Asymptotic Distribution of Isolated Nodes in Secure Wireless Sensor Networks under Transmission Constraints

The Eschenauer-Gligor (EG) key predistribution is regarded as a typical approach to secure communication in wireless sensor networks (WSNs). In this paper, we establish asymptotic results about the distribution of isolated nodes and the vanishing small impact of the boundary effect on the number of isolated nodes in WSNs with the EG scheme under transmission constraints. In such networks, nodes are distributed either Poissonly or uniformly over a unit square. The results reported here strengthen recent work by Yi et al.


Introduction
A wireless sensor network is composed of a collection of wireless sensors distributed over a geographic region.A wireless sensor network can be an integral part of military command, control, communication, computing, intelligence, surveillance, reconnaissance, and target system.It has been the subject of intense research in recent decades.Asymptotic analysis, valid when the number of nodes in the network is large enough, has been useful for understanding the characteristic of the network.
In many applications, a large number of wireless sensors are independently and uniformly deployed in the sensor field.They can be deployed by dropping from a plane or delivered in a missile.To model such a randomly deployed wireless sensor network, it is natural to represent the sensor nodes by a finite random point process over a network square area S. In addition, due to the short transmission range of communication links, two wireless sensors can build a link if and only if they are within each other's transmission range.Assume all  sensors have the same transmission range of radius   , then the induced network topology is a   -disk graph in which two nodes are joined by an edge if and only if their distance is at most   .This model is proposed by Gilbert [1] and referred to as a random geometric graph, denoted by (,   , S).
However, in many applications, a wireless sensor network is composed of low cost sensors.Due to the limited capacity, traditional security schemes and key management algorithms are too complicated and not feasible for such a system.The Eschenauer-Gligor (EG) [2] key predistribution scheme is a widely recognized way to secure communication.In this scheme, in a WSN with  sensors and sensor set  = {V 1 , V 2 , . . ., V  }, the EG scheme independently assigns a set of   distinct cryptographic keys, which are selected uniformly at random from a pool of   keys, to each sensor node; the set of keys of each sensor is called the key ring and is denoted by   for sensor V  .The EG scheme is denoted by a random key graph (,   ,   ) in [3][4][5][6], in which an edge exists between two nodes V  and V  if and only if they possess at least one common key; that is,   ∩   ̸ = 0.In a WSN using the EG scheme under transmission constraint, two sensors V  and V  establish a direct link between them if and only if they share at least one key and their distance is no greater than   .We denote the event establishing this direct link by   .If we let the graph (,   , S) model such a WSN, it is obvious to see (,   , S) is the intersection of random key graph (,   ,   ) and random geometric graph (,   , S) with  nodes uniformly distributed over a square S; namely,  (,   , S) =  (,   ,   ) ∩  (,   , S) , where   represents the parameters   ,   , and   together.
Let V ∈ S be the location of point V  .A direct link   exists in (,   , S) if both of the following two conditions are satisfied: V − V      ≤   ; where ‖ ⋅ ‖ represents the Euclidean norm.
We let   be the probability of key sharing between two sensors and note that   is also the edge probability in random key graph (,   ,   ).It holds that Clearly if   < 2  , then   = 1.If   ≥ 2  , as shown in previous work [5][6][7], we have If   ≥ 2  , by [8], it further holds that By [9], (5) implies that if  2  /  = (1), then We will use ( 5) and ( 6) throughout the paper.Let   be the probability that a secure link exists between two sensors in the WSN with EG scheme under practical constraint; obviously,   is the edge probability in (,   , S). (6).The secure wireless sensor network with nodes uniformly distributed is modeled by graph (,   , S), in which two nodes have an edge if their Euclidean distance is at most   and have a secure link if their key rings have at least one common key.Compared to the work done by Yi et al. [10], in this paper, by using a different method, we establish that the number of isolated nodes in the secure wireless sensor network has an asymptotic Poisson distribution whether the  nodes are induced by a uniform point process or Poisson point process.

It holds that 𝑝
To model transmission constraints, we use the popular disk model [11], and under the disk model, two nodes are directly connected if and only if their Euclidean distance is smaller or equal to a given threshold   , where parameter   is termed as the transmission range.
The rest of the paper is organized as follows.Section 2 reviews related work.Section 3 comparatively studies the distribution of isolated nodes in WSNs employing the EG scheme with nodes Poissonly or uniformly distributed over a unit square S. Through the study, it shows that under certain conditions the impact of boundary effect on the number of isolated nodes is negligible.Finally, Section 4 summarizes conclusions and discusses prospects of establishing the distribution of isolated nodes and tighter connectivity thresholds for secure wireless sensor networks under improved conditions.

