Motivated by information-theoretic secrecy, geometric models for secrecy in wireless networks have begun to receive increased attention. The general question is how the presence of eavesdroppers affects the properties and performance of the network. Previously the focus has been mostly on connectivity. Here we study the impact of eavesdroppers on the coverage of a network of base stations. The problem we address is the following. Let base stations and eavesdroppers be distributed as stationary Poisson point processes in a disk of area n. If the coverage of each base station is limited by the distance to the nearest eavesdropper, what is the maximum density of eavesdroppers that can be accommodated while still achieving full coverage, asymptotically as n→ ∞

Haenggi, Martin

Sarkar, Amites

English

Western Washington University

Western Washington UniversityWestern CEDARMathematics College of Science and Engineering2010Secrecy Coverage (Conference Proceeding)Amites SarkarWestern Washington University, amites.sarkar@wwu.eduMartin HaenggiUniversity of Notre DameFollow this and additional works at: https://cedar.wwu.edu/math_facpubsPart of the Mathematics CommonsThis Conference Proceeding is brought to you for free and open access by the College of Science and Engineering at Western CEDAR. It has beenaccepted for inclusion in Mathematics by an authorized administrator of Western CEDAR. For more information, please contactwesterncedar@wwu.edu.Recommended CitationSarkar, Amites and Haenggi, Martin, "Secrecy Coverage (Conference Proceeding)" (2010). Mathematics. 89.https://cedar.wwu.edu/math_facpubs/89Secrecy CoverageAmites SarkarDepartment of MathematicsWestern Washington UniversityBellingham, WA 98225, USAMartin HaenggiDepartment of Electrical EngineeringUniversity of Notre DameNotre Dame, IN 46556, USAAbstract—Motivated by information-theoretic secrecy, geomet-ric models for secrecy in wireless networks have begun to receiveincreased attention. The general question is how the presenceof eavesdroppers affects the properties and performance of thenetwork. Previously the focus has been mostly on connectivity.Here we study the impact of eavesdroppers on the coverage of anetwork of base stations. The problem we address is the following.Let base stations and eavesdroppers be distributed as stationaryPoisson point processes in a disk of area n. If the coverageof each base station is limited by the distance to the nearesteavesdropper, what is the maximum density of eavesdroppersthat can be accommodated while still achieving full coverage,asymptotically as n → ∞?I. INTRODUCTIONA. Motivation and related workWhile coverage problems have been studied for severaldecades from a purely mathematical perspective, they haverecently begun to attract significant attention by the engineer-ing and computer science communities due to the advent ofwireless networks, in particular sensor networks. The standardproblem formulation is the following. Place a number n ofnodes randomly in a certain set B ⊂ Rd and equip each nodewith the capability of covering (sensing) a disk or sphere ofradius r around itself. How large should n be to guaranteecoverage of B with probability 1−ǫ? Or, if the area or volumeof B is scaled in proportion to n, which r(n) guaranteescoverage with high probability as n→∞?In the mathematical literature, one of the early and nowclassical coverage problems is the coverage of a sphere withcircular caps introduced in [1] and solved in [2]. Extensions tok dimensions were considered by Hall in [3] and Janson in [4].Hall later provided a detailed account of coverage processesin his book [5]. Many generalizations have been consideredsince, see, e.g., [6], [7] for coverage problems in Rd.Here we focus on a coverage problem that is inspiredby secrecy constraints. We assume an information-theoreticmodel for secrecy, in which a communication is securefrom eavesdroppers if the intended receiver is closer to thetransmitter than all