Coverage of Space in Boolean Models Author(s): Rahul Roy Source: Lecture Notes-Monograph Series, Vol. 48, Dynamics & Stochastics (2006), pp. 119-127 Published by: Institute of Mathematical Statistics Stable URL: http://www.jstor.org/stable/4356366 . Accessed: 15/06/2014 04:49 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . Institute of Mathematical Statistics is collaborating with JSTOR to digitize, preserve and extend access to Lecture Notes-Monograph Series. http://www.jstor.org This content downloaded from 188.72.127.119 on Sun, 15 Jun 2014 04:49:18 AM All use subject to JSTOR Terms and Conditions
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Coverage of Space in Boolean ModelsAuthor(s): Rahul RoySource: Lecture Notes-Monograph Series, Vol. 48, Dynamics & Stochastics (2006), pp. 119-127Published by: Institute of Mathematical StatisticsStable URL: http://www.jstor.org/stable/4356366 .
Accessed: 15/06/2014 04:49
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp
.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].
.
Institute of Mathematical Statistics is collaborating with JSTOR to digitize, preserve and extend access toLecture Notes-Monograph Series.
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? Institute of Mathematical Statistics, 2006 DOI: 10.1214/074921706000000158
Coverage of space in Boolean models
Rahul Roy1 *
Indian Statistical Institute
Abstract: For a marked point process {{x?,S?)?>\} with {x{ G ? : i > 1} being a point process on ? ? Rd and {S? ? Rd : i > 1} being random sets consider the region C = U?>i(x? + Si). This is the covered region obtained from the Boolean model {(xi + Si) : i > 1}. The Boolean model is said to be completely covered if ? ? C almost surely. If ? is an infinite set such that s -?- ? ? ? for all s ? ? (e.g. the orthant), then the Boolean model is said to be eventually covered if t 4- ? ? C for some t almost surely. We discuss the issues of coverage when ? is U.d and when ? is [0, oo)d.
1. Introduction
A question of interest in geometric probability and stochastic geometry is that of
the complete coverage of a given region by smaller random sets. This study was
initiated in the late 1950's. An account of the work done during that period may be
found in Kendall and Moran (1963). A similar question is that of the connectedness
of a random graph when two vertices u and ? are connected with a probability pu~v
independent of other pairs of vertices. Grimmett, Keane and Marstrand (1984) and
Kalikow and Weiss (1988) have shown that barring the 'periodic' cases, the graph is almost surely connected if and only if 5Z?p? = oo.
Mandelbrot (1972) introduced the terminology interval processes to study ques- tions of coverage of the real line R by random intervals, and Shepp (1972) showed
that if S is an inhomogeneous Poisson point process on R x [0, oo) with density measure ? ? ? where ? is the Lebesgue measure on the x-axis and ? is a given measure on the ?/-axis, then U(x>y)es(a;,2; + y) ? R almost surely if and only if
f0 dxeyip(f^?(y ? x)p(dy)) = oo. Shepp also considered random Cantor sets de-
fined as follows: let 1 > t\ > t<? > ... be a sequence of positive numbers decreasing to 0 and let V\,p2,. ?. be Poisson point processes on R, each with density ?. The
set V :? R \ (U* UxeVi (%,% + U)) is the random Cantor set. He showed that V
has Lebesgue measure 0 if and only if J^? U = oo. Moreover, P(Vr = 0) = Oorl
according as 5Z^=i n~2 e??{?(?? ?-l? tn)} converges or diverges. In recent years the study has been re-initiated in light of its connection to percola-
tion theory. Here we have a marked point process {(xiy S?)?>i} with {xi : i > 1} be-
ing a point process on ? ? Rd and Si ? -Rd being random sets. Let C = U?>i (x?~\-S?) be the covered region of the Boolean model {(xi + Si) : i > 1}.
The simplest model to consider is the Poisson Boolean model, i.e., the process
{xi : i > 1} is a stationary Poisson point process of intensity ? on Rd and
Si = [0,pi]d, i>l (1)
*This research is supported in part by a grant from DST. 1 Statistics and Mathematics Unit Indian Statistical Institute, 7 SJS Sansanwal Marg, New
are d-dimensional cubes the lengths of whose sides form an independent i.i.d collec- tion {pi : i > 1} of positive random variables. (Alternately, ??'s are d-dimensional
spheres with random radius pi.) In this case, Hall (1988) showed that
Theorem 1.1. C = Rd almost surely if and only if Epd = oo.
