First critical probability for a problem on random orientations in $G(n,p)$

We study the random graph $G(n,p)$ with a random orientation. For three fixed vertices $s,a,b$ in $G(n,p)$ we study the correlation of the events $a \to s$ and $s\to b$. We prove that asymptotically the correlation is negative for small $p$, $p<\frac{C_1}n$, where $C_1\approx0.3617$, positive for $\frac{C_1}n<p<\frac2n$ and up to $p=p_2(n)$. Computer aided computations suggest that $p_2(n)=\frac{C_2}n$, with $C_2\approx7.5$. We conjecture that the correlation then stays negative for $p$ up to the previously known zero at $\frac12$; for larger $p$ it is positive.


Introduction
Let G(n, p) be the random graph with n vertices where each edge has probability p of being present independent of the other edges. We further orient each present edge either way independently with probability 1 2 , and denote the resulting random directed graph by G(n, p). This version of orienting edges in a graph, random or not, is natural and has been considered previously in e.g. [1,2,3,5].
Let a, b, s be three distinct vertices and define the events A := {a → s}, that there exists a directed path in G(n, p) from a to s, and B := {s → b}. In a previous paper, [2], we showed that, for fixed p, the correlation between A and B asymptotically is negative for p < 1 2 and positive for p > 1 2 . Note that we take the covariance in the combined probability space of G(n, p) and the orientation of edges, which is often referred to as the annealed case, see [2] for details. We say that a probability p ∈ (0, 1) is critical (for a given n) if the covariance Cov(A, B) = 0. We have thus shown in [2] that there is a critical probability 1 2 + o(1) for large n. (Moreover, this is the largest critical probability, since the covariance stays positive for all larger p < 1.) We also conjectured that for large n, there are in fact (at least) three critical probabilities when the covariance changed sign. Based on computer aided computations we guessed that the first two critical probabilities would be approximately 0.36 n and 7.5 n . In this note we prove that there is a first critical probability of the conjectured order, where the covariance changes from negative to positive, and thus there must be at least three critical probabilities. Our theorem is as follows. This research was initiated when all three authors visited the Institut Mittag-Leffler (Djursholm, Sweden). Theorem 1.1. With p = 2c n and sufficiently large n, the covariance Cov(A, B) is negative for 0 < c < c 1 and positive for c 1 < c < 1, where c 1 ≈ 0.180827 is a solution to (2 − c)(1 − c) 3 = 1. Furthermore, for fixed c with 0 ≤ c < 1, In fact, the proof shows that (1.1) holds uniformly in 0 ≤ c ≤ c for any c < 1; moreover, we may (with just a little more care) for such c write the error term as O(c 4 n −4 ). This implies that for large n, the critical p ≈ 2c 1 /n is indeed the first critical probability, and that the covariance is negative for all smaller p > 0. Remark 1.2. In a random orientation of any given graph G, it is a fact first observed by McDiarmid that P(a → s) is equal to P(a ↔ s) in an edge percolation on the same graph with probability 1/2 for each edge independently, see [5]. Hence the events A (and thus B) have the same probability as P(a ↔ s) in G(n, p/2). With p = 2c/n it is well known that for c < 1 this probability is c (1−c) n −1 + O(n −2 ), see e.g. [4]. Hence the covariance in (1.1) is of the order O(P(A) P(B)/n).
The outline of the proof is as follows, see Sections 2 and 3 for details. Let p := 2c/n, where c < 1. Let X A := #{a → s} be the number of paths from a to s in G(n, p) and X B := #{s → b}. (In the proof below, for technical reasons, we actually only count paths that are not too long.) We first show that, in our range of p, the probability that X A ≥ 2 or X B ≥ 2 is small, and that we can ignore these events and approximate Cov(A, B) by Cov(X A , X B ). The latter covariance is a double sum over pairs of possible paths (α, β), where α goes from a to s and β goes from s to b, and we show that the largest contribution comes from configurations of the following two types: Type 1: The two edges incident to s, i.e the last edge in α and the first edge in β, are the same but with opposite orientations; all other edges are distinct. See Figure 1.
