Proof of Theorem 5.1. Define the auxiliary process yN(t) via yiN(t) :=vi(0)− ∑
j:j̸=i
∫ t 0
rij(¯xN(u))du+ ∑
j:j̸=i
∫ t 0
rji(¯xN(u))du
+ ∑
h∈S0
∑
j∈S1
∫ t u=0
pijFj(t−u)rhj(¯xN(u))du
− ∑
h∈S0
∑
j:j∈S0
∫ t u=0
pjiFi(t−u)rhi(¯xN(u))du (5.8)
fori∈ S.
Then
|x¯Ni (t)−vi(t)| ≤ |x¯Ni (t)−yiN(t)|+|yNi (t)−vi(t)| for anyi∈ S.
Denote
DiN(T) = sup
t∈[0,T]
|x¯Ni (t)−yiN(t)|
We estimate∥yN(t)−v(t)∥by
|yiN(t)−vi(t)| ≤ ∑
j:j̸=i
∫ t 0
|rij(¯xN(u))−rij(v(u))|du+ ∑
j:j̸=i
∫ t 0
|rji(¯xN(u))−rji(v(u))|du
+ ∑
h∈S0
∑
j∈S1
∫ t u=0
pijFj(t−u)|rhj(¯xN(u))−rhj(v(u))|du
+ ∑
h∈S0
∑
j:j∈S0
∫ t u=0
pjiFi(t−u)|rhi(¯xN(u))−rhi(v(u))|du
≤ZR
∫ t 0
∥xN(u)−v(u)∥du
where
Z :=|S0|+|S0|+|S0| · |S1|+|S0|2
and∥.∥is the maximum norm onRS. We aim to show thatDNi (T)→0 in probability asN → ∞for eachi∈ S; once we have that, we have
∥¯xN(t)−v(t)∥ ≤max
i∈S DNi (T) +ZR
∫ t 0
∥¯xN(u)−v(u)∥du (5.9)
and an application of Gr¨onwall’s lemma ([19], page 498) readily yields
∥x¯N(t)−v(t)∥ ≤max
i∈S DNi (T) exp(ZRT), proving the theorem.
It now remains to show that for eachi∈ S,DNi (T)→0 in probability asN→ ∞.
To see this note that:
DiN(T)≤ |vi(0)−x¯Ni (0)|+ ∑
j:j̸=i
sup
t∈[0,T]
∫ t 0
rij(¯xN(u))du− 1 NPij
( N
∫ t 0
rij(¯xN(u))du)
+ ∑
j:j̸=i
sup
t∈[0,T]
∫ t 0
rji(¯xN(u))du− 1 NPji
( N
∫ t 0
rji(¯xN(u))du) +
+ ∑
h∈S0
∑
j∈S1
pij ∫ t
u=0
Fj(t−u)rhj(¯xN(u))du−
−
∫ t u=0
1 (
Tji
Phj(N∫0zrhj(¯xN(u))du)≤t−z ) 1
NdPhj
( N
∫ z 0
rhj(¯xN(u))du)
+ ∑
h∈S0
∑
j∈S0
pji ∫ t
u=0
Fi(t−u)rhi(¯xN(u))du−
−
∫ t u=0
1 (
Tij
Phi(N∫0zrhi(¯xN(u))du)≤t−z ) 1
NdPhi (
N
∫ z 0
rhi(¯xN(u))du)
(5.10)
fori∈ S
The first term in (5.10) converges to 0 in probability per our assumptions. The second and third terms in (5.10) are essentially the same; they are handled in the following lemma.
Lemma 5.2. For anyi, j∈ S
sup
t∈[0,T]
∫ t 0
rji(¯xN(u))du− 1 NPji
( N
∫ t 0
rji(¯xN(u))du) →0
almost surely as N→ ∞.
Remark. The lemma states almost sure convergence; this makes sense because the coupling pro-vided by the Poisson representation puts the PGSMP for different values ofN in the same probability space. That said, convergence in probability is enough for our purposes.
