A NNALES DE L ’I. H. P., SECTION B
T ERENCE C HAN
Indefinite quadratic functionals of gaussian processes and least-action paths
Annales de l’I. H. P., section B, tome 27, n
o2 (1991), p. 239-271
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239-
Indefinite quadratic functionals of Gaussian processes and least-action paths
Terence CHAN
Statistical Laboratory, University of Cambridge,
16 Mill Lane, Cambridge CB2 ISB, U.K.
Vol. 27, n° 2, 1991, p. -271. Probabilités et Statistiques
ABSTRACT. - A method for
finding
the law of certain indefinite qua- dratic functionals of Gaussian processes ispresented.
This involves estab-lishing
aSpectral
Theorem forcompact operators
which are"selfadjoint"
with
respect
to a certain class of indefinite scalarproducts
andproving
aversion of Mercer’s Theorem for their
integral
kernels. Someapplications
are
given,
inparticular
anapplication
tochanging
the law of a randomharmonic oscillator
according
to theprinciple
of least action. Some rela- tions between the formulae established here and certain formulae forFeynman path integrals
are also discussed.Key words : Gaussian process, quadratic functional, random harmonic oscillator.
RESUME. 2014 On
presente
une methode pour calculer la loi de certaines fonctionnellesquadratiques
indefinies de processusgaussiens.
Dans cebut,
on etablit un théorème
spectral
pour desopérateurs compacts qui
sont«
auto-adjoints
» parrapport
avec une certaine classe deproduits
scalairesindefinis,
d’ou on deduit une version du théorème de Mercer pour leurs noyaux. On donnequelques applications,
enparticulier
uneapplication
au
changement
de loi d’un oscillateurharmonique aleatoire,
selon leprincipe
de moindre action. On traite deplus
les relations entre des formules etablies ici et certaines formules pour lesintégrales
de cheminsde
Feynman.
Classification A.M.S. : 60 G (60 F, 60 H, 8 1 C).
Annales de I’Institut Henri Poincaré - Probabilités et Statistiques - 0246-0203
Vol. 27/91/02/239/33/$ 5,30/© Gauthier-Villars
Let
Zt
be a continuous Gaussian process(in JR~).
We think ofZt
as afunction in a suitable Hilbert space with inner
product ( ., . ~
and weare interested in the
problem
ofcalculating
the law ofquadratic
functionals of Z of the formwhere S is a selfadjoint oounaea operator. in particular, we present a method for
calculating
theLaplace
transform(here,
E~ denotesexpected
value for the process Z withZo = z).
Initially,
we consideronly
centred Gaussian processes(so ZQ=0).
Laterwe will use the results for centred Gaussian processes to calculate the
Laplace
transform forgeneral starting point
z.Our motivation for
considering
thisproblem
is thefollowing.
Let(Xr, Vt)
be an Ornstein-Uhlenbecktype
model of asimple
harmonicoscillator: ’
1
where
B~
is a Brownian motion. We are interested in least-actionpaths
ofsuch a process. Since the
Lagrangian
of such a harmonic oscillator is(Vf - X~)/2,
the action of asimple
harmonic oscillator isprecisely
of the/*T
_____
form
( 1. 1 ) with f, g,>
= forf,
g in the Hilbert spaceL2 ([0, r], C2)
and theoperator
S is definedby
There are many other situations in which one is interested in the law of a
quadratic
functional of a Gaussian process and we shall consider a few of these as well.In the case where S is a
positive-definite
operator, there isalready
amethod for
evaluating
theLaplace
transform(1 .2).
Webegin by briefly reminding
the reader how this works. Forsimplicity
we take S = I.Let c
(s, t)
denote the covariance kernel for the process Z withZo
= 0:Annales de I’Institut Henri Poincaré - Probabilités et Statistiques
For a continuous process the kernel c
( . , . )
isjointly
continuous. Define the covarianceoperator
Cby
acting
on functionsfe
L 2([0, r], (Cd-valued
Lfunctions).
