final report - probability.ca

Theorem A.10. There exists a unique probability measure µ on (H, B (H)) with mean a, covariance Q and Fourier
transform
ˆµ (h) = exp
{i 〈a, h〉 − 1 }
2 〈Qh, h〉 , h ∈ H. (A.3)
We denote µ by N a,Q , and if a = 0 we simply write N Q .
Proof. We check that the restriction of l 2 of the product measure µ, defined by (A.2) fulfills the necessary properties.
First note that by (A.2) we have ˆ
∣ ∣ x
2 µ (dx) = traceQ + |a| 2 .
(A.4)
H
Assume for simplicity, that Ker (Q) = {0}, and λ 1 ≥ λ 2 ≥ · · · ≥ λ n ≥ · · · . Let (P n ) be the sequence of projectors
defined by (A.1), and let h ∈ H. Since |〈x, h〉| ≤ |x| |h|, and |x| µ (dx) is finite by (A.4), the dominated
´H
convergence theorem thus implies
ˆ
ˆ
〈x, h〉 µ (dx) = lim 〈P n x, h〉 µ (dx) .
H
n→∞
H
However,
ˆ
H
〈P n x, h〉 µ (dx) =
=
=
n∑
ˆ
k=1
H
n∑
ˆ
h k
k=1
x k h k µ (dx)
R
x k N ak ,λ k
(dx k )
n∑
h k a k = 〈P n a, h〉 → 〈a, h〉 ,
k=1
as n → ∞. Therefore, the mean of µ is a. Likewise, in order to determine the covariance of µ, we fix y, z ∈ H and
write
ˆ
ˆ
〈x − a, y〉 〈x − a, z〉 µ (dx) = lim 〈P n (x − a) , y〉 〈P n (x − a) , z〉 µ (dx) .
n→∞
Furthermore,
ˆ
H
H
〈P n (x − a) , y〉 〈P n (x − a) , z〉 µ (dx) =
H
=
=
n∑
ˆ
k=1
H
n∑
ˆ
y k z k
k=1
(x k − a k ) 2 y k z k µ (dx)
R
(x k − a k ) 2 N ak ,λ k
(dx k )
n∑
y k z k λ k = 〈P n Qy, z〉 → 〈Qy, z〉 ,
as n → ∞. Hence, the covariance of µ is indeed Q.
Finally, for any h ∈ H,
ˆ
ˆ
e i〈x,h〉 µ (dx) = lim e i〈Pnx,h〉 µ (dx)
H
n→∞
H
n∏
ˆ
= lim e ix kh k
N ak ,λ k
(dx k )
n→∞
k=1
= lim
n∏
k=1
R
e ia kh k − 1 2 λ kh 2 k
n→∞
k=1
= lim
n→∞ ei〈Pna,h〉 e − 1 2 〈PnQh,h〉
= e i〈a,h〉 e − 1 2 〈Qh,h〉 .
This verifies that the Fourier transform of µ is given by (A.3).
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