A NNALES DE L ’I. H. P., SECTION B
R OSS P INSKY
M ICHAEL S CHEUTZOW
Some remarks and examples concerning the transience and recurrence of random diffusions
Annales de l’I. H. P., section B, tome 28, n
o4 (1992), p. 519-536
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519-
Some remarks and examples
concerning the transience and
recurrenceof random diffusions
Ross PINSKY
(1)
Michael SCHEUTZOW
Technion - Israel Institute of Technology, Haifa, Israel
Technische Universitat Berlin, Allemagne
Vol. 28, n° 4, 1992, p. -536. Probabilités et Statistiques
ABSTRACT. - Let
k (t)
be anergodic
Markov chain onE== {1,2,
...,~}
and let
for
k =1, 2, ... , n.
Then for each realization of the Markov
chain, Lk(t)
maybe
thought
of as a timeinhomogeneous
diffusion generator. The process it generates,X (t) = X (t; k ( . )),
will be called a random diffusion. Wegive examples
such that eachLk
generates apositive
recurrent(transient)
diffusion but such that the random diffusion is a. s. transient
(positive recurrent).
We alsogive
some results andexamples
for transience andrecurrence in the
special
case that(X (t), k (t))
is a reversible process.Finally,
we consider the effects ofspeeding
up orslowing
down thejump
rate of the Markov chain.
(1) This work was supported by the Fund for the Promotion of Research at the Technion.
Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques - 0246-0203
1. INTRODUCTION
Let
k (t),
be anergodic
Markov chain on the finite state spaceE={ 1, 2, ...,~}
with generator Gand,
for eachk E E,
define on Rd thesecond order
elliptic
operatorThen for each realization of the Markov
chain, Lk(t)
may be
thought
of as a timeinhomogeneous
diffusion generator(with
coefficients
aij (x; k (t, (0))
and~(~; k (t, co)).
We willalways
assume thatk) ~
is continuous andpositive
definite and thatk), i =1, 2, ..., d,
are bounded on compacts andmeasurable;
it then follows from[8]
that generates aunique time-inhomogeneous
diffusion pro-cess X
(t)
= X(t; k ( . )). (We
allow for thepossibility
ofexplosion;
if theprocess
explodes
thenX (t; k ( . ))
is defined up to theexplosion time.)
Wecall such a process a diffusion in a random
temporal
environment orsimply
a random diffusion. Note that in fact X(t)
may bethought
of asthe first component of the Markov process
(X (t), k (t))
E Rd x Egenerated by
L + G.( L
+ G acts on smooth functionsf (x, k)
defined on Rd x Eby
B
(L + G) ( f ) (x,
Fromthis,
the compact-ness of E and the
ergodicity
ofk (t),
it is easy to see that the random diffusion X(t)
is almostsurely
recurrent(transient)
if(X (t), k (t))
is recur-rent
(transient).
In[6],
necessary and sufficient conditions were obtained for the transience or recurrence of a certain class of random diffusions.In certain cases, it turned out that each
Lk generated
a transientdiffusion,
but that the random diffusion was recurrent. This was because the direc- tions of transience of the various
Lk’s
cancelled one another out. Thequestion
was then raised whether the random diffusion mustnecessarily
be recurrent if all the
Lk’s
are recurrent generators.In section
2,
we construct two somewhatsurprising examples
onR~,
where the
topology plays
no role. In the firstexample,
we construct twopositive
recurrent generators such that the random diffusion obtainedby switching
from one generator to the other at random times is transient.Then we construct two transient generators such that the random diffusion constructed from
switching
from one to the other at random times ispositive
recurrent.These
examples
couldeasily
be constructed on Rby defining everything
symmetrically
on R - . The reason we choose to consider R ± isthat,
withregard
toexample
two, it is easy to construct two transient generatorswhose drifts
point
inopposite
directions and which get cancelled out in the random diffusion. Thetopology
of R + does not allow for suchcancellation.
It is
usually
difficult to establish transience or recurrence results for random diffusions and the above discussion shows that ingeneral
onecannot deduce transience or recurrence from the
corresponding
behaviorof the
Lk’s. However,
it turns out that if the Markov process.(X (t), k (t))
is
reversible,
then the situation is moreorderly.