We denoteby ( . , . )
the usual innerproduct
onL2:
Strictly speaking,
weought
to considercomplex-valued
functions. However since C is anintegral operator
with realkernel,
it actsseparately
on thereal and
imaginary parts of f,
so forpractical
calculations we needonly
consider real-valued functions. As is
well-known,
the covarianceoperator
C withjointly
continuousintegral
kernel is apositive, selfadjoint,
trace-class
operator
on the Hilbert space ~ =L2 ([0, ’r], (we
do not assumepositive-definiteness
for C since it may ingeneral
have non-trivial nullspace). Therefore,
it has a countable set of realnon-negative eigenvalues
with
corresponding
orthonormaleigenfunctions
cpn. Mercer’s theorem states that the continuousintegral
kernel of apositive selfadjoint compact operator
admits aneigenfunction expansion
of the formFrom this we see that the process Z admits a series
expansion
of the formwhere §1, ~2 - - -
are i. i. d.N(0, 1)
random variables.Substituting
theexpansion ( 1 . 5)
for Z into( 1. 2)
we see that theLaplace
transform may be calculated asFurthermore,
we know that theproduct
at(1.6)
converges since C is trace-class(which implies that ~ (0).
We can then obtain anexplicit
formula in terms of a for the
product
at(1.6) by
means of the Hadamard factorisation theorem.(But
more on thislater.)
The aim of the present work is to extend this method to
operators
S which are indefinite and obtaineigenfunction expansions
which are anal-ogous to
( 1 . 4)
and( 1 . 5).
Vol. 27, n° 2-1991.
suppose that ~ has a bounded inverse. We can detine an indefinite scalar
product (., .)
on ~fusing
theselfadjoint operator J = S -1:
inner
product
and( . , . )
to denote the indefinite scalarproduct.]
Notethat
( . , . )
isnon-degenerate
i. e.(x, y) = 0
for allimplies
x= 0.
Let K be the
operator
on H definedby
mis is men an integral operator wnn integral Kernel ~~, ~.
The
operator
K isJ-selfadjoint,
thatis, selfadjoint
withrespect
to the scalarproduct ( . , . );
indeedanu me same calcultation snows maL iL is aiso j-positive:
Because K is
compact,
itsspectrum
consists of a countable set ofeigenva-
lues of finite
multiplicity. Let "An
be theeigenvalues
of K withcorresponding eigenfunctions
en. Because K isJ-selfadjoint, the eigenfunctions e’l
areJ-orthogonal [(ei, e~) = 0
if~7].
Theproof
of this is identical to the usualproof
from Hilbert spacetheory. Moreover, (K f, /)==( C/, f)
= 0only
iffe
kerC,
since from standard Hilbert spacetheory
we haveand Therefore
(K en, en) _ ~,n (em en)
= C en, >0,
so(en, en) ~ 0 and Àn
is real with sgn(~,n)
= sgnen).
In otherwords,
eigenspaces corresponding to positive (resp. negative) eigenvalues are positive (resp. negative) definite. (1 . 8)
This is a
particular
case of ageneral
result which is true for a muchbigger
class of
positive operators (see Chapter
VII ofBognar [1]).
For the moment, let us
ignore
mathematicalrigour
andbriefly
describehow we would like to
proceed. By analogy
with(1.4),
we would like to express the kernel k(s, t)
aswhence we would obtain a series
expansion analogous
to(1.5)
of the form.-
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
The
Laplace transform (1 . 2)
may then be calculatedby using
Notice that
J en
areprecisely
theeigenfunctions
of the usualadjoint K* = C J -1 of K,
so theeigenfunction expansion (1.9)
should come as nosurprise
to anyone familiar with the so-calledspectral representation
forfinite-dimensional matrices which are
diagonalisable
but notnecessarily symmetric.
2. RIGOROUS TREATMENT OF HEURISTIC CALCULATIONS IN SECTION 1
It turns out that
by
far thebiggest
hurdle on the way to ananalogue
of Mercer’s Theorem which would
give
theeigenfunction expansion ( 1 . 9)
is in
proving
ananalogous Spectral
Theorem forcompact J-selfadjoint operators.
Much of thepresent theory
of indefinite scalarproduct
spaces andspectral theory
forselfadjoint operators
is due to Krein. While Krein and others haveproved
the existence ofspectral
functions for variousoperators (e. g.
see the referencesgiven
inBognar [ 1 ]),
it is incredible that thesimplest
case - that of acompact selfadjoint operator
with countablespectrum -
does not seem to have been considered and asyet
there is noanalogous
result to thesimplest
of all thespectral
theorems.However, assuming
that such aspectral
theoremholds,
thatis, assuming
theeigen- functions en
of K span theimage
ofK,
it isquite
easy togeneralise
Mercer’s Theorem to the case of
J-selfadjoint operators.
