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3.3 Non-constant jump rates

3.3.3 The contracting distance

Markov processes with random switching

inequalities, we have E

eR0τn(Ls)ds1{τn>nr−1(1+)}

≤E

he2ni1/2P

τn > nr−1(1 +)1/2

≤E

he2ni1/2eθ0nr−1(1+)E

heθ0τni1/2

≤exp −n 2 ln

1− 2%

r

+ ln 1− θ0 r

!

+ θ0(1 +) r

!!

, whereθ0 ≥0. Note that in the previous inequality, we have supposed thatr >2%. Letγ ∈(0,1), we setθ =θ0 =γr. We can find a largerand a smallγverifying

ln (1 +γ)−(1−)γ−(1−)r−1% >0, and

ln

1−2%

r

+ ln (1−γ) +γ(1 +)>0.

and thus there existC0 >0andε >0such that

X

n≥1

E

eR0τn(Is)ds

X

n≥1

C0eεn<+∞, thus concluding the proof by combining this with (3.16) and (3.17).

Remark 3.3.9. If all Markov processes contract, then the proof simplifies considerably. Indeed, if for alliF, one hasα(i)≥ζ >0, then one has

X

n≥1

E

eR0τn(Ls)ds

X

n≥1

E

heζτniX

n≥1

r r+ζ

!n

= r

ζ <.

Markov processes with random switching

3.3.3.1 Definition ofde

Here, we build a distancede: (E×F)×(E×F)→[0,1]such that there existt >0andα ∈(0,1) verifying

d(x,e y)<1 ⇒ ∀tt, Wd˜(δxPt, δyPt)≤αd(x,e y). (3.19) wherex = (x, i)andy = (y, j)belong to E ×F. Since we can say nothing when i 6= j, we will taked(x,e y)constant equal to1in this case. Wheni=j we want to use Assumption3.1.3to prove a decay. But it is more useful to “decrease the contraction” of the underlying Markov semigroup. More precisely, by Jensen inequality, Assumption3.1.3gives

Wdq(µPt(i), νPt(i))≤e(i)tWdq(µ, ν),

for allt≥0,q∈(0,1]and every probability measuresµ, ν. Finally, we definedeby d(x,e y) = 1i6=j+1i=jδ−1dq(x, y)∧1,

whereδ >0will be determined later. Now, if a coupling(Xt,Yt)t≥0 = ((Xt, It),(Yt, Jt))t≥0starting from(x,y), verifiesd(x,e y)<1, thenI0 =J0 =i=j. So, we will try to build our coupling in such a way thatIandJ remain equal for as long as possible. More precisely, if we set

T = inf{s≥0|Is6=Js}, (3.20)

then we will prove that there existsK >0and a choice of coupling such that P(T <∞)≤Kd(x, y).

3.3.3.2 Construction of our coupling

Here, we fixx = (x, i), y = (y, j)in Eand we let t > 0. Let r ≥ 0and(Nt)t≥0 be a Poisson process of intensity r with Nt = Pn≥01{τnt} and τn = Pnk=1Ek for a family (Ek)k≥0 of i.i.d.

exponential variables as before andτ0 = 0. We assume thatr ≥ 2¯a, i.e. that isr is bigger than the jump rates ofIorJ. As in the proof of Lemma3.3.8and Theorem3.1.4, we give the construction of our coupling(X,Y)at the jump times ofN. Letn ∈ {0, .., Nt}, we consider the following dynamics:

– IfIτn 6=Jτn thenXsandYs evolve independently for everys∈[τn, τn+1t).

Markov processes with random switching

– IfIτn =Jτn then by Assumption3.1.3, we can coupleXandY in such a way that E

hd(Xτn+1t, Yτn+1t)|Gτn

ieρ(Iτn)(τn+1tτn)d(Xτn, Yτn), whereGn=σ{(Xτn,Yτn),(τk)k≥0}.

At the jump times ofN the situation is different sinceI orJ may jump. We will optimise the chance thatI andJ jump simultaneously. For eachn∈N, we cut[0,1]in four partsI0n, I1n, I2n, I3nin such a way that

λ(I0n) = 1 r

X

jF

(a(Xτn, Iτn, j)−a(Yτn, Iτn, j))+, λ(I1n) = 1

r

X

jF

(a(Yτn, Iτn, j)−a(Xτn, Iτn, j))+, λ(I2n) = 1

r

X

jF

a(Xτn, Iτn, j)∧a(Yτn, Iτn, j), λ(I3n) = 1− 1

r

X

jF

a(Xτn, Iτn, j)∨X

jF

a(Yτn, Iτn, j),

whereλis the Lebesgue measure and(x)+ = max(x,0). Let(Un)n≥0 be a sequence of i.i.d. random variables uniformly distributed on[0,1], we coupleI andJ at the jump times as follows:

– ForUnI0n,Ijumps, butJ does not jump.

– ForUnI1n,J jumps, butIdoes not jump.

– ForUnI2n,IandJ both jump simultaneously to the same location.

– ForUnI3n,IandJ both stay in place.

The second components,X andY, do not jump. Finally, we also coupleXandYwith a continuous Markov chainLwhich only depend toU andN and which verifies

t≥0, ρ(It)≥α(Lt).

This Markov chainLis constructed as in the proof of Lemma3.3.8.

