( ) ( ) ( )
E (W (αf)− αW (f ))2 = E W (αf)2 + E α2W (f)2 − 2E (W (αf)αW (f))
= α2|f|2 + α2|f|2 − 2α (αf,f) = 0.
Why is
{W (h) : h ∈ H }
a subspace? This is obvious because W is linear. Why is it
closed? Say W
(h )
n
→ f ∈ L^{2}
(Ω)
. This requires that
{h }
n
is a Cauchy sequence. Thus
h_{n}→ h and so
( 2) [ ( 2) ( 2)]
E |f − W (h)| ≤ 2 lni→m∞ E |f − W (hn)| + E |W (hn)− W (h)|
( 2) 2
= 2nli→m∞ E |W (hn)− W (h)| = 2nl→im∞ |hn − h|H = 0
and so f = W
(h)
showing that this is indeed a closed subspace. ■
Next is a technical lemma which will be of considerable use.
Lemma 63.6.3Let X ≥ 0 and measurable. Also define a finite measure onℬ
(ℝp)
∫
ν (B ) ≡ XX (Y )dP
Ω B
Then let f : ℝ^{p}→ [0,∞) be Borel measurable. Then
∫ ∫
f (Y )XdP = f (y )dν(y)
Ω ℝp
where here Y is a given measurable function with values in ℝ^{p}. 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 = X_{D} and replace f
(Y )
with X_{Y−1(B)
}. Then
∫
X X −1 dP = P (D ∩ Y −1(B))
Ω D Y (B)
∫ ∫
ℝp XB (y)dν(y) ≡ ν(B ) ≡ Ω XDXB (Y )dP
∫ ( − 1 )
= XDXY −1(B )dP = P D ∩ Y (B)
Ω
∫ ∫ ∑m ∑m ∫
sn (y )dν(y) = ckXBk (y)dν(y) = ck XBk (y )dν(y)
ℝp ℝpk=1 k=1 ℝp
∑m
= ckP (D ∩Y −1(Bk))
k=1
∫ m ∫ m
s (Y )X dP = ∑ c X X (Y )dP = ∑ c P (D ∩ Y −1(B ))
Ω n D k=1 k Ω D Bk k=1 k k
which is the same thing. Therefore,
∫ ∫
sn(Y )XDdP = sn(y)dν(y)
Ω ℝp
Now pass to a limit using the monotone convergence theorem to obtain
∫ ∫
f (Y )X dP = f (y)dν(y)
Ω D ℝp
Next replace X_{D} with ∑_{k=1}^{m}d_{k}X_{Dk}≡ s_{n}
(ω )
, a simple function.
∫ ∑m m∑ ∫
f (Y ) dkXDk dP = dk f (Y) XDkdP
Ω k=1 k=1 Ω
∑m ∫
= dk ℝp f (Y) dνk
k=1
where ν_{k}
(B)
≡∫_{Ω}X_{Dk}X_{B}
(Y )
dP. Now let
∫ ∑m ∫
νn(B ) ≡ dkXDk XB (Y) = snXB (Y )dP
Ωk=1 Ω
It is indexed with n thanks to s_{n}. Then
m ∫ m
ν (B) = ∑ d X X (Y )dP = ∑ d ν (B)
n k=1 k Ω Dk B k=1 k k
Hence
∫ ∫ ∑m ∑m ∫
f (Y )sndP = f (Y ) dkXDkdP = dk f (y)dνk
Ω Ω k=1 k=1 ℝp
∫ ∑m ∫
= f (y) dkdνk = f (y)dνn
ℝp k=1 ℝp
(sndP = dνn so to speak.)
Then let s_{n}
(ω)
↑ X
(ω)
. Clearly ν_{n}≪ ν and so by
the Radon Nikodym theorem dν_{n} = h_{n}dν where h_{n}↑ 1. It follows from the
monotone convergence theorem that one can pass to a limit in the above and
obtain
∫ ∫
f (Y)XdP = f (y)dν ■
Ω ℝp
The interest here is to let f
(Y )
≡ e^{λ⋅Y} so f
(y)
= e^{λ⋅y}. To remember this, XdP = dν in
a sort of sloppy way then the above formula holds.
Lemma 63.6.4Each e^{W}
(h)
is in L^{p}
(Ω)
for every h ∈ H and for every p ≥ 1. Infact,
∫ ∫
( W (h))p W(ph) 12|ph|2H
Ω e dP = Ω e dP = e .
In addition to this,
∑n W (h )k W (h) p
--k!-- → e in L (Ω, ℱ,P) , p > 1
k=0
Proof:It suffices to verify this for all positive integers p. Let p be such an
integer. Note that from the linearity of W,
( W(h))
e
^{p} = e^{pW(h)
} = e^{W(ph)
} and
so it suffices to verify that for each h ∈ H,e^{W(h)
} is in L^{1}