Let usbegin by doing
so.First,
somesimple preliminary
results.Throughout
this section we usethe
symbols
1 and * to denote the usualorthogonal complement
andadjoint respectively,
while thesymbols (1)
and(*)
denoterespectively orthogonal complement
andadjoint
withrespect
to(.,.).
It is easy tosee that the
following relationship
holds between the two different kinds ofadjoint:
The
proof
of thefollowing
verysimple
result is left to the reader.LEMMA 2.1. - Let M be a
subspace of
ThenCOROLLARY 2 , I . -
If ~
M is a closedsubspace of ~,
then (-1.) = M.Vol. 27, n° 2-1991.
Proof -
so (-L)=(Here
wehave used the
analogous
classical result for from Hilbert spacetheory.)
0The result in
Corollary
2.1 isproved
in muchgreater generality
inBognar [ 1 ] .
LEMMA 2.2. - Let A : H ~ H be a linear operator. Then the
following properties
hold:(i )
ker A =(Im A *~)~1> ; (ii)
kerA~*~ _ (Im A)~1j ;
,(iii)
Im A =(ker A~*~)~1~ ; (iv) 1m A *> = (ker A)~1~.
Proof. - (i )
Given x E kerA,
we have.... ...F .I. ,., . ....
so ker A c Im A (*) On the other hand if v E Im A (*)
-L),
then for all y E~,
_ .. ,. -... ~ _ . _
The
proofs
of(ii)-(iv)
are identical to theproofs
from the usual Hilbert spacetheory, provided
we have the result ofCorollary
2.1. DIn
particular,
if A isJ-selfadjoint
then kerA (1-)
= Im A.However,
unless ker A isnon-degenerate [i. e.
for ally e ker A imply x = 0],
it is not true thatNow obviously, unless 1B. nas a non-generate nuu space [so that tne direct sum
decomposition (2 . 1 )
holds forK],
itseigenfunctions
cannotpossibly
span its range. We shall return to this
point
later. For now, let us prove therequired analogue
of Mercer’s Theorem for aJ-selfadjoint operator satisfying
the necessary conditions.THEOREM 2.3. - Let J be a bounded
selfadjoint
operator on the Hilbert space ~c° =L 2 ([0, ’t], ~d)
and let( . , . )
be anindefinite
scalarproduct defined
via J as in
(1 . 7)
and let K beJ-positive J-selfadjoint integral
operator withintegral
kernelk (s, t)
which is continuous on[0, [0, T].
LetÀn,
n =
0, 1 ,
... denote the(real) eigenvalues of K
with associatedJ-orthogonal eigenfunctions
en.Suppose
thatspan {
= Im K. Then the kernel k(s, t)
admits the
eigenfunction expansion
1
1 ne series converges uniformty in norm (i. e. tne natural norm Jor square
matrices)
on[0, ’t] X [0, ’t].
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
Proof -
Since the kernelk (s, t)
is real andcontinuous,
theeigen-
functions are real and continuous. Let
and let be the
integral operator
withintegral
kernelkn- By calculating
for
fEkerK
and1,
... it is easy to see that for Hencekn(t, t)
is aJ-positive
matrix for each t:[Here, kn (t, t) v
is a valued-function in ~f soJ kn (t, t)
makessense.]
Hence
Since
J ej (t) (J e f (t))~1~
ispositive-definite
in the usual sense, wehave,
forall v E
[Rd
Recall that
~,~ (ej, e~) ~ 0,
so each term in the above sum ispositive.
Hencethe above
inequality implies
Schwartz’s
inequality implies,
foreach v,
w e fRHence ior fixed s ana 8>u, tnere exists im sucn that ior n > m 4 m anu
each v,
for some constant A. Therefore for each fixed s and for
each v,
we have that
Vol. 2~. n° 2- 1 991 .
converges
absolutely
anduniformly
in t. Since the sum(2.2)
is bilinear in v, w, there exists amatrix k
such thatnn
wmcn oy virtue 01 ine unnorm convergence ana the conunuity 01 ej(.), is continuous in t for fixed s.
Consider now
since
Iff=en
for some n, theright-hand
side isHence
by
virtue of thecontinuity
ofk(s, .)-R(s, .),
we havek (s, ~)=~(~ ~
for all s, t.We now prove the uniform convergence in s and t of the
eigenfunction expansion
fork (s, t).
For all v E!Rd
we havenon-decreasing sequence of
continuous functionsconverging pointwise
tothe continuous function
t) v.
