Download Baker, Charles R.; (1971)Zero-one laws for Gaussian measures on Banach space."

This research was supported in part by the Air Force Office of Scientific Research under Contract AFOSR-68-14l5 and by the Office of Naval Research under Contract N00014-67-A-032l-0006 (NR042-69).
-e
ZERO-oi~E ~!S FOR GAUSS INJ r1EAsURES
ON BANAQi SPACE
Charles R. Baker
Department of Statistics
Uni versi ty of North Cam Una at Chape Z Hi ZZ
Institute of Statistics Mimeo Series No. 785
November, 1971
ZERO-oNE
lAws
FOR GAUSS IAN l''1EAsURES
ON BANACH SPACE
Charles R. Baker
ABSTRACT
B be a real separable Banach space t
Let
Borel a-field of
If
G
B [B]
€
~
B,
and
B [B]
-e
1
for the case where
a Gaussian measure on the
the completion of the Borel a-field under
is a subgroup, we show that
due to Kallianpur and Jain.
~(G) =
~
~
~(G)
=0
or 1,
~.
extending a result
Necessary and sufficient conditions are given for
G is the range of a bounded linear operator.
These results are then applied to obtain a number of
0-1
statements for the
sample function properties of a Gaussian stochastic process.
The zero-one law
is then extended to a class of non-Gaussian measures, and applications are
given to some non-Gaussian stochastic processes.
2
L
INTRODUCTION
Kallianpur [1] and Jain [2] have proved the following result.
a complete separable metric space,
on
T,
and
B[X]
that
P
function
X sets generated by sets of the form
tl, •.. ,t
n
E T
and
C a Borel set in
is a Gaussian probability measure on
K and zero mean, and that
Hilbert space of
K.
Let
BO[x]
B[X]
n
R •
Suppose
with continuous covariance
X contains the reproducing kernel
denote the completion of
B[X]
With these assumptions, Kallianpur and Jain have shown that
for every
T be
X a linear space of real-valued functions
the a-field of
{x: (x(tl), ••• ,x(tn»EC},
Let
BO[x]-measurable subgroup of
under
P(G)
=0
P.
or 1
X.
This result does not immediately apply to the case of a Gaussian measure
on the Borel sets of an arbitrary real separable Banach space, even in the case
of
·e
L,
P
since on
We first show
L
p
the co-ordinate maps
1T
t
'.
X
-+
xt
are not well-defined.
that this zero-one law holds for Gaussian measures
on a Banach space, and without the assumption of zero mean.
Some necessary
and sufficient conditions for the alternatives are also given.
is extended to a class of non-Gaussian measures.
The
0-1
law
A number of applications are
given on path properties of Gaussian stochastic processes and some nonGaussian processes.
As noted in [1], Cameron and Graves [3] first considered this problem for
the case of Wiener measure.
In another direction, Pitcher [4] essentially
proved the following result.
separable Hilbert space
for
Sand
T
H
Let
~
be a zero-mean Gaussian measure on a real
with covariance operator
Hilbert-Schmidt linear operators.
K.
Then
Suppose
~[range(T)]
Our interest in this problem was motivated by Pitcher's work.
result to any bounded operator
law holds for any measurable
T,
K
= TST*
=0
or 1.
We extended his
and conjectured from this that a sioilar 0-1
linear manifold in,, Bilbert space.
'~Je
are indebted
to the appearance of [1] for the key step in completing our proof (Lemma 2,
below), and to [2] for enabling us to extend our result to subgroups.
3
2.
ilEFIN IT! ONS
B will denote a separable Banach space, with norm
Hilbert space with inner product (.,.).
Borel a-field of A sets.
A Gaussian measure
All linear spaces are defined over
IF is the characteristic function of the set F.
~
on B[B] is by definition a probability measure
such that all bounded linear functionals on
B are Gaussian with respect
S* is the space of all bounded linear functionals on
~.
separable
For a given topological space A, B[A] will denote the
the real numbers.
to
11·1 I, H a
B• We will
usually not distinguish between Hand H*•
·e
3.
THE ZERO-(}JE 1A1 FOR BANAa-I SPACE
In this section we obtain a zero-one law for any Gaussian measure
on a real and separable Banach space.
In particular, the measure need
not have zero mean element.
We first obtain the zero-one law for
H.
Let
~l
be a Gaussian
measure on B[H] with null mean element and covariance operator K;
= f (x,u)<x,v)d~l(x),
all u,v in H. K is linear, bounded, nonH
negative, self-adjoint, trace-class, and uniquely determines ~l [5]. Since
(Ku,v)
H is a linear topological space,
A is a Borel set, any z
~(A)
= ~l{x:
{e }, n
n
x + m
= 1,2, •••
€
A}.
€
{x: x + z
€
A}
is a Borel set in H if
H. We define the Gaussian measure
~
~
on B[H] by
has covariance operator K and mean element m.
will denote a complete orthonormal set of eigenvectors
for K; we require only that
I IKen II
# 0 for at least one value of n.
4
\l • Let
We will obtain the zero-one law for
B[H]
\l.
under
For any
v
\l{x: x+Kv EA},
H,
in
and
= exp{<x,v>
Comparing
L2 [\l])·
G E B [H].
\l
define the Gaussian measure
by
\lv
=
\l (A)
v
are mutually absolutely continuous
[d\l /d\l] - l
and have Radon-Nikodym derivative
(=
be the completion of
The following result is well-known.
Lemma 1 [6]:
lv(x)
H,
G is a subgroup of
B [H]
\l
Since
definition of
\1
v
with
given by
v
~I IK~vl 12 }.
-(m,v) -
12v
v
12v'
\lv ~ \l,
lv E L 2 [H,B\l[H],\l]
it is clear that
= B\1[H],
B\1v[H]
= L2 [\1V],
L2 [\1]
{x: x+Kv E A} E B [H]
\1
we obtain that
and from the
A E B [H].
\1
for
Lemma 2:
-e
0, l-l , vEH} is dense in L [\1]·
2
v
Proof: Suppose IHg(x)d\l(x) = I Hg(x)[l-lv(x)]d\1(x) =
for all v in
H, some g in L [\1]· Then IHg(x)exp{(x-m,v)-~1 I~vl ,2}d\1(x) = 0, all
2
vEH. Let v - L~ai e , where the {a. } are scalars. Proceeding as in the
i
1
N
N
proof of Lemma 3 of [1], one obtains E[gIB[H ]] = a.e. d\1, where B[H ] is
°
°
the a-field generated by
f : x + <x,e >,
n
n
smallest a-field such that
E[gIB[H]]
= ° a.e.
d\l,
f
n
= 1, ••. ,N.
is measurable for all
n
and thus
= ° a.e.
g(x)
e,
is the
one obtains
n
d\1(x)
B[H]
Since
(see Appendix 1 for
details) •
Lemma 3:
For any
are scalars such that
\1(F)
Proof:
If
0,
\1 (Fa )
n
>
that
= 1,
all
l
a v
n
n,
FEB [H],
\1
lim
n
\l(F) = 0,
=
IFl a
(x) + 1
n
a
n
= 0.
then
\l(F
(x)d\l(x).
v
as
we see that
a
n
{l
+
a v
F
let
0,
Then
a v
n
a v
n
= {x:
lim
) = \1
n
a v
n
x+a Kv E F},
n
\1(F
a v
Hence
{IF1
a v}
n
{a }
n
= \l(F).
n
(F) = 0,
Noting that
la v
since
\l
~
\la v
If
n
is positive for each
n
for all
xEH,
and that
f.P
H-a
n
v
(x) d\1 (x)
} is uniformly integrable with respect to \l
n
[7, p. 18].
)
where
is uniformly integrable,
5
Ip(x)la v(x)
Ip(x)
+
a
as
n
fHIp(x)la v(x)d~(x)
~(p)
+
+
0,
as
a
n
n
Lemma 4: If
~(G) >
all
+
n
x
H,
in
and this implies
O.