Related Work
With regard to the sensor distribution, we consider that  nodes are independently and uniformly deployed in unit square area S. The disk model induces a random geometric graph [4,[12][13][14][15] which is denoted by (,   , S); an edge exists between two sensors if and only if their Euclidean distance is no more than   .Extensive research has been done on random geometric graphs.The connectivity of random geometric graphs has been studied by Dette and Henze [16], Penrose [17], and others [12,[18][19][20].For a uniform  point process over a unit area square S, Dette and Henze [16] showed that, for any constant , if   = √(ln  + )/, then graph (,   , S) has no isolated nodes with probability  − asymptotically.Eight years later, Penrose [14] established that if a random geometric graph induced by a uniform point process or Poisson point process has no isolated nodes, then it is almost surely connected.Besides the overall connectivity, some applications are concerned with whether there exists a giant connected component.Continuum percolation is a useful theorem in analyzing threshold phenomena.Ammari and Das [21] focused on percolation in coverage and connectivity in three-dimensional space and found out whether the network provides long distance multihop communication.
For random key graph (,   ,   ), Godehardt and Jaworshi [22] focused on the distribution of the number of isolated vertices in (,   ,   ).Some partial results concerning the connectivity of random key graphs were given in [6,7,23].In [5], Rybarczyk gave asymptotic tight bounds for the thresholds of the connectivity, phase transition, and diameter of the largest connected component in random key graphs for all ranges of   .Other related works regarding (,   ,   ) model have been reported.
For example, Bloznelis et al. [24] treated the evolution of the order of the largest component.Connectivity and communication security aspects of (,   ,   ) in various important settings are also studied in [7,25,26].Although some properties of secure WSNs with the EG scheme have been extensively studied in [4-7, 13, 27], most research [5][6][7]27] unrealistically assumes unconstrained sensor-to-sensor communications; that is to say, any two sensors can communicate regardless of the distance between them.
Recently, there is interest in random graphs in which an edge is determined by more than one random property, that is, intersection of different random graphs.The intersection of Erdős-Rényi random graph (, ) [28] and random geometric graph (,   , S) has been of interest for quite some time now.Recent work on such random graphs is by [20,29] where connectivity properties and the distribution of isolated nodes are analyzed.And the intersection of random graphs (, ) and random key graph (,   ,   ) is considered in [30].Such a graph is constructed as follows: a random key graph (,   ,   ) is first formed based on the key distribution and each edge in this graph is deleted with a specified probability.
In addition to random key graphs and random geometric graphs, the Erdös-Rényi graph (, ) [28] has also been extensively studied.An Erdös-Rényi graph (, ) is defined on a set of  nodes such that any two nodes establish an edge independently with probability .As already shown in the literature [5][6][7]27], random key graph (,   ,   ) and Erdös-Rényi graph (, ) have similar connectivity properties when they are matched through edge probability; that is,   = .Hence, it would be tempting to exploit this analogue and conclude that the distribution of isolated nodes in (,   , S) ((,   ,   ) ∩ (,   , S)) is similar to that of (, ) ∩ (,   , S) in [20], whether the  nodes are distributed Poissonly or uniformly on a unit square S.

Main Result
In this section, we study the expected number of isolated nodes in WSNs with the EG scheme under transmission constraints with nodes either Poissonly or uniformly on a unit square S. The number of isolated nodes is a key parameter in the analysis of network connectivity.A necessary condition for a network to be connected is that the network has no isolated nodes, and this may be possibly true for the intersection of random key graph and random geometric graph.
In order to obtain the distribution of isolated nodes in (,   , S), we prove the same result for its Poissonized version, graph  Poisson (,   , S), where the only difference between  Poisson (,   , S) and (,   , S) is that the node distribution of the former is a homogeneous Poisson point process with intensity  on a unit square S while that of the latter is a uniform  point process.