Proof of Lemma 5.2. By the Lipschitz-condition, 0 ≤ ∫t
0rijN(¯xN(u))du≤RT for any t ∈[0, T] and thus
sup
t∈[0,T]
∫ t 0
rji(¯xN(u))du− 1 NPji
( N
∫ t 0
rji(¯xN(u))du) ≤ 1
N sup
s∈[0,RT]
|Pc(N s)−N s|,
which goes to 0 almost surely by the functional strong law of large numbers (FSLLN) for the Poisson process ([65], Section 3.2).
What remains is to prove that the last two terms in (5.10) go to 0 in probability.
Before proceeding, we change the notation a bit. h, iandjwill be fixed from now on until the end of this section. We will thus drop them from notation and useF =Fj, r=rhj, Tk =TkjiandP =Phj in (5.10) for the first of the last two terms in (5.10) (and, correspondingly,F =Fi, r=rhi, Tk =Tkij andP =Phi for the last term in (5.10)).
We will also use the shorthand
JN(t) =P (
N
∫ t 0
r(¯xN(u))du )
for the (measure generated by the) Poisson process.
Using the above notation, either of the last two terms in (5.10) (without the finite summations
∑
h∈S0
∑
j∈S1·and the bounded constantpji) simplifies to ∫ t
0
F(t−u)r(¯xN(u)) du−
∫ t 0
1(
TJN(u)≤t−u) 1
NdJN(u)
(5.11)
We note that ∫ t
0
1(
TJN(u)≤t−u) 1
NdJN(u)−
∫ t 0
F(t−u)r(¯xN(u)) du ≤ ∫ t
0
F(t−u)1
NdJN(u)−
∫ t 0
F(t−u)r(¯xN(u)) du +
∫ t 0
1(
TJN(u)≤t−u) 1
NdJN(u)−
∫ t 0
F(t−u)1
N dJN(u)
. (5.12)
The first term on the right hand side will be dealt with in Lemma 5.3 and the second in Lemma 5.4.
We have some more preparations first. From Lemma 5.2 we already have that sup
t∈[0,T]
1 NP
( N
∫ t 0
r(¯xN(u))du )
−
∫ t 0
r(¯xN(u))du →0
almost surely asN → ∞. As a direct consequence of this, we also have sup
s,t∈[0,T]
1 NP
( N
∫ t s
r(¯xN(u))du )
−
∫ t s
r(¯xN(u))du →0 almost surely since
sup
s,t∈[0,T]
∫ t s
·
= sup
s,t∈[0,T]
∫ t 0
· −
∫ s 0
·
≤2 sup
t∈[0,T]
∫ t 0
· Also as a preparation, we have
sup
t∈[0,T]
∫ t 0
r(¯xN(u))du≤ sup
t∈[0,T]
∫ t 0
R∥x¯N(u)∥du≤ sup
t∈[0,T]
Rt=RT
independent of N, again using ∥¯xN∥ ≤ N and r(¯x) ≤ R∥x¯∥. Lemma 5.2 then also implies
1 N
∫t
0dJN(u)≤RT+εN, whereεN →0 almost surely asN → ∞.
Lemma 5.3.
sup
t∈[0,T]
∫ t 0
F(t−u)1
NdJN(u)−
∫ t 0
F(t−u)r(¯xN(u))du →0
almost surely as N→ ∞. Proof. Letε >0 be fixed. Write
F(t−u) =gt,ε(u) +ht,ε(u),
whereg=gt,εis a piecewise constant function with 0≤g(u)≤1 and∥h∥∞≤ε. Their exact definition is as follows. Take the ε,2ε, . . . quantiles of F(t−u) (F(t−u) as a function ofu is nonincreasing between 0 and 1); that is, letuk = inf{u: F(t−u)≤kε}. Some of theseuk’s may be equal ifF has discontinuities. The number of distinct quantiles is certainly no more than⌈ε−1⌉, independent ofN andt.