0, then G ~ range(K).
Proof: We apply a proof used to establish Theorem 1 of [2]. Suppose
~(G)
> l/s,
s
o ~ k ~ s,
Ku~G.
a positive integer.
by
Por each
n,
This implies that
(s+l)~(G) ~
= G,
G(n)o
G(n)
I:=o
=0
~(G)
Proof: Suppose
-_
v
(G) =
~{x:
G(n)
j
= 1,2, •••
define
G(n)k'
1 ~ k ~ s,
x+(slkn)-lKuEG },
j # k
are disjoint if
where
(see [2]).
~(G(n)k) ~ 1, all n, and hence by Lemma
3
or 1.
~(G) >
O.
= ~(G),
x+KvEG}
We note that
and
k
= {x:
n
a contradiction.
1,
Lemma 5:
~
G(n)k
Por
From Lemma
and
1 J. {l-l , vEH} ,
Span{l}
@
fHIG(x)[l-lv(x)]d~(x)
L2[~]
so that by Lemma 2
v
decomposition
4, G ~ Kv for all vEH, so that
Span{l-l , vEH}.
v
= 0,
all
v
has the orthogonal
IG(x) = constant a.e.
Hence,
H.
in
d~(x).
Corollary: If ~(G) = 1, then G ~ range(K~).
Proof: Let
~
,
~' v (A) = ~{x: X+K~VEA}.
then
~(G)
=
1~~'
~
Now let
B [B]
~
v
(G)
~'
v
.....
= 0,
~
[6],
so that
B~t U1]
v
defined by
=B
~
[H].
If
K~V~G,
a contradiction.
be a Gaussian measure on
be the completion of
subgroup of
B[H]
be the Gaussian measure on
v
B[B]
under
B[B]
~,
with mean element
m,
GEB [B]
and suppose that
~
let
is a
B.
Theorem 1:
~(G)
=0
or 1.
Proof: We can construct a separable Hilbert space H such that the e1ements of
B[B]
B constitute a dense linear manifold in
= BnB[H];
see [8].
Let
H,
BEB[H],
v be the Gaussian measure on
and
B[H]
defined by
6
v(A) = l.I(BnA).
sets in
B[B]
be the completion of
v
of l.I-measure zero belong to
is clear that
group in
B [H]
Let
B [H]
v
B,
G
then
B [B].
H,
suIt now follows from Lemma 5 and the definition of
v.
1.I
Remarks: 1. Theorem
ment.
G is a sub-
Hence, if
1.I
G is a subgroup in
€
Since all
v.
under
and have v-measure zero, it
B[H]
contains all elements of
B [8],
B[H]
G
€
B [H].
The re-
v
1 holds for Gaussian measures with non-zero mean ele-
This does not appear to be obvious if one has only the zero-one law for
zero-mean Gaussian measures.
2.
If one limits attention to manifolds, there are several ways to prove
Lemma 4.
KV~G.
Inl
·e
+
For example, let
Then
00.
G
n
Also,
and
l.I(Gn )
= {x: xt~v
n
Gn
G are disjoint if
m
= l.I v / n (G) ,
uniformly integrable.
Now suppose
such that
while
l.I(G)
n < 0,
€
G}
n
~
n
m,
and the family
l.I(G)
>
O.
= ±1,±2, •••
so that
l.I(G)
n
{l vn
/}
For any
0
and suppose
+
0
as
is easily seen to be
>
0,
there exists
f G l -v / n (x)dl.l(x) = 1.I-v / n (Gn ) = l.I(G).
n
This
n
contradicts the uniform integrability of
4.
{l vn
/ }.
f'1EAsURABLE SUBGROUPS
The following two lemmas can often be used to show that a given subgroup
is measurable.
Lemma 6 [9]: Let T be a map from a complete separable metric space Ml
into a complete separable metric space
B[Ml]/B[M ]
Z
measurable.
Then
T[A]
€
M2• Suppose T is one-to-one and
B[M Z]
when
A
€
B[M l ].
7
In our applications, we deal with a bounded linear operator
two Banach spaces.
T between
The following lemma extends Lemma 6 (Kuratowski's theorem)
NT
to the case where
(the null space of
T)
contains elements other than the
null element.
Lemma 7:
flexive.
Let
Suppose
T[A] E B[B ]
2
Proof:
B
l
B2 be two separable Banach spaces, with Bl
and
T: B
l
B2 is a bounded linear operator. Then
+
A E B[B ].
if
l
See Appendix 2.
We note that Lemma 7 can be used if
p
Bl
is
lp
or
Lp
(Lebesgue) for
E (1,00).
5.
·e
re-
NECESSARY AND SUFFICIENT CONDITIONS FOR ~(G) = 1
~(G)·
There are many problems in which one will wish to determine if
for some measurable subgroup
ditions for
=1
~(G)
Theorem 2:
Theorem 2 gives necessary and sufficient con-
for a large class of subgroups.
Suppose
a Gaussian measure on
G.
T: H + B is a bounded linear operator.
B[B],
with mean element
m,
If
~
is
then
(a)
~[range(T)]
=
0 or 1.
(b)
~[range(T)]
=
1 if and only if there exists a Gaussian measure v
on B[H] such that
(c)
Let
HI
~[A]
= v{x:
TXEA} for all A in B[B].
be a separable Hilbert space containing
linear manifold and such that
be the Gaussian measure on
AEB[H ].
l
1
Then
~[range(T)]
B E B[H],
B[H ]
l
=1
B[B]
defined by
B as a dense
= BnB[H].
~l
Let
~l[A]. ~[BnA],
if and only if m E range(T) and
there exists a covariance operator
S
in
H
such that
K
l
= TSTl *,
8
where
Kl
~l
is the covariance operator of
restriction of
T*
to
and :1*
is the
HI'
Proof: Part (a) follows from Theorem 1 and Lemma 7. To prove (b), we
first note that if such a measure
~[range(T)]
Thus, suppose
v·
and
{x: x=Tv,
..l.
iJm
Yx = 0
:
J.
v € A n
{x: xfrange(T)} •
hence
Y is
~;
~{X:
Yx J. N }.
v(A)
if
T
Define a map
xfrange(T).
measurable.
=
v(H)
~[range(T)]
range(T*);
Y: B
For
O€A,
y-l(A)
In both cases,
T
If
OfA.
= ~{xIYxe:A}.
any element of
·e
if
N J.r
B[B]/B[H]
from
= 1.
H by
+
then
Yx
=v
y-l(A) = T[A
Y thus induces a measure
=
~{x:Yx
= 1.
€ H} = 1;
To see that
= ~{x:
Thus
v{y: (y,u)<k}
= ~{x:
(Yx,u)<k}
~{x:
lim u (TYx) <k}
nn
= ~{x:
lim u (x)<k},
n n
n NTJ.]
u
on
v
B[H]
is Gaussian, let
B*
c
such that
limn(YX,T*un)<k}
u
T*u
n
+
be
u.
=
range(T) = {x: TYx=x} •
since
= Tv
also,
v
{u }
n
then there exists
x
by Lemma 7, and
B[B],
belongs to
if
=
y-l(A)
AEB[H],
= 1.
~[range(T)]
exists, then obviously
v
As the
a.e. limit of a sequence of Gaussian random variables, the random variable
fu: x
(x,u)
+
bility measure.
~{x:
To see that
~
= ~{x:x e: Anrange(T)} =
x€A}
To prove (c), we note that
if
~1
~[range(T)]
= vaT
-1
v.
is Gaussian with respect to
= 1,
• Let
let
v
Sand y
We first note that
Ty
= m,
v is thus a Gaussian proba-
is induced from
~{x:TYx
v
by
T,
one notes that
€ A} - v{x:Tx € Ale
~[range(T)]
=1
<~ ~l[range(T)]
be the Gaussian measure on
B[H]
= 1.
