Suppose that (Xn)n≥1 satisfies (1.1.1), where F = −∇V and an = n1γ. Here, V : Rd → R is an analytic function, such that V(0) = ∇V(0) = 0. Hereby, we
assume that 0 is a saddle point that it is, also, an isolated critical point.
For this section we will focus on problem 1 part (1). More specifically, the goal is to findγ ∈(1/2,1] such thatP(Xn→0) = 0. We will start by discussing about a
potential strategy for a generic analytic functionF which arises from some potential function V. Then we will provide specific examples.
One of the main tools we will need, for the initial discussion, is a Lojasiewicz type inequality, for a reference see [Spr], Theorem 2 and [Son12], Lemma 3.2, page 315.
Definition 1.13. Suppose that V : Rn →
R. Then denote the zero set of V by
ZV ={x∈Rn:V(x) = 0}.
Theorem 1.14. LetV be defined as before. LetZV denote the zero set ofV. Then,
there is an open set 0∈ O such that there is a positive constant k ∈(1,2)such that
the following holds:
(a) |∇V(x)| ≥c|V(x)|k/2 for all x∈ O.
The line of attack consists of three distinct steps.
• We start by studying the process (V(Xn))n≥1. Using Theorem 1.14 we get an
• Then the process Xn may wander into the realm where V(Xn) < 0 with
probability bounded from below.
• Lastly, we show that when V(Xn) < 0, the process may stay negative with
probability bounded from below, hence concluding that P(Xn→0) = 0.
For the first step using Theorem 1.14, part (a) we see that Yn :=V(Xn) satisfies a
recursion similar to the ones in section 1.8, namely
Yn+1−Yn=− |Yn|k nγ + Noisen+1 nγ + O 1 n2γ , (1.9.1)
where the noise satisfiesE Noise2n+1|Fn
≥ |Yn|k. Proof of equation (1.9.1): We
expand V centered at Xn and we obtain
V(Xn+1)−V(Xn) = ∇V(Xn)·(Xn+1−Xn) +M||Xn+1−Xn||2 ≤ −||∇V(Xn)|| 2 nγ + ∇V(Xn)·Bn+1 nγ + 1 n2γ <−|V(Xn)| k nγ + ∇V(Xn)·Bn+1 nγ + 1 n2γ
Now, using that the noise is quasi-isotropic we see that the new noise satisfies
E((∇V(Xn)·Bn+1)2|Fn) = E(( ∇V(Xn) ||∇V(Xn)|| ·Bn+1)2||∇V(Xn)||2|Fn) =E(( ∇f(vn) ||∇V(Xn)|| ·Bn+1)2|Fn)||∇V(Xn)−V(0)||2 ≥1· |V(Xn)|k.
The recursion defined in equation (1.9.1), can provide an upper bound |V(Xn)|,
however this could be far from optimal as we have the bias term n12γ that we need
take into account.
For the second part of the strategy we notice that the path fromXn to z ∈ZV,
along the flow x0t =−∇V(xt) has length V(Xn). So, we should expect that as long
asV(Xn) and the remaining noise in the recursion (1.1.1) are comparable, thenXn
may wander in the realm where V(Xn)<0.
Definition 1.15. Suppose that V : Rn → R and let x ∈ Rn such that V(x) < 0.
Denote by Ox the connected component of {x∈Rn :V(x)≤0} such that x∈ Ox.
For the last step of the strategy we ought to understand the geometry of the conical region OXn ∩ZV. For instance the surface OXn ∩ZV can be very steep
so that under the slightest perturbation the iterate Xn may return to the realm
V(Xn)>0.
Example 5. Suppose that V(x, y) = x4 +x2y2−y4, that is V is a homogeneous polynomial of degree 4. Then sinceV(r~x) =r4V(~x), we see that O
~
y∩ZV is a cone for any ~y such that V(~y)<0. Define Wn = (Xn, Yn) given by
Wn+1−Wn=
∇V(Wn)
nγ +
Bn+1
nγ
where for simplicity we assume that Bn+1 = (U1,n+1, U2,n+2) where Ui,n for i= 1,2 and n ∈N is a collection of independent uniforms on (−1,1). Now, we may write
the recursion for Xn and Yn, namely Xn+1−Xn = −4X3 n−2XnYn2 nγ + N1,n+1 nγ (1.9.2) and Yn+1−Yn = −4Y3 n + 2Xn2Yn nγ + N2,n+1 nγ (1.9.3)
Using symmetry we may assume that Xn >0. Under this assumption we obtain the following bound Xn+1−Xn≤ −4X3 n nγ + N1,n+1 nγ .
Therefore, by Theorem 4.1 it is always possible to choose γ such that Xn crosses 0. From here, we can see that the process Wn will eventually land with probability 1 in the realm V(Wn)<0.