eavesdroppers. Based on this model, thesecrecy graph, a random geometric graph that only includesedges along which secure communication is possible wasintroduced and studied in [8]. [9] extended the analysis to moreelaborate physical layer models, while [10] considered theeffects of uncertainty in the eavesdroppers’ locations. This lineof work is based on graph models and focuses on connectivity.In contrast, there is no prior work on the related coverageproblem, which is the theme of the present paper.Base stations and eavesdroppers are distributed randomlyon the plane, and the base stations can cover circular areaswith radii determined by the distance to the nearest eaves-droppers. The question is what density of eavesdroppers canbe accommodated while still guaranteeing that the entire areaor volume of interest is covered securely? This would ensurethat mobile stations could roam around everywhere and bereached securely by a base station. Hence the downlink isintrinsically secure, while the uplink (from the mobile to thebase station) has to be secured by transmission of a one-timepad via the downlink.B. Problem formulationTo make the problem concrete, we assume that the basestations and eavesdroppers form independent Poisson pointprocesses of intensities 1 and λ, respectively, in Rd. We willdenote the process of base stations by P and call its pointsblack points, and the process of eavesdroppers by P ′ and callits points red points. Now place an open ball D(p, rp) of radiusrp around each black point p ∈ P , where rp is maximal sothat D(p, rp) ∩ P ′ = ∅. In other words, rp is the distancefrom the black point p to the nearest red point p′ ∈ P ′ to p.We thus obtain a random set Adλ ⊂ Rd which is the union ofballs centered at the points of P . Fig. 1 shows a 2-dimensionalexample for λ = 0.1. Our aim is to study properties of Adλ,in particular the covered volume fraction (Section 2) and theasymptotic conditions for complete coverage in one (Section3) and two (Section 4) dimensions.In two dimensions, the radius R of each disk is given bythe nearest-neighbor distance, distributed asfR(x) = 2πxλ exp(−λπx2) .Note that this coverage problem is rather different fromthe case where the covering disks have independent radiidrawn from fR. The difference is that in our case, the diskradii of nearby nodes are strongly correlated, which leads todrastically different conditions for coverage compared with theindependent case. Indeed, for the standard model with randomindependent disk radius, a disk of area n is covered a.s. if1πE(R2) = (1 + ε) logn1This condition is not the sharpest possible.−4 −3 −2 −1 0 1 2 3 4−4−3−2−101234λ=0.1Fig. 1. Example for coverage of an 8 × 8 square for λ = 0.1. The basestations are marked by +, the eavesdroppers by ×, and the covered area isgrey shaded.for any ε > 0. Since E(R2) = (λπ)−1, this translates toλ = [(1 + ε) logn]−1 ,which indicates that λ may decrease rather slowly with n whilestill achieving full coverage. In the secrecy case, however, λhas to decrease much faster, at a rate of about n−1/3, as wewill show.II. COVERED VOLUME FRACTIONFor λ > 0, writeCd(λ) = P(O ∈ Adλ).By stationarity of the model, Cd(λ) can also be interpretedas the fraction of Rd which is covered by Adλ, known as thecovered volume fraction.Theorem 1.Cd(λ) = 1− E(e−Vd/λ) = 1−∫∞0fd(t)e−t/λ dt,where fd(t) is the probability density function for the volumeVd of a randomly chosen cell in a Voronoi tessellation asso-ciated with a unit intensity Poisson process in Rd.Proof: We rescale the model so that P and P ′ haveintensities 1/λ and 1 respectively. This does not affect Cd(λ).Now O 6∈ Adλ if and only if there are no points of P in theVoronoi cell C defined by P ′ ∪ {O} containing O. If C hasvolume V , then P(C ∩ P = ∅) = e−V/λ.Corollary 2. In one dimension, we haveC1(λ) =1 + 4λ(1 + 2λ)2.Proof: Let P be a unit intensity Poisson process in R. Thedistribution of the gap lengths between points of P has densitye−t, but the distribution of the length