The class of random diffusions for which(X (t),
k(t))
is reversible is described in thefollowing proposition
whoseproof
isgiven
in section 3.PROPOSITION l. - In addition to the
ellipticity assumption
madeabove,
assume that
(x; k)
E C2(Rd)
andbi (x; k)
E c1(Rd) for
eachk=1, 2,
..., n.Then the Markov process
(X(t),k(t))
is reversibleif
andonly if
G isreversible and the
Lk
are all reversible with a common(~ finite)
invariantdensity;
that is, theLk
have theform
The measure with respect to which the process is reversible is the
product
measure x p on Rd X
E,
where p is the measure with respect to which G is reversible.n
In the reversible case, define
a (x) _ ~
a(x; k)
where{ ~k ~ k =1
andk=1 a
(x; k)
are as inProposition 1,
and defineThus L is the operator obtained
by averaging
the coefficients ofLk
with respect to the invariant measure ofk (t).
We have thefollowing
theorem.THEOREM 1. - Assume that the Markov process
(X (t), k (t))
is reversible.(i) If
at least oneof
theLk’s
is transient, then the randomdiffusion
isa. s. transient.
(ii)
Allof
theLk’s
arepositive
recurrentif
andonly if
the randomdiffusion
is a. s.
positive
recurrent.(iii ) If
theaveraged
operator L is recurrent, then the randomdiffusion
isa. s. recurrent.
The diffusion
generated by
g YL - 1 ~ .
k 2 a(x; ( ~ )
+ a(x; ( ~ ) Q Q( ) (x) . V
ispositive
recurrent if andonly
if it does notexplode
and dx 00.Therefore,
in the reversible case, if all theLk’s
are recurrent, thenthey
will either all be null-recurrent or all
positive
recurrent. Theorem 1 then leaves open twoquestions:
( 1 )
If all theLk’s
arenull-recurrent,
must the random diffusion necess-arily
be recurrent?(2)
Does the transience or recurrence of L determine that of the random diffusion? Thatis,
if L is transient must the random diffusionnecessarily
be transient?
We will answer each of these
questions
in thenegative
with anexample.
The reversible case will be studied in section 3.
In section
4,
we consider the random diffusionX (t) corresponding
tothe Markov process
(X (t), k (t)) generated by
L + ÀG,
where ~, is apositive
parameter.Thus,
thelarger (smaller)
Àis,
the faster(slower)
theswitching
rate is. We discuss the behavior of
k),
the invariantprobability density
for(X (t),
k(t)) (assuming
that(X (t),
k(t))
ispositive recurrent)
as X - 0 and X - oo .
2. TWO EXAMPLES
Example
l. -L1
andL2
arepositive
recurrent, but the randomdiffusion
is transient.
Before
constructing
theexample,
weprovide
a bit of intuition. We will construct two diffusions on R + with reflection at 0 with generatorsL - 1
kd 2 k = 1 , 2.
The idea is toconstruct bi
such that it is2 dx 2
dx
> > 1periodic
on R + withlong
low mesasinterspersed
withdeep
canyons such that theintegral of b 1
over aperiod
isnegative,
which makesL 1 positive
recurrent.
Then b2
is chosen to bebi
shiftedby
half aperiod.
Ineffect,
the chances are
high
that when the medium switches from 1 to 2 the process is close to thenegative
side of a canyonof b1
and afterswitching
feels the
positive
drift ofb2,
and therefore moves to the next canyon ofb2
in thepositive direction,
and so on.Let a denote a
positive
number to be chosen later. Chooseb 1°‘~ (x)
E(R) satisfying
Now define
~(jc)=~(jc+1),
x e R. We consider the diffusion generators= -
20142014 +
~2014,
k =1,2,
on R + with reflection at 0.[This corresponds
2 dx dx
to the
boundary
conditionu’ (0) = 0.]
Thedensity
of the invariant measurecorresponding
to with theboundary
conditionu’ (0) = 0
satisfies theadjoint equation [[«> v (x) = 0
for and(i/-~)(0)==0,
and isgiven by
Since
Jo
exp1 (x)
oo for k=1, 2,
it follows that an invari-0 0
ant
probability
measure exists for and thus generates apositive
recurrent diffusion.
We will show that for a
sufficiently large,
the Markov processgenerated by
L(03B1)+G istransient,
whereG=( )
andL(«) B L(03B1)
= (L(03B1)1 L(03B1)2
As noted in theintroduction,
it then follows that the random diffusionX(t)=X(t,k(.))
is almostsurely
transient.To prove the transience of
(X (t), k (t)),
it isenough
to find anre[0, oo)
and a function Ve C °°
([r, oo ) X ~ 1,2 }) satisfying
To see
this,
let forYE[O, oo).