Letg be the piecewise constant function
g(u) =F(t−uk) if u∈(uk−1, uk],
sog(u)≤F(t−u). The choice ofuk’s guarantees thath(u) =F(t−u)−g(u)≤ε.
Then we can write ∫ t
0
F(t−u)1
NdJN(u)−
∫ t 0
F(t−u)r(¯xN(u))du ≤ ∫ t
0
g(u)1
NdJN(u)−
∫ t 0
g(u)r(¯xN(u))du + ∫ t
0
h(u)1
NdJN(u)−
∫ t 0
h(u)r(¯xN(u))du
Sinceg is piecewise constant, ∫ t
0
g(u)1
NdJN(u)−
∫ t 0
g(u)r(¯xN(u))du = 1
N
⌈∑ε−1⌉ k=1
g(uk) (
JN(uk)−JN(uk−1)−
∫ uk
uk−1
N r(¯xN(u))du )
≤ 1
N
⌈∑ε−1⌉ k=1
g(uk)
(
JN(uk)−JN(uk−1)−
∫ uk uk−1
N r(¯xN(u))du )≤
1 N
⌈∑ε−1⌉ k=1
JN(uk)−JN(uk−1)−
∫ uk uk−1
N r(¯xN(u))du ≤
⌈∑ε−1⌉ k=1
sup
s,t∈[0,T]
1 NP
( N
∫ t s
r(¯xN(u))du )
−
∫ t s
r(¯xN(u)) =
⌈ε−1⌉ · sup
s,t∈[0,T]
1 NP
( N
∫ t s
r(¯xN(u))du )
−
∫ t s
r(¯xN(u))du →0 almost surely asN → ∞sinceεis independent ofN.
Since∥h∥∞≤ε, we have ∫ t 0
h(u)1
NdJN(u)−
∫ t 0
h(u)r(¯xN(u))du ≤ ∫ t
0
h(u)1
NdJN(u) +
∫ t 0
h(u)r(¯xN(u))du ≤ ε
N ∫ t
0
dJN(u) +ε
∫ t 0
r(¯xN(u))du
≤ε(2RT +εN), independent oft (withεN →0 almost surely asN → ∞).
Lettingε→0 proves sup
t∈[0,T]
∫ t 0
F(t−u)1
NdJN(u)−
∫ t 0
F(t−u)r(¯xN(u))du →0
almost surely asN → ∞. Lemma 5.4.
sup
t∈[0,T]
1 N
∫ t 0
1(
TJN(u)≤t−u)
dJN(u)−
∫ t 0
F(t−u)dJN(u) →0 almost surely as N→ ∞.
Proof. Letε >0 be fixed. Also fixtfor now. We want to prove P
(1 N
∫ t 0
1(
TJN(u)≤t−u)
dJN(u)−
∫ t 0
F(t−u)dJN(u) > ε
)
is exponentially small inN via Azuma’s inequality [11, 2]. Once we have that, we can apply Borel–
Cantelli lemma (see e.g. [16] Chapter 2.3) to conclude that for any fixedϵ, the above event happens only finitely many times, which is equivalent to almost sure convergence to 0. To apply Azuma, we need to write the above integral as a martingale with bounded increments. The measure dJN(u) is concentrated on pointsuwhere P has an arrival atN∫u
0 r(¯xN(z))dz. Let we denote these random points byu1, u2, . . . (in fact, the whole sequence depends on the value ofN; we will keepN fixed as long as we use this sequence, and not includeN in the notation). The integral only has contributions from these points; it is natural to write (using a slightly different notation)
Sl:= (1(T1≤t−u1)−F(t−u1)) +· · ·+ (1(Tl≤t−ul)−F(t−ul)) MN :=P
( N
∫ t 0
r(¯xN(z))dz )
so that ∫ t
0
1(
TJN(u)≤t−u)
dJN(u)−
∫ t 0
F(t−u)dJN(u) =SMN. We first resolve the difficulty thatMN is in fact random.