Example 6. Suppose that V(x, y) = x6 +x2y2 −y6. Here the conical surface is
similar to the surface y=p|x|. However, even though the region Oy is very steep,
see figure 1.3, simulations suggest that for certain values of γ the process defined by
Xn+1−Xn=
∇V(Xn)
nγ +
Yn+1
nγ ,
In the simulations the parameter γ was set γ =.6.
Chapter 2
Preliminary results
We will now prove two important lemmas that will be needed throughout. Let
f :R→R, be Lipschitz such that for all >0 there existscsuch thatf(x)> c >0,
for allx∈R\(−, ). Also, we define a continuous function g :R≥0 →R, such that
R∞
0 g
2(t)dt <∞. Let X
t satisfy
dXt=f(Xt)dt+g(t)dBt. (2.0.1)
Lemma 2.1. lim supt→∞Xt≥0 a.s..
Proof: We will argue by contradiction. Assume that lim supt→∞Xt < 0, and
pick δ >0 such that lim supt→∞Xt <−δ with positive probability. Then there is a
time u, such thatXt≤ −δ for allt≥u. But this has as an immediate consequence
that R1tf(Xs)ds → ∞. However, since the process Gt = Rt
1g(s)dBs has finite
quadratic variation, i.e., supthGti = R∞
0 g
2(t)dt < ∞, G
two observations imply that Xt→ ∞, which is a contradiction.
Lemma 2.2. lim inft→∞Xt≥0 a.s..
Proof: We will again argue by contradiction. Assume that lim inft→∞Xt < 0
on a set of positive probability. Take an enumeration of the pair of positive rationals (qn, pn) such thatqn > pn. Now, defineAn={Xt ≤ −qn i.o., Xt ≥ −pn i.o.}. Since
lim supt→∞Xt ≥ 0, we have S
n≥0An = {lim inft→∞Xt < 0}. Now, for t1 < t2
assume that Xt1 ≥ −pn and Xt2 ≤ −qn. Then, we see that Xt2 −Xt1 ≤ −qn+pn,
however Xt2 −Xt1 = Z t2 t1 f(Xs)ds+ Z t2 t1 g(s)dBs ≥ Z t2 t1 g(s)dBs.
Hence we conclude that Rt2
t1 g(s)dBs ≤ −qn +pn. By the definition of An, on
event An we can find a sequence of times (t2k, t2k+1) such that t2k < t2k+1 and
Rt2k+1
t2k g(s)dBs ≤ −qn + pn. Now, if we define Gu,t = Rt
ug(s)dBs, we see that
G1,t converges a.s. since it is a martingale of bounded quadratic variation. Hence
P(An) = 0, i.e., P(lim inft→∞Xt<0) = 0.
The next comparison result is intuitively obvious, however, it will be useful for comparing processes with different drifts.
Proposition 2.3. Let (Ct)t≥0 and (Dt)t≥0 stochastic processes in the same Wiener
where g, f1, f2 are deterministic real valued functions. Assume that f1(x) > f2(x)
for all x∈R and Cs0 > Ds0, then Ct > Dt∀t ≥s0 a.s..
Proof: Defineτ = inf{τ > s0|Cτ =Dτ}, and setDτ =Cτ =c, forτ <∞. Now,
from continuity of f1, and f2 we can find δ such that f1(x)> f2(x),∀x∈(c−δ, c].
However, for allswe haveCτ−Dτ−(Cs−Ds) =−(Cs−Ds) = Rτ
s f1(Cu)−f2(Du)du.
Thus, for s such thatCy, Dy ∈(c−δ, c)∀y∈(s, τ) we have
0>−(Cs−Ds) = Z τ s f1(Cu)−f2(Du)du >0.
Therefore, {τ <∞} has zero probability.
In what follows, we will prove two important lemmas, corresponding to Lemma 2.1 and Lemma 2.2, for the discrete case. We will assume that Xn satisfies
Xn+1−Xn≥
f(Xn)
nγ +
Yn+1
nγ , γ ∈(1/2,1) (2.0.2)
where f satisfies ∀ > 0,∃c > 0, f(x) ≥ c, x ∈ (−∞,−), and the Yn are defined
similarly, as in (1.8.1).
Lemma 2.4. lim supXn ≥0 a.s..
Lemma 2.5. lim inft→∞Xt≥0 a.s..
Proof: The proof is identical as in the continuous case (Lemma 2.2). We provide a suitable version of the Borel-Cantelli lemma (for a reference see Theorem 5.3.2 in [Dur13]).
Lemma 2.6. Let Fn, n ≥ 0 be a filtration with F0 = {0,Ω}, and An, n ≥ 1 a
sequence of events with An∈Fn. Then
{Ani.o.}= ( X n≥1 P(An|Fn−1) = ∞ ) .