2014 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

for some m and σ2 showing that h◦X is normally distributed.Next suppose h ◦ X is normally distributed whenever h ∈ E ′ and L (X) = µ. Then

letting F be a Borel set in R, I need to verify

µ(h−1 (F)

)=

1√2πσ

∫F

e−|x−m|2

2σ2 dx.

However, this is easy because

µ(h−1 (F)

)= P

(X−1 (h−1 (F)

))= P

((h◦X)−1 (F)

)which is given to equal

1√2πσ

∫F

e−|x−m|2

2σ2 dx

for some m and σ2. This proves the lemma.Here is another important observation. Suppose X is as just described, a random vari-

able having values in E such that L (X) = µ and suppose h1, · · · ,hn are each in E ′. Thenfor scalars, t1, · · · , tn,

t1h1 ◦X + · · ·+ tnhn ◦X

= (t1h1 + · · ·+ tnhn)◦X

and this last is assumed to be normally distributed because (t1h1 + · · ·+ tnhn) ∈ E ′. There-fore, by Theorem 61.6.4

(h1 ◦X , · · · ,hn ◦X)

is distributed as a multivariate normal.Obviously there exist examples of Gaussian measures defined on E, a Banach space.

Here is why. Let ξ be a random variable defined on a probability space, (Ω,F ,P) which isnormally distributed with mean 0 and variance σ2. Then let X (ω)≡ ξ (ω)e where e ∈ E.Then let µ ≡L (X) . For A a Borel set of R and h ∈ E ′,

µ ([h(x) ∈ A]) ≡ P([X (ω) ∈ [x : h(x) ∈ A]])

= P([h◦X ∈ A]) = P([ξ (ω)h(e) ∈ A])

=1

|h(e)|σ√

2π

∫A

e− 1

2|h(e)|2σ2 x2

dx

because h(e)ξ is a random variable which has variance |h(e)|2 σ2 and mean 0. Thus µ isindeed a Gaussian measure. Similarly, one can consider finite sums of the form

n

∑i=1

ξ i (ω)ei

where the ξ i are independent normal random variables having mean 0 for convenience.However, this is a rather trivial case.