Thus,
such that
be the covariance operator and mean element of
since
I H (x,u)ld~l(x) = I H(x,T 1*u)dv(x),
all
v.
u
1
in
HI'
where
product on
H.
(0'·)1
is the inner product on
Now, for all
u,v
in
HI'
(K1u,v)1
= I H(x-y,T 1*u)(x-y,T l *V)dv(x) = (TSTl*u,v)l.
if
K
1
define
= TST 1*,
v
S
a covariance operator in
to be the Gaussian measure on
and mean element
y.
The map
T: H + HI
HI and (0,0) is the inner
B[H]
Hence,
H,
= I H (x-m,u)l(x-m,v)ld~l(x)
1
= TST l *.
Kl
and m = Ty
for
Conversely,
y€H,
with covariance operator
induces from
v
we
S
a Gaussian measure
9
having covariance operator TST1 * and mean element m;
on
~
Gaussian measure on
HI
since a
is uniquely specified by its covariance operator and
mean element, this: measure must be
Some applications of Theorem 2 are given in the next section.
it may be of interest to note here the following.
measure on
that
B[H],
with covariance operator
~{x:JC+m € range(K~)} ..
arbi trary Gaussian
example, if
~
lJ,
for all
0
This property is not shared, for
m€H.
~l{x:x-~€ range(T)}
.. 1,
where
then
~l
is the translate of
~
and T: If + H
~ [range (T)]
€
is bounded and linear, then
range(T)}
=0
for all
m€H.
this holds for all subgroups in
For
1
for
lJl[range(T)] .. 0,
However, it is clear from Theorem 2 that if
~{x:X+m
lJ (G) .. O.
~[range(T)]:II
is a zero-mean Gaussian measure such that
H,
is a Gaussian
Part (c) of Theorem 2 shows
by all subgroups G € B[ H] such that
T a bounded and linear map in
·e
K.
~
Suppose
However,
~
by
m(range(T) •
m,
is a zero-mean Gaussian measure
.. 0
<=0;>
It would be interesting to know whether
B[H].
If so, the method used to prove
Theorem 1 shows that a similar result would hold for any subgroup in
B[B].
Part (c) of Theorem 2 is an extension of a result by Pitcher [4], who
proved the following.
~
Suppose
with covariance operator K.
is a zero-mean Gaussian measure on
Suppose K
= TST,
where
T and
S
are linear,
bounded, non-negative, self-adjoint, and Hilbert-Schmidt operators in
T strictly positive.
and only if
that
K
Then
~[range(T)]
S is trace-class.
= TST,
that
range(T)
= 0 or 1,
and
B[H],
~[range(T)]"
H, with
1
if
His method of proof requires the assumptions
= range(S) = H~
and that
~
is self-adjoint
La'J.d compact.
The following theorem generalizes (a) and (b) of Theorem 2 to two Banach
spaces.
10
Theorem 3:
~
T: Bl
that
Suppose
B
2
+
T
on
B[B l ]
Proof:
to-one.
Then
~[range(T)]
Moreover,
v
is one-to-one and
B[B ].
2
Gaussian measure on
sure
~
is a bounded linear operator.
reflexive, or else
or 1.
such that
Suppose that either
= Bl *.
range(T*)
range(T)
B[B ],
2
E
~
there exists a probability measure
B
3
T
= Bl *
range(T*)
v
;nduces
•
Since
on
~
T
-1
and
B[B ]
l
f rom
Define
such that
B ;
range(T)
exists ([10], p. 44).
belonging to
range(T*);
easy to see that
v{x: y(x) < k}
y(x)
tion that
y
say
y
= Tl*v l , VI
= Il{x:
= T*v.
If
is Gaussian with respect to
v,
has a norm-
Since
be any element of
B1 *
is not the null element, it is
B : y(Tl
3
E
vI
all
v
-1
y
to
x) < k}
B •
3
Now
= Il l {x:
range(T*).
E
vl(x) < k}.
The assump-
B1* thus implies that v is a Gaussian measure.
For the case where
suppose that
y
y
the restriction of
yeT-Ix) < k} = Il {x
l
range(T*) =
Let
and the
be the restriction of
~l
is thus a Gaussian measure.
2
as defined
i s Gauss~an.
.
v
then by the Hahn-Banach theorem
Hence,
v
is one-
v(A) =
We s h ow
v.
Let
preserving extension to
T
exists, it is obvious that
denote the separable Banach space consisting of
norm.
=0
is induced from v by T.
= 1.
11 [range (T)]
and that
is
be a
~[range(T)]
and
It is clear that we need only prove the existence of
"{X·,
T-lx E A},
~
~
Let
B
l
= 1 if and only if there exists a Gaussian mea-
First consider the case where
in the theorem when
Let
is a real separable Banach space, and
B. (i=1,2)
Bl
ll[range(T)]
is reflexive, with
= 1.
T not necessarily one-to-one,
The proof of Lemma 7 (Appendix 2) shows that
one can imbed
B
as a dense Borel-measurable linear manifold in a separable
Hilbert space
H.
in such a way that
and
T
i
=v
B[B ]
i
C
B[H ],
i
B[B ]
i
= Bi
n B[H ],
can be extended by continuity to a bounded linear operator
It is clear that
Yx
~
if
range(T)
x = T v
1
and
= Tl[B l ]
v.L
NT ' while Yx =
1
.1
NT
1
0
]. Define Y: H2
if
x
~
i
T l :H l +H 2 •
+
range(T ).
l
HI by
Define
11
e
~
on B[B 2 ] by
~
(A)
= ~(B2nA).
The procedure used to prove (b) of Theorem 2
shows that there exists a Gaussian measure v
on B[H ] such that
l
,
(A)
~
=
A} for A in B[B ]. v(B ) = v{x: Tlx €B } = ~ (B ) = 1. Let va
Z
2
2
l
* then u(x) is a
be the restriction of v to B[B ]. If u € B * belongs to HI'
l
l
v{x: Tlx
€
Gaussian random variable with respect to vo' since v
Hence va
is Gaussian.
provided H * is dense in 3 *. Suppose there exists f in 8 **
1
l
1
such that f(v) = 0 for all v in H *. Since 3 is dense in HI and 8 is
1
1
l
is Gaussian
** ~ **
**
1 ' 0 1 is dense in HI and hence f must be
the null element of Bl ** • Thus H1* is dense in 8 *' so that va is Gaussian,
1
isometrically isomorphic to B
with
~(A)
= vO{x:
Tx
€
A}
for A
€
B[B Z].
The remainder of the proof is obvious.
Theorems 2 and 3 extend the following well-known result:
measure on B[B ], and T: 3
1
-e
1
~
If
~
is a Gaussian
B is bounded and linear, then the measure on
2
B[B 2 ] induced from ~ by T is also Gaussian.
Thus, if B is reflexive, and T- l
l
exists, then Theorem 3 shows that for a Gaussian measure
~
on B[B 2 ],
~[domain(T-l)] = a or 1; if ~[domain(T-l)] = 1, then the measure induced from
~
by T-1 is Gaussian.
if and only if
by T.
If T-1 does not exist, we still have that
~[range
there exists a Gaussian measure on B[B ] from which
I
~
( T)]
=1
is induced
12
o. PfPLICATIONS TO GAUSSIAN rEASURES AND GAuSSIAN STOCHASTIC PROCESSES
In this section we apply the preceding theorems to obtain a number
of zero-one statements for Gaussian measures and Gaussian stochastic
processes.
Statements regarding analytical properties of sample functions
are contained in Theorems 4, 7, and 8.
Theorems 5 and 6 deal with the
convergence of certain series, and with integrability properties of the
sample functions of measurable Gaussian stochastic processes.