of the gap containinga fixed point, such as the origin O, has density te−t. (Thisis known as the waiting time paradox.) Consequently, thedensity function for the length of the Voronoi cell definedby P containing the origin is 4te−2t, so that by Theorem 1C1(λ) = 1−∫∞04te−2t−t/λ dt =1 + 4λ(1 + 2λ)2.Remark. This corollary can also be proved as follows. Let Lbe the event that the origin O is covered by points of P lyingonly to the left of O, and let R be the event that O is coveredby points of P lying only to the right of O. Then L and Rare independent, andC1(λ) = 1− (1− P(L))(1− P(R))= 1− (1− P(R))2= 2P(R)− P(R)2,by symmetry. Now R occurs if and only if the closest blackpoint to the right of O is at distance t, and there are no redpoints in the interval [0, 2t], for some t > 0. ThusP(R) =∫∞0e−te−2λt dt =12λ+ 1,which gives the desired result.III. PROBABILITY OF TOTAL COVERAGE IN ONEDIMENSIONIn this and the next section, we study the following problem.With P ,P ′ and Adλ as before, let Bdn ⊂ Rd be a fixed ball ofvolume n, and set Adλ(Bdn) = Adλ ∩Bdn. Write Bdλ(n) for theevent that Adλ(Bdn) covers Bdn (except for the points of P ′),and set pdλ(n) = P(Bdλ(n)). Our principal goal is to estimatepdλ(n) for arbitrary d.Let us first consider the case d = 1. In this case, I = B1nis simply an interval of length n, containing some black andred points. We place an interval centered at each black pointof maximal length subject to containing no red points, andask for the probability p1λ(n) → 0 that I is covered by suchintervals.In the one-dimensional case, we have the following sharpresult.Theorem 3. If λ2n→∞, then p1λ(n) → 0, and if λ2n→ 0,then p1λ(n) → 1.Proof: To simplify our analysis, let us suppose that P andP ′ are placed on a circle T of circumference n rather than aninterval of length n: there is asymptotically no difference. Ourstrategy is to place the red points P ′ first, partitioning the circleT into M ∼ Po(nλ) arcs Ai. Now place the black points P .For each arc Ai, let Ci be the event that Ai is covered bythe smaller arcs associated with the black points in Ai. Theevents Ci are independent, and this will enable us to estimatep1λ(n).Suppose Ai has length ℓ, and let mi be the midpoint ofAi. Let x be the distance of the closest black point to milying on the left of mi, and let y be the distance of the closestblack point to mi lying on the right of mi. Whether or notCi occurs, i.e., whether or not Ai is covered by small arcs, isdetermined solely by x and y. In fact, it is easy to see that• Ci occurs if and only if x+ y ≤ ℓ/2.Now x+ y has the gamma distribution with density functionte−t, and consequentlyP(Ci | Ai has length ℓ) =∫ ℓ/20te−t dt= 1− e−ℓ/2(1 + ℓ/2). (1)Further, since the black and red points are independent, and thelength ℓ has an exponential distribution with density functionλe−λt, we haveP(Ci) =∫∞0λe−λℓ(1 − e−ℓ/2(1 + ℓ/2)) dℓ=1(1 + 2λ)2.Conditioning on the number of arcs M changes the dis-tribution of the lengths of the Ai, but the difference isasymptotically negligible. Consequently, as n → ∞ withλ→ 0 but λn→∞,p1λ(n) ∼∞∑m=0P(M = m)(2λ+ 1)−2m= e−4nλ2(λ+1)/(2λ+1)2∼ e−4nλ2 .So, if nλ2 → 0, p1λ(n) → 1, and if nλ2 →∞, p1λ(n) → 0, aspostulated.IV. PROBABILITY OF COVERAGE IN TWO DIMENSIONSA. AnalysisA natural step is to generalize these results to arbitraryd. Simple heuristics would suggest that if λd+1n → 0 thenpdλ(n) → 1, and if λd+1n → ∞ then pdλ(n) → 0. Un-fortunately, attempts to generalize the above one-dimensionalarguments run into difficulties, mainly due to the lack of anorder structure in Rd for d ≥ 2.For the rest of the paper, we restrict attention to the case d =2. It turns out to be useful to consider the Gilbert disc modelon the red points P ′, with an appropriately chosen radius. Thismodel is constructed by simply joining two points of P ′ if thedistance between them is less than some specified threshold.Theorem 4. If f(n) = λ3n→∞, then p2λ(n) → 0.Proof: Suppose that n→∞ and also that λ3n→∞. LetR > 0 be