63.6.1 Lots Of Independent Normally Distributed Random Variables
You can use the Kolmogorov extension theorem to prove the following corollary. It is
Corollary 58.20.3 on Page 6510.
Corollary 63.6.1 Let H be a real Hilbert space. Then there exist real valued random
for h ∈ H such that each is normally distributed with mean 0 and for
is normally distributed and
is an orthogonal set of vectors of H, then
random variables. Also for any finite set
is normally distributed.
Corollary 63.6.2 The map h → W
is linear. Also,
is a closed
subspace of L2
Proof: This follows from the above description.
a subspace? This is obvious because
is linear. Why is it
closed? Say W
→ f ∈ L2
This requires that
is a Cauchy sequence. Thus
hn → h
and so f
showing that this is indeed a closed subspace.
Next is a technical lemma which will be of considerable use.
Lemma 63.6.3 Let X ≥ 0 and measurable. Also define a finite measure on
Then let f : ℝp → [0,∞) be Borel measurable. Then
where here Y is a given measurable function with values in ℝp. Formally, XdP = dν.
Note that Y is given and X is just some random variable which here has nonnegative
values. Of course similar things will work without this stipulation.
Proof: First say X = XD and replace f
Now let sn
and let sn
is a Borel set.
which is the same thing. Therefore,
Now pass to a limit using the monotone convergence theorem to obtain
Next replace XD with ∑
k=1mdkXDk ≡ sn
a simple function.
It is indexed with n thanks to sn. Then
Clearly νn ≪ ν
and so by
the Radon Nikodym theorem dνn
where hn ↑
1. It follows from the
monotone convergence theorem that one can pass to a limit in the above and
The interest here is to let f
. To remember this, XdP
a sort of sloppy way then the above formula holds.
Lemma 63.6.4 Each eW
is in Lp
for every h ∈ H and for every p ≥
In addition to this,
Proof: It suffices to verify this for all positive integers p. Let p be such an
integer. Note that from the linearity of W,
so it suffices to verify that for each h ∈ H,eW
is in L1
. From Lemma
In using this lemma,
would be 0 because by the construction,
Consider the last claim. It is enough to assume p is an integer.
This converges to 0 for each ω
because it says nothing more than that the nth
term of a
convergent sequence converges to 0.
The second term clearly converges to 0 as n →∞
. Consider the first term. To simplify,
Then this term reduces to