Thenby
Ito’sformula,
for n>x>r,Since
the bk
arebounded, clearly
oo as n - oo . Thus we obtainIt remains to find a V
satisfying properties
1-3 above. To dothis,
wefirst choose
W 1
andW2 ([0,1]) satisfying
Now define
Clearly
V E COO(R + )
and conditions( 1 )
and(2)
on V are satisfied for r=0.To
verify
condition(3),
we consider two cases.Case 1. -
[x] is
even and k =1 or[x]
is odd and k = 2. Thenand
Thus, by
condition(iv) above, (Lk°‘~
+G)
V(x, k) >_
0 for all x ? 0 and k =1 , 2 provided
a issufficiently large.
Case 2. -
[x]
is odd and k =1 or[x]
is even and k = 2. Thenand
For
X-[X]~[1 3,2 3], W2
is constant and henceby
condition(vi)
above.For
From conditions
(iii)
and(vi)
and the fact thatW2
issmooth,
it followsfor
- 1 and y=2 3.
Thususing
condition
(v),
it follows that for all andk =1, 2 provided
a issufficiently large.
Thiscompletes
theproof.
Example
2. -L1
andL2
are transient, but the randomdiffusion
ispositive
recurrent.For this
example
we switch the direction of the drift in theprevious example.
LetBk°‘~ (x) _ - bk°‘~ (x), k =1, 2,
wherebk°‘~ (x)
is as inExample
1.We consider the diffusion generators g
Lj~=-2014-+B~2014,
kk= 1, 2
on2 dx2 k
dx
’[0, oo)
with reflection at 0. Since~0 exp ( / - 2 r B(03B1)k (x)
00, A; =1,2,
it follows that the
L(03B1)k
are transient generators[2, Chap. 9].
We will show that for asufficiently large,
the Markov process(X (t),
k(t)) generated by
L(0152) + G ispositive
recurrent, where G=1
- 1 1
andL(03B1) = (L(03B1)1 L(03B1)2)
. Todemonstrate
positive
recurrence it isenough
to find a V E Coo(R +
x{ 1, 2 })
satisfying
and
To see that this is a sufficient condition for
positive
recurrence, let Vsatisfy (i )
and(ii )
aboveand,
say,Then
by
Ito’sFormula,
for x > c,and
letting
t ~ ~, one obtains for x > c.By
Khasminskii’s construction[4], Ex, x
tc oo isequivalent
topositive
recurrence.Define
Wi, W2
E C °°([0,1 ], R)
with thefollowing properties:
Now define
A calculation similar to that in
example
one shows that V satisfieslim +
G)
V(x, k) = -
~, if a issufficiently large.
3. THE REVERSIBLE CASE
For the processes we are
considering,
a necessary and sufficient condi- tion forreversibility
is that there exist a reference measure v(dx, k) ~ 0
such that
We
begin
with theProof of Proposition
1. - We must show that theequality (3 .1 )
holdsfor all if and
only
if L is of the formspecified
in thestatement of the
proposition
and G is reversible. One direction is clear: if Jl is the invariantprobability
measure of G and is the commoninvariant
density
of theLk’s,
then(3 .1 )
holds with v(dx, k)
= dx.Now assume that
(3 .1 )
holds. First we will show that for allkEE,
Let and assume that
Eo
is notempty. Let
ko E Eo
and fork~k0 in (3 .1 ).
Then the lefthandside of
(3 .1 )
vanishes and therighthand
sideequals
Thus,
this must vanish forall fk0,
gkEC~c (Rd),
all k E E -Eo
and allko
EEo.
This
implies
thatGk,
ko =0,
for k E E-Eo
andk0 ~ Eo
which contradicts theergodicity
of k(t).
By fixing
ako E E
andtaking g~ = fk = 0
for one sees that eachLk
must bereversible,
thatis, Lk
must be of the formin which case the
density
with respect to whichLk
is reversible is e2Qk.Therefore,
v is of the form v(dx, k)
= y~ dx where y~ > 0 for all k e E.Without loss of
generality,
we may assume thatQk(O) = 0
for all k E E.For v of this
form,
we havefor
cj° {Rd).
Thus for(3 . .1)
tohold,
we needfor all
(Rd).