Expected Number of Isolated Nodes in 𝐺 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 (𝑛, 𝜃 𝑛 , S).
In graph  Poisson (,   , S), let   denote the event that node V  is isolated, and let    (V  ) denote the intersection of S and the disk centered at position V ∈ S with radius   .When node V  is at position V , the number of nodes within area    (V  ) follows a Poisson distribution with mean    (V  ), and to have an edge with V  in graph  Poisson (,   , S), a node not only has to be within    (V  ) but also has to share at least a key with node V  .Then the number of nodes neighboring to V  at V follows a Poisson distribution with mean      (V  ), and the probability that such number is 0 is equal to  −  |   (V  )| .Integrating V over S, then the probability that the node V  is isolated is given by Theorem 1. Suppose that  2  /  = (1),   = (1/ ln ), and  nodes are Poissonly distributed on a unit square S with the maximum transmission radius  2  ⋅  2  /  = (ln  + )/ for some constant .Then the expected number of isolated nodes in   (,   , S) converges asymptotically to  − as  → ∞.
Proof.Let  denote the number of isolated nodes in graph  Poisson (,   , S).By (7), we know P(  ) = ∫ S  −  |   (V  )| V  holds.To compute P(  ) based on S, we divide S in a way similar to that by Li et al. [35] and Wan and Yi [15].Specially, S is divided into S 0 , S 1 , S 2 , and S 3 , respectively, as illustrated in Figure 1 (note that   = (1)).S 0 consists of all points each with a distance greater than   to its nearest edge to S, whereas S 3 is a square of size   ×  at the four corners of S. We further divide S\{S 0 ∪S 3 } into S 1 and S 2 as follows.In S\{S 0 ∪S 3 }, S 1 contains points whose distance to the nearest edge of S is no greater than   /2, while the remaining area is S 2 .Then the expected number of isolated nodes is given by The unit square S and its divisions S 0 , S 1 , S 2 , and S 3 .
The four summands in (8) represent, respectively, the expected number of isolated nodes in the central area S 0 , in the boundary area along the four sides of S, and in the four corners of S. In the following analysis, we will show that the first term approaches  − as  → ∞, and the remaining terms approach 0 an  → ∞.

Corollary 2.
In graph   (,   , S) under conditions  2  /  = (1),   = (1/ ln ), and  2  ⋅( 2  /  ) = (ln +)/ with  = (ln ) or a constant , then for any constant  > 0 such that () = (  ) holds.Now we examine the impact of boundary effect on the number of isolated nodes in  Poisson (,   , S) since the network area in our analysis is a square.The square accounts for the real-world boundary effect whereby some transmission region of a sensor close to the network boundary may fall outside the network field.In contrast, the torus eliminates the boundary effect.The analysis of impact of the boundary effect is done by comparing the number of isolated nodes in  Poisson (,   , S) and the number in a network with nodes Poissonly distributed on a unit torus S  with a pair of nodes V  , V  separated by a toroidal distance ‖V  − V ‖  .Denote such network on a unit torus by   Poisson (,   , S  ).The following Theorem can be established.Proof.Let   denote the number of isolated nodes in graph   Poisson (,   , S  ), and let     (V  ) denote the intersection of S  and the disk centered at position V ∈ S  with radius   .The probability that node V  is isolated is Using the condition  2  ⋅ ( 2  /  ) = (ln  + )/, we obtain On the basis of Theorems 1 and 3, using the coupling technique, the following theorem can be obtained. ⋅ ( 2  /  ) = (ln  + )/ for some constant .Then the number of isolated nodes in   (,   , S) due to the boundary effect converges asymptotically to 0 as  → ∞.
Proof.Comparing Theorems 1 and 3, it is noted that the expected numbers of isolated nodes on a unit torus S  and on a unit square S, respectively, asymptotically converge to the same constant  − as  → ∞.Now we use the coupling technique [36] to construct the connection between  and   .Consider a graph   Poisson (,   , S  ), and the number of isolated nodes in that graph is   .Remove each connection of the above graph with probability We further note that only connections between nodes near the boundary will be affected.Denote the number of newly appearing isolated nodes by   ; namely,   is the number of isolated nodes due to the boundary effect; it is straightforward to show that   is a nonnegative random integer.Further, such a connection removal process results in a random network with nodes Poissonly distributed with density .That is, a random network on a unit square with boundary effect is included.The following equation result holds: By Theorems 1 and 3 and the above equation, it can be shown that lim Due to the nonnegativity of   , lim →∞ P (  = 0) = 1. (33)

Distribution of the Number of Isolated Nodes
in   (,   , S).In this subsection, we analyze the distribution of the number of isolated nodes in  Poisson (,   , S).For this purpose, we give some definitions.Let Poi() be a Poisson random variable with parameter .
Let Γ be a finite set of indices and let (  ) ∈Γ be a family of random indicator variables.We say (  ) ∈Γ are positively related (see [37]), if, for each  ∈ Γ, there exist random indicator variables (  ) ∈Γ\{} with the distributions such that   ≥   for every  ̸ = .A useful result obtained by Stein-Chen method is the following.
Proof.The triangular inequality for the total variation distance implies By a coupling argument [39] and Theorem 3, we have Combining this with (36), we now only need to prove lim First, we claim that (   )  =1 are positively related.To see this, define where , where   represents the key ring which is adjacent to Poisson (,   , S  ).Hence, we have For every  ̸ = , if    = 1 then    = 1.Consequently, we get    ≥    .
We now consider the asymptotic distribution of the number of isolated nodes in  Poisson (,   , S).From Theorem 4, lim →∞ P(  = 0) = 1 holds, and using Slutsky's Theorem [40], the following result on the asymptotic distribution of  can be readily obtained.

