A non-linear model of trading mechanism on a financial market

0.17 MB·English

by  N. Vvedenskaya

A non-linear model of trading
mechanism on a financial market
2
1
0 N.Vvedenskaya1 , Y.Suhov1,2,3, V.Belitsky3
2

n
a
J Abstract

2 We introduce a prototype model in an attempt to capture some as-
2 pects of market dynamics simulating a trading mechanism. The model

description startswith adiscrete-space, contiinuous-timeMarkovprocess
]
R describing arrival and movement of orders with different prices. Wethen
perform a re-scaling procedure leading to a deterministic dynamical sys-
T
tem controlled bynon-linear odinarydifferential equations(ODEs). This
.
n allows us to introduce approximations for the equilibrium distribution of
i the model represented by fixedpoints of deterministic dynamics.
f
-
q
[ 1 Introduction

1
v This paper proposes a model that takes into account, in a rather stylized form,
0 some aspects of automated trading mechanisms adopted in modern financial
8
markets, in particular, the dynamics of the limit order book.
5
In short, a limit order book keeps records of arivals, movements and depar-
4
. tures of market participants (traders) who declare their trading positions. An
1
arriving trader may wish to buy or sell at a certain price, and can move his
0
declared price when time progresses. If the declared price is met by a trader
2
1 withthe oppositeintention,atradeisrecorded: this mayleadtodisappearance
: of one or both participants from the market, due to exhaustion of their offers.
v
For a detailed description of some common limit order book models and their
i
X
applications, see [2, 3, 4] and references therein.
r ———————————————–
a
1Institute for Information Transmission Problems, RAS, Moscow, Russia
2DPMMS,University ofCambridge,andSt John’s College,Cambridge,UK
3IME, Universidade de Sao Paulo, Sao Paulo, Brazil
1
In the current paper we present a somewhat different model, including el-
ements of queueing behavior of arrived offers; this model is studied by using
techniques of asymptotic analysis. An earlier account of this work (in its pre-
liminary version) can be found in [1]. Compared with [1], in the present text
weadoptacontinuous-timesettingforthebasicMarkovprocess: itclarifiesthe
meaning of the main parameters of the model and shortens the proof of some
of our main results.
The model under consideration is a prototype; at this stage it does not aim
to take into account all possible aspects that can be viewed as defining, either
theoreticallyorpractically. Instead,we optedfor asimplified descriptionwhich
leads to some straightforward,yet instructive, answers.
Ourmodeldiffersfromknownmodelsofthelimitorderbookinanumberof
aspects. Arguably,itcanbeasubjectofcriticism(asanumberofotherproposed
models). In particular, the strategic behavior of the model in its current form
(and some further details) do not quite match existing mechanisms governing
electronictradingonfinancialmarkets. Nevertheless,themodelshowsacertain
amount of flexibility, and covers a borad range of situations. Its mathematical
advantage is that in the scaling limit under consideration, it leads to a single
fixed point.
Our scaling limit is based on suppositions that (i) the number of market
participants is large, (ii) during a very short time period only part of them
makesadecisionofperformingatradeormaikingamovealongthepricerange,
and(iii)theprobabilityforanygivenparticipanttomakesuchdecisionissmall.
This makes it natural to change, in a suitable manner, parameters of original
Markov process.
After rescaling, a limiting dynamical system emerges, with a determinis-
tic behavior described by a system of non-linear ordinary differential equa-
tions. The rescaling techniques greatly simplify the structure of the model,
andthis phenomenonextends farbeyondbasicexampleslike the currentproto-
type model. As we mentioned earlier, the present paper focuses on a simplified
model,with‘minimal’numberofconstantparameters,wheresomeoftechnically
involved issues are absent.
A similar approach is commonly used in the literature on stochastic com-
munication networks; see, e.g., [5, 6, 7, 8] and [9]. We also find similarities,
as well as differences, with models proposed (in a different context) in a recent
paper [10]; analogies with [10] could be useful for the aforementioned purpose
of defining the prices that are appropriate for trades.
In the next section we describe the underlying Markov process. In Section
3 the rescaling of the process is presented and the main results are stated and
the proofs are given. The last section contains concluding discussion of various
aspects of the model.
2
2 The underlying Markov process
Therationaleforthemodelsbelowisasfollows. Weconsiderasingle-commodity
marketwherepricesmaybeatoneofN distinctlevels(say,c <c <...<c ,
1 2 N
although the exact meaning of these values is of no importance here).