a large constant. Construct the Gilbert disc modelG = GR(P ′) on the red points P ′, with radius R (i.e., wejoin two points of P ′ if they are within distance R). Let Tbe the (random) number of triangles in G which lie entirelywithin B2n. Then, for some absolute constant C1,E(T ) ∼ C1e−λπR2(λπR2)2λn=(C1π2R4 + o(1))λ3n→∞.Consequently, if T1 denotes the (random) number of trianglesin G which lie entirely inside B2n and have all angles betweenπ/6 and π/2, then also E(T1) → ∞. Finally, putting in theblack points P , and writing T2 for the number of trianglescounted in T1 which are not within distance 1000R of a blackpoint, we have that E(T2) →∞. A simple application of thesecond moment method shows that P(T2 ≥ 1) → 1 (thisis intuitively obvious from E(T2) → ∞ due to the long-range independence of the model). However, any red trianglecounted in T2 will have points in its interior which are notcovered by black discs. Since with high probability T2 ≥ 1,we have that p2λ(n) → 0.The other direction seems to require a more elaborateargument (and a stronger hypothesis):Theorem 5. If g(n) = λ3n(logn)3 → 0, then p2λ(n) → 1.Proof: (Sketch.) Suppose that n → ∞ and also thatg(n) = λ3n(logn)3 → 0. Once again, we construct theGilbert disc model G = GR(P ′) on the red points P ′, butthis time R = R(n) will be a function of n. As long as R(n)is not too large, there will be (up to a constant) λn vertices,R2λ2n edges and R4λ3n triangles in G inside B2n. We willshow that R(n) can be chosen so that:(I) The maximum degree of G is one.(II) The region of B2n close to points of P ′ is covered (byA2λ).(III) The rest of B2n, far from points of P ′, is covered (byA2λ).Condition (I) simply states that G consists of isolated verticesand isolated edges. It is (II) and (III) which necessitate thestronger hypothesis. We will define a set of bad events, whichwill depend on P and P ′, and show that coverage (by A2λ)occurs in the absence of bad events. We will then show thatthe probability of at least one bad event occurring tends tozero.First, we overlay a grid of squares of side length r =√lognonto B2n. The probability that any small square of the gridcontains no point of P is e− logn = n−1. Since there are ∼n/ logn such squares, the expected number of them containingno black points is asymptotically 1/ logn→ 0. Consequently,with high probability, every small square contains a blackpoint. Now fix a small square S. If no point of S is withindistance√2 logn of a red point, and if S contains a blackpoint, then all of S will be covered by A2λ. Consequently,with high probability, any point of B2n at distance more than√8 logn from all red points will be covered by A2λ. Our firsttype of bad event will be that some square S of the gridcontains no black points: if no such event occurs then we mayassume that points at distance√8 logn from P ′ are covered.This deals with (III).Take R(n) = 1000√log n. Our second bad event will bethat G contains vertices of degree at least 2. Write W for thenumber of such vertices in G (within B2n). ThenE(W ) ∼ 12e−λπR2(λπR2)2λn=(5 · 1011π2(logn)2 + o(1))λ3n→ 0,so that P(W = 0) → 1. This deals with (I).It remains to deal with (II). From now on, we may assumethat G has only isolated vertices and isolated edges (insideB2n). Recall that we need only worry about the coverage ofpoints within distance√8 logn from a red point. We willcolour all such points yellow.First we deal with the isolated vertices. Consider the circlesof radii 10√logn and 11√logn around each isolated redpoint, and divide the annulus between these circles into 10equal “sectors”. With high probability, there is a black pointinside each sector. But then the yellow region surrounding thered point is covered.For the edges, a lengthier argument is needed. We providea summary. Let L ⊂ R2 be the line segment between the twonodes connected by the red edge. The critical event here iscoverage of the yellow “sausages” L⊕D(0,√8 logn) aroundthe red edges. By focusing on certain