Henceby
Dini’s Theorem the series convergesuniformly
in t.For 8>0,
there exists N such that forn > m >_ N
n ~
application Ul s inequanty as before then gives LnaL mere exists N such that for n
> m >_
Nwhence it follows that
U n ~ t)
Next we prove that the
eigenfunctions
of the operator K from theprevious
section span theimage
ofK,
so that our formal calculations at(1.9)-(1.11)
are indeed valid. In view of themotivating example (1.3)
and in view of the fact that
operators
defined viasymmetric
matrices asin that
example
are aparticularly simple
class ofselfadjoint
operators,Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
one would
ideally
like to prove ageneral analogue
of theSpectral
Theoremfor
J-selfadjoint operators,
where J is defined via asymmetric matrix,
which would allow us to do the calculations at
(1.9)-(1.11)
for any n-dimensional Gaussian process. It is therefore a
great disappointment
anda source of immense frustration
doing
this hasproved
for the moment tobe
intractably
difficult. Here we prove the existence of aspectral expansion
for scalar
products
of the formwhere S is a compact
selfadjoint operator.
For a Gaussian processZt,
weshall then be able to find the law of any
quadratic
functional of the form~Z, (I + S) Z ) .
The
key
result is thefollowing
theorem due to Krein.THEOREM 2.4. - Let H be a compact
selfadjoint
operator on a Hilbert space.Suppose
H is either a trace-class or a Hilbert-Schmidt operator. Let S be a compact operator and suppose B is a compactperturbation of
H inthe sense that B =
(I
+S)
H.Suppose further
that B does not vanish on anynon-zero element
of
the closureof
itsimage.
Then the root vectorsof
Bspan the
image of
B.This theorem is a
special
case of a moregeneral
one(Theorem V.S.1)
in
Gohberg
and Krein[5].
In
general,
we cannothope
that theeigenvectors
of anon-selfadjoint ("selfadjoint"
is taken to have the usual Hilbert spacemeaning here) operator
will span theimage
of thatoperator;
we need to consider other root vectors aswell,
as in Theorem 2.4 above.However,
forJ-selfadjoint operators (J
here is any boundedoperator
via which an indefinite scalarproduct
isdefined)
we have thefollowing
result.LEMMA 2.5. - Let A be a non-zero
eigenvalue of
aJ-positive
andJ-selfadjoint
operator B. Then the associated root space coincides with theeigenspace:
Proof -
SinceÀ # 0
the associatedeigenspace
is eitherpositive-definite
or
negative-definite by
the result(1. 8).
Now suppose that for someinteger
rc >__ 2
and some functionf,
we have butThen f1 :
-(B-03BB I)n-1 f is
a non-zero element of theeigenspace
associatedwith 03BB and we have
which contradicts the definiteness of the
eigenspace
associated with À. DVol. 27, n° 2-1991.
We are now in a
position
to prove our main result in thissection,
whichis that the calculations
( 1. 9)-( 1. 11 )
are valid if thequadratic
functional( 1.1 )
arises as acompact perturbation
of theidentity.
--
THEOREM 2.6. - Let
Zt
be a continuousRd-valued
centred Gaussian process with co-variance kernel c(s, t)
= [E(ZS Z).
Denoteby
C theintegral
operator with kernel c
(s, t)
and suppose that C is apositive-definite.
Let Sbe a compact
selfadjoint
operator. Then(2 . 3) for
a1 /"AmiD’
where is thelargest (positive) eigenvalue
and"AmiD
is the smallest(negative) eigenvalue of (I
+S)
C and the determinant isdefined
as theproduct of eigenvalues.
Proof -
Since S iscompact,
the non-zeroeigenvalues
of(I + S)
are bounded away from zero, so if 0 is in thespectrum
of I + S it is in thepoint spectrum. Moreover,
thenull-space
of I + S is at most finite-dimensional. So
by discarding
a finite-dimensionalsubspace
of wemay assume we are
working
on a Hilbert space on which(I + S) -1
1exists and is bounded. Let K =
(I + S) C
and let J =(I + S) -1.
Define ascalar
product ( . , . )
via J in the usual way as in(1 . 7).
Theoperator
K hasintegral
kernelk (s, t) = J -1 c ( . , t) (s),
that is theoperator J -1 acting
on the function c
(., t)
for fixed t,with
the result evaluated at s.Moreover,
we know from Section 1 that K is
J-selfadjoint
andJ-positive,
and that ifand en
arerespectively
theeigenvalues
andeigenfunctions
ofK,
then the are real andthe en
areJ-orthogonal.