The marketis operating in continuous time t∈R where R =[0,∞). (As
+ +
wasmentionedabove,the earlierversion[1]usedamorecumbersomediscrtete-
time versionof the underlying process.) At a given time t∈R , there are b (t)
+ i
traders prepared to buy a unit of the commodity at price c and s (t) traders
i i
prepared to sell it at this price, which leads to vectors
b(t)= b (t),...,b (t) , s(t)= s (t),...,s (t) ∈ZN.
1 N 1 N +
Here and below Z(cid:0)+ = {0,1,...} s(cid:1)tands for(cid:0)a non-negative (cid:1)integer half-lattice
and ZN for the non-negative integer N-dimensional lattice orthant. The pair
+
(b(t),s(t))representsastateofaMarkovprocess{U(t)}thatwillbethesubject
of our analysis.
Suppose that, for given k = 1,...N and t ∈ R , we have that b (t) ≥
+ k
s (t)>0 then eachof the sellers gets a trade at a givenrate ρ >0 and leaves
k T
the market, together with one of the buyers. Therefore, both values b (t) and
k
s (t) decrease at rate ρ . In addition, each one among s (t) sellers (i) quits
k T k
the market at rate ρ >0 or (ii) moves to the price level c at rate ρ >0,
Q k−1 M
if k > 1. Similarly, every buyer among the b (t) buyers (i) quits the market at
k
the same rate ρ as above or (ii) moves to the price level c , again at rate
Q k+1
ρ , provided that k <N.
M
Symmetrically, if s (t) ≥ b (t) > 0 then each of the buyers gets a trade
k k
at rate ρ and leaves the market, together with his seller companion. The
T
remaining traders at the price level c proceed in a manner as above.
k
Further, when k = N, a buyer leaves the system with rate ρ +ρ . Simi-
Q M
larly, for k=1, a seller leaves the system with rate ρ +ρ .
Q M
Finally, a random Poissonflow of new exogenousbuyers arrivesat the price
levelc ; the rateofthis arrivalequalsλ >0. Similarly,a Poissonrandomflow
1 B
of new sellers arrives at the price level c ; the rate of this arrival is λ >0.
N S
As usually,standardindependence assumptionsareinplace. This generates
the aforementionedMarkov process U(t) with trajectories (b(t),s(t)) , t∈
R .
+
(cid:8) (cid:9) (cid:8) (cid:9)
Theorem 1 For any values of parameters λ , ρ and ρ , the process
B/S Q/M T
{U(t)}isirreducible,aperiodicandpositiverecurrent. Therefore, ithasaunique
set of equilibrium probabilities π = π b,s : b,s ∈ ZN , and for any initial
+
stateU(0) (deterministic or random(cid:16)), t(cid:0)he d(cid:1)istribution of t(cid:17)he random state U(t)
at time t converges weakly to π as t→∞:
lim P U(t)=(b,s) =π b,s .
t→∞
(cid:16) (cid:17) (cid:0) (cid:1)
3
Proof of Theorem 1. Irreducibilityandaperiodicityoftheprocessisevident.
Positive recurrence follows from the following observation. The (random) time
thatagiventrader(abuyeroraseller)spendsinthesystem,i.e.,thetimefrom
his arrival till exit, is majorized by a sum of N independent exponential vari-
ables. Therefore,the process {U(t)} canbe majorized,in a naturalfashion, by
anM/M/∞queueing process. But the latter is knownto be positive recurrent.
The remaining assertions of Theorem 1 are standard. N
Despite a concise description, the detailed pattern of behavior of process
{U(t)} is rather complex, particularly for large values of N. For instance,
consider the differences between vectors b(t′) and b(t) and between s(t′) and
s(t), on a time interval (t,t′) where 0 < t < t′. The increments for the entries
b (·) and s (·) for 1≤k ≤N −1 are captured by the following equations:
k k
b (t′)=b (t)+iM (t,t′)−n (t,t′)−iM(t,t′)−iQ(t,t′),
k k k−1 k k k
and
s (t′)=s (t)+jM (t,t′)−n (t,t′)−jM(t,t′)−jQ(t,t′).
k k k+1 k k k
HereiM(t,t′)isthenumberofbuyerswhomovewithintimeinterval(t,t′)from
k
level k to k +1 and jM(t,t′) that of sellers who move from level k to k−1.