In addition to the results on Gaussian measures and processes
presented in this section, Theorem 10 of Section 7 contains results on
the convergence of sequences and series of Gaussian random variables.
We use L or L [T] to denote the set of equivalence classes of all
p
-e
p
real-valued Lebesgue-measurable functions f on T (an interval) such
that
fTlftlPdt < ~, together with the n~rm
I If I Ip =
-/
[fTlftIPdt]~ p.
We do not distinguish between a function and its equivalence class.
l
p
x = (x l ,x 2 ' ••• )
together with the norm IIxl Ip = [L n Ixn IP]~/p.
denotes the set of all sequences of real numbers
such that
L
n
Ixn IP
<
~,
We use either C or CrT] to denote the space of real-valued functions
defined and continuous on the compact interval T, together with the
norm
I If II = sup
t
E T
Iftl.
13
Theorem 4.
~
C[T].
Suppose
= {x:
ACP[T]
any fixed
(2)
(a. ,x) ,
a.
[a,b].
denote the real separable Banach space with elements
P
C[T].
= ACP[T],
(2)
Cn[T]
I Ixl I = I~=o
+
in
x
C[T]
P
be defined by
Noting that
P
SUPt€T
IX t (i)l.
> 1
s-1 f =
=
where
x(s)ds,
range(S)
is a linear manifold and that
=0
or 1.
The natural injection of
and
C[T],
)J(AC P)
(f(a),f(l»
Thus,
Cn[T]
Cn[T]
into
C[T]
is
is a Borel-measurable
and the result follows.
= 1,
differentiation induces from
that
= a+f at
t
II(a,x)11
is a real separable Banach space under the norm
linear manifold in
If
S(a.,x)
range(S)
~(ACP[T])
one sees that
a bounded, linear, and one-to-one map.
.tit
and with norm defined by
L [T],
is a one-to-one bounded linear map, so that by Lemma 6
S
is a Borel set in
range(S)
x is n-times continuously differentiable on T}.
S: Q
Let
zero or one:
derivative},
p
a real scalar,
1a.1+llxll·
p
~-measure
and have
x is absolutely continuous with L
Let
(1)
B[C[T]]
p€[l,oo)
= {x:
Cn[T]
Proof:
=
is a Gaussian measure on the Borel a-field of
The following sets belong to
(1)
T
~
for
induces a Gaussian measure on
then Theorem 3 shows that the operation of
a Gaussian measure on
~
f
€
AC P ,
B[Qp]'
S
L
p
.
For this, one notes
as in the proof of Theorem 4.
and projection into
L
p
This
gives a Gaussian
(1)
€ A} for A€ B[L].
B[L ], \) (A) = ].l{x € AC P : x
p
P
The following theorems deal with various properties of the sample functions
measure
\)
on
of Gaussian stochastic processes.
real-valued.
Lemma 8:
In each case the process is assumed
We first establish a lemma
Suppose
(X t ) ,
t€T
essentially used by Pitcher [4].
(a compact interval), is a measurable
Gaussian stochastic process on a probability space
(n,B,p).
If almost all
14
sample paths of
(X )
t
Gaussian measure
belong to
on
~
is the sample path of
evaluated at
e.
-+
L
{y:
duces a probability measure
~
L ,
S/B[L]
on
every bounded linear functional on
X(w)
>
belongs to
form a neighborhood base
0,
for
A€B[L].
Hence,
P
X in-
By a result of Doob [11, pp.64-65].
p
p
where
X(w)€A},
= X-l{x: I Ixl Ip<E}
£
B[L l.
L
induces a
t
measurable, one notes that
p
X-l[A]€e
P
(X)
w.
I Iyl Ip <£},
at zero for the norm topology in
then
= p{w:
~(A)
{w: [!T1Xt(w)IPdt]1/p<E}
Since sets of the form
that
is
p
P < 00,
S
defined by
p
(X )
t
1
p
B[L],
Proof: To see that X: n
by Fubini's theorem
L [T],
so
is Gaussian with respect to
p is a Gaussian measure.
The problems treated in Theorem 5 have a long history (see ell, and the
references given there).
The proof is remarkably simple, given Theorem 1.
Theorem 5: Suppose (X t ) , t€T
(a compact interval), is a measurable
Gaussian stochastic process on a
covariance function
(X ) ,
t
sample path of
(1)
(2)
K(t,s)
probability space
and mean element
evaluated at
w.
met).
Let
(n,e,p),
with
denote the
X(w)
Then
P{w: X(w)€L } = 0 or 1, any fixed p ~L
p
Let {en } denote a set of functions continuous on T,
with
{f }
n
p{w: X(w)€L } = 1, then
Z
p{w: Ien (t)!Tfn (s)X s (w)ds converges uniformly} = 0 or 1.
Proof: As shown in [12], one can define a function g€L l such that
2
gX(w)€L a.e. dP(w); e.g., get) = 1 if K(t,t)+m (t) S 1, and get) =
l
2
-1
2
[K(t, t)+m (t)]
i f K(t,t)+m (t) > 1. Let (Y t) = (gtXt); (Y ) is a meast
urable Gaussian process with almost all sample paths in L • Thus Y induces
l
from P a Gaussian measure ~ on B[L ]. Now let G: Ll -+ Ll be defined
l
any set of functions belonging to
L •
2
If
IN
by
Gv
= gv,
{w: X(w)€L }
l
g
as above.
= {w:
G is bounded, linear, and one-to-one.
Y(w)€range(G)}.
Hence,
{w: X(w)€L } € e,
l
and
Clearly,
15
p{w:
X(w)€L l } = ~y{range(G)} = 0 or 1.
For
and that the natural injection of
p > 1,
L
P
one uses the fact that
into
is a bounded,
linear, and one-to-one operator.
The set of
(2)
A=
w for which the series converges uniformly is
~~1 U~l nm>n>M nt€S {w: IL~=n+l ek(t) IT fk(s) Xs(w)dsl
is a countable dense subset of
T.
But
A = X- 1 [D],
1
< -}
N '
where
s
where
D = nun
n
{x: IL~=n+l ek(t) IT fk(s) xsdsl < .!}
N~l M~l m>n>M
N • D is obt€S
viously in B[L ], since {x:
IIk€I Yk IT fk(s)xsdsl < .!}
2
M € B[L 2 ] for any finite
index set I and scalars {Y }. D is also a linear manifold. Hence,
k
P[A] = ~X[D]
=0
~X
or 1,
the measure induced from
P by
X.
This completes
the proof.
Part (1) of Theorem 5, for zero-mean (X ), has been previously
t
announced by B. Rajput. Part (2) contains the orthogonal series
.
~
(Karhunen-Loeve) representation of (X
t
)
as a special case.
It is shown
in [13,14] that the orthogonal series converges uniformly with probability
one when (X
t
)
has continuous paths.
The following theorem generalizes Part (2) of Theorem 5, and
gives the analogue of Part (1) for l
Theorem 6:
p
spaces.
Let {Z }, n = 1,2, ••• be a family of jointly Gaussian
n
random variables on a probability space (n,8,p).
Let {e } be a set of
n
functions defined and continuous on the interval T.
1 ::; p <
Define, for
00,
A = {w: IN1 Z (w)e (t) converges uniformly on T},
n
n
B = {w: INl[z n (w)]Pen (t) converges absolutely and uniformly on T},
P
C
P
= {w: I~lzn(w)IP
< oo}.
Then A, B , and C each belongs to 8, each has probability one or zero,
p
p
and P[C ] = 1 if and only if
p
I~[E(zn2(w»]P/2 <
00
•
16
Proof:
g
Let
n
.. n- 1 if E Z; 2 (w)
n
:5
and
1,
g .. n- 1[E Z 2(w)]-1
n
.
n
then Y(w) = (Y (w),Y (w), ••• ) € i a.e.
