Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission.


 Clifford Washington
 1 years ago
 Views:
Transcription
1 On the Probability of Covering the Circle by Rom Arcs Author(s): F. W. Huffer L. A. Shepp Source: Journal of Applied Probability, Vol. 24, No. 2 (Jun., 1987), pp Published by: Applied Probability Trust Stable URL: Accessed: 22/11/ :31 Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms Conditions of Use, available at JSTOR's Terms Conditions of Use provides, in part, that unless you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, you may use content in the JSTOR archive only for your personal, noncommercial use. Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission. JSTOR is a notforprofit service that helps scholars, researchers, students discover, use, build upon a wide range of content in a trusted digital archive. We use information technology tools to increase productivity facilitate new forms of scholarship. For more information about JSTOR, please contact Applied Probability Trust is collaborating with JSTOR to digitize, preserve extend access to Journal of Applied Probability.
2 J. Appl. Prob. 24, (1987) Printed in Israel C Applied Probability Trust 1987 ON THE PROBABILITY OF COVERING THE CIRCLE BY RANDOM ARCS F. W. HUFFER,* Florida State University L. A. SHEPP,** AT& T Bell Laboratories Abstract Arcs of length lk, 0 < lk < 1, k = 1, 2, *, n, are thrown independently uniformly on a circumference W having unit length. Let P(11, 12, * * *,) be the probability that W is completely covered by the n rom arcs. We show that P(I?, 12,*., 14) is a Schurconvex function that it is convex in each argument when the others are held fixed. COVERAGE PROBABILITIES; SCHURCONVEX; GEOMETRICAL PROBABILITY 1. Introduction Suppose arcs of length lk, 0 < k <, 1 _k _ n, are thrown independently uniformly on a circumference W having unit length. Let P(1l, 12, ***, n) be the probability that W is completely covered by the n rom arcs. Stevens (1939) has explicitly evaluated P(11, 12, **, I) in the case 11 = 12 =. = I4. It seems hopeless to give a simple formula for the case of general arc lengths. In this paper we examine qualitative properties of the function P(l, 12, *, In). In particular, we show that (1.1) P(1l, 12,.., ) is Schurconvex, (1.2) P(ll, 12, *, In) is convex in each argument (keeping the others fixed). These results are proved in Sections 2 3. There is an extensive literature concerned with rom arcs coverage problems. Much of this literature deals exclusively with the case of equal arc Received 16 July 1985; revision received 18 February * Postal address: Department of Statistics, Florida State University, Tallahassee, FL , USA. Research supported by Office of Naval Research Contract N C ** Postal address: AT & T Bell Laboratories, 600 Mountain Avenue, Murray Hill, NJ 07974, USA. 422
3 On the probability of covering the circle by rom arcs 423 lengths. A few of the papers concerned with coverage of the circle by arcs of different lengths are Shepp (1972), Siegel Hoist (1982), Janson (1983) Huffer (1986). Shepp works with infinitely many arcs having fixed lengths given by some sequence {l,}. In the last three papers, the arc lengths are taken to be independent rom variables. A survey of some of the coverage literature may be found in Solomon (1978). Our main result is the Schurconvexity of P which was proposed as a conjecture by Frank Proschan. Let {xi, }) {y i},.. be collections of arc lengths with xl > x2 *** > xn Yl > Y2 * * *? y Yn. Schurconvexity says the following: If then ii k k x _< Yi for 1l k_ n1 ii ii n y i = P(x,, x2,.., Xn) n ii Y P(yl, Y2,' *', Yn) Loosely speaking, if xl, x2, * *, xn are 'less spread out' or 'more nearly equal' than are Y1, Y2, *, Yn, then P(x1, x2,..*, X) ' P(1, y2,., Yn). Schurconvexity allows us to compute explicit bounds in some cases. For instance, P(l, 12,, In) > P(l,, *,) where I = n ~ 1i Stevens' formula gives P(I, 1,, )= z.. (l) k(n)(1kl) kn See Marshall Olkin (1979) for further information on Schurconvexity. One can also consider covering the circle by n romly placed sets (not necessarily arcs). A result due to Wschebor (1973) states roughly that among sets with measures 11, 12, ***,,, the coverage probability is largest when all the sets are arcs (intervals). Our result (1.1) shows that among arcs with ?? + In = c, greater inequality in the lengths leads to a greater probability of coverage. 2. Proof of Schurconvexity Assume without loss of generality that To demonstrate Schurconvexity it suffices to show that