k
Next,iQ(t,t′)isthenumberofbuyerswhoquitthesystemduringinterval(t,t′)
k
from level k and jQ(t,t′) the number of sellers who quit the system from level
k
k. Finally, n (t,t′) is the number of buyers and sellers who got a trade over
k
(t,t′) at level k. All listed quantities are non-negative integer-values random
variables. For k = 1 the structure of the expression is similar, with the term
iM (t,t′) being replacedby i (t,t′)≥0, while for k =N the term jM (t,t′) is
k−1 B k+1
replaced by j (t,t′)≥0; both i (t,t′) and j (t,t′) being distributed according
S B S
to a Poisson law with mean λ (t′−t).
B/S
In the simplest case of a market with one price level (N = 1), the pro-
cess {U(t)} is a continuous-time random walk on the two-dimensional lattice
quadrant Z2, where
+
b(t′)=b(t)+i (t,t′)−n(t,t′)−i (t,t′)
B Q
and
s(t′)=s(t)+j (t,t′)−n(t,t′)−j (t,t′).
S Q
This already makes analytical representations for the invariant distribution π
rather complicated; cf. [11] and references therein.
3 Scaling limit
The complexity of the time-dynamics and of the equilibrium distribution π for
process U(t) makesitdesirabletodevelopefficientmethodsofapproximation.
In this paper we focus on one such method based on scaling the parameters of
(cid:8) (cid:9)
the process (including states and time-steps).
4
The re-scaling procedure is as follows: we fix values γ > 0, β > 0, α > 0,
λ >0 and λ >0 and set:
B S
γ β α
ρ = , ρ = , ρ = . (1)
T Q M
L L L
In addition, we re-scale the states and the time: pictorially,
b s t
k k
x ∼ , y ∼ , τ ∼ . (2)
k k
L L L
Formally, denoting the Markov process generated for a given L by U(L), we
consider the continuous-time process
1
V(L)(τ)= U(L) τL), τ ≥0. (3)
L
Let RN denote a positive orthant in N(cid:0) dimensions. Suppose we are given
+
a pair of vectors (x(0),y(0)) ∈ RN × RN where x(0) = (x (0),...,x (0)),
+ + 1 N
y(0) = (y (0),...,y (0)). Consider the following system of first-order ODEs
1 N
for functions x =x (τ) and y =y (τ) where τ >0 and 1≤i≤N :
k k k k
x˙ = λ − β+α x −γmin x ,y ,
1 B 1 1 1
x˙k = αxk−1(cid:16)− β+(cid:17) α xk−γm(cid:2)in xk(cid:3),yk , 1<k ≤N,
(4)
y˙k = αyk+1− (cid:16)β+α(cid:17)yk−γmin (cid:2)xk,yk(cid:3), 1≤k <N,
y˙N = λS− β+(cid:16)α yN(cid:17)−γmin xN(cid:2),yN ,(cid:3)
(cid:16) (cid:17)
(cid:2) (cid:3)
with the initial data
x (0)≥0, y (0)≥0, 1≤k ≤N. (5)
k k
The fixed point x∗,y∗ of system (4) has
(cid:0) x∗ =(cid:1)(x∗,...,x∗ ) and y∗ =(y∗,...,y∗ )
1 N 1 N
where x∗ and y∗ give a solution to
k k
λ = β+α x∗+γmin x∗,y∗ , (1′)
B 1 1 1
αx∗ = (cid:16)β+α(cid:17)x∗ +γmin (cid:2)x∗,y∗(cid:3), 1<k ≤N, (2′)
k−1 k k k
(6)
αy∗ = (cid:16)β+α(cid:17)y∗+ γmin(cid:2)x∗,y∗(cid:3), 1≤k <N, (3′)
k+1 k k k
λS = (cid:16)β+α(cid:17)yN∗ +γmin (cid:2)x∗N,yN∗(cid:3) . (4′)
(cid:16) (cid:17) (cid:2) (cid:3)
(In (6) we noted individual equations by addition signs that are used below).
Bothsystems(4)and(6)arenon-linear. However,thenon-linearity‘disappears’
at a local level which greatly simplifies the analysis of these systems.
In Theorems2 and3below, we use the distance generatedby the Euclidean
norm in RN ×RN.