The condition that C is
positive-definite implies
that it has trivial null- space. It is clearthat,
since C does not vanish on any non-zero function and since we areinsisting
that(I
+S)
also hasonly
trivial null space(by discarding
a one-dimensionalsubspace
ifnecessary),
thedecomposition (2.1)
holdstrivially.
Therefore K =(I + S) C
is acompact perturbation
ofa
selfadjoint operator
which vanishesonly
at 0 on ImK,
and hence theeigenfunctions en
span Im Kby
Lemma 2.5 and Theorem 2.4.By
Theorem
2.3,
theintegral
kernelk (s, t)
has anexpansion
of the formHence Zt admits an expansion
/2014:20142014
Annales de Henri Poincaré - Probabilités et Statistiques
where 03BEn
are i. i. d.N(0, 1)
random variables. The result(2.3)
nowfollows. D
The
assumption
that C ispositive-definite
is used in theproof
ofTheorem 2.6
only
to show that the direct sumdecomposition (2. 1)
holdsfor K: in this case
(2 .1 )
holdstrivially.
Ingeneral,
if C has non-trivialnull-space,
Theorem 2.6 still holds if we can prove(2 . 1 ) directly.
We conclude this section
by briefly discussing
how onemight
calculate(1.2)
for a non-zerostarting point
z. So suppose now thatZo = z
and letThen
Zt:=Zt-m(t)
is a centred Gaussian process. We write+2(~(.),
+(~(.), {I+S)m(.)~. (2.6)
We know how to handle the first and last terms in
(2.6) (the
last term isjust
a deterministicfunction). By
means of theeigenfunction expansion (2.4)
for the centred Gaussian processZ,
we may write the second term in(2 . 6)
as[Recall
that J =(1+S) ~.] Putting
thistogether
with the contributions of~
from the first term in(2.6)
and thencompleting
the square, we seethat the
Laplace
transform of(2. 6)
may be written as the infiniteproduct
of
Laplace
transforms of non-centralx-squared distributions,
the non-central
x-squared replacing
the centralx-squared
distribution in(2. 5).
Perhaps
a better idea of how such a calculation works inpractice
canbe
gained
from the detailedexample
in the next section.3. DETAILED CALCULATIONS FOR A SPECIFIC EXAMPLE We return to our
motivating example
mentioned in Section 1. Let Z =(X, V) satisfy
the SDEwith initial condition
Z~=z= (~ ~).
Consider first the caseZo=0.
Forfixed 1, we wish to find the law of
,.-
where J is the matrix
Vol, 27, 11a 2- 1 991 .
Notice that J is both
symmetric
andorthogonal.
We cannotapply
Theorem 2.3
directly
since J = I on an infinite-dimensional invariant sub- space and so is not acompact perturbation
of theidentity. However,
the process Z does have thefollowing
crucialproperty:
/*t
Bearing
in mind theproperty (3 .4),
it is easy to showusing integration by parts
and Fubini’s Theorem that~-
where A is tne compact posmve-aeimite selfadjoint operator defined oy
~T
1 ne operaior y - A.) nas elgenvames
.,
Let
C1 (s,
be the covanance kernel ot V, which detmes anintegral operator C 1
in the usual way. Set Now let(I - A) - 1
and define an indefinite scalarproduct ( . , . ) 1 on H
1 viaJ 1
as before. We can thereforeapply
Theorem 2.3 to thequadratic
func-tional
(3 . 5).
Inparticular,
we haveand
where 03BDn
and fn
are theeigenvalues
andeigentunctlons
ofK1.
However, from thepoint
of view ofpractical calculations,
this would meandoing
calculations which involve the
composition
of twointegral
operators as well as the inverse of anintegral
operator, and ingeneral
one wouldexpect
this to beconsiderably
harder to handle thanworking
with theoperator
K = J C where J is the matrix at(3 . 3).
In any case the calculations( 1 . 9)-( 1 .11 )
for a matrix J are morenatural,
since as we havealready
said at the
beginning
of thissection,
one would like to be able to do calculationsalong
the lines of(1.9)-(1.11)
for ageneral
Gaussian process notnecessarily satisfying (3.4),
and ourapproach
here is in many ways somewhatunsatisfying.
Annales de I’Institut Henri Poincaré - Probabilités et Statistiques
Because of
(3 . 4),
it turns out that we can in factapply
the results(2 . 3)- (2. 5)
to the case where J is the matrix at(3 . 3)!
