Trivariate distributions do not
characterize the texture of a
random set
C. Lantuéjoul
[email protected]
Centre de Géostatistique
Ecole des Mines de Paris
Outline
Presentation of three spatial stochastic models
– Gaussian excursion
– Poisson tessellation
– Dead leaves model
Comparing their multivariate distributions
1
Gaussian excursion
2
Gaussian excursion
Basic ingredients:
– a standardized gaussian random function Y with covariance C.
– a numerical value λ.
Definition:
The excursion of Y at level λ is the random function
Xλ(x) =
+1 if Y (x) ≥ λ
−1 if not
3
Examples at various levels
Gaussian random function (gaussian covariance) and its excursions at levels
λ = −1, −0.5, 0, 0.5 and 1.
4
Relationship between covariance functions
2
π
C(h)
exp
0
2
−λ
1+x
√
dx
1 − x2
h ∈ IRd
0.0
−1.0
−0.5
C(lambda,h)
0.5
1.0
Cλ(h) =
Z
−1.0
−0.5
0.0
0.5
1.0
C(h)
Covariance of excursions at levels 0, ±0.5, ±1, ±1.5 and ±2 versus covariance of the
underlying Gaussian random function
5
Can the covariance at 0 be exponential?
C0(h) =
2
arcsin C(h)
π
2
e−|h|
0.0
0.2
0.4
C(h)
0.6
0.8
1.0
C0(h) = e−|h| ⇐⇒ C(h) = sin
π
0
1
2
3
4
5
h
Is h −→ sin
π −|h|
2e
a function of positive type?
6
More about this function
Main property:
π −|h|
sin 2 e
divisible covariance function, i.e. it can be
is an infinitely
written as exp −γ(h) where γ is a variogram.
Sketch of the proof:
Start from the representation of the sine function as an infinite product:
!
«
„
∞
−2|h|
Y
π
π −|h|
e
= e−|h|
sin
1−
e
2
2
4n2
n=1
and take the logarithm. After some calculations, one gets
ln sin
„
π −|h|
e
2
«
=−
∞
X
ζ(2n) “
n=1
n4n
2n|h| − 1 + e
−2n|h|
”
,
and the proof is completed by showing that the function h −→ |h| − 1 + e −|h| is a
variogram.
7
Simulation of a gaussian random function with
exponential sine covariance function
Bochner theorem:
There exists a probability density function f (spectral density) such that
Z
π
ei <h,ω> f (ω) dω
sin e−|h| =
2
IR2
Remark:
Let Ω ∼ f , and√let Φ ∼ U([0, 2π[) be independent. Then the random
function x −→ 2 cos < Ω, x > + Φ is standardized with covariance C.
Algorithm:
(i) generate independently n spectral vectors ω1, ..., ωn ∼ F and n phases
φ1, ..., φn ∼ U([0, 2π[);
√ n
2X
cos < ωi, x > + φi for each x ∈ IR2.
(ii) return y(x) = √
n i=1
8
Spectral density
By Bochner theorem, and owing to the fact that it is square integrable, the
spectral density f can be written as
Z
π
1
−<ω,h>
−|h|
2
e
dh
ω
∈
I
R
e
sin
f (ω) =
(2π)2 IR2
2
2
0.10
f(ω)
0.15
0.20
0.25
4 n=0 (2n)!
1
(2n + 1)2 + |ω|2
32
0.05
f (ω) =
∞
n 2n
X
(−1) π
1
0.00
Explicitly
0.0
0.5
1.0
1.5
2.0
2.5
3.0
ω
9
Simulation of the spectral density
Put also fk (ω) =
that
k
n 2n
X
1
(−1) π
4 n=0 (2n)!