More generally, Meester and Roy (1996) obtained
Theorem 1.2. For {xi : i > 1} a stationary point process, if Epf = oo then C = Rd almost surely.
The above results relate to the question of complete coverage of the space Rd. Another question which arises naturally in the Poisson Boolean model is that
of eventual coverage (see Athreya, Roy and Sarkar [2004]). Let {xi : i > 1} be a stationary Poisson process of intensity ? on the orthant R+ and the Boolean model is constructed with random squares 5? as above yielding the covered region C = Ui>i[xi(l),Xi(l) + pi] ? ??? ? [xi(d),Xi(d) + pi\. In that case, P(R% CC)=0, however we may say that R+ is eventually covered if there exists 0 < t < oo such that (t, oo)d ? C. In this case, there is a dichotomy vis-a-vis dimensions in the
coverage properties. In particular, while eventual coverage depends on the intensity ? for d = 1, for d > 2 there is no such dependence.
Theorem 1.3. For d = I,
(a) ifO < I := \imm?x-+00xP(pi > ?) < oo then there exists 0 < ?? < } < oo such that
P\(R+ is eventually covered by C) = { "
if X > ?0; c
(b) ifO<L:= limsupx_+00xP(pi > x) < oo then there exists 0 < ? < ?? < oo such that
Pa(R+ is eventually covered by C) = <
[1 ifX>Xi\
(c) if limx_>oo xP(p\ > x) = oo then for all X > 0, R+ is eventually covered by C almost surely (??);
(?) z/limx_oo xP(p\ > x) = 0 then for any X > 0, R+ is not eventually covered
by C almost surely (??).
Theorem 1.4. Let d>2, for all X > 0.
(a) P\(^+ w eventually covered by C) = 1 whenever lim inf x?oc %P(p\ > x) > 0.
(b) P\(R+ is eventually covered by C) = 0 whenever limx_00xF(pi > x) = 0.
In 1-dimension for the discrete case we may consider a Markov model as follows:
Xi, Xi,... is a {0,1} valued Markov chain and Si :? [0, p?], i = 1,2,... are i.i.d. intervals where pi is as employed in (1). The region ?J(i + Si)lxi=\ is the covered
region. This model has an interesting application in genomic sequencing (Ewens and Grant [2001]). U Pij = P(Xn+\ = j \ Xn = i), i,j = 0 or 1, denote the transition
probabilities of the Markov chain then we have
Theorem 1.5. Suppose 0 < ???,??? < 1?
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(a) /// := lim inf j^00jP(pi > j) > 1, then P{C eventually covers N} = 1 when-
ever ? *T? > ?/?? Pio+Poi '
(b) // L := limsupj-^oo jP(pi > j) < oo, then P{C eventually covers N} = 0
whenever n p?* < 1/L. Pio+Poi '
Molchanov and Scherbakov (2003) considered the case when the Boolean model
is non-stationary. For a Poisson point process {xi : i > 1} of intensity ?, we place a d-dimensional ball B(xi,pih(\\xi\\)) centred at Xi and of radius p?/i(||ic?||) where
Pi is as before and h : [0, oo) ?> (0, oo) is a nondecreasing function. Let
C = UZiB(xi,Pih(\\xi\\))
denote the covered region. Let p^ denote the volume of a ball of unit radius in d / , \l/d
dimensions and take ho(r) = ? jj^ l?Sr )
Theorem 1.6. Suppose Epd+ri < oo for some ? > 0 and h is as above.
The result in (a) above cannot be translated into an almost sure result because
of the lack of ergodicity in the model.
2. Complete coverage
We now sketch the proofs of Theorems 1.1, 1.2 and 1.6.
Let V = [0, l]d \ C denote the 'vacant' region in the unit cube [0, l]d;
E{i{V)) = E? 1{i jg not covered}dx
= exp(-??>?) (2)
where ? stands for the d-dimensional Lebesgue measure. Hence, if Epd < oo then
E(i(V)) > 0 and so P([0, l]d ? C) < 1.