LEMMA 3 .1. - Let Z =
(X, V)
be the Gaussian processsatisfying (3 , 1)
and let J be as at
(3 . 3).
Denoteby
C the covariance operatorof
Z and letK = J -1
C = J C. Denoteby (en , ef)
theeigenfunction of K corresponding
to
eigenvalue Ån-
ThenProof -
We denoteby c (s, t)
the covariance matrix ofZ:c(~~)=E(Z~Z~).
Because of theproperty (3 . 4),
Fubini’s Theorem allows us to writec (s, t)
in terms of c 1(s, t):
Write
J(t)
=(fA (t), (t))
and consider functions of the form g(s)
=C f(s) .
We have _
Differentiating
the aboveexpression
withrespect
to s shows that.J
and
consequently
theeigenfunctions en exactly
reflect thephase-space
structure of Z:
j
In the
present
context, Lemma 3.1implies
thatsince we have the obvious
boundary
condition en(0)
= O.LEMMA 3.2. - Let
Z, J, K, Àn
and e,~ be as is Lemma 3.1. The(2 . 4)
and(3 . 6) give
the sameexpansion for V,
and inparticular
Theorem 2.3 holdsfor
thequadratic functional (3 . .2).
Vol. 27, n° 2-1 991 .
Proof - Comparing
the twoexpansions (2.4)
and(3.6)
forV,
we see that we must prove---_~ ;r fit ~ ,, "
where we nave written en= (e, en). Because 01 ( ) Lne equation
lUl’ the en component
An
integration by parts
shows thatwe calculate
Annales de Henri Poincaré - Probabilités et Statistiques
But the last two terms on the
right-hand
side areprecisely
those on theleft-hand
side,
soHence the
equation
reads for theef component
’l’heretore
is an
eigenfunction
1corresponding
toeigen-
value This proves the result
(3 . 8)
andhence
the lemma. DWe can now
proceed
to find the law of(3.2).
The SDE(3.1)
can besolved
explicitely by observing
thatis a local
martingale:
Vol. 27, n° 2-1991.
Hence the covariance matrix c
(s, t)
may be calculated asWe now find the
eigenvalues
ot K = C[recall
that we take J to be the matrix at(2.4)].
We haver~
Differentiating
the aboveexpression
with respect to sgives
- ~ . B
Annales de Henri Poincaré - Probabilités et Statistiques
Differentiating again gives
Now notice that
so
therefore,
from(3 . 10)
and(3.11)
we see thatand also from
(3 . 9)
and(3.11)
27, n° 2-1991.
Adding (3 .12)
and(3 .13) yields
If we write
as nerore, men 14) simply says
which we know already trom tne previous section. using tne equation
(3.15)
andintegration by parts
we obtain-- ,-
Hence
Adding (3.9)
and(3.11)
we then obtain thefollowing
set ofinhomogen-
eous ODEs for the
eigenfunctions:
" / V ~B,Y~~B Y~~.Y~~ ~ / v ~~~~B v
[from (3 . 9)] en (0)
= 0.By taking
a suitable linear combination of(3 . 9)
and
(3. 10),
we see that the otherboundary
condition must beinus me two natural oounuary conditions lor me ODEs (3. 1/) are
-x/nB2014.,v/fnB 2014.n /Q 1 ’7 .-. B
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
Equations (3 . 17 a)-(3.
17b)
have the samehomogeneous
solution and the functions(of s) en (L)
cos(t - s)
and -en (L)
sin(L - s)
areparticular
sol-utions of
(3 .17 a)
and(3 . 17 b) respectively.
We need to consider three cases.
Firstly
if or?~n - l, (x
+1 )/?~,~
> 0 so the solution to(3 . 1 7)
iswhere
[using (3.17 c)
and(3 . 15)]
Finally,
theboundary
condition(3. 17 d) gives
thefollowing equation
forthe
eigenvalues ~:
"[We
are of courseonly
interested in thepositive
solutions of(3 . 19). J Next,
for -1~0, (~+ 1 )/~,,~ o
so the solution to(3 . 17)
is~~ ... _ ~ _ r ~~ .. ~ ~. _
wnere J
Again,
theboundary
condition(3 . 17 d) gives
rise to anequation
for~,,~
as follows:
For sin
(3 . 21 )
may be written asVol. 27, n° 2-1991.