Distribution of the Number of Isolated Nodes in
(,   , S).We derive the distribution of isolated nodes in (,   , S) by using standard Poissonization technique [11,14].The idea is that the result about the distribution of isolated nodes for (,   , S) follows once we establish the result with Poissonization, that is to say, once we obtain the distribution of isolated nodes for  Poisson (,   , S). See the following lemma for rigorous argument.Lemma 8. Suppose that  2  /  = (1),   = (1/ ln ), and  nodes have the same maximum transmission radius  2  ⋅ ( 2  /  ) = (ln  + )/ for some constant .Then with  denoting ⌈ −  1/2+ 0 ⌉, where  0 is an arbitrary constant with 0 <  0 < 1/2, then node V  is isolated in (,   , S) if and only if V  is an isolated node in   (,   , S).
Proof.We will use the standard de-Poissonization technique [11,14,18] to prove Lemma 8. Let  denote the number of nodes in graph  Poisson (,   , S); clearly  follows a Poisson distribution with mean , for any positive , and from Chebyshev's inequality, we have Without loss of generality, we take  −  1/2+ 0 as an integer.
With  = ( − )/√, substituting  =  −  1/2+ 0 into (46), we get Hence,  − 2 1/2+ 0 <  <  holds almost surely.When  < , we construct a coupling C between graph (,   , S) and graph  Poisson (,   , S); let graph (,   , S) be the result of adding  −  nodes uniformly distributed on S to graph  Poisson (,   , S).Let   be the node set of  Poisson (,   , S); clearly,   is a subset of , where  is the node set of (,   , S).In addition, it is straightforward to see that the edge set of  Poisson (,   , S) is also a subset of that of (,   , S).Then under coupling C, graph  Poisson (,   , S) is a subgraph of (,   , S).
Let   ,   denote the set of isolated nodes in (,   , S) and  Poisson (,   , S), respectively.To prove Lemma 8, we show It is straightforward to see By (47), we will prove P(  ̸ =   ) = (1) if we can derive To prove (48) and (49), with −2 Then there exists at least one node V  in  \   such that V  and V  are neighbors in (,   , S). Due to | \   | =  −  < 2 1/2+ 0 , noting that   is the edge probability in (,   , S), then with   denoting the number of isolated nodes in  Poisson (,   , S), as an easy consequence of the union bound, In order to apply Corollary 2 to  Poisson (,   , S), for  sufficiently large, the following condition holds: We demonstrate (51) in view of Second, in order to prove (49) with  − 2 1/2+ 0 <  < , we consider event that   \   ̸ = 0 occurs if and only if there exists at least one node V  such that V  ∈   and V  ∉   .With V  ∈   , then V  is isolated in (,   , S), which along with V  ∉   leads to V  ∉   and V  ∈  \   .Let  denote the probability that a node is isolated in (,   , S); it follows via a union bound that P [(  \   ̸ = 0) ∩ ( − 2 1/2+ 0 <  < )] ≤ 2 1/2+ 0 ⋅ .Noting that the number of isolated nodes in a network is a nonnegative integer, the following result can be obtained as a consequence of Theorems 7 and 9. Notice that, in formulating this result, we drop the assumption that  originally is a constant and allow it instead to be -dependent.

Conclusion and Future Work
Yi et al. [10] considered that a wireless ad hoc network consists of  nodes distributed independently and uniformly in a unit disk D or a unit square S. They used Brun's sieve to show that, for graph (,   , D) or (,   , S), if  2  ⋅ ( 2  /  ) = (ln  + )/ and  2  /  = (1/ ln ), the number of isolated nodes asymptotically follows a Poisson distribution with mean  − .Pishro-Nik et al. [34] also obtained such result on asymptotic Poisson distribution with condition  2  /  = (1/ ln ) generalized to  2  /  = Ω(1/ ln ).In this paper, we discuss the distribution of isolated nodes in WSNs employing the widely Eschenauer-Gligor key