5.1. Bounds on Percolation on Finite Boxes
The basis for all the proofs of the previous statements is Lemma 9 below which determines large connected components on finite boxes. For integers and we define the box of size and lower left corner x by , and we abbreviate . Let be the largest connected component in box (with a fixed deterministic rule if there is more than one largest connected component in ).
Lemma 9 Assume and .
Choose with and let .
For every and there exists such that for all where is the largest connected component in box .
Sketch of proof of Lemma 9. This lemma corresponds to Lemma 2.3 of [
22] in our model. Its proof is based on renormalization arguments which only depend on the fact that
and that the edge probabilities are bounded from below by
for any
. Using that
for all
, a.s., we see by stochastic dominance that the renormalization holds also true for our model. Renormalization shows that for
m sufficiently large, the probability of {
contains at least a positive fraction of
vertices that are connected within a fixed enlargement of
} is bounded by a multiple of the probability of the same event but on a much smaller scale. To bound the latter probability we then use the fact that the model is percolating, and from this we can conclude Lemma 9. We skip the details of the proof of Lemma 9 and refer to the proof of Lemma 2.3 of [
27] for the details, in particular, the bound on
in our homogeneous percolation model (see proof of Lemma 2.3 in [
27]) also applies to the inhomogeneous percolation model.
Although the above lemma does not allow the connected component
to have size proportional to the size of box
, it is useful because it allows to start a new renormalization scheme to improve these bounds. This results in our Theorem 6 and is done similar as in
Section 3 of [
11]. For the proof of Theorem 6 we use the following lemma which has two parts. The first one gives the initial step of the renormalization and the second one gives a standard site-bond percolation model result. Once the lemma is established the proof of Theorem 6 becomes a routine task.
Let denote the largest connected component in box (with a fixed deterministic rule if there is more than one largest connected component in ). For , we say that boxes and are pairwise attached, write , if there is an occupied edge between a vertex in and a vertex in .
Lemma 10 - (a)
Assume and .
Choose such that .
For each and there exist and an integer such that for all .
- (b)
[Lemma 3.6, [11]] Let and consider the site-bond percolation model on with sites being alive with probability and sites are attached with probability where and .
Let be the size of the largest attached cluster of living sites in box .
For each there exist ,
and such that holds for all whenever and .
Proof of Lemma 10 (a). We adapt the proof of Lemma 3.5 of [
11] to our model. Fix
and
. Choose
such that
note that this differs from choice (3.13) in [
11]. Lemma 9 then provides that there exists
such that for all
For the choice
the first part of the result follows. For the second part we choose
. For
and
we have upper bound, using that
for all
, a.s.,
a.s., where the latter no longer depends on the weights
. For our choices of
δ and
ρ, Equation (
2) implies
This shows the second inequality of part (a). For part (b) we refer to Lemma 3.6 in [
11].
Proof of Theorem 6. The proof follows as in Theorem 3.2 of [
11], we briefly sketch the main argument. Choose the constants
,
,
,
and
as in Lemma 10, and note that it is sufficient to prove the theorem for
, where
and
m is chosen (fixed) as in Lemma 10 (a). In this set up
can be viewed as a disjoint union of
for
. There are
such disjoint boxes. We call
alive if
and we say that disjoint
and
are pairwise attached if their largest connected components
and
share an occupied edge. Part (a) of Lemma 10 provides that
is alive with probability exceeding
r and
and
are pairwise attached with probability exceeding
for living boxes
and
with
(note that in this site-bond percolation model the attachedness property is only considered between living vertices because these form the clusters). For any
, let
be the event that box
contains a connected component formed by attaching at least
of the living boxes. On event
we have for the largest connected component in
thus, the volume of the largest connected component
in box
is proportional to the volume of that box and there remains to show that this occurs with sufficiently large probability. Part (b) of Lemma 10 and stochastic dominance provide (note that we scale
from Lemma 10 (a) to the site-bond percolation model on
in Lemma 10 (b))
Choosing
provides
This finishes the proof of Theorem 6.
Proof of Corollary 7. The proofs of
and
of Corollary 7 follow completely analogous to the proofs of Corollaries 3.3 and 3.4 in [
11] (note that Lemma 10 (a) replaces Lemma 3.5 of [
11] and Theorem 6 replaces Theorem 3.2 of [
11]).
5.2. Proof of Continuity of the Percolation Probability
The key to the proofs of the continuity statements is again Lemma 9.
Proof of Theorem 3. Note that and imply that . Therefore, there exists with . For these choices of we have a unique infinite cluster , a.s., and we can apply Lemma 9.
