60.8 Gaussian Measures For A Separable Hilbert Space
First recall the Kolmogorov extension theorem, Theorem 58.2.3 on Page 6238 which is
stated here for convenience. In this theorem, I is an ordered index set, possibly infinite,
even uncountable.
Theorem 60.8.1(Kolmogorov extension theorem) For each finite set
J = (t1,⋅⋅⋅,tn) ⊆ I,
suppose there exists a Borel probability measure, ν_{J} = ν_{t1}
⋅⋅⋅
t_{n}defined on the Borel sets of∏_{t∈J}M_{t}where M_{t} = ℝ^{nt}such that if
where if s_{i} = t_{j}, then G_{si} = F_{tj}and if s_{i}is not equal to any of the indices, t_{k}, thenG_{si} = M_{si}. Then there exists a probability space,
(Ω,P,ℱ )
and measurable functions,X_{t} : Ω → M_{t}for each t ∈ I such that for each
Then for the index set equal to ℕ the measures satisfy the necessary consistency
condition for the Kolmogorov theorem above. Therefore, there exists a probability space,
which shows the random variables are independent as well as normal with mean 0 and
variance 1. This proves the Lemma.
A random variable X defined on a probability space
(Ω,ℱ, P)
is called Gaussian
if
∫ --1- 2
P ([X ∈ A ]) = ∘--1---- e−2σ(v)2(x−m(v)) dx
2πσ(v)2 A
for all A a Borel set in ℝ. Therefore, for the probability space
(X, ℬ(X ),μ)
it is natural
to say μ is a Gaussian measure if every x^{∗} in the dual space X^{′} is a Gaussian random
variable. That is, normally distributed.
Definition 60.8.3Let μ be a measure defined on ℬ
(X)
, the Borel sets of X, aseparable Banach space. It is called a Gaussian measure if each of the functions in thedual space X^{′}is normally distributed. As a special case, when X = U a separable realHilberts space, μ is called a Gaussian measureif for each v ∈ U, the function u →
(u,v)
_{U}is normally distributed. That is, denoting this random variable as v^{′}, it follows for A aBorel set in ℝ
∫
λ ′(A ) ≡ μ([u : v′(u) ∈ A ]) = ∘-1-- e− 2σ(1v)2(x− m(v))2dx
v 2π σ(v)2 A
in case σ
(v)
> 0. In case σ
(v)
= 0
λv′ ≡ δm (v)
In other words, the random variables v^{′}for v ∈ U are all normally distributed on theprobability space
(U, ℬ(U) ,μ )
.
Also recall the definition of the characteristic function of a measure.
Definition 60.8.4The Borel sets in a topological space X will be denoted by ℬ
(X )
. Fora Borel probability measure μ defined on ℬ
(U)
for U a real separable Hilbert space, defineits characteristic function as follows.
∫
ϕμ(u) ≡ ^μ (u) ≡ ei(u,v)dμ (v) (60.8.31)
U
(60.8.31)
More generally, if μ is a probability measure defined on ℬ
(X )
where X is a separableBanach space, then the characteristic function is defined as
∫
∗ ∗ ix∗(x)
ϕμ(x ) = ^μ (x ) ≡ U e dμ (x)
One can tell whether μ is a Gaussian measure by looking at its characteristic function.
In fact you can show the following theorem. One part of this theorem is that if μ is
Gaussian, then m and σ^{2} have a certain form.
Theorem 60.8.5A measure μ on ℬ
(U)
is Gaussian if and only if there exists m ∈ Uand Q ∈ℒ
(U )
such that Q is nonnegative symmetric with finite trace,
∑ (Qe ,e ) < ∞
k k k
for a complete orthonormal basis for U, and
i(m,u)− 12(Qu,u)
ϕμ(u) = ^μ(u) = e (60.8.32)
(60.8.32)
In this case μ is called N
(m, Q)
where m is the mean and Q is called the covariance. Themeasure μ is uniquely determined by m and Q. Also for all h,g ∈ U
Proof: First of all suppose 60.8.32 holds. Why is μ Gaussian? Consider the random
variable u^{′} defined by u^{′}
(v)
≡
(v,u)
. Why is λ_{u′} a Gaussian measure on ℝ? By the
definition in 60.8.31,
∫ ∫ ∫
itu′(v) it(v,u) ix
U e dμ (v) ≡ U e dμ(v) = ℝe dλu′ (x)
∫ i(v,tu) it(m,u)− 1t2(Qu,u)
= e dμ(v) = e 2
U
and this is the characteristic equation for a random variable having mean
(m, u)
and
variance
(Qu,u)
. In case
(Qu,u )
= 0, you get e^{it(m,u)
} which is the characteristic
function for a random variable having distribution δ_{(m,u)
}. Thus if 60.8.32 holds, then u^{′} is
normally distributed as desired. Thus μ is Gaussian by definition.