Fixk,
mEE andchoose fj~0 for j~k
andfor j#m. Substituting
this in(3 . 2)
allows us to conclude thatLetting
x = 0 shows that G is reversible and thaty={y~}
is amultiple
ofFurthermore, Qm
wheneverGkm > o.
Since G isirreducible,
it follows thatQi=Q2=
...= Qn-
Thiscompletes
theproof.
In the reversible case, transience or recurrence can be characterized
by
a
simple
variationalprinciple.
and for any functionm n
v={vk}
onE, Then vGv~~0
withequality
if andk= 1
only
if v is amultiple
of the vector 1. In the reversible case, this follows from theRayleigh-Ritz
formula and the factthat, by ergodicity,
zero is asimple eigenvalue. (In
fact the result holds even in the nonreversiblecase.)
The
quadratic
form associated with L + G on any domainD,
whereT
IS
where
THEOREM
(Ichihara [3]).
- LetThen
Àm
is monotonedecreasing
and the reversible process(X (t), k (t))
isrecurrent
if
limÀm
= 0 and transientif
lim~,m >
O.m - m
Remark. - Ichihara
actually proved
thistheorem,
for operators of theform 1 ~ ~
a(x) V,
but it is easy to extend it to the L + G where L is as2
( ) ~
Yabove
(see
also[7]).
We can nowgive
theProof of
Theorem l. - Definefor
k =1, 2, ... , n
and and definePart
(i ). By
Ichihara’stheorem, Lk
is recurrent or transientaccording
to whether lim
03BBkm=0
or lim03BBkm>0. Thus, by assumption,
there existsa
ko
such that lim03BBk0m
> 0.However,
we haveJm (u) >_ Jlko Jk0m (uko)
for anyu = (ul,
...,un), because - (
u Gu ~ >_
o.We conclude that and thus lim
X > 0. By
Ichihara’s theo-rem
again,
it follows that(X (t), k (t))
is transient and thus the random diffusion is almostsurely
transient.Part
(ii ).
It is easy to see that the random diffusionexplodes
if andonly
if at least one of theLk’s
generates anexplosive
diffusion. Sinceexplosion precludes
recurrence, we may assume that the random diffusion and the diffusionsgenerated by
theLk’s
are allnonexplosive.
For reversiblenonexplosive
processes,positive
recurrence isequivalent
to theintegrability
of the measure with respect to which the
semigroup
and generator are reversible.(For diffusions,
this follows fromChapter
31 in[9].
Itcan easily
be extended to random
diffusions.) Thus,
under theassumption
of noexplosion,
all of theLk’s
generatepositive
recurrent diffusions if andonly
if and the random diffusion is
positive
recurrent if andonly
ifBy
Ichihara’sTheorem,
L is recurrent or transientaccording
to whetherlim lim
~>0.
We havewhere the
equality
follows because G u = 0 if ul = u2 = ...= un.
Since
by assumption,
limÅm
=0,
we obtain limÀm
= 0 and the randomdiffusion is almost
surely
recurrent.We now turn to the two
examples
as mentioned in the introduction.Example
3(Reversible case). - L1
andL2
are null recurrent, but the randomdiffusion
is transient.Before
constructing
theexample,
weprovide
a bit of intuition. ChooseQ - 0 (that is,
constant invariantdensity)
andchoose a 1
such that it hasbroad and very
high
mesasinterspersed
with very narrow canyons with al almost zero at the bottom of the canyons. As xincreases,
theseproperties
are to become more and more extreme.Now a2
is chosen to have the same form but such that each canyonof a2
issignificantly
closerto the next canyon of ai to the left than to the next canyon
of al
to theright.
Since the canyons are verydeep, L1
andL2
are(null)
recurrent.Starting
atlarge
x on a mesa, the distribution will almostimmediately
become almost uniform over the mesa
and,
aslong
as the medium does notswitch,
it will remain like that for along
time. Therefore the random diffusion is morelikely
to increase than to decrease between two successiveswitches. The
following
construction is based on this intuitivedescription although
it does not follow it in every detail.Let =(1 -1 -1
and consider the generatorson the half line
[0, w)
with reflection at 0. The recurrence ofL~, i =1, 2,
isequivalent
to the conditionAs in
(2 .1 ),
in order to show that the random diffusion istransient,
it isenough
to find functionsV l’ V 2 E Coo ([0, (0)) satisfying
and
Solving (3 . 4)
for ai,i =1, 2, gives
and
where Cl and c~ are
arbitrary
constants. We will setc~ = e~ = ~ .