5
Theorem 2 (a) For any initial data (x(0),y(0)) with x (0) ≥ 0, y (0) ≥ 0,
i i
1 ≤ i ≤ N for all τ > 0 there exists a unique solution (x(τ),y(τ)), to problem
(4)–(5) and x (τ)>0,y (τ)>0.
i i
(b) As τ → ∞, the solution approaches a fixed point which is a unique
solution to system (6):
dist x(τ),y(τ) , x∗,y∗ →0. (7)
h(cid:0) (cid:1) (cid:0) (cid:1)i
Proof of Theorem 2, (a). Obviously, a unique solution exists for sufficiently
small τ > 0. Suppose that x = max x and max x < x . Then x˙ ≤ 0.
i k k i i
k k<i
Similarly if y = max y and max y < y then y˙ ≤ 0. Also, x˙ < 0 if
j k k j j 1
k k>j
x > λ /(β +α) and y˙ < 0 if y > λ /(β +α). Moreover, x˙ ≥ 0 when
1 B N N S 1
x = 0 and y˙ ≥ 0 when y = 0. Thus the components of the solution x (τ)
1 N N i
and y (τ) are non-negative and uniformly bounded. By standard constructions
i
of the ODE theory, a unique solution (x(τ),y(τ)) exists for all τ > 0, and
x(τ),y(τ)∈RN. N n o
+
Before proving assertion(b), we discuss severalproperties of the solutionto
(4)–(5). The followingProposition1indicatesthatthe solutionto(4)possesses
a kind of the min/max principle.
Proposition 1 Suppose that we have two initial points, x = (x ,...,x ) and
1 n
x′ =(x′,...,x′ )∈RN such that for some k =1,...,N,
1 n +
x ≤x′, and y ≥y′. (8)
k k k k
Then for the solutions (x(τ),y(τ)), (x′(τ),y′(τ)), to (4), with the initial con-
ditions (x(0),y(0))=(x,y) and (x′(0),y′(0))=(x′,y′), for all τ >0,
x (τ)≤x′(τ), and y (τ)≥y′(τ). (9)
k k k k
Similarly if x ≥x′ and y ≤y′ then the corresponding solutions obey x (τ)≥
k k k k k
x′(τ) and y (τ)≤y′(τ), for all τ >0
k k k
Proof of Proposition 1. It is sufficient to consider the situation where the
strict inequalities take place. Suppose that (9) holds strictly for τ < τ and
0
fails at τ = τ . For instance, let x (τ ) = x′(τ ) and assume that k > 1 is the
0 k 0 k 0
minimal index for which such an equality takes place.
If y (τ ) > y′(τ ) then x˙ (τ ) < x˙′(τ ) and the above equality x (τ ) =
k 0 k 0 k 0 k 0 k 0
x′(τ ) is impossible. The other case is considered in a similar manner. N
k 0
A corollary of proposition 1 is
Proposition 2 If x˙ (0) ≥ 0 and y˙ (0) ≤ 0 for all 1 ≤ k ≤ N then x (τ)
k k k
increases and y (τ) decreases in τ, for all τ > 0 and 1 ≤ k ≤ N. Similarly if
k
x˙ (0)≤0 and y˙ (0)≥0, 1≤k ≤N, then x (τ) decreases and y (τ) increases
k k k k
in τ.
6
Proof of Proposition 2. Itagainsufficestoassumethatthestrictinequalities
holdtrue: x˙ (0)>0andy˙ (0)<0. Then,forasmallδ >0: x (δ)>x (0)and
k k k k
y (δ)<y (0).
k k
Set x′(0) = x (δ), y′(0) = y (δ). The coefficients of equations do not
k k k k
depend on τ, therefore x′(τ) = x (τ +δ) and y′(τ) = y (τ +δ) for all τ > 0.
k k k k
By Proposition 1, x (τ +δ) = x′(τ) ≥ x (τ) and y′(τ +δ) = y′(τ) ≤ y (τ),
k k k k k k
for all τ > 0 and 1 ≤ k ≤ N. As δ may be arbitrarily small, the assertion of
Proposition 2 is valid. N
Proof of Theorem 2, (b). Given x(0)=(x (0),...,x (0)) and
1 N
y(0) = (y (0),...,y (0)) ∈ RN, let (x(τ),x(τ)) be the solution to (4), (5).