Y .. g z·
l
2
2
n
n n'
dP(w). Let n = {w: Y(w)€i }. Y: Q + i
induces a probability measure ~Y
l
2
l
2
on B[i 2 ]; ~Y is Gaussian, since if LI~12 < 00, L~Yk is a Gaussian random variable as the a.e. limit of a sequence of Gaussian random variables. If
\' n
-2' {I x 12 :5 sup I x 12 L\' n -2: :5 2 II x 11 2 • Hence we
Ln IxIP<oo,
p ~ 1, then L
n
n
n
"Il'm
n
p
by (G x) ..
define the bounded 9 linear, and one-to-one operator G : ip + i
2
p n
p
gnxn for x€ip. Define the continuous function en' by e~'(t) .. g n -Pen (t) '
otherwise.
Let
and define
e"
n
l
A .. y- {x€i 2 :
l
Bp = y- {x€i 2 :
by
e "(t) .. g -1 e (t).
n
n
n
L~ xn en"(t) converges uniformly on T},
L~ Ixnl P en:(t) converges absolutely and cniformly on T}, and
Cp" y-l[range(G )].
Using the method of proof for part (b) of Theorem 5, one
p
sees easily that
-e
'/a + bl P
:5
Note that
A, is in Band P(A) .. 0 or 1. - Similarly, using the inequality
2P (la1 P + fbI P ), P(B )
p
il,nd that P (C ) = 0 or 1.
p
=0
or 1.
It is clear that C belongs toB.
p
Finally, P.(C ) .. 1 if and only 1'f the paths of (Zn)
p
induce a Gaussian measure on B[i ]. the results of [15, 16] show that this occurs
p
if and only if
~ [EZ 2(w)]p/2
n
n
<
00
•
For the next result, we make the following definitions.
set of functions on
cn eq [T]
and
T
that are equal a.e. dt
AC P [T]
eq
are defined similarly.
Theorem 7: Suppose (X t ), t€T
to an element of
(1)
(2)
(3)
B and have
{w: X(w)€C [T]}
eq
n
{w: X(w)€C
[TD
eq
{w: X(w)€AC P [TD.
eq
is the
CrT];
(a compact interval), is a measurable
Gaussian stochastic process on a complete probability space
following sets belong to
Ceq [T]
(n,B,p).
probability zero or one:
The
17
Proof: It is clear that Ceq [T], Cn eq [T]
Ll[T].
From Theorem 5, either almost all paths of
or almost all do not belong to Ll[T].
l
then X- [A]€8 for all subsets A of
AC P [T]
eq
and
(X )
t
are subsets of
belong to
Ll[T],
If almost all paths lie outside
Ll ,
L , since (n,8,p) is complete.
l
Thus, to prove the theorem, it is sufficient to show that the sets indicated
in (1)-(3) of the theorem are elements of
one or zero, when almost all paths of
Thus, suppose that
p{w: X(w)€L }
l
Gaussian measure induced on
be the map taking
fTlxtldt
Thus
S
x
in
T SUPt€T Ixtl,
range(W) € B[L l ]
range(W)
=C
~
[T],
B[L ]
l
by
8,
and that each set has probability
(X )
t
belong to
= 1,
and let
~X(A)
(X );
t
Ll[T].
~X
be the
= p{w:
X(w)€A}.
C into its equivalence class in
W is bounded and linear;
X-l(C
~
)€8
W
Since
W is also one-to-one.
~X[range(W)]
by Lemma 6, and hence
this shows that
L •
l
Let
and that
=0
or 1.
P[X-l(C
~
)]
Since
=0
or 1.
To prove the remainder of the theorem, we note that in the proof of Theorem
.
~
4 both Cn[T]
and
Defining
+
AC Peq
W: C
= W[ACP ] ,
L
l
ACP[T]
were shown to be Borel linear manifolds in
as above, noting that
Cn
eq
= W[Cn]
CrT].
and that
the remainder of the proof is clear.
The preceding theorems contain various zero-one laws for Gaussian processes.
The problem of determining necessary and sufficient conditions for
the alternatives is usually more difficult.
rection are given in Theorem 2.
Some general results in this di-
The following theorem is an application of
Theorem 1 an1 Theorem 2.
Theorem 8: Suppose that (X t ) , t€(-oo,oo), is a measurable mean-square
continuous Gaussian process on a complete probability space (n,8,p). Suppose that
(X ) is stationary with rational spectral density function
t
mean. Let
f
and with zero
18
A
= {w:
~
X(w) is absolutely continuous on (-00,00),
with derivative belong:lng
to L2 [T] for every compact interval T}
A - {w: X(w) is equal a.e. (Lebesgue) to an element of A}.
eq
Then,
and
P(X-l[A ])
eq
or 1.
x-leA
(2)
(X ) is separable, then
t
l
X- [A]€8, and P(X-l[A]) = 0 or 1.
(3)
P(X-l[Aeq ]) ... 1
eq
]€s,
=0
(1)
If
separable~.
2 f(A)dA < 00
foo-A0
0'
if and only if
If
is
then this condition is also necessary and sufficient for
P(X-l[A]) ... 1.
Proof: A function is absolutely continuous on (-00,00) if and only if it
is absolutely continuous on every compact interval.
X-leA ]€8
eq
·e
Let
The fact that
and has probability zero or one thus follows from Theorem 7.
T be any fixed compact interval.
gral operators in
L [T]
having kernels
2
R(t,s)
J~oo
...
Let
and
R.r
and
R(t,s)
f(A) exp(iA(t-s»
RO(t,s),
defined by
dA,
2 -1
-00 (A +1)
exp(iA(t-s»
...
be the inte-
ROT
OO
f
dA •
Using a result of Hajek [17, Sec. 7], one can show that
W trace-class if and only if
the range of
(R~OT)
f:00 (A 2+l)f(A)dA
consists of all elements of
dt to an absolutely continuous function with
concludes that
P(X-l[A ]) ... 1
eq
(n,8,p)
c
A,
In [18]
L [T]
2
L [T]
it is shown that
that are equal a.e.
derivative.
2
if and only if
To prove the statements regarding
x-leA]
00.
<
for
foo-00
A2f(A)dA
<
Hence, one
00.
we first note that
f:
2
]. Hence, if
= 00, then X-1 [A]€8, since
oo A f(A)dA
eq
:I.s complete, and P(X-l[A]) = o. To complete the proof, it is suf-
x-leA
ficient to show that
f:
oo
2
A f(A)dA < 00
and
(X )
t
separable imply that almost
19
e
(X )
t
all paths of
are absolutely continuous.
This has been proved by Doob
[11, pp. 535-537].
As
noted in the proof of the theorem, this result is an extension in one
(X ) has absolutely contint
F the spectral distribution function of
direction of a result of Doob, which states that
uous paths if
(X t ) ,
and
f:
(X t )
oo
2
A dF(A)
<
00,
How~ver,
separable [11, pp. 535-537].
the result given here
(X ) has a spectral density, and this
t
spectral density must be rational. The requirement that (X ) be Gaussian is
t
is restricted to the case where
not assumed by Doob; the process need only be separable, measurable, meansquare continuous, and wide-sense stationary.
Under these same assumptions,
with the process also assumed to have a spectral density function, which is
rational, the proof of Theorem 7 can be used to show that almost all sample
-e
path derivatives belong to L [T] for all compact intervals
2
co
2
f -00 A f(A)dA < 00; i.e., normality is not required.
Theorem 8 is proved for zero-mean Gaussian processes.
mean function
m,
A
eq
(or
to have probability one is that
each compact interval
T,
sufficient condition for
oo
1~(A)12 A2dA
<
00
where
~
[19],
A,
oo
J
-00
if separability of
2
A f(A)dA
ffir(t)
to be in
where
<
= met),
and
00
t€T,
1.range(R~OT)
(X t )
is assumed)
~
ffir€range(R OT)
for
= 0,
A
~(t)
t4T.
is that
is the fourier transform of
We discuss an application to information theory.
