RANDOM FIELDS AND THEIR GEOMETRY

22 1. Random fields
Since indicator functions are in K, we can define a process W on RN by
setting
W (λ) = Θ −1 � �
(1.4.22)
� (−∞,λ] ,
where (−∞, λ] = �N j=1 (−∞, λj]. Working through the isomorphisms shows
that W is a complex ν-noise and that �
RN exp(i〈t, λ〉) W (dλ) is (L2 (P) indistinguishable
from) ft. ✷
There is also an inverse 22 to (1.4.19), expressing W as an integral involving
f, but we shall have no need of it and so now turn to some consequences
of Theorems 1.4.3 and 1.4.4.
When the basic field f is real, it is natural to expect a ‘real’ spectral
representation, and this is in fact the case, although notationally it is still
generally more convenient to use the complex formulation. Note firstly that
if f is real, then the covariance function is a symmetric function (C(t) =
C(−t)) and so it follows from the spectral distribution theorem (cf. (1.4.16))
that the spectral measure ν must also be symmetric. Introduce three 23 new
measures, on R+ × R N−1 , by
ν1(A) = ν(A ∩ {λ ∈ R N : λ1 > 0})
ν2(A) = ν(A ∩ {λ ∈ R N : λ1 = 0})
µ(A) = 2ν1(A) + ν2(A)
We can now rewrite24 (1.4.16) in real form, as
�
(1.4.23) C(t) =
cos(〈λ, t〉) µ(dλ).
R+×R N−1
There is also a corresponding real form of the spectral representation
(1.4.19). The fact that the spectral representation yields a real valued processes
also implies certain symmetries 25 on the spectral process W . In particular,
it turns out that there are two independent real valued µ-noises,
22 Formally, the inverse relationship is based on (1.4.22), but it behaves like regular
Fourier inversion. For example, if ∆(λ, η) is a rectangle in R N which has a boundary of
zero ν measure, then
W (∆(λ, η)) = lim
K→∞ (2π)−N
Z K Z K
. . .
−K −K
NY e
k=1
−iλktk − e−iηktk f(t) dt.
−itk
As is usual, if ν(∆(λ, η)) �= 0, then additional boundary terms need to be added.
23Note that if ν is absolutely continuous with respect to Lebesgue measure (so that
there is a spectral density) one of these, ν2, will be identically zero.
24There is nothing special about the half-space λ1 ≥ 0 taken in this representation.
Any half space will do.
25To rigorously establish this we really need the inverse to (1.4.19), expressing W as
an integral involving f, which we do not have.