We consider the same site-bond percolation model on
as in Lemma 10 (b). Choose
,
and
and define the model as follows: the following events are independent and every site
is alive with probability
and sites
are attached with probability
. For given
we choose the parameters
such that there exists an infinite attached cluster of living vertices, a.s., which is possible (see proof of Theorem 2.5 in [
22]).
The proof is now similar to the one of Theorem 6. Choose
such that
. From Lemma 9 we know that for all
m sufficiently large and any
where
denotes the largest connected component in
. The latter events define alive vertices
x on the lattice
(which due to scaling is equivalent to the above aliveness in the site-bond percolation model on
). Note that this aliveness property is independent between different vertices
. Attachedness
, for
, is then used as in the proof of Theorem 6 and we obtain in complete analogy to the proof of the latter theorem
where in the last step we used the choice of
ρ and the fact that
. Since
we get percolation and there exists an infinite cluster
, a.s., which implies
. Of course, this is no surprise because of the choice
with
.
Note that the probability of a vertex
being alive depends only on finitely many edges of maximal distance
(they all lie in the box
) and therefore this probability is a continuous function of
λ and
α. This implies that we can choose
and
so small that
where
is the measure where for occupied edges we replace parameters
λ by
and
α by
. As above we obtain, note
,
Since
we get percolation and there exists an infinite cluster
, a.s., which implies that
. This finishes the proof of Theorem 3.
Proof of Theorem 5. We need to modify Proposition 1.3 of [
23] because in our model, edges are not occupied independently induced by the random choices of weights
.
(i) From Theorem 3 it follows that
for all
, which proves continuity of
on
.
(ii) Next we show that
is left-continuous on
, that is,
To prove this we couple all percolation realization as
λ varies. This is achieved by randomizing the percolation constant
λ, see [
23,
28]. Conditionally given the i.i.d. weights
, define a collection of independent exponentially distributed random variables
, indexed by the edges
, which have conditional distribution
compare to Equation (
1). We denote the probability measure of
by
in order to distinguish this coupling model. We say that an edge
is
ℓ-open if
, and we define the connected cluster
of the origin to be the set of all vertices
which are connected to the origin by an
ℓ-open path. Note that we have a natural ordering in
ℓ,
i.e., for
we obtain
. Moreover for
, the
λ-open edges are exactly the occupied edges in this coupling (note that the exponential distribution Equation (
4) is absolutely continuous). This implies for
By countable subadditivity of
and the increasing property of
in
ℓ we have
Moreover, the increasing property of
in
ℓ provides
. Therefore, to prove Equation (
3) it suffices to show that
Choose
. Since there is a unique infinite cluster for
, a.s., there exists an infinite cluster
on the set
. If the origin belongs to
then the proof is done. Otherwise, because
is a subgraph of
, there exists a finite path
π of
λ-open edges connecting the origin with an edge in
. By the definition of
λ-open edges we have
for all edges
. Since
π is finite we obtain the strict inequality
. Choose
and it follows that
. This completes the proof for the left-continuity in
λ.
(iii) Finally, we need to prove right-continuity of
on
. For integers
we consider boxes
centered at the origin, see also Theorem 6. We define the events
,
i.e., the connected component
of the origin leaves box
. Note that
is the decreasing limit of
as
. Therefore, it suffices to show that
is a continuous function in
λ. We write
to indicate on which parameter
λ the probability law depends. We again denote by
the connected component of the origin connected within box
, see Corollary 7. Then, we have
Choose
, then we have for all
and all
We bound the two terms on the right-hand side of Equation (
5).
(a) First we prove that for all
there exists
such that for all
This is done as follows. For
we define the following events
This implies for
that
Moreover, note that
is decreasing in
,
We prove Equation (
6) by contradiction. Assume that Equation (
6) does not hold true,
i.e.,
. Then the first lemma of Borel-Cantelli implies
The latter implies that the degree distribution
has an infinite mean. This is a contradiction to Theorem 2.2 of [
12] saying that for
the survival function of the degree distribution has a power-law decay with rate
which provides a finite mean. Therefore, Equation (
6) holds true.
(b) For all
and all
there exists
such that for all
Note that
only contains finitely many edges of finite distance. Therefore, continuity in
λ is straightforward which provides Equation (
7).
Combining Equations (
6) and (
7) provides continuity of
in
λ for all
n, see also Equation (
5). Therefore, right-continuity of
follows. This finishes the proof of Theorem 5.
5.3. Proofs of the Graph Distances
In this section, we prove Theorem 8. Statement (a) of Theorem 8 is proved in Theorems 5.1 and 5.3 of [
12], the lower bound of statement (b1) is proved in Theorem 5.5 of [
12]. Therefore, there remain the proofs of the upper bound in (b1) and of the lower bound in (b2) of Theorem 8.
