobtain the equality:

E

π

[τ

∗

∧ τ

n

+ g

a,b

(π

τ

∗

∧τ

n

)] = e

V (π, y)

(2.56) for all

(π, y) ∈ [0, 1] × R

, whi h together with (2.54) dire tly imply the desired

assertion.



A knowledgments. A part of theseresults waspresented atthe Conferen e

'Kol-mogorovandContemporaryMathemati s'held atMos owStateUniversityinJune

2003. The author thanks Hans RudolfLer he and Goran Peskir for their interest.

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