which has no solution unless
When (~. 23) holds,
(~. 22)
has a uniquepositive
solution.The third case we need to consider is
when Àn
== 2014 1. In this case unless_~*~ _ ! ~-~ _ rB ~^~ ~~t~ B.
the
equations (3 . 17)
have no solution.Notice that for L
~/2,
weonly get positive eigenvalues
as we wouldexpect,
since we know from theprevious
section that theoperator
K ispositive-definite
forRecall from Section 1 that
(for Zo
=0)
- ¿t’ 18- 1 -~
We are now in a
position
to calculate theproduct
oo
We know that tor
/~>U,
we have1/(~2014 1) where xn
arethe
positive
solutions of(3 .19).
However even for -1~0,
we couldobtain 03BBn by solving
anequation
of the form(3.19)
since(3.21)
may be written as(3 . 19)
if we let xn = +1 )/~,n
bepurely imaginary,
in whichcase we still have
03BBn=1/(x2n-1). Also,
sincewe see tnat (3.19) is consistent with j3 . 24) even m the case 03BBn=-1. Thus
irrespective
of the value ofÀn,
it is valid to saythat Àn
solves(3.19)
with~=1/(~-1).
Now the roots occur at
oc = -1 /~,n =1- x~,
which because ofequation (3.19),
coincide with the real roots ofn~~~ /i -
Suppose
nowthat y
has othercomplex
roots andlet ç ==
x+ iy
E Csatisfy
. / t-’B
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
If either x=0 or
y = 0
then thisjust
reducesrespectively
to theequations (3.19)
and(3.21),
so we suppose that neither of xand y
iszero. For sin i ~
0,
the aboveequation
may be written asWriting
this in terms of the real andimaginary
partsgives
Since x~ 0
and y~
0 we can divide these twoequations respectively by x and y
togive
and, putting ~=2r~ ~=2r~
thisimplies
thatwhich is
impossible,
sinceand
equality
holdsonly
whenx’ == y’ == O.
This shows that the functionf (ac)
hasonly
those real roots which we have described and no others.Also,
since sin
(r x)/x ~
Tas x ~ 0,
thefunction f
has a removablesingularity
ata = and so we may
regard f as
anintegral
function.Moreover,
since thefunctions sin
Jç
and cosJç
are both of order1/2, f ’ (oc)
has order1/2
and
since /(0)==1,
the Hadamard factorisation theorem(see
Titchmarsh
[8]) gives
and we
have finally
arrived at the formulaIt is worth
emphasising again
that(3 . 26)
isonly
valid for certain values of a[viz.
those for andthat,
when there arenegative
aswell as
positive eigenvalues,
the admissible values of a must lie in an Vol, 27, n° 2-1991.interval around
0,
viz.where is the
largest (positive) eigenvalue
and is the smallest(negative) eigenvalue. Moreover,
notice from(3.19)
that as iincreases,
both and increase in
modulus,
so that(3.27) gives
a smallerand smaller admissible range of a.
Let us now see how to calculate the
Laplace
transform( 1 . 2)
for a non-zero
starting point v). Let Z
be defined as in Section 2 with/ .-. ~, ~ t B
We deal with each of the terms in
(2.6)
in turn. The first term may be dealt withby
means of(3.25).
The last term iseasily
calculated to be[Note
that(3.28)
isprecisely
the action of the classical least-actionpath starting
at(x, v).]
Consider now the second term in(2.6) which,
as wehave seen, may be written as in
(2.7).
Theintegral
on theright-hand
sideof
(2.7)
may be evaluatedby integration by parts:
- A
If we now normalise the
eigenfunctions
so thate) (I) - 1,
this thengives
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
where we have
put
In our calculation
(3.30)
we have used the formula for theLaplace
transform of a non-central
x-squared
distribution with 1degree
of freedomand
non-centrality parameter (by
which we mean the square of themean of the
corresponding
Normaldistribution),
with theLaplace
transform
being
evaluated at theargument
For the moment we donot bother to calculate the scalar
product (em en) because, except
for somespecial
valuesof T,
the calculation is rather messy.For a as in
(3.27),
let G(a, r)
denote thenon-negative
function(The
above sum caneasily
be shownby
direct calculation toconverge.)