1 N +
Considertwoadditionalsolutions,(x′(τ),y′(τ)) and(x′′(τ),y′′(τ)), to(4)with
x′(0)=0, y′(0)=max λ /(α+β), maxy ,
k k S i
i
h i
and
x′′(0)=max λ /(α+β), maxx (0) , y′′(0)=0.
k B i k
i
h i
By Propositions 1 and 2
x′(τ)≤x (τ)≤x′′(τ), y′(τ)≥y (τ)≥y′′(τ). (10)
k k k k k k
Further, x′(τ) increases, while y′(τ) decreases in τ. By the same token, x′′(τ)
k k k
decreases and y′′(τ) increases in τ. Therefore, both pairs (x′(τ),y′(τ)) and
k
(x′′(τ),y′′(τ)) tendtolimits asτ →∞,whicharefixedpoints,i.e., solutionsto
Eqns (6). By (10), any solution to (4), (5) eventually lies between these limits.
To finish the proof of the theorem, we have to show that the solution (x∗,y∗)
to Eqn (6) is unique.
For convenience, we state the corresponding assertion as Lemma 1.
Lemma 1 For any values λ ,α,γ and β ≥ 0 there exists a unique solution
B/S
to Eqns (6).
Proof of Lemma 1. To start with, note that every solution to (6) has
x∗ >x∗ >...>x∗ , y∗ <y∗ <...<y∗ .
1 2 N 1 2 N
Therefore, if x ≤ y then x < y , 1 < k ≥ N and if x ≥ y then x >
1 1 k k N N k
y , 1<≥k <N .
k
Itisconvenientto introduceauxiliaryvariablesv ,w ≥0intermsofwhich
k k
Eqns (6) will be treated. Geometrically the idea is as follows: we start with
Eqn(6)(1) : λ =(α+β)v +min[v ,w ] and watchhow this relationbetween
B 1 1 1
v , w istransformedby(6)(2),(6)(3) torelationbetweenv , w ,thentorelation
1 1 2 2
between v , w etc.
3 3
The locus of points (u,v)∈R2 where
v,w >0, λ =(α+β)v+min[v,w]
B
7
coincides with a continuous broken line, L ⊂ R2, formed by two pieces: 1) a
1 +
vertical ray L(0) emitted from the point o(0) lying on the bisectrix where
1 1
o(0) = λ (α+β+γ),λ (α+β+γ)
1 B B
(cid:0) (cid:14) (cid:14) (cid:1)
and2) a line segmentL(1) of a negative slope −(α+β)/γ, joining o(0) with the
1 1
point o(1) on the horizontal axis:
1
o(1)1= λ (α+β),0 .
1 B
(cid:0) (cid:14) (cid:1)
See the figure below.
w
(0)
L
2
L(0)
1
λ
/γ (1)
B L
2 M
N
(1)
(2) L
L 1
2
λ /(α+β) v
λ /(α+β+γ) B
B
The passage from values v , w to v , w generates a 1 − 1 map acting on
1 1 2 2
L . The image of L under the map (v ,w ) 7→ (v ,w ) is another continuous
1 1 1 1 2 2
brokenline, L ⊂R2, formedby threepieces: 1)averticalrayL(0) issuedfrom
2 + 2
point
o(0) = λ α (α+β+γ)2,λ α ,
2 B B
2) a line segment L(1) joinin(cid:0)g the(cid:14)points o(0) and (cid:14)o(1(cid:1)) where o(2) lies on the
2 2 2 2
bisectrix and has both co-ordinates equal to
λ α(α+β+γ)
B
,
(α+β+γ) (α+β)(α+β+γ)2+α2γ
and3)alinesegmentL(2) joini(cid:2)ngo(1) withthepointo(2) on(cid:3) thehorizontalaxis:
2 2 2
λ α
o(2) = B ,0 .
2 (α+β)2
(cid:18) (cid:19)
8
The slopesdw/dv ofsegments L(1) andL(2) arenegative,butthe slope flattens
2 2
when we pass from L(1) to L(2).
2 2
In the above figure, L and L are shown on the same (v,w)-plane R2; in
1 2
this figure line L lies to the left of L . (The third broken line, M present in
2 1 N
the figure is explained below.)
Asimilarpicturepersistswhenweiterate,i.e.,passfrom(v ,w )to(v ,w )
2 2 3 3
and so on. At step k (v ,w ) 7→ (v ,w ) where again the map is 1 - 1,
k k k+1 k+1
(v ,w ) belongs to a continuous broken line L in R2 formed by k+2
k+1 k+1 k+1 +
pieces. One piece, L(0) , is a vertical ray while the k +1 others, L(1) , ...,
k+1 k+1
L(k+1), are line segments of negative slopes, the slope flattens when we pass
k+1
from L(1) to L(k). The last segment, L(k+1), joins a point on the bisectrix and
k k k+1
a point on the horizontal axis.