P (1
N ∫ t
0
1 (
TJeN(u)≤t−u
)
dJN(u)−
∫ t 0
F(t−u)dJN(u) > ε
)
= P(
SMN
N > ε
)
=
∑∞ l=0
P( Sl
N
> ε, MN =l )
≤
2RT N∑
l=0
P( Sl
N > ε
) +
∑∞ 2RT N+1
P(MN =l) =
2RT N∑
l=0
P( Sl
N > ε
)
+P(MN >2RT N).
The sum was cut at 2RT N because MN is stochastically dominated by a Poisson distribution with parameterRT N, soP(MN >2RT N) is exponentially small due to Cram´er’s large deviation theorem (see e.g. Theorem II.4.1 in [18]):
P(MN >2RT N)≤e−RT N(2 ln 2−1).
(The Cram´er rate function of the Poisson-distribution with parameterλisI(x) =xln(x/λ)−x+λ.) To apply Azuma to each of the terms P(SNl> ε)
, we also need to check that Sl is indeed a martingale with bounded increments. To set it up properly as a martingale, note that {ul} is an increasing sequence of stopping times, so the filtration {Fl} is well-defined; Fl contains all the
information known up to timeul, including the values of all of the non-Markovian clocks that started by the timeul.
Sl has bounded increments, since
|1({Tl≤t−ul)−F(t−ul)| ≤1.
The last step to apply Azuma is that we need to check thatSl is a martingale with respect toFl. It is clearly adapted, and
E(1(Tl+1≤t−ul+1)|Fl) =E(E(1(Tl+1≤t−ul+1)|Fl, ul+1)|Fl) = E(P(Tl+1≤t−ul+1|Fl, ul+1)|Fl) =E(F(t−ul+1)|Fl)
shows that it is a martingale as well. (In the last step, we used the fact thatul+1 is measurable with respect toσ{Fl∪{ul+1}}whileTl+1 is independent from it.)
We have everything assembled to apply Azuma’s inequality (Theorem 5.2 in [11]):
P(|Sl−E(Sl)|> λ)≤2e−λ
2 2l
and thus (consideringE(Sl) = 0)
2RT N∑
l=0
P( Sl
N > ε
)
≤
2RT N∑
l=0
2e−ε
2N2 2l ≤
≤2RT N·2e−ε
2N2
4RT N = 4RT N e−ε
2N 4RT.
In the last inequality, we estimated each term in the sum by the largest one, which is forl= 2RN. The estimate obtained is
P (1
N ∫ t
0
1(
TJN(u)≤t−u
)dJN(u)−
∫ t 0
F(t−u)dJN(u) > ε
)
≤ 4RT N e−ε
2N
4RT +e−RT N(2 ln 2−1).
Remember that t was fixed; we need to upgrade this estimate into an estimate that is valid for supt∈[0,T](.) before applying Borel–Cantelli lemma. We do this by partitioning the interval [0, T] into N subintervals uniformly, and then controlling what happensat the partition points andbetween the partition points separately. For the former, we apply the previous estimate. Let
tm:= mT
N , m= 0,1, . . . N,
then
P (
max
0≤m≤N
1 N
∫ tm 0
1(
TJN(u)≤t−u
)dJN(u)−
∫ tm 0
F(t−u)dJN(u) > ε
)
≤ (N+ 1)
(
4RT N e−ε
2N
4RT +e−RT N(2 ln 2−1)) ,
which is still summable.
Now we turn our attention to the intervals [tm, tm+1]. Since
∫ t 0
1(
TJN(u)≤t−u
)dJN(u) and
∫ t 0
F(t−u)dJN(u)
are both increasing int, we only have to check that neither of them increases by more than εN over an interval [tm, tm+1].