The next task is to suppose μ is Gaussian and show the existence of m,Q which have
the desired properties. This involves the following lemma.
Lemma 60.8.6Let U be a real separable Hilbert space and let μ be a probabilitymeasure defined on ℬ
(U )
. Suppose for some positive integer, k
∫
|(x,z)|kdμ (x) < ∞
U
for all z ∈ U. Then the transformation,
∫
(h1,⋅⋅⋅,hk) → (h1,x)⋅⋅⋅(hk,x)dμ (x ) (60.8.36)
U
(60.8.36)
is a continuous k−linear form.
Proof:I need to show that for each h ∈ U^{k}, the integral in 60.8.36 exists. From this
it is obvious it is k− linear, meaning linear in each argument. Then it is shown it is
continuous.
First note
k k
|(h1,x)⋅⋅⋅(hk,x)| ≤ |(h1,x)| + ⋅⋅⋅+ |(hk,x)|
This follows from observing that one of
|(hj,x)|
is largest. Then the left side is smaller
than
|(hj,x)|
^{k}. Therefore, the above inequality is valid. This inequality shows the
integral in 60.8.36 makes sense.
I need to establish an estimate of the form
∫
|(x,h)|kdμ (x) < C < ∞
U
for every h ∈ U such that
||h||
is small enough.
Let
{ ∫ }
U ≡ z ∈ U : |(x,z)|kdμ(x) ≤ n
n U
Then by assumption U = ∪_{n=1}^{∞}U_{n} and it is also clear from Fatou’s lemma that each U_{n}
is closed. Therefore, by the Bair category theorem, at least one of these U_{n0} contains an
open ball, B
(z0,r)
. Then letting
|y|
< r,
∫ ∫
|(x,z0 + y)|kdμ(x), |(x,z0)|kdμ (x ) ≤ n0,
U U
∫ k (k+2 ) k k
|(x,y)| dμ ≤ 2 ∕r n0 ||y|| ≡ C ||y||.
U
Thus by Holder’s inequality,
∫ ∏k ( ∫ )1∕k
|(h1,x)⋅⋅⋅(hk,x)|dμ(x) ≤ |(hj,x)|kdμ(x)
U j=1 U
∏k
≤ C ||hj||
j=1
This proves the lemma.
Now continue with the proof of the theorem. I need to identify m and Q. It is
assumed μ is Gaussian. Recall this means h^{′} is normally distributed for each h ∈ U. Then
using
is a continuous linear transformation and so by the Riesz representation theorem, there
exists a unique m ∈ U such that
∫
(h,m )U = U (h,x)dμ(x)
Also the above says
(h,m )
is the mean of the random variable x →
(x,h)
so in the
above,
m (h) = (h,m ) .
U
Next it is necessary to find Q. To do this let Q be given by 60.8.34. Thus
∫
(Qh,g) ≡ ((x,h) − (m, h))((x,g)− (m,g))dμ (x)
∫U
= U (x− m, h)(x − m,g)dμ (x)
It is clear Q is linear and the above is a bilinear form (The integral makes sense because
of the assumption that h^{′},g^{′} are normally distributed.) but is it continuous? Does
(Qh, h)
= σ^{2}
(h)
?
First, the above equals
∫ ∫
(x,h)(x− m, g)dμ− (m,h)(x− m, g)dμ(x)
∫ U U
= (x,h)(x− m, g)dμ (60.8.37)
U