Conversely, Ep\ = oo implies E(i(V)) = 0 and thus by stationarity, E(?(Rd \ C)) = 0. Using the convexity of the shapes 5? we may conclude that P(C = Rd) ? 1.
Here the Poisson structure was used to obtain the expression (2); for a general
process we need to extract, if possible, an ergodic component of the process and show that the Boolean model obtained from this ergodic component covers the entire space when Epd = oo. To this end let {x? : i > 1} be an ergodic point process with density 1. Let Dn = [0,2n/d]d and En = {there exists x? in the
annulus Dn+1 \ Dn such that Do ? (xi + Si)}. Also let Am be the event that m is the first index such that #{z : x i G Z)n+i \Dn} > <z2n for all n > m and for some
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To prove Theorem 1.6 (a) we study the case when h(r) = llf/dh0(r) and lnEpd > 1 + d for some d > 0. It may be easily seen that for ? 6 Zd, P{z + (-1/2, l/2]d g C) < ???{-??(^)}, where Rz = {(r,x) ? [?,??) ? Rd : ? + (-l/2,l/2]d ?
??(#, rh(x)) and ? is the product measure of the measure governing p\ and Lebesgue measure. From the properties of ho it may be seen, after some calculations, that
given e > 0, there exists r such that for ||z|| > r, p(Rz) > (1 -
e)nd(h(z))dEpd. Thus we obtain, for some constant K,
S PU + (-1/2, l/2]d % C} < ? S exp{-d(l + d)(1 - e) log(||*||)} zGZd ^6Zd
= /^ ? ||z||-d(l+*)<l-e) < oo
zGZd
whenever (1 -f ?)(1 - e) > 1. Invoking the Borei-Cantelli lemma we have that ? -I- (?1/2, l/2]d g C occurs for only finitely many z G Zd. Using this we now
complete the proof of Theorem 1.6 (a). The proof of Theorem 1.6(b) is more delicate and we just present the idea here.
For d > 2, if we place points in a spherical shell of radius n7, such that the interpoint distances are maximum and are of the order of n^ where 0 < ? < 7 and 7 > 1 then the number of points one can place on this shell is of the order of n7-^^-1). Let Vn be the event that one such point is not covered by C. It may be shown that there is a choice of 7 and ? such that infinitely many events Vn occur with probability 1. For d = 1, the same idea may be used and, in fact, the proof is much simpler.
3. Eventual coverage
We discretise the space R^ by partitioning it into unit cells {(ij,...,id) + (0, l]d :
iii- ? ? iid = 0,1,...} and call a vertex i := (i\,..., id) green if x G i -h (0, l]d for some point x of the point process. Consider two independent i.i.d. collections of random variables {p\} and {p[} where the distribution of p\ and p\ are identical
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to that of 2 + [max{pi,...,Pn}\ and max{0, [max{pi,...,p?v}\ -
1} respectively; here ? is an independent Poisson random variable with mean ? conditioned to be 1 or more. We now define two discrete models, an upper model and a lower model: in both these models a vertex i is open or closed independently of other vertices, and the covered region for the upper model is U{i open}(i 4- [0,p?*]), and that for
the lower model is U{i open}(i + [0,/>(]). Observe that the eventual coverage of the Poisson model ensures the same for the upper model and eventual coverage of the lower model ensures the same for the Poisson model. Thus it suffices to consider the eventual coverage question for a discrete model, as in Proposition 3.1 below, and check that the random variables pu and pl satisfy the conditions of the proposition.
We take {X{ : i G Nd} to be an i.i.d. collection of {0,1} valued random variables with ? := P(X\ = 1) and {p\ : i G Nd} to be another i.i.d. collection of positive integer valued random variables with distribution function F(= 1 ?
G) and inde-
pendent of {X\ : i G Nd}. Let C := U{ieNd|xl=1}(i + [0, p\\d). We first consider
eventual coverage of Nd by X\.
Proposition 3.1. Let d>2 and 0 < ? < 1.