(Nt)' t€T
If the process has
then application of Theorem 2 shows that the necessary and
sufficient condition for
J:
T whenever
Suppose that
ffir'
(St)'
(compact interval), are measurable zero-mean Gaussian stochastic
processes on a probability space
(Q,e,p).
We assume that
(St)
and
(Nt)
are statistically independent, and that almost all sample paths belong to
for each process.
Let
denote the Gaussian measures
respectively; e.g.
~S(A)
=
20
p{w: S(w)eA},
~
the sample path of
S(w)
be the Gaussian measure induced on
mutual information (AMI) of
(St)
evaluated at
B[L2xL2]
and
by
(St+Nt)
(St,St+Nt).
if
Let
llS,S+N
The average
is defined as [20]
=
AMI (S,S+N)
w.
dllS,S+N (x,y)
and equal to
+00
otherwise.
This quantity is of much
interest in many Communication Theory problems, Where, it represents the average
information about the "signal"
noise"
(St+Nt).
AMI(S,S+N)
~
<
00
·e
obtained by observing "signal plus
Hajek [17] has shown that for the case considered here,
<~ KS = ~~G~~ for a trace-class operator G, where KS and
denote the covariance operators of
2 above, one sees that
of
(St)'
(St)
AMI(S,S+N) <
belong to the range of
00
(St)
and
(Nt).
From this and Theorem
if and only if almost all sample paths
~~. ?his result has been previously
obtained by Pitcher [4].
7
I
ZERO-oNE LAWS FOR NON-GAUSS I N..J MEASURES
Our results so far involve only Gaussian measures.
Gaussian measure on
that
o or
II
and
ll'
B[B],
and
II v
However, if
is any probability measure on
are mutually absolutely continuous, then obviously
1 for any subgroup
G belonging to
B [B].
II
II
B[B]
is a
such
ll'(G)
=
In this section, we consider
21
a class of non-Gaussian measures for which this holds.
~
on the following lemma.
Lemma 9 [21]: Suppose
~x @ ~
Let
{(x,y): x+y€A}
~X+y
measure
v
in
B,
~2
and
~x
belongs to
a meas ure
is absolutely
are two probability measures on
~y
(BxB, B[B]xB[B]).
denote product measure on
B[B]xB[B]
~X+y(A)
on B[B] by
~l
means that
~2'
continuous with respect to
B[B].
~l«
The notation
Our results are based
for all
A€B[B].
= ~X0~y{(X,y):x+y
Define a probability
€A},
and for each fixed
~X+v (A) = llX{x: x+v EO A}.
llX+v by
The set
We then have the
following results:
(1)
If
llX+Y «
(2)
If
llX «
(3)
If
llX
~
a.e. dlly(y),
then
~X+y «
llX+Y a.e. dlly(Y),
then
llX «
~x
d~(y),
llX+Y a.e.
then
We thus consider Gaussian measures
defined as in the lemma, where
~y
case where
to imbed
space
H,
K,
then
~
llX+Y
~X+Y'
and the class of measures
need not be Gaussian.
quence of this lemma is the following.
has a covariance operator
llX
llX
llX'
If
llX+Y
An obvious conse-
B is a Hilbert space, so that
~y[range(K~)] = 1. For the
if
~X+y '" ~X
llX
B is not a Hilbert space, one can use the methods of Kuelbs [8]
B as a dense measurable linear manifold in a separable Hilbert
and work with the covariance operator of the extension of
~X.
There are thus obvious extensions of the results of the preceding section.
Two
such extensions are given below.
n = 1,2, •••
Theorem 9: Suppose {Zn}' n = 1,2, ••• and {V},
n
are
two families of real random variables defined on a probability space
(Q,S,P).
{V}.
n
Let
Suppose that
K(n,m)
{Zn}
is Gaussian, and that
denote the covariance of
denote the correlation of
V
n
and
V.
m
Let
Z
{Zn}
and
n
{e}
n
functions. defined and continuous on the interval T.
is independent of
Z,
m
and let
R(n,m)
be any set of real-valued
Define
0
e
A'
B'
C'
l.~
= {w:
p
p
(z
n (w)+Vn (w»
= {w: l.~ (Z n (w)+Vn (w»p
= {w: l.~ Izn (w)+Vn (w)I P
where
converges absolutely and uniformly on T}
e n (t)
'
< oo},
is a real scalar,
p
converges uniformly on T},
e n (t)
p
~
1.
g = (gl,gZ' ••• ) be a sequence of real numbers such tha t gi > 0, all i,
,00
2
2
and li=l gi E[Zi (w)] < 00. Suppose that one of the following conditions is
Let
satisfied:
(1)
(l.i giVi(w)xi)z
k,
(Z)
x€iZ'
all
for a fixed random variable
k(w) li;j x.x.K(i,j)
1.' J
a.e. dP(w);
For some matrix operator
R' = K"~SK'~
1
l'
by
'-
~
(3)
l.
KZ(i,j)
i~j K(i,i)K(j,j)
~
Kii
such that
where the matrix operators
= gigjR(i,j)
R'(i,j)
(a)
S in iZ
0,
~
Kjj
<
0)
and
Kl(i,j)
R(i,i)
and
~
and K
l
R'
<
00,
are defined
= gigjK(i,j).
such that
i,j
(summation over
1
l.i Sii
for all
kK(i,i)
i
and some
finite scalar k,
(b)
V i 2 (w)
,
li K( i,i ) <
00
a.e. dP(w).
A', B'p' and C'p
Then,
zero, and
P(C'p) = 1
Proof: Let
(gi)
induces from
P
each has probability one or
'r
liLE(Zn 2 (w)] P/2
if and only if
<
.00.
be defined as in the theorem, and set
Then Y(w)
gnZn(w).
a,
all belong to
= (Y l (w),Y 2 (w), •.• )
~Y
a Gaussian measure
Yn(w) =
is in iZ a.e. dP(w),
B[iZ].
on
~Y
and hence
has covariance oper-
condition implies
= gigjK(i,j). Consider condition (1) of the theorem. This
that V'(w) = (glVl(w),g2VZ(w), .•• ) belongs to iZ a.e.
dP(w) ,
V'
ator
Ky,
Ky(i,j)
and thus
induces from
P
a probability measure
~'
on
Moreover, condition (1) is a necessary and sufficient condition that
range(Ky~)
duced on
for almost all
B[iZ]
by
Y+V'.
w
[~].
Since
Let
~l
k
B[iZ].
V' (w)
€
be the probability measure in-
~'[range(K2y)]
= 1,
~l ~~,
by Lemma 9.
23
A', B'
The assertions of the theorem regarding
Theorem 6.
For example, defining
P(A' )
e "(t)
n
= gn-1
C'
and
p
now follow from
p
e (t),
n
=
()
'IN
~l { X€~2:
l 1 x~ e~(t)
converges uniformly on T}
=
1
0
~y { x€~z:
l'IN
converges uniformly on T } ,
because the latter quantity is
e~(t)
x
n
0
or
1,
~l ~ ~.
by Theorem 6, and
Condition (2) implies condition (1) [18].
KyZ(i,j)
Condition (3a) of the theorem is equivalent to
2
and
~
gi R(i,i)
kKy(i,i),
2
gi R(i,i) ~ kKy(i,i)
all
i,
some finite
k.
The assumption that
V' (w)€lZ a.e. dP(w),
implies that
V'i(w)
Kuelbs has shown [8] that the first part of (3a) implies that
~'y
are mutually absolutely continuous, where
B[lZ]
trix
Vi 2 (w)
I i ---'-'---,K(i,i)
for almost all
w,
~'y
is the Gaussian measure on
but with diagonal covariance ma-
R' ,
the range of
First,
V'(w)
V'(w)
To see this, we note that if
12) eigenvectors of R' ,
and
hi '
eignevectors corresponding to the zero eigenvalues of
R' ,
In
Hence,
I k j hi' hi' Vk'(w)Vj'(w)dP(w)
k
j
V'(W)~hi'}
all
h.'.