Sincei =1, 2,
we obtain the conditionThe condition
(3.3)
in terms ofVt and V 2
now reads asThus to
complete
theexample,
we must findfunctions V1
andV 2 satisfying (3 . 5 a)-(3 .
5e).
We
begin
withLEMMA 1. - Let
bk,
ek,dk,
ek, k =0, 1,
be constantssatisfying bo co do eo
andb1,
ci,d1, e 1 > Q. Define A=c0-b0 4
andB=-.
Then there exist
F,
G E Coo([0, 1])
such thatProof. -
Obvious.We now construct
V1 and V2
as follows. For i=1,2,
..., letand
Clearly, ti + 1- ti -1
> 0 andm~ > 0
for alli =1, 2,
...Define
V 1
andV 2
on[0, such
thatand
For
.x E [ti, ti +
1] define
where F and G are as in Lemma 1 with
~o=/~ do = gi,
eo =gi + m;,
di
= el = ~f+1. Note that theassumptions
of thelemma are satisfied and that
A,
B in the lemma coincide withAi, B~.
It isobvious that
Vi, V2
E Coo([0, oo))
andsatisfy
3 . 5(a), (b)
and(c).
We now consider
3 . 5 (d). By observing
where(V1- V2) (x) changes sign,
it suffices to show3 . 5 (d )
forx=ti+1 2, i= 1 , 2,
...For i even,
For i
odd,
Furthermore,
Hence, for i even,
and for i
odd,
This
gives
3 . 5(d).
The
inequality
and an
analogous
one forExample
4(Reversible case). -
Theaveraged
operator L is transient but the randomdiffusion
is recurrent.1 d d , d
(-1 1)
Let
- 1 d d
on R for k =1 2 and let G = 1 1 LetLk
.2dx
akd dx+
akQ
dx 1 - 1- - 1 d -d - d
Then the
averaged
operator L isgiven b y
L= 1 d
ad
+a ’ Q d
where2dx dx dx
a x ) 1 ( a 1 + a 2) ( x).
Recall thatLk(L)
is recurrent if andonly
if2
Choose
Q satisfying j
and{gn}~n=-~ be a
sequence ofJ 2014 oo
functions defined on
[0,1) satisfying
Define
and define Then
akEc1(R)
fork= 1, 2
and it followsfrom the
properties
and ofQ
and from anapplication
of theSchwartz
inequality
on the function thatLk
is recurrent fork =1, 2
and that L is transient.(In fact,
it turns out that L isexplosive.) Furthermore,
the invariantdensity
of L + G ise2 Q ~x~ ~k
which isintegrable.
Thus as
long
as(X(t), k(t))
isnonexplosive,
then in fact it ispositive
recurrent and
consequently
the random diffusion is almostsurely positive
recurrent. To prove that
(X (t), k(t))
isnonexplosive,
note that sinceL1
and
L2
are recurrent,X (t)
can notexplode
whenk(t)= 1
nor whenk
(t)
= 2. We have thusgiven
anexample
where eachLk
isrecurrent,
L is transient and the random diffusion is recurrent.4. THE DEPENDENCE OF L + À G ON ?~
The
parameter ~,
can have an effect on transience or recurrence, but not in the reversible case.Indeed,
from the variationalprinciple
of theprevious section,
it is easy to see that L + ~, G generates either a recurrent random diffusion for all ~, > 0 or a transient one for all ~>0. To see that this is not true ingeneral,
consider[6, Corollary
2 in the case8=0]
withG
replaced by
~,G. In this section we will discuss thefollowing question
for the
general
notnecessarily
reversible case.Assuming
that(X (t),
k(t))
is
positive
recurrent for alllarge X
or all small~,,
how does the invariantprobability
measure behave as.
The case X - 00 is an
example
of the"averaging principle".