The proof of the upper bound in Theorem 8 (b1) follows from the following proposition and the fact that
is a continuous function. The following proposition corresponds to Proposition 4.1 in [
11] in the homogeneous long-range percolation model.
Proposition 11 Let and and .
For each and each ,
there exists such that holds for all with .
Sketch of proof of Proposition 11. We only sketch the proof because it is almost identical to the one in [
11]. Definition 1 and
Figure 1 of [
11] define for
a hierarchy of depth
connecting
x and
y as the following collection of vertices:
is a hierarchy of depth
connecting
x and
y if
- (1)
and ,
- (2)
and for all and ,
- (3)
for all and and such that the edge between and is occupied,
- (4)
each bond specified in (3) appears only once in .
The pairs of vertices
and
are called gaps. The proof is then based on the fact that for large distances
the event
of the existence of a hierarchy
of depth
m that connects
x and
y through points
which are dense is very likely (
m appropriately chosen), see Lemma 4.3 in [
11], in particular formula (4.18) in [
11] (where the key is Corollary 7
). On this likely event
, Lemma 4.2 of [
11] then proves that the graph distance cannot be too large, see (4.8) in [
11]. We can now almost literally translate Lemmas 4.2 and 4.3 of [
11] to our situation. The only changes are that in formulas (4.16) and (4.21) of [
11] we need to replace
of [
11]’s notation by
λ in our notation and we need to use that the weights
are at least one, a.s. We refrain from giving more details.
There remains the proof of the lower bound in (b2) of Theorem 8. We use a renormalization technique which is based on a scheme introduced by [
13]. Choose an integer valued sequence
,
, and define the box lengths
as follows: set
and for
,
Define the
n-stage boxes,
, by
For
, the children of
n-stage box
are the
disjoint
-stage boxes
We are going to define
good n-stage boxes
, note that we need a different definition from Definition 2 of [
13].
Definition 12 (good
n-stage boxes)
Choose and fixed. Lemma 13 Assume .
For all there exist and a constant such that for all and all ,
Proof of Lemma 13. Let
and
be two independent random variables each having a Pareto distribution with parameters
and
. For
we have, using integration by parts in the first step,
where the last step follows by distinguishing between the cases
,
and
. Choose
so large that
which, together with the above calculations, implies that for all
and
with
,
where the second inequality holds for all
with
large enough. It follows that for all
, using
,
Hence, for an appropriate constant
and for all
with
sufficiently large,
which finishes the proof of Lemma 13.
Lemma 14 Assume .
For ,
,
and sufficiently large we have This lemma is the analog in our model to Lemma 1 of [
13] and provides a Borel-Cantelli type of result that eventually the boxes
are good, a.s., for all
n sufficiently large.
Proof of Lemma 14. We prove by induction that
is summable. Choose
and set
. For
sufficiently large we obtain by Lemma 13,
where the last step holds true for
sufficiently large. Because
has only one child (because
) we get for
sufficiently large
For the induction step we note that
n-stage box
is not good if at least one of the
translations
,
, fails to have property (a) or (b) of Definition 12. Using translation invariance and Lemma 13 we get for all
and for all
sufficiently large, set
,
Note that the event in the probability above ensures that there are at least two children
and
of
that are not good and are separated by at least Euclidean distance
. Therefore, using
,
, the two boxes
and
are well separated in the sense that the events
and
are independent. Note that for the latter we need to make sure that
and
are disjoint for all
, which is the case because
and
have at least distance
. The independence implies the following bound
It follows that there is
such that for all for all
and
large enough
and we can choose
so large that Equation (
10) holds true also for all
. We claim that for all
sufficiently large and all
,
which will imply Lemma 14 because the right-hand side is summable. Indeed, Equation (
11) is true for
by Equations (
8) and (
9). Assuming that Equation (
11) holds for all
with
we get, using Equation (
10),
where the last step follows since
and
.
The following lemma is the analog of Proposition 3 of [
13] and it depends on Lemma 2 of [
13] and Lemma 14. Since its proof is completely similar to the one of Proposition 3 of [
13] once Lemma 14 has been established we skip this proof.
Lemma 15 (Proposition 3 of [
13])
Choose for .
There exists a constant such that for every n sufficiently large, if for every the n-stage box is good and for every the l-stage boxes centered at are good, then if satisfy then .
Proof of Theorem 8 (b2). Lemma 14 says that, a.s., the l-stage boxes are eventually good for all . Moreover, from Lemma 15 we obtain the linearity in the distance for these good boxes which says that, a.s., for n sufficiently large and we have .