2
1
(2n + 1)2 + |ω|2
32
and observe
f2k+1 < f2k+3 < · · · < f < · · · < f2k+2 < f2k < · · · < f0
f0
Algorithm:
(i) generate ω ∼ f0 and u ∼ U. Set k = 0;
(ii) set k = k + 1;
(iii-i) if uf0(ω) > fk (ω) and k even, then goto (i);
(iii-ii) if uf0(ω) > fk (ω) and k odd, then goto (ii);
(iii-iii) if uf0(ω) ≤ fk (ω) and k even, then goto (ii);
(iii-iv) if uf0(ω) ≤ fk (ω) and k odd, then return ω.
f 2k
f2k+2
f
f2k+3
f2k+1
ω
10
Simulation of an excursion
11
Poisson tessellation
12
Parametrization of a line
p
p
α
0
0
α
π
L(a,p)
Equation of a line:
x cos α + y sin α = p
0 ≤ α < π direction
−∞ < p < +∞ location
13
Poisson line process
A Poisson line process is parametrized by a Poisson point process on
[0, π[×IR.
Poisson point process intensity: λ
14
Poisson tessellation
Basic ingredients:
– a network of Poisson lines with intensity λ;
– a mark distribution uniform on {−1, +1}.
Definition:
The random function X obtained by assigning polygons independent marks.
15
Covariance function of a Poisson tessellation
Basic property:
D
The number of lines hitting a convex domain D follows a Poisson distribution
with mean λp(D)
(p(D) = perimeter of D)
Covariance function:
C(h) = E{X(x)X(x + h)} = P {x, x + h ∈ same polygon} = e−2λ|h|
The Poisson tessellation has an exponential covariance function with scale
factor (2λ)−1. If λ = 12 , then C(h) = e−|h|.
16
Dead leaves model
17
Dead leaves model
t
0
x
y
18
Dead leaves model
Basic ingredients:
– a homogeneous Poisson point process P on IRd×] − ∞, 0[;
– a family (A(p), p ∈ P) of independent and identically distributed compact
random sets, or leaves;
– marks on leaves. They are independent and uniform on {−1, +1}.
Definition:
The value X(x) assigned to each point x ∈ IRd is the mark of the most
recent leaf that covers it.
19
Examples of dead leaves models
The three models have the same Poisson intensity and the same mean leaf area. Leaves
are squares, disks and Poisson polygons.
20
Covariance function of the dead leaves model
Geometric covariogram of a leaf:
h
Lh
L
K(h) = E
Z
IR2
1x∈L 1x+h∈L dx
Covariance function:
C(h) = E{X(x)X(x + h)} = P {x, x + h ∈ same leaf} =
K(h)
2K(o) − K(h)
21
Can the covariance function be exponential?
2K(o)
K(h)
−|h|
=e
⇐⇒ K(h) =
2K(o) − K(h)
1 + e|h|
This is possible if L is a uniform planar cross-section of a random ball, the
radius of which satisfies
sinh r 1
r>0
P {R ≥ r} =
3
r cosh r
22
Comparison between distribution
23
The three random sets considered
Gaussian excursion
Poisson polygons
Dead leaves model
24
Autodual random set
Definition:
The random set X is said to be autodual if X and its complement X̄ have
the same spatial distribution.
Property:
If X is autodual, then its multivariate distributions of order 2n + 1 are given
by those of order 2n.
Example:
If X is autodual, then its trivariate distributions are completely specified by
its bivariate distributions.
25
Proof of the example
Let A, B and C three events over X. By autoduality, complementation
and inclusion-exclusion, we have
PA∩B∩C
= PĀ∩B̄∩C̄
= PA∪B∪C
= 1 − PA∪B∪C
= 1 − (PA + PB + PC − PA∩B − PA∩C − PB∩C + PA∩B∩C )
Hence
PA∩B∩C
1
= (1 − PA − PB − PC + PA∩B + PA∩C + PB∩C )
2
26
Questions
Questions are raised assuming that the three random sets have their
statistical properties specified by their spatial distibution.
– How can these three random sets be statistically made distinct?
– How much information is required for this?
– At what order does this take place?
– How to quantify the information conveyed by all nth order multivariate
distributions?
– How large is the nth order information compared to that of the spatial
distribution?
27