Ifwe
put G (ex, 0)
= 0 for all a thenG (., .)
isactually jointly
continuous.Thus, by
means of(3.26)
and(3.30)
we have now arrived at the formula*2014 ~ ti - ~ 2014*
4. SOME CONNECTIONS WITH FEYNMAN INTEGRALS In this section we show how the results obtained so far may be inter-
preted
in the context of results onFeynman integrals,
inparticular
theCameron-Martin formula and the translation formula of
Elworthy
andTruman
[4]
forFeynman integrals.
We first recall the essential
ingredients
fordefining
theFeynman path integral.
Let H be a Hilbert space ofpaths
with innerproduct (., . )H’
The basic class of
Feynman-integrable
functionals are those functionalsVol. 27, n° 2-1991 .
f:
H - C which are Fourier transtorms otcomplex
measures on H withbounded variation. We denote this class of functionals
by F (H);
thusf E ff (H)
if andonly
iffor some
complex measure f.
TheFeynman integral
will bedenoted
by (See [4]
and the many referencesgiven
there for how suchintegrals
areconstructed.)
In
[4], Elworthy
and Truman obtained thefollowing
Cameron-Martin formula for theFeynman integral:
THEOREM 4 . 1. - Let L : H - H be a trace-class and
selfadjoint (with
respect
to ( . , . )>n!)
linear operator such that(I
+L)
is abijection.
Letg : H -~ C be
defined by
where
(H).
Then g isFeynman-integrable
andHere det is the Fredholm determinant and the index
ind (T)
of anoperator
T withcountably
manyeigenvalues
of finitemultiplicity,
is thenumber of
negative eigenvalues
ofT,
countedaccording
tomultiplicity.
We also have the
following
translation formula which isanalogous
to thefamiliar Cameron-Martin-Girsanov
change
of drift formula for Brownian motion. For a EH,
let ga: H --+ C be definedby ga (y) = g (y + a), g
E %(H).
Then the functional
)n}~(-)
isFeynman integrable
andF(exp{~~}~(.))=exp{-~- ~/2}P~). (4 . 3)
Return now to our
original setting
and continue with the same notation established in theprevious
sections. We define a Hilbert space ofpaths
Hto be the space of
absolutely
continuous functionsf
on[0, T] satisfying
the
boundary
conditions[These boundary
conditionscorrespond
to(3 .17 c)-(3 .
17d)].
The innerproduct
on H isgiven by
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques
(Recall
thatC 1
is the"velocity"
covarianceoperator
as defined in Section 2and ( . , . )
is of course the same asever.)
Let L be the operatorL = C1 (I - A).
Notice that L isthe ( >-adjoint
ofK1
and is( )H-selfadjoint.
Inparticular,
L has the sameeigenvalues
asK1.
Observe now that the action of the harmonic oscillator may be written
as
/*~-
so the
Laplace
transform of the action as at(3.25)
isjust
Moreover the
product
on theright-hand
sideof
(3 . 25) ~ ( 1 + a,~,n) -1~~
isnothing
but since L is has acontinuous kernel and so the Fredholm determinant is the same as the characteristic determinant. Notice also that the range of admissible values of a ensures that
(I
+ aL)
ispositive-definite,
so its index is zero.Therefore,
the formula
(3.26)
is the "Cameron-Martin formula" for Wienerintegrals
which is the exact
analogue
of the formula(4.2)
forFeynamn integrals (with fm 1).
It is also clear from the way(3.32)
was arrived at that itcorresponds exactly
to the translation formula(4. 3)
forFeynman
inte-grals.
The formula(4.2)
is itselfanalogous
to the classical Cameron- Martin formula[2]
for the Wienerintegral
of linear transformations of Brownian motion. In fact Cameron and Martin[2]
were able to use theirformula to obtain the
Laplace
transform of11 B;
ds.Pushing
theanalogy
withFeynman integrals
a littlefurther,
observe that while we do not have anexplicit representation for ( . , . >H
as we donot know what
C 1 is explicitly,
we do have a naturalreproducing
kernelfor H: this is
simply
the covariance kernel c1 (s, t),
forby definition,
ithas the
property
thatThese connections become even more
transparent
if we consider the verysimple example
of Brownian motion started at 0 and run up to time 1.Let G (s, t) = s n t, s,
t E[0, 1],
be the covariance kernel. The covariance operator haseigenvalues
and
eigenfunctions
cpnsatisfying
cpn(0)
= 0 andcpn ( 1 ~
= O. The natural Hil- bert space ofpaths
H is the space ofabsolutely
continuous functions on[0, 1] satisfying
these sameboundary conditions,
with innerproduct
VoL 27, n° 2-1991.