Letmbe fixed. We handle the two integrals separately. First, for
∫ t 0
F(t−u)dJN(u), we have
∫ tm+1
0
F(tm+1−u)dJN(u)−
∫ tm
0
F(tm−u)dJN(u) =
∫ tm 0
F(tm+1−u)−F(tm−u)dJN(u) +
∫ tm+1 tm
F(tm+1−u)dJN(u)≤
∫ tm 0
F(tm+1−u)−F(tm−u)dJN(u) +
∫ tm+1 tm
1dJN(u).
The second term is equal toJN(tm+1)−JN(tm). By the Lipschitz-condition, this is stochastically dominated byZ ∼Poisson(RT) since the length of the interval isT /N, and thus
P (1
N
∫ tm+1 tm
F(tm+1−u)dJN(u)> ε )
≤P (Z
N > ε )
=P (Z
ε > N )
.
Note that the right hand side is summable inN, its sum being equal to the expectation of⌈Z
ε
⌉. To estimate the other term, note that
u∈[tl−1, tl] =⇒F(tm+1−u)−F(tm−u)≤F(tm+1−tl−1)−F(tm−tl) = F(tm+1−tl−1)−F(tm+1−tl) +F(tm+1−tl)−F(tm−tl),
which gives
∫ tm
0
F(tm+1−u)−F(tm−u)dJN(u) =
∑m l=1
∫ tl tl−1
F(tm+1−u)−F(tm−u)dJN(u)≤
∑m l=1
∫ tl tl−1
F(tm+1−tl−1)−F(tm−tl)dJN(u) =
∑m l=1
(F(tm+1−tl−1)−F(tm−tl))(JN(tl)−JN(tl−1)).
We use two things here: the fact that (JN(tl)−JN(tl−1)) is stochastically dominated by Poisson(RT) and the fact that the sum
∑m l=1
(F(tm+1−tl−1)−F(tm−tl)) =
∑m l=1
F(tm−l+2)−F(tm−l) = F(tm+1) +F(tm)−F(1)−F(0)≤2
is telescopic. This means that the whole sum can be stochastically dominated by Poisson(2RT) (note that the number of clocks starting at each interval is not independent, but because of the Lipschitz-condition, we may still use independent Poisson variables when stochastically dominating the sum).
Using the notationZ ∼Poisson(RT) again, we get that
∑∞ N=1
P (2Z
N > ε )
=
∑∞ N=1
P (2Z
ε > N )
≤2RT ε + 1.
(In fact,P(2Z
ε > N)
goes to 0 superexponentially inN.) The last term to estimate is the increment of
∫ t 0
1(
TJN(u)≤t−u
)dJN(u).
betweentmandtm+1, e.g. the number of clocks expiring betweentm andtm+1.
Partition the clocks according to which interval [tl−1, tl] they started in. The number of clocks starting in [tl−1, tl] is stochastically dominated byZ∼Poisson(RT) by the Lipschitz-condition, and for each such clock, the probability that it goes off in [tm, tm+1] is less than or equal toF(tm+1)−F(tl−1).
This implies that the number of the clocks starting in [tl−1, tl] and going off in [tm, tm+1] is stochasti-cally dominated byWm,l∼Poisson(RT(F(tm+1)−F(tl−1))). The total number of clocks going off in [tm, tm+1] is stochastically dominated by Poisson(RT∑m
l=1(F(tm+1)−F(tl−1))), where the familiar telescopic sum appears in the parameter. (Once again, the Lipschitz-condition was used implicitly.) So the total number of clocks going off in [tm, tm+1] is stochastically dominated by Poisson(2RT), which means we arrive at the also familiarP(2Z
ε > N)
value, which we already examined and proved
to be summable inN.
Putting it together, we get that
P (
sup
t∈[0,T]
1 N
∫ t 0
1(
TJN(u)≤t−u)
dJN(u)−
∫ t 0
F(t−u)dJN(u) > ε
)
≤CN,ε
where
∑∞ N=1
CN,ε<∞,
so the Borel–Cantelli lemma gives almost sure convergence asN → ∞.
With Lemmas 5.2-5.4 finished, the proof of Theorem 5.1 is complete.