It followsfrom Khasminskii
[5] (see
also[ 1 j, Chap. 4,
Sect.3)
that the randomdiffusion X
(t)
convergesweakly
to the diffusiongenerated by
theaveraged
operator
where( )
is as defined in section 3.Assuming
that L ispositive
recurrentand that L+ XG is
positive
recurrent forlarger
it also follows from Khasminskii that the invariant measures VÀcorresponding
to L + ÀG con-verge
weakly
to theproduct
measure onR d X E,
where v is theinvariant
probability
measure for L.For A
0,
we will prove thefollowing
theorem.THEOREM 2. -
(i )
Thepositive
recurrenceof
all theLk’s
is a necessarybut not a
sufficient
conditionfor
thetightness o_, f ’ ~ vÀ}
as X -+ 0.-
is
tight as Â. -+ 0,
then w - lim v~(dx, k) =
vk(dx),
x - o
k =
1,2,..., ~
where v~ is the invariantprobability measure-for Lk.
Proof -
Assumethat {v~}
istight
as A -+ 0 and be asequence such that lim
An = 0
and limvÀn
= v. Note that themarginal
onM -~ 00
E of VÀ
is y
for all ~>0. It is easy to see that the measureP Àn
onC
([0, oo), Rd
xE),
inducedby
the processgenerated by
L+ ~
G with initialdistribution
v~~,
convergesweakly
as n -+ oo to the measuren
P --_
E
where6~
is the delta measure on C([0, E)
atA;(~)=~ 0~~oo
andPk
is the measure onC([0,
inducedby
theprocess
generated by Lk
with initial distribution v(dx, k).
It also follows that P is astationary
measure withmarginal
v. Butgiven
the form ofP,
it is clear that this can occuronly
if eachLk
ispositive
recurrent and ifv
(dx, k) =
~,k vk(dx),
where vk is the invariantprobability
measure forLk.
It remains to show that the
positive
recurrence of all of theLk’s
is nota sufficient condition for the
tightness of {v~}
as ~ -~ 0. Return to the firstexample
in section 2.Replace
the generator L + Gby
L+~G. Foreach fixed
0~1,
one can find anappropriate (X = (X (À)
for which therandom diffusion is transient
despite
the fact thatL i
andL2
arepositive
recurrent. If we could
pick
one a to worksimultaneously
for all 0 A 1,we would be done.
Running through
theproof
reveals that aproblem
occurs
only
in case 2 withx-[x]e 0, - U ( -. 1
where we obtainSince
W’(z)
andW"(z)
vanishon -
~z~-,
and since from the construc- tionW"2(z)
can not benonnegative
in a lefthandneighborhood ofz=2014,
it follows that no matter how
large
ais, (z) +1 2W"2 (z)
will benegative
for
z -
andsufficiently
close Thus as X -0,
in order that(4.1)
remain
nonnegative,
we mustpick
(xlarger
andlarger.
Thus we amendexample
one as follows. Let be a sequence ofpositive
numberssuch that lim rtj= 00. For
each j,
letk= 1, 2,
be defined as inexample
one and chosen so thatsup j 2
Fork= 1,
2define
bk(x)=b(03B1j)k(x),
if[x]=j,j=0, 1,
..., and defineLk= +.
’2 dx dx Now for each
~>0,
the V ofexample
one willsatisfy (2.1)
withinstead of L~ + À G and r
sufficiently large depending
on À. To see thisnote that in
(4.1),
if[x] = j,
then rtj instead of a will appear in(4.1)
andthus for any fixed À > 0
(4.1)
will remainnonnegative
forsufficiently large
REFERENCES
[1] M. I. FREIDLIN, Functional Integration and Partial Differential Equations, Princeton University Press, Princeton, 1985.
[2] A. FRIEDMAN, Stochastic Differential Equations, Vol. 1, 1975, Academic Press.
[3] K. ICHIHARA, Some Global Properties of Symmetric Diffusion Processes, Publ. RIMS Kyoto Univ., Vol. 14, 1978, pp. 441-486.
[4] R. Z. KHASMINKII, Ergodic Properties of Recurrent Diffusion Processes and Stabilization of the Solution of the Cauchy Problem for Parabolic Equations, Proc. Theory and Appl., Vol. 5, 1960, pp. 179-196.
[5] R. Z. KHASMINSKII, On the Averaging Principle for Stochastic Differential Equations, Kybernetika, Academia, Praha, Vol. 4, 1968, pp. 260-279 (Russian).
[6] M. A. PINSKY and R. G. PINSKY, Transience and Recurrence for Diffusions in Random
Temporal Environments, Annals of Probability (to appear).
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[9] S. R. S. VARADHAN, Diffusion Problems and Partial Differential Equations, Tara Institute, Bombay, 1980.
(Manuscript received July 4, 1991.)