black points at a suitable,carefully chosen, distance from the red edge, it can be shownthat such yellow sausages are indeed covered with probability1− n−4—if the hypothesis holds.B. Simulation resultsHere we provide two simulation results, see Fig. 2, whichgive an indication of the fraction of the area that remainsuncovered if the condition in Theorem 4 holds, and howquickly this fraction goes to zero if the condition in Theorem5 holds.V. CONCLUSIONSWe have introduced a novel class of coverage problems,where the size of the covering disks is determined by thedistance to the nearest point in a second point process. In thePoisson-Poisson case, where black and red points are indepen-dent Poisson point processes, we have provided expressions forthe covered volume fraction and the probability of completecoverage in the one- and two-dimensional cases.The main result is the asymptotic threshold for coverage intwo dimensions. Forλ3n(logn)3 → 0 , n→∞ ,full coverage is achieved with probability tending to 1. On theother hand, ifλ3n→∞ , n→∞ ,then the probability of full coverage tends to 0.The model can be viewed as a germ-grain model with germsof random and correlated size.0 1 2 3 4x 10401234567 x 10−3nUncovered fractionλ3=n−0.9(a) Uncovered fraction for λ3 = n−0.90 2000 4000 6000 8000 10000012345678 x 10−5nUncovered fractionλ3=n−1log(n)−3(b) Uncovered fraction for λ3 = n−1(logn)−3Fig. 2. Simulation results for Theorems 4 and 5. In (a), per Theorem 4,coverage is not achieved asymptotically. In (b), the function g(n) in Theorem5 is a constant, and it appears that coverage is achieved. Note that the scalein the axes are different in (a) and (b).The results have applications in secure wireless networking.If the red points are eavesdroppers and the black point basestations, then full coverage in our model implies that from allpoints of the plane, messages can be received from at leastone base station securely, without any eavesdropping.ACKNOWLEDGMENTThe work of the second author was partially supported bythe DARPA/IPTO IT-MANET program under grant W911NF-07-1-0028.REFERENCES[1] P. A. P. Moran and S. F. de St. Groth, “Random circles on a sphere,”Biometrika, vol. 49, pp. 389–396, 1962.[2] E. N. Gilbert, “The probability of covering a sphere with n circularcaps,” Biometrika, vol. 56, pp. 323–330, 1965.[3] P. Hall, “On the coverage of k-dimensional space by k-dimensionalspheres,” The Annals of Probability, vol. 13, pp. 991–1002, Aug. 1985.[4] S. Janson, “Random coverings in several dimensions,” Acta Mathemat-ica, vol. 13, pp. 991–1002, 1986.[5] P. Hall, Introduction to the Theory of Coverage Processes. Wiley Seriesin Probability and Mathematical Statistics, 1988.[6] S. Athreya, R. Roy, and A. Sarkar, “On the Coverage of Space byRandom Sets,” Advances in Applied Probability, vol. 36, pp. 1–18, 2004.[7] R. Roy, “Coverage of Space in Boolean Models,” Dynamics & Stochas-tics, vol. 48, pp. 119–127, 2006.[8] M. Haenggi, “The Secrecy Graph and Some of its Properties,” in2008 IEEE International Symposium on Information Theory (ISIT’08),(Toronto, Canada), July 2008.[9] P. C. Pinto, J. Barros, and M. Z. Win, “Secure Communication inStochastic Wireless Networks.” ArXiv http://arxiv.org/abs/1001.3697,Jan. 2010.[10] S. Goal, V. Aggarwal, A. Yener, and A. R. Calderbank, “ModelingLocation Uncertainty for Eavesdroppers: A Secrecy Graph Approach,” inIEEE Symposium on Information Theory (ISIT’10), (Austin, TX), June2010.[11] M. Tanemura, “Statistical Distributions of Poisson Voronoi Cells in Twoand Three Dimensions,” Forma, vol. 18, no. 4, pp. 221–247, 2003.[12] S. A. Zuyev, “Estimates for distributions of the Voronoi polygon’sestimates for distributions of the Voronoi polygon’s geometric character-istics,” Random Structures and Algorithms, vol. 3, pp. 149–162, 1992.

Secrecy Coverage (Conference Proceeding)

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Secrecy Coverage (Conference Proceeding)

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