Type 2: α and β contain a common subpath with the same orientation, but all other edges are distinct. See Figure 2. If (α, β) is of Type 1, then α and β cannot both be paths in G(n, p), since they contain an edge with opposite orientations. Thus each such pair (α, β) gives a negative contribution to Cov(X A , X B ). Pairs of Type 2, on the other hand, give a positive contribution. It turns out that both contributions are of the same order n −3 , see Lemmas 3.2 and 3.3, with constant factors depending on c such that the negative contribution  Figure 2. Configurations of Type 2 (i, j ≥ 0, k, l, m ≥ 1).
from Type 1 dominates for small c, and the positive contribution from Type 2 dominates for larger c.
Open problem 1.3. It would be interesting to find a method to compute also the second critical probability, which we in [2] conjectured to be approximately 7. 5 n . (The methods in the present paper apply only for c < 1.) Even showing that the covariance is negative when p is of the order log n n is open. Moreover we conjecture that (for large n at least) there are only three critical probabilities, but that too is open.

Proof of Theorem 1.1
We give here the main steps in the proof of Theorem 1.1, leaving details to a sequence of lemmas in Section 3.
By a path we mean a directed path γ = v 0 e 1 · · · e v in the complete graph K n . We use the conventions that a path is self-avoiding, i.e. has no repeated vertex, and that the length |γ| of a path is the number of edges in the path.
We let Γ be the set of all such paths and let, for two distinct vertices v and w, Γ vw be the subset of all paths from v to w.
If γ ∈ Γ, let I γ be the indicator that γ is a path in G(n, p), i.e, that all edges in γ are present in G(n, p) and have the correct orientation there. Thus Let I A and I B be the indicators of A and B. Note that the event A occurs if and only if α∈Γas I α ≥ 1, and similarly for B.
It will be convenient to restrict attention to paths that are not too long, so we introduce a cut-off L := log 2 n and let Γ L vw be the set of paths in Γ vw of length at most L. Let i.e, the numbers of paths in G(n, p) from a to s and from s to b, ignoring paths of length more than L.
Write X A = I A + X A and X B = I B + X B , where I A and I B are the indicators for the events X A ≥ 1 and X B ≥ 1 respectively, so that Similarly, since where Lemma 3.5 shows that the last three terms are O(n −4 ). Hence, it suffices to compute Lemmas 3.2 and 3.3 yield the contribution to this sum from pairs (α, β) of Types 1 and 2, and Lemma 3.4 shows that the remaining terms contribute only O(n −4 ). Using (2.2)-(2.4) and the lemmas in Section 3 we thus obtain is negative for c = 0 and has two real zeros, for example because its discriminant is −283 < 0, see e.g. [6]; a numerical calculation yields the roots c 1 ≈ 0.180827 and c 2 ≈ 2.380278, which completes the proof.

Lemmas
We begin with some general considerations. We assume, as in Theorem 1.1, that p = 2c/n and 0 ≤ c < 1.
Consider a term Cov(I α , I β ) in (2.4). Suppose that α and β have lengths α and β . Furthermore, suppose that β contains δ ≥ 0 edges not in α (ignoring the orientations) and that these form µ ≥ 0 subpaths of β that intersect α only at the endvertices. (We will use the notation β \ α for the set of (undirected) edges in β but not in α.) The number αβ of edges common to α and β (again ignoring orientations) is thus β − δ. By (2.1), E I α = (c/n) α and E I β = (c/n) β .
(i) If α and β have no common edge, then I α and I β are independent and (3.1) Cov(I α , I β ) = 0.
(ii) If all common edges have the same orientation in α and β, then (iii) If some common edge has different orientations in α and β, then E(I α I β ) = 0 and We denote the falling factorials by (n) := n(n − 1) · · · (n − + 1). Note that the total number of paths of length in Γ vw is (n − 2) −1 := (n − 2) · · · (n − ), since the path is determined by choosing − 1 internal vertices in order, and all vertices are distinct.
