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# Full text

Title
Mapping without the sun
Author
Zhang, Jixian

80
• then two Poisson points, namely yi and 2/2, fall in ¿(xo).
They generate
X 2 (x 0 ) — S(yi) U 6(1/2) = 8(h), in medium grey,
(6)
and h(x 0 ) = {yi} U {2/2}; (7)
• then again a new Poisson realization generates one point,
z\ in ¿(2/1), and another Poisson realization the three points
-22,1,-22,2, and 22,3 in ¿(2/2), hence
^3(2:0) = ¿(zi)U[5(z2,i) U S(z 2 ,2) U ¿(-22,3)] ,0-.
= 8(h), in light grey.
and h(xo) = {zi}u[{z 2 ,i} U {22,2} U {22,3}]- (9)
The (n + 1) step is obtained by the two induction relations
X n+1 (I 0 ) = 8[I n (I 0 )} (10)
/n+l(/o) = 0[I n (Io)} /iix
= u{(S(xi, n ) n Ji, n ), Xi, n e in, Ji, n e J(0)}
We observe that the fire of tomorrow is determined by points seats
stemming from the zone which bums today. What burnt yester
day, before yesterday, etc., has no longer importance
Since the two spreads I n and X n are RACS, we shall character
ize them by their functionals Q n (K) (Matheronl975), (Stoyan
et al. 1995) . In the RACS X n , the primary grain only is inde
pendent of step n, since at each time n, the intensity of the seats
is different. This circumstance simplifies the theoretical study of
the time evolution. Also, it suggests to find induction relations
between Q n (K) and Q n +i(K) that reflect the two definitions
by induction of Relation (10) and (11).
2.2 Characteristic functional
Relation (1) allows us to take for Io a point initial seat, xo say,
of dilate X\ = 8(xo), and whose intersection of the dilate with
Poisson points J provides the first random set h — 8(x0) fl J.
The distribution of a RACS X is characterized by the probabili
ties
Q(K) = Pr{K C X c } (12)
as K spans the class the compact sets of R 2 (Matheronl975),
(Molchanov2005). For the random spread X n (xo) this func
tional Qn(K | xo) is the probability that K misses the n th spread
X n (xo) of initial seat xq. It satisfies the following induction re
lation between steps n and n + 1 (Serra2005)
Qn+i(K | xo) = exp - f 0(dy)[ 1 - Q n (K \ y)\. (13)
J¿(xo)
Each step involves an exponentiation more than the previous one.
For example, for the first two steps we find that
Q 2 (K) = exp-0[C(jFOn5(®o)], (14)
Qo(K) = exp - [ 0(dy)[l-e~ e[aK)n,5(!/)1 ] (15)
J <5(xo)
where £ is the reciprocal function of (5, i.e.
y € C,(x) if and only if x £ 8(y) x,y £ E.
(16)
The seat spread I n+ 1 satisfies the same induction relation (13) as
the fire spread X n +i- The only change holds on the first term, for
which it suffices to replace C(K) by K in Relation (14).
Figure 4: a) the weight u < 1, then g(n) —> 0, as n —+ 00. b) the
weight u > 1, then g(n) —+ p, as n —* 00.
2.3 Order two dependency
A Markov type assumption of order one underlies the random
spread model, since it suffices to know the n th seats for construct
ing the (n+l) th ones. This assumption just allowed us to achieve
the formal calculus of the functionals Q n and R n . But it does not
prevent us from the comeback of new seats on already burnt ar
eas, after a few steps. The successive spreads may turn around
¿(xo), as depicted in Figure 3.
The trouble can partly be overcome by a Markov assumption of
order two, i.e. by making depend the (n + 2) th seats of both the
(n + l) th and n th steps. We can impose for example to keep the
second seats if they fall in the 8(xi),Xi (E h but not in ¿(xo),
and so on. The new version of the induction relation (13) involves
now two successive terms
Qn+2(K I xo)
= exp {- f s(xo) 9(dy)[ 1 - Q n +i(K I xo \ y)]}
where
Q n +i(K I xo I y) = exp - [ Q(dz)[ 1 - Q n (K \ z)\
Jô(y)\S(x0)
2.4 Spontaneous extinction
The fire which stems from the point seat xo may go out, spon
taneously, after one, two, or more steps. The description of this
phenomenon does not involve any particular compact set K. De
note by g(n | xo) the probability that the fire extinguishes after
step n. This event occurs after the first step, when no Poisson
point falls inside set <5(xo), hence when
y(l | xo) = 1 - exp -0[<5(x o )].
For a spontaneous extinction after step n+l, we find an induction
relation similar to Equation ( 13)
g(n + l I x 0 ) = 1 -exp{ - [ 6(dy)g(n\y)}. (17)
J S(x 0)
Suppose for the moment that 6 and Ô are translation invariant.
The extinction probabilities no longer depend on xo, y, etc., and
reduce to g(n + 1 | xo) = g(n + 1), g(n \ y) = g(n),
etc..Similarly, the integral u = f S(zj 9(dx) is independent of z £
R d , so that
g(n + 1) = 1 — exp —ug(n). (18)
As n —► 00, the behavior of g depends on u. If u < 1, which cor
responds to Figure 4a, then g(n) —► 0, i.e. the fire extinguishes