At the end of this process we obtain a continuous broken line L , the locus
N
ofpoints(v ,w ). Ournextstepistoconsiderthe intersectionofL andM
N N N N
where M is the locus where
N
λ =(α+β)w +γmin[v ,w ].
S N N N
Moreprecisely,M isacontinuousbrokenlineformedbyahorizontalrayissued
N
from the point
λ λ
S S
, ,
α+β+γ α+β+γ
(cid:18) (cid:19)
and a line segment joining this point with the point
λ
S
0,
α+β
(cid:18) (cid:19)
lying on the vertical axis. Cf. the figure.
We want to check that the point of intersection is always unique: it yields
to unique solution to (6).
Line L may intersect the horizontal part of M . That means that there
N N
exist a solution to (6) where x∗ > y∗ and x∗ > y∗ (this case is not presented
1 1 N N
onourfigure). Forthe proofoflemmaitisneededto showthatinthis caseL
N
cannotintersectthesloppypartofM ,wherey >x . Forthatitissufficient
N N N
toprovethatdw dv onL \L(N),thepartofL abovethebisectrix,issteeper
N N N
than −γ (α+β), the slope of the segment of line M .
(cid:14) N
Onthe otherhandifL doesnotintersectthe horizontalpartofM it has
N N
(cid:14)
to intersect the sloppy part of M and it is needed to show that in this case
N
such intersection is unique. Here again it is sufficient to show that dw dv on
L \L(N) is steeper than −γ (α+β).
N N (cid:14)
To prove the assertion of the lemma we show that dw dv on L \L(N) is
(cid:14) N N
always steeper than −γ (α+β). In fact, it suffices to verify that
(cid:14)
dw(cid:14) γ
N <− on Li , 1≤i<N. (11)
dv α+β N
N
9
Any segment L(i), 1<i<k, maps onto segment L(i) , segment L(k) maps
k k+1 k
ontotwosegments: L(k) andL(k+1). TheslopeofsegmentL(i) ,i=1,...,k+1
k+1 k+1 k+1
is steeper than that of L(i) and the slope of L(k+1) is steeper than that of L(k).
k k+1 k
More,the slopes of L(i), 0<i≤N, are steeper then the steep segmentof M .
N N
To show that consider three cases of the map L →L :
k k+1
1) v >w , v >w ;
k k k+1 k+1
2) v >w , v <w ;
k k k+1 k+1
3) v <w , v <w .
k k+1 k+1 k+1
In these cases we have, the following equations
A (α+β+γ)dw
k+1 k
,
α2 dv
ddwvkk++11 =(α+αβ2+γ)2ddwvkk, k (12)
Bk(α+β+γ)dwk,
α2 dvk
where, respectively, 
(α+β)v +γw (α+β+γ)w
k+1 k+1 k
v = , w = ,
k k+1
α α
 (α+β+γ)v (α+β+γ)w
k+1 k
vk = (α+β+αγ)v , wk+1 = (α+βα)w +γv,
k+1 k k
v = , w = .
k k+1
α α
 wk+1 vk
Here A =α+β+γ and B =α+β+γ .
k+1 k
v w
k+1 k
Therefore for N >2
dw (α+β+γ)N−1dw (α+β+γ)N−1α+β
N 1
> =
dv αN−1 dv αN−1 γ
N 1
(cid:12) (cid:12) (cid:12)dw γ (cid:12)
(cid:12) (cid:12) >(cid:12) = . (cid:12)
(cid:12) (cid:12) (cid:12)dv MN α+β (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
For N = 2 we hav(cid:12)e th(cid:12)e middle case in (12) with k = 1, k+1 = 2. Here
again the needed inequality takes place. In fact, dw1 = −α+β. By using (12),
dv1 γ
we get that for N =2,
dw (α+β+γ)2dw
2 =− 1 on L1.
du α2 du 2
2 2
This finishes the proof of Lemma 1. N
Theorem 3 Suppose thatthere-scaledinitial states convergein probability: for
any ǫ>0,
1
lim P dist U(0),(x(0),y(0)) ≥ǫ =0.
L→∞ L
(cid:18) (cid:20) (cid:21) (cid:19)
10