4 424 F. W. HUFFER AND L. A. SHEPP (2.1) P(,l, 1213,, l,n) P(I, + e, 12  e, 13,., In) for all sufficiently small e > 0 (see A.2.b on p. 55 of Marshall Olkin (1979)). An outline of the proof follows. To prove (2.1) we must compare two situations: one in which the first two arcs have lengths 11, 12 the other in which these arcs have lengths 1, + e, 12  e. To do this we construct these two situations on the same probability space then condition on the positions of the first two arcs. Having fixed the first two arcs, we examine the conditional probability of covering the remaining part of the circle by the remaining n  2 arcs with lengths 13,?., In. It is easy to hle the case when the first two arcs overlap. When the first two arcs are disjoint, a more complicated proof is needed. For this case we use a reflection argumento express the coverage probability inequality in the form given in (2.5). A further conditional argumenthen reduces this to a consequence of the lemma given at the close of this section. For notational convenience, we parameterize the circle by using the real line modulo 1. A real number x corresponds to a unique point on ' which we denote by [x]. [x] = [y] if only if x  y is an integer. An interval (a, b) on the line corresponds to a unique arc on W which we denote by [(a, b)]. Let Xi, X2,... X, be independent uniformly distributed on (0, 1). Define l = [(X1, XI + 14)], 2 = [(X2, X2 + 12)], S; = [(XI, Xl + 1l + E)], 2 = [(X2 + E, X2 + 12)] n e = U i3 [(Xi,x,+,)]. Formula (2.1) is equivalento (2.2) P{ C Sl U d2 U J} <P{ 6" C ' U a2 JUf}. To prove (2.2) we show that (2.3) P{ c,1 u a2 U I X,i, X,) } P{' c.\ U 2 U 1 Xl, x2} almost surely. There are three cases to consider: (i) [X2] E 1,, (ii) [X2]d1l [X2 + 12] E, (iii) si, r n 2 = 0 In case (i) we have Si U ds2 C d U 2. In case (ii) both il U d2 d; U 2 are arcs on ' the length of 1, U d2 is less than or equal to the
5 On the probability of covering the circle by rom arcs 425 length of d/[ U.2/. The reader can easily verify these facts. Thus (2.3) holds in both cases (i) (ii). The remainder of the proof deals with case (iii) so that in the following we take XI X2 to be fixed (nonrom) values such that /i, n r42 0. This implies./1 n 2/ = 0. Let V V' denote the complements of li U s2./4 U.2 respectively. We can now rewrite (2.3) as (2.4) Pt V c X} _ P{ V' C X}. V consists of two disjoint arcs separated by a distance of 12. V' consists of disjoint arcs having the same lengths but situated closer together (separated by 12  e). Let A, B, C C*be the disjoint arcs given in Figure 1. Both C C* have length e. With this notation V =A U B U C V' =A U B U C*. Let A* be the reflection of A about the diagonal through the midpoint of B. Noting that the reflection of C is C* the reflection of B is B, symmetry yields so that (2.4) is equivalento Since P{A u B U C c } = P{A* u B U Cc X} P{A U B U Cc Y}) P{A* U B U Cc Y}. P{A u B u C c =P{A UB u B C c cpa U,,C t} by symmetry P{A U B C XW} = P{A* U B C Jf}, this may be rewritten as P{A U B c Jr, cg r} > P{A* u B c r, C4 0} B Figure 1
6 426 F. W. HUFFER AND L. A. SHEPP or equivalently (2.5) P{A c li B /c, Ct } > P{A* CY lb c, C e1 }. In proving (2.5), in addition to conditioning on the event {B c ar, C?t Xj}, we shall assume we know which of the rom arcs intersect B the exact position of these arcs. More precisely, we are given (B C XW, C? J') the value Xi for all i such that [(Xi, Xi + 1/)] intersects B. For convenience we use y. to denote this conditioning. To prove (2.5) complete the proof of our tlt/orem it suffices to show that (2.6) P{A c JF} > P{A* c }. Let D be the union of all the rom arcs [(X,, Xi + 1,)] which intersect B. Given F, we know that D is an arc containing B that one of its endpoints (call it Y) must lie in C. We can ignore the case where D is very large has both endpoints in Cbecause in that case both conditional probabilities in (2.6) are equal to 1. At this point we shall assume that e (which is the length of C) is less than the minimum of 13, 14,, **,. Let ( be the collection of those rom arcs which do not intersect B, that is, = {i: i 3 B n [(,, Xi + l)] = 0}. Given 3F, the arcs in i cannot intersect D n C, for this would lead to complete coverage of C. Thus, given 3F, the arcs in r are uniformly independently distributed inside the arc extending from Y to Z (where Z is the endpoint of B on the side not adjoining C) which we denote by E. Various items introduced in the previous paragraphs are pictured in Figure 2. The diagram depicts A A* as being disjoint, but this will not always be the case. B Figure 2