~
.
= 1,
each
hi"
Noting that
=0
for each
so that
hi'.
range(R').
{w: V' (w)·frange(R')}
is in the null space
Suppose
N of
Q
Q,
Qii
= O.
and
are
{hi}
are those
then
p{w: V'(w) is not orthogonal to hi'}
thogonal to hi'}, we have that P{w:V'(w)€range(R')}
:::>
belongs to
belongs almost surely to the closure of
= gigjR(i,j).
R'ij
the complete orthonormal (in
,
[21] •
We now show that if this sum is finite
and if condition (3a) is satisfied, then
k
r,ange(Q)
and
[V'i(w)]2
= Ii ---~~
Ky(i,i) •
range(Q2) a.e. dP(w).
= 0,
~y
~ is that Q~ and K~Y have the same range space
and
Note that
p{w:
= giVi(w).
A necessary condition for mutual absolute continuity
Q,
~' y
of
~Y'
having the same mean element as
< 1
Ii;j Ky(i,i)Ky(j,j)
= Uh
,{w: V' (w) is not or-
=
Next we show that
i
1.
Then the element
Ik,j e
ij
eikR'(j,k)
= <Sij'
since
24
R'(i,i)
e
~
space of
R' •
i,
and
= Ky(i,i).
Qii
Hence
range(R') c range(Q).
e
such that
i
Hence,
e
is in
i
v' (w)Erange(Q)
(= range(~»
for almost
,l;
then xErange(Q) if
Ii
all
w.
We now note that if xEl
and x Erange(Q),
Z
x 2
and only if I -!- <~. Hence, the assumption that
QU
i
i,
,l;
E range(Q ) a.e. dP(w).
Ky~i,i) < ~ a.e. dP(w)
Ii
imply that
\'
KZ(i,j)
< 1
li,j K(i,i)K(j,j)
The assumption that
= range(Ky,l;).
range(Q,l;)
R(i,i) ~ k!<y(i,i),
all
[V' (w)]2
and the assumption that
that
is in the null
ei
N span N , so that
Q
Q
Q is contained in the null space of R' • This implies
The elements
the null space of
that
all
k!<y(i,i),
V'(w)
implies
Thus, the assumptions of (3) imply that
V'(w)
€
range(Ky~) a.e. dP(w). Hence, assumption (3) implies assumption (1), and the
proof is
co~pleted.
Remark: It is clear that
Corollary: Suppose (X t ) and (Y t ) ,
.~
tET
R(i,i) < ~
\'
if
li K(i,i)
(a bounded interval), are
(n,S,p).
measurable stochastic processes on the complete probability space
Suppose that
(X )
t
sample paths in
and that
(Y )
t
(Y )
t
that
and
L [T],
2
are independent, that both have almost all
(X )
t
is Gaussian with covariance operator
has covariance operator
R
and mean element
K,
set of functions defined and continuous on
{y}
T.
two sets of real scalars, and define, for any
Bp "
= {w:
Let
p
m.
and let
be a set of complete orthonormal eigenfunctions for
A"
•
n
and
Let
K,
{e }
n
{f}
n
{A}
be any
be any
n
€ [l,~),
IN1 A [(X(w)+Y(w),e \j+Y ] f (t) converges uniformly on T},
n
n
n
n
= {w: I~
on T},
A:«X(W)+Y(w),en)+yn)p fn(t) converges absolutely and uniformly
and
cp " = {w: I~l
IA
where (.,.)
denotes the
n
I PI(x(w)+Y(w),e n )+Y n IP
for all
n
<
~},
L inner product.
Z
and some finite scalar
(Re ,e )+(m,en )2 ~
n n
<Y(w),e n>2
k, and if I n <Ke ,e > < ~
If
n
a.e. dP(w),
then
A", B"
p
and
C"
p
each belongs to
S,
n
each has probability
25
one or zero, and
p[e "]
p
=1
if and only if
Proof: We note that (Ken,e m> = 0 if n f m;
condition (3) of Theorem
7 is thus satisfied for the random variables
<Y(w),e >2
\ _
n
Ln <Ke ,e > < 00 a.e. dP and '(Ren,en ) ~
n n
finite), are given in [ 21], where a number of examples
Sufficient conditions for
k ( Ken,e n ),
all
n
(k
are also presented.
For instance, if
tegra1 operator with kernel either
Y(w)
on T
belongs to
k
range (K 2 )
T
=
[O,b],
exp(-alt-sl),
if and only if
Y(w)
to an absolutely continuous function with
b <
00,
and
a
0,
or
>
K is the inb-It-sl,
then
is equal a.e. (Lebesgue)
L [T]
2
derivative.
We also
note that the corollary has straightforward extensions to processes whose
sample paths belong (a.s.) to
L
p
for any
p~[l,oo).
The following theorem gives additional results on the convergence of
sequences and sums of random variables.
Theorem 10.
Let {Z }, n
n
= 1,2, •• ,
and {V}, n
n
= 1,2, ••• ,
or real random variables on a probability space (n) S, P).
be two families
Suppose
Let {gn} be a set
that {Z } is Gaussian, and that {Z n lis independent of {V}.
n
n
g ZE [Z Z] < 00 L: g 2 < 00, and gi > 0 for i = 1,2, ••
of real scalars such that I: nn
n
' nn
(1) T:te, fpllovitJ,S sets have IJro0P..bility zero
A
1
= {w:
or one:
Z (w) converges to some real scalar}
n
AZ = {w: Zn (w) converges to k}, where k is a fixed real number.
(2) Suppose that any of the conditions (1)-(3) of Theorem 9 is satisfied for
{g } defined as above.
n
or
one~
Then the following sets have probability zero
26
Z (w) + V (w) converges},
n
n
Z (w) + V (w) converges to k}.
n
n
Moreover, P(B i )
=1
if and only if P(A )
i
Define yew) by Yi(w)
Pro0f:
=
g.Z.(w).
~
~
convergent sequences of real numbers x
=
(xl' x , ••• ).
Z
I ([22], p.
II xii = suplx
n
n
Z
Z
Li(gixi)Z S (Lig )I Ixl I • Thus we can define
i
to-one operator G: ~ ~tz by (Gx)i = gixi. We
]J
yew)
range(G)}.
£
o
on B[-t.. Z]' Al
assertion
for i
= 1,2.
Let c denote the space of all
space under the norm
{w:
= 1,
68).
c is a separable Banach
Suppose x £~.
Then
a bounded, linear, and onesee that {w: Z(w) £~}
=
Now we note that Y induces from P .a Gaussian measure
= Y-1 [range(G)],
and peAl)
=
]J(range(G», proving the
for AI.
For the set A2 , we consider the space =0 of real sequences that are
convergent to zero.
We note that w
£
A if and only if Zn(w) - k
2
~o is a separable Banach space under the norM
I Ixl I = sMPlxnl
~
o.
([22], p. 68).
Defining the operator G: ~o ~t'2 by (Gx)i = gixi' one sees that G is bounded,
linear, and one-to-one.
Let yew) be defined by Yi(w)
= gi(Zi(w)
is in t 2 a. e. dP (w), thus induces a Gaussian measure on B[i 2] •
Z (w) - k
n
~
0 if and only if yew)
£
range (G).
- k). yew)
Moreover,
The remainder of the proof
is clear.
We now consider B •
l
gi[Zi(w) + View)].
Theorem 9
Define yew) by Yi(w)
= giZi(w)
and Q(w) by Qi(w)
=
Under the assumption that one of the conditions (1) -(3) of
. is satisfied, both yew) and Q(w) belong to t 2 for almost all w,
and thus induce probability measures 1.ly and ]JQ on
9 also shows that
of the previous result
The assertion, on B
1
for AI.