Proof. J A is the indicator of the event that there is a path in G(n, p) from a to s, and that every such path has length > L = log 2 n. Thus, 0 ≤ J A ≤ α∈Γas, |α|>L I α and thus, using (2.1) and the fact that there are (n − 2) −1 ≤ n −1 paths of length in Γ as , Proof. Let the path α from a to s consist of i + 1 edges, where the last edge is the first in the path β of length j + 1 from s to b, see Figure 1. The paths must not share any more edges, but could have more common vertices. Here i, j ≥ 0 and i + j ≥ 1 since a = b. Let R i,j be the number of such pairs of paths, for given i and j. If j ≥ 1, the paths are determined by the choice of i distinct vertices for α and then j − 1 distinct vertices for β; if j = 0, then i ≥ 1 and the paths are determined by the choice of i − 1 distinct vertices for α. Order is important so, for i, j ≤ L, with a minor modification if j = 0, and summing over all such pairs (α, β) gives by A pair (α, β) of paths of Type 2 must contain a directed cycle containing s, from which there are m ≥ 1 edges to a vertex x to which there is a directed path of length i ≥ 0 from a. The cycle continues from x with k ≥ 1 edges to a vertex y, which connects to b via j ≥ 0 edges. The cycle is completed by l ≥ 1 edges from y to s, see Figure 2.
Let R i,j,k,l,m be the number of such pairs (α, β) with given i, j, k, l, m. The path α is determined by i + k + l − 1 distinct vertices and given α, if j ≥ 1, then the path β is determined by choosing m + j − 2 vertices; if j = 0 then b lies on α, so α is determined by choosing i + k + l − 2 vertices, and then β is determined by choosing m − 1 further vertices. Reasoning as in the proof of Lemma 3.2 we have Due to our cut-off, we have to have i + k + l ≤ L and j + k + m ≤ L, but we may for simplicity here allow also paths α, β with lengths larger than L; the contribution below from pairs with such α or β is O(c L ) = O(n −99 ). Summing over all possible configurations gives i,j≥0, k,l,m≥1 Proof. Consider pairs (α, β) with some given α , δ, µ. The path α, which has α − 1 interior vertices, may be chosen in ≤ n α−1 ways. The 2µ endvertices of the µ subpaths of β \α are either b or lie on α, and given α, these may be chosen (in order) in ≤ ( α +2) 2µ ways. The δ − µ internal vertices in the subpaths can be chosen in ≤ n δ−µ ways. They can be distributed in δ−1 µ−1 (interpreted as 1 if µ = δ = 0) ways over the subpaths. The path β is determined by these endvertices, the sequence of δ − µ interior vertices in the subpaths between these endvertices and which vertices belong to which subpath; hence the total number of choices of β is ≤ δ−1 µ−1 ( α + 2) 2µ n δ−µ . For each such pair (α, β), we have by (3.1)-(3.3) |Cov(I α , I β )| ≤ (c/n) α+δ . Consequently, the total contribution to |Cov(I α , I β )| from the paths with given α , δ, µ is at most We consider several different cases and show that each case yields a contribution O n −4 , noting that we may assume that β > δ, since otherwise α and β are edgedisjoint, and thus Cov(I α , I β ) = 0 by (3.1).
(iv) µ ∈ {1, 2} and α and β have some common edge with opposite orientations: In this case, (3.3) applies, and δ−1 µ−1 ≤ δ ≤ β . Thus, if we let αβ = β − δ ≥ 1 be the number of common edges in α and β, then the total contribution to |Cov(I α , I β )| for given α , β , µ, αβ (which determine δ = β − αβ ) is at most, in analogy with (3.5) but using (3.3), For fixed µ, the sum of (3.10) over α , β ≥ 1 and αβ ≥ 3 − µ is O n −4 , so we only have to consider 1 ≤ αβ ≤ 2 − µ. In this case we must have µ = 1 and αβ = 1 (and δ−1 µ−1 = 1); thus α and β have exactly one common edge, which is adjacent to one of the endvertices of β. If the common edge is adjacent to s, we have a pair (α, β) of Type 1, see Figure 1; we may thus assume that the common edge is not adjacent to s. Then, β ≥ 2 and the common edge is adjacent to b, which implies b ∈ α. Given α , the number of paths α that pass through b is ( α − 1)(n − 3) α−2 , since b may be any of the α − 1 interior vertices. The choice of α fixes the last interior vertex of β (as the successor of b in α), and the remaining β − 2 interior vertices may be chosen in ≤ n β −2 ways. The total contribution from this case is thus at most and summing over α and β we again obtain O n −4 .