7 On the probability of covering the circle by rom arcs 427 Define A,=A D, A*=A*D Then P{A C O = U [(Xi, + )]. iec J.y} = P{Ar C,r} P{A* C t} j } = P{A*C Jr}. If D intersects A*, then A* c A* so that (2.6) holds. Now consider the case when D A* are disjoint so that A* = A*. Because A A* have the same length A is closer than A* to the midpoint of E, the following lemma shows that P{A c Y} ) P{A* C Y,r Since Ar A ra we w} have P C P(A C atr} so that again (2.6) holds. This completes the proof. It remains to present prove the lemma needed above. This will be done independently of the previous material. Lemma. Romly place k intervals of length 01, 02,' *, Ok inside the interval (0, L). The left endpoint of the interval of length Oi is uniformly distributed on (0, L  60). Fix a value b in (0, L). For 0 _ 2 L  b define Q(A) to be the probability that (A, i + b) is completely covered by the k rom intervals. Q(A) is nondecreasing for 0 _ < (L  b)/2. That is, for an interval of fixed length, the probability of complete coverage increases as the interval is moved toward the center of (0, L). Proof. The k rom intervals will be denoted I, I2, *., Ik. Let B= (A, i + b) with A <(Lb)/2. Choose e small enough so that ; + e (L  b)/2. Instead of comparing the coverage probabilities of the intervals B = (A, + b) (A + e, A + e + b), we shall consider only B transform the intervals II, I2,., Ik uniformly distributed in (0, L) into intervals I*, I2*,. *, Ik uniformly distributed in ( e, L  e). The intervals I* will be constructed so that (2.7) if Bc U Ii, thenb C U I*. i= i=l k k
8 428 F. W. HUFFER AND L. A. SHEPP The lemma follows immediately from (2.7). Before describing the construction we need some preliminaries. Let C= U {Ii: Ii?(0, b + 2A))}. Let 9 be the event that B n Cis nonempty there exists j such that B n C c Ij Ij n (L , L) # 0. Note that (O, b + 2l) (L  e, L) are disjoint. We use R to denote the operation of reflection about the midpoint of B. The interval B is centered inside (0, b + 2A) so that if Ii C (0, b + 2A) then also R(I,) c (0, b + 2A). An interval Ii = (c, d) is transformed into J* in ( e, L  e) according to this rule: if d >L e, then = (d  L, d  L +(d  c)); if Ii c (0, b + 2A)! occurs, then i = R(Ii); otherwise, take It = Ii. To verify that this construction works, condition on the exact positions of those intervals Ii for which I,i(0, b + 2)). Conditionally, the remaining rom intervals will be independently uniformly distributed inside (0, b + 22). The reader is left to complete the argument. A crucial observation is that p(b n I.)  (B n IP) for all i where # denotes Lebesgue measure. 3. Convexity in each argument We now prove (1.2). Let Xl, X2,, X, be independent uniformly distributed on (0, 1). Choose y z satisfying 0 <y < z <y + z < 1. Define, = [(X1, X X2 +)], [(X + Y, X1 + Y + Z)], 3= [(X, + y, x, + )], 4 = [(X, X1 + y + Z)], n W= U i2 [(, x, + )]. Here we use the bracket notation described early in Section 2. For 1 _ i _ 4 define J, to be the indicator of the event { C Ri U C). Because X, n 92 = a3 Xl U  S2= 4, it is easily seen that Taking expected values leads to J1 + J2 ' J3 + J4. P(z, 12,'", In) _< :?[P(z  y, 12,, * n) + P(z + y, 12, * *, In)] This suffices to show that P(z, 12, * *, In) is convex in z when 12, ***, 1, are held fixed. The argument above is essentially that used by Huffer (1987).
9 On the probability of covering the circle by rom arcs 429 Acknowledgements The authors wish to thank Frank Proschan for conjecturing (1.1), Andrew Odlyzko for remarks which were helpful in proving (1.1), Doug Critchlow for a suggestion which improved the proof of the lemma. References HUFFER, F. W. (1987) Inequalities for the M/G/oo queue related shot noise processes. J. Appl. Prob. 24(4). HUFFER, F. W. (1986) Variability orderings related to coverage problems on the circle. J. Appl. Prob. 23, JANSON, S. (1983) Rom coverings of the circle with arcs of rom lengths. In Probability Mathematical Statistics: Essays in Honour of CarlGustav Esseen, ed. A. Gut L. Holst. Department of Mathematics, Uppsala University, Sweden. MARSHALL, A. W. AND OLKIN, I. (1979) Inequalities: Theory of Majorization Its Applications. Academic Press, New York. SHEPP, L. A. (1972) Covering the circle with rom arcs. Israel J. Math. 11, SIEGEL, A. F. AND HOLST, L. (1982) Covering the circle with rom arcs of rom sizes. J. Appl. Prob. 19, SOLOMON, H. (1978) Geometric Probability. SIAM, Philadelphia, PA. STEVENS, W. L. (1939) Solution to a geometrical problem in probability. Ann. Eugenics 9, WSCHEBOR, M. (1973) Sur le recouvrement du cercle par des ensembles places au hasard. Israel J. Math. 15, 111.