B~Z].
The proof of Theorem
is an
obvious consequence
27
The assertions on B follow
2
in the same way.
As a corollary of this theorem, we note that if {Zn }, n
is a family of Gaussian random variables, then n
with probability one or zero.
-1 n
= 1,2, •••
i[Zt(w) - EZ i ]
~
0
Similarly, an L~ Zi(w) converges with
probability one or zero, and an L~ Zi(w) ~ k with probability one or zero,
0..
for any set of scalars
{a }.
n
ACKNOWLEDGEMENT
The author thanks S. Cambanis for helpful discussions on Theorem 5 •
......
.~.:.
.,""'~,
28
10
J'PPENDIX
PROOF OF LE~1J'AA
2
In this appendix, we supply the missing steps in the proof of Lemma 2, as
outlined in the proof of Lemma 3 of [1].
Lemma 2 above; we suppose that
I:
=f H g(x)exp{2~
g
The notation is as in the proof of
L2[~]'
€
ai(x,ei)}d~(X) = 0
and that
for any N and scalars (al, ••• ,aN).
Now
I:
= I E[g(x)exp{2~ai(x,ei)}
N
IB[H ]]
d~(x)
H
= I eXP{2~ai(x,ei)}
H
N
The conditional expectation E[g(x)1 B[H ]]
N
from the definition of B[H ].
N
can be written as E[g(x)1 B[H ]
qN: x
--
+
«x,el),···,(x,eN»
E[g(x)1 B[HN]] d~(x),
= h[(x,el>' ••• '(x,~)] = h[qN(x)],
where
N
and h is B[RN]/B[R] measurable, since B[H ]
is generated by sets of the form
{x: qN(x)
€
A},
A
€
N
B[R ].
N
Since qN is B[H]/B[R ] measurable, the usual change of variable formula
gives
I
H
h[qN(x)]exp{ r~ ai(x,e i )} d~(x)
=I N
h(rl,··,rN)exp{
R
2~airi}dP(rl,···,rN)
N
l
where P is the Gaussian measure on B[R ] having covariance matrix K , K~j
/
>_ <
> and mean vector
\Kei,e
j - 0ij Kej,e j
11
/
)
m,
mi _
- \m,e
i •
N
family of probability measures {Q} on B[R ], defined by
=
We now have a
a
dQa(rl,.··,rN)
where a
=
= C(a)
(al' •••
'~)
exp{
2~airi} dP(rl, ••• ,rN)
N
is any element of R , and C(a) is a normalizing
By Theorem 1, p. 132 of [23], the family {Qa} is complete; i.e.,
N
N
IN
f(x)dQ (x) = 0, all a in R , implies f(x) = 0 a.e. dQ a (x), all a in R •
R
a
factor.
-
29
= ~{x:
h[q~(x)]=O}
= ~{x:
h[(x,el), ••• ,(x,eN)]}~ 0, we see that
h[(x,el), ••• ,(x,eN)] = 0 a.e.
d~(x), or, E[g(x)IB[~]] = 0 a.e. d~(x). By
the martingale convergence theorem ([7], pp. 84-85),
exists and is equal to
containing
B[HN]
E[g(x)IB[H]],
for all
N.
E[g(x)IB [H]] = 0 a.e. d~(x),
~
--
Thus
since
B[H]
E[g(x)IB[H]]
and thus
li~E[g(x)IB[tli]]
is the smallest a-field
=
0 a.e. d~(x),
g(x) = 0 a.e. d~(x).
30
;\PPENDIX
2
D
PROOF OF
LEM'1A 7
Let B be the separable Banach space consisting of range(T) and
3
the B norm. Define T : B1~ B by T x = Tx. Since range(T) is dense in
1
2
3
1
*-1
~
B3' Tl
exists (t 10 ], p.44). Let {xn}' n = 1,2, ••• be a dense subset
of B • If there exists u in B * such that U(T x ) = 0 for all n, then
l
3
1 n
*
[T ](x ) = 0 for all n; since {xn} is dense in B and T *-1 exists, this
l
n
l
l
implies that u is the null element of B * • Hence, {Tx } is dense in B •
3
3
n
For each xn' define an element F in B3* as follows. If II TXn 11 2 = 0,
n
Fn is the null element.
II.Fn II = 1 and
F (Tx)
n
n
IITx
n
11
2
•
Let {a } be a set of real scalars such that a
n
n
=0
while otherwise a n > 0, and Ln a n = 1. Define a norm
2
2
1 I
11
I lui 1
on
8
by
3
3
3 = Ln ahFn (u). This norm is obviously weaker than
2
2
2
the II • 11 2 norm, since I ITul 1
~
sRP Fn (TU) ~ I ITul 1 2 • Let H3 denote
3
if
--
=
I
ITxnl 1 2
= 0,
the completion of B in the I 1.11
3
c
B[H ], so that B[B ] = B
3
3
Consider T1*F.
n
II Txnl 1 2
Hence
=
3
n
II T *Fn II
1
3
norm.
Kuelbs has shown [8] that B[B 3 ]
B[H ].
3
:;
liT II since II Fn II
= 1.
Moreover,
ITl*Fn(Xn )I ~ II Tl*Fn I I II xn 11 1 , so that IITII ~ IIT/Fnll.
I IT1*Fnl I = I ITI I,
each Fn •
Since 8 1 is reflexive, there exists an
B1 such that Ilyn 11 1 = 1 and T1*Fn (yn ) = I ITI I ([23], p.103).
l
Define G = I ITI l- T1*F ; IIG II = 1 and G (y ) = 1. Choose Ln in B *,
1
n
n
n
n n
n = 1,2, ••• such that IlLn II = 1 and Ln (xn ) = I Ixn I I. Let {Sn },
element y
n
in
be a set of strictly positive real scalars such that
an'- inner product (.'.)1 on 81 by
L
S L (u)L (v).
n n n
n
(u,v)l
=
~L
L S
n n
=~.
a G (u)G (v)
n n n
n
Define
+
Let H be the completion of B under the norm obtained
l
l
from this inner product.
Using the results of [8], we conclude that
31
2
Consider T as a map from HI into H3 • For u in 8 , one has I ITlul 1 3 =
l
1
2
LnanFn2(Tlu} S 21 ITI 1 <u,u)1. Thus Tl is a bounded linear operator from 81
(cHI) into H , and can be extended by continuity to a bounded linear operator
3
on HI.
Denote this extension as T2 •
We show that T maps Borel sets in HI into Borel sets in
2
range (T Z)
is in B[H ], since TZ[H ]
3
l
= T2 [NT!
Z
H3• Note that
] and T2 is a one-to-one bounded
linear map of the separable Hilbert space NT! (with the HI inner product) into
2
H3 • Lemma 6 thus implies that TZ[NT~] is in B[H ].
3
in HI.
We have T2 [A]
must belong to B[H3] •
Finally, let A
= TZ[A
n NT
!
2
].
!
Now let A be any Borel set
!
!
Since AnNT is in B[NT ], T2 [AnNT ]
J
Z
2
2
j
B[B ]. Then A e; B[fl ], and T [A] = T[A]. T[A] is thus
I
2
l
a Borel set in H , and since B[B ] = B nB[H ], this implies that T[A] € B[B 2 ].
3
2
z 3
This concludes the proof.
e;
32
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FOOTNOTES
AMS Subject Classifications:
Primary 28A40, 60G15, 60G17;
Secondary 60G30.
Key Phrases:
Gaussian measures, Gaussian stochastic processes,
Zero-one laws
1.
This research was supported in part by the Air Force Office of Sci-
entific Research under Contract AFOSR-68-1415 and by the Office of Naval Research under Contract N00014-67-A-0321-0006 (NR042-69).
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