(a) ?/limj-.oo jG(j) = 0 then PP(C eventually covers Nd) ? 0,
(b) ?/liminfj^oo jG(j) > 0 then PP(C eventually covers Nd) ? 1.
We sketch the proof for d = 2. For i, j G ? let A(i, j) := {(i,j) & C}. Clearly,
P(A(k,j) ? A(i,j)) = P(A(k - i,j))P(A(i,j)) for k > i,
i.e., for each fixed j the event A(i, j) is a renewal event. Thus, if, for every j > 1,
S^? P(A{i,j)) = oo then, on every line {y = j}, j > 1, we have infinitely many z's for which (i, j) is uncovered with probability one and hence Nd can never be
eventually covered.
To calculate Pp(A(i, j)) we divide the rectangle [l,i] ? [1, j] as in Figure 1. For
any point (k, /), 1 < k < i ? j and 1 < / < j, in the shaded region of Figure 1, we ensure that either X(k,i) = 0 or p(k,i) ^ & + 3 ~ 1? The remaining square region in
Figure 1 is decomposed into j sub squares of length t,\ <t < j -\ and we ensure that for each point (k, I) on the section of the boundary of the sub square t given by the dotted lines either X(kti) = 0 or P(k,i) ^ t. So,
Since am < oo for all m > 1, both the numerator and the denominator in the fraction above are finite. Moreover, each term in the sum of the numerator is less than the corresponding term in the sum of the denominator; yielding that the fraction is at most 1. Hence, for 0 < a < 1 as chosen earlier
^^ < (l-pG(jo-rm)r+m+l Im
< a.
This shows that S?=? 7m < oc and completes the proof of part (b) of the propo- sition.
It may now be seen easily that pu and pl satisfy the conditions of Proposition 3.1 and thus Theorem 1.4 holds.
4. Markov Model
The relation between the Poisson model and the discrete model explained in Section 3 shows that Theorem 1.3 would follow once we establish Theorem 1.5. In the setup of the Theorem 1.5, for each A: G ? let Ak '= {k & C}. To prove Theorem 1.5(a) we show that ]T^ P(Ak) < oo and an application of the Borei-Cantelli lemma
yields the result, while to prove Theorem 1.5 (b) we show that S*:^(^*) = ???
However, the Ak's are not independent and hence Borei-Cantelli lemma cannot be
applied. Nonetheless using the Markov property one can show that P(Ak ? Ai) ?
P(Ak-i)P(Ai) and therefore, A^s are renewal events; so by the renewal theorem, if S^? P(Ai) = ?? then Ai occurs for infinitely many ?'s with probability one.
For k > 1, let P0(Ak) = P(Ak \ ?? = 0) and Px(Ak) = P(Ak \ ?? = 1). The
following recurrence relations may be easily verified
P0(^fc4i) = PooPo(Ak)+poiPi(Ak) (7)
Pi(4fe4i) = Fik-VfaoPo^+PnPi^k)}. (8)
We use this to prove Theorem 1.5(b) first. Let *o(s) = S?1*;0 Po(Ak)sk and
^i(s) = S??at? Pi(Ak)sk denote the generating functions of the sequences
{Po(Ak) : k > ko} and {P\(Ak) : k > ko} respectively, where k0 is such that for a given e > 0 and C = L + 6>0 (where L is as in the statement of the theorem), fco+(l-C)>0, Po(Ako)>0, Pi(Ako)>0, andF(fc-l)>l-^ for k > **,. Such a ko exists by the condition of the theorem.
Using the recurrence relations (7) and (8) we obtain
^'1(s)P(s)>Q(s)B(s) + R(s), (9)
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Now for s < 1, ?W = Tj?^-h t^T H- 1 qM f?r?r, for some real numbers ' P(s) X-Poos X-s l-s(l-poi-Pio) '
D,E,F. It may now be seen that #i(l) = oo whenever E > 0. Also the recurrence
relations show that F?(1) = oo whenever F?(1) = oo. A simple calculation now
yields that E > 0 if and only if ??"1 < ^ =
2?7 ? Since ? is arbitrary, we obtain
Theorem 1.5(b). The proof of Theorem 1.5(a) is similar.
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