(v) µ ∈ {1, 2} and all common edges in α and β have the same orientation: The edge in β at s does not belong to α (since it would have opposite orientation there), so one of the µ subpaths of β outside α begins at s. If µ = 1, or if µ = 2 and b / ∈ α, then (α, β) is of Type 2, see Figure 2 (j = 0 and j ≥ 1, respectively). We may thus assume that µ = 2 and b ∈ α. As in case (iv), given α , we may choose α in ( α − 1)(n − 3) α−2 ≤ α n α−2 ways. The µ = 2 subpaths of β outside α have 4 endvertices belonging to α; one is s and the others may be chosen in ≤ 3 α ways. For any such choice, the remaining δ − 2 vertices of β may be chosen in ≤ n δ−2 ways. The total contribution for given α and δ is thus, using (3.2), at most Proof. We only need to consider paths in Γ L , which is assumed throughout the proof. Define Y A := X A 2 , the number of pairs of (distinct) paths from a to s, and similarly Further, let Z A := X A 3 , the number of triples of (distinct) paths from a to s. Then . Finally we will show that Cov(Y A , X B ) = O(n −4 ) finishing the proof of the first part of the lemma.
If all common edges of the three paths have the same direction, E(I α 1 I α 2 I α 3 ) = c n l 1 +δ 2 +δ 3 , otherwise it is 0, so we need only consider paths with the same direction.
, where α 1 , α 2 and α 3 are three distinct paths from a to s and β is a path from s to b. We need only consider paths where all common edges have the same direction, as E(I α 1 I α 2 I α 3 I β ) = 0 otherwise.
where α 1 and α 2 are two distinct paths from a to s and β 3 and β 4 are two distinct paths from s to b. As above, we need only consider paths where all common edges have the same direction. As before, α 1 and α 2 are described by l 1 = |α 1 | ≥ 1, δ 2 = |α 2 \ α 1 | ≥ 1, the number of edges in α 2 not in α 1 , and µ 2 ≥ 1, the number of subpaths they form that intersect α 1 in (and only in) the endvertices. Then β 3 is described by δ 3 = |β 3 \ (α 1 ∪ α 2 )|, the number of edges in β 3 not in α 1 or α 2 , and µ 3 , the number of subpaths they form with no interior vertices in common with α 1 , α 2 . Similarly, β 4 is described by δ 4 = |β 3 \ (α 1 ∪ α 2 ∪ β 3 )| ≥ 0, the number of edges in β 4 not in α 1 , α 2 or β 3 and µ 4 ≥ 0, the number of subpaths they form which intersect α 1 , α 2 , β 3 in (and only in) the endvertices. Note that µ 3 ≥ 1 for every non-zero term, as otherwise the first edge in β 3 from s would be the last edge in one of the α paths, and therefore would have opposite direction.
The number of choices for the α paths are the same as in (ii) and given those, and δ 3 , µ 3 , δ 4 , µ 4 , the β paths can be chosen in at most n δ 3 −µ 3 · (l 1 + δ 2 − µ 2 + 1) 2µ 3 · δ 3 −1 , where the last factor is an upper bound for the number of ways β 4 can choose different sections from the α paths and β 3 .
Case 3: µ 2 + µ 3 = 2, µ 4 = 0. This can only occur if µ 2 = µ 3 = 1. Thus, β 3 starts with an edge not in any of the α paths and, as this is its only excursion it must end up at one of the α paths and follow it to b (if β 3 were to go straight to b without coinciding with any of the α paths then β 4 would have to do the same, so that β 3 = β 4 ). β 4 must start as β 3 until it encounters an α path and must have the possibility to chose a different path to b than β 3 along the α paths. This means that both α paths must pass through b and that they only differ somewhere between a and b. Thus, see Figure 3, there must be three vertices x (possibly x = a), y (possibly y = x) and z (possibly z = b) between a and b, so that both α paths pass in order a, x, y, z, b, s, and both β paths pass in order s, x, y, z, b. Both the two α paths and the two β paths follow different subpaths between y and z. Let the number of edges between a and x be i ≥ 0, between x and y be j ≥ 0, between y and z be k ≥ 1 and l ≥ 1 for the two possibilities (with k + l ≥ 3), between z and b be m ≥ 0, between s and x be r ≥ 1 and between b and s be t ≥ 1.