Remark 3.3.10. This coupling is not quite Markovian since, between times τn andτn+1, it already uses information about the pair (Xt, Yt) at time τn+1. However, in many situations to which our results apply there exists a Markovian coupling with generatorL(i)which minimises the Wasserstein distance for each of the underlying processes. In this case, we can make our coupling Markovian with

Markov processes with random switching

generator

Lf(x,y, n) =L(i)f(x,y, n) + X

kF

(a(x, i, k)−a(y, j, k))+f((x, k),y, n+ 1)

+ X

kF

(a(y, j, k)−a(x, i, k))+f(x,(y, k), n+ 1)

+ X

kF

a(x, i, k)∧a(y, j, k)f((x, k),(y, k), n+ 1) +

rX

kF

a(x, i, k)∨a(y, j, k)

f(x,y, n+ 1)−rf(x,y, n).

3.3.3.3 The distancedeis contracting forP

In this subsection, we show that the distancededefined above is indeed contracting for the coupling constructed in the previous subsection. This is formulated in the following result.

Lemma 3.3.11. Let (Xt,Yt)t≥0 be the coupling of the previous section. Under the assumptions of Theorem3.3.2, we can chooserandδin such a way that

tt, E

h

d(Xe t,Yt)iγd(x,e y),

for someγ ∈(0,1)andt >0, and allx,yE×F verifyingd(x,e y)<1.

Proof. Recall that sinced(x,e y) < 1one hasI0 =J0 and thatT, defined in (3.20), denotes the first time of separation ofI andJ. Using Lemma3.2.6, there existq ∈(0,1]andC, η >0such that

E

h

d(Xe t,Yt)i≤E

1{T=∞}

1

δdq(Xt, Yt) +1{T <+∞}

≤ 1 δE

e

Rt

0(Ls)ds

E[dq(x, y)] +P(T < +∞).

Ceηtd(x,e y) +P(T <+∞). Here, we have used the fact that

E

h1{T=∞}dq(Xt, Yt)i≤E

h1{TτNt}e(LτNt)(tτNt)dq(XτNt, YτNt)i

≤E

h1{TτNt}e(LτNt)(tτNt)E

hdq(XτNt, YτNt)|Gn

ii

≤E

"

1{TτNt−1}e

Rt

τNt−1(Ls)ds

dq(XτNt−1, YτNt−1)

#

≤E

eR

t

0(Ls)ds

E[dq(x, y)].

Markov processes with random switching

It remains to obtain a bound onP(T <+∞). SinceI andJ can only jump whenN jumps,T can be finite only if it is one of the jump times ofN. So, we set

An ={T =τn}={Tτn and Iτn 6=Jτn}.

By Assumption3.1.1, we have

P(An) = P({UnI0nI1nI3n} ∩ {Tτn})

≤E

"

21{Tτn}P

jF |a(Xτn, Iτn, j)−a(Yτn, Iτn, j)|

r

#

≤E

"

21{Tτn}P

jF |a(Xτn, Iτn, j)−a(Yτn, Iτn, j)|

r

!q#

≤ 2qκq

rq E[d(Xτn, Yτn)q]≤ 2qκq rq E

eq

Rτn

0 α(Ls)ds

d(x, y)q. Hence

P(T <∞) = X

n≥1

P(An)≤ 2qκq

rq d(x, y)qX

n≥1

E

eq

Rτn

0 α(Ls)ds

.

Now, similarly to the proof of Lemma3.3.8, ifr is large enough then there existC0 > 0andε > 0 verifying

X

n≥1

E

eq

Rτn

0 α(Ls)ds

X

n≥1

C0eεn=: ˜C <+∞.

Combining these bounds, we obtain the estimate E

h

d(Xe t,Yt)iCeηt+(2κ)qC˜ rq δ

!

d(x,e y).

First makingδsufficiently small and then takingtlarge enough, we thus obtain the announced result.

3.3.4 Bounded sets are d-small

e

Here, we prove that if a set is bounded then it isd-small.e

Lemma 3.3.12. Under the assumptions of Theorem 3.3.2, if SE ×F is of bounded diameter in the sense that

R= sup{d(x, y)|x,yS}<+∞, then there existt, t >0such thatSisd-small fore Pt, for allt∈[t, t].

Markov processes with random switching

Proof. Let x = (x, i) and y = (y, j) be two different points of S. By the assumptions of Theo- rem 3.3.2, there exists i0F such that ρ(i0) > 0. Let (Xt)t≥0 and (Yt)t≥0 be two independent processes generated by (3.2) and starting respectively fromxandy. Let us denote

τin = inf{t ≥0|It =Jt=i0} and τout = inf{tτin |It6=i0orJt 6=i0}. For everyb, c >0such thatb > c, we define

pc,b(x,y) = P(τin < c, τout > b) .

By Assumptions3.1.1and3.1.2, we havepc,b(x,y)>0. Using the fact thatais bounded, a coupling argument shows thatpc,b is lower bounded by a positive quantity which only depends oniandj. We then obtain the bound

E

h

d(Xe t,Yt)i≤E

h1{τin<c,τout>b}d(Xe t,Yt)i+ 1−pc,b(x,y)

≤1−pc,b(x,y)1−δ−1e%ceρ(i0)td(x, y)

≤1−pc,b(x,y)1−δ−1e%ceρ(i0)tR,

where% was defined in (3.18). There existc > 0and t > c such that1−δ−1e%ceρ(i0)tR > 0.

SinceF is finite, we can furthermore boundpc,b from below by the minimum over all i, jF, and the result follows for anyb > t andt ∈(t, b).

Remark 3.3.13. One can see from this proof that it is not necessary that the jump rates are lower bounded, as in Assumption3.1.2. Indeed, we need that, for each i, jF, the jump times of I are stochastically smaller than a variable which does not depend of the dynamics ofX.