Consistent Criteria in Metric Space
Z.Zerakidze, M.Mumladze, L. Eliauri, L. Aleksidze
Gori univetsity
[email protected], [email protected],
[email protected], [email protected]
Abstract - In this paper, we propose the consistent criteria
of hypothesis verification with zero probability of an errors of
any kind. We prove necessary and sufficient conditions for the
existence such consistent criteria in metric space.
Key words: consistent criteria, metric space, measure.
Let ( E , S ) be a measurable sample space with a given
family of probability measures
{ h , h  H } where H is
a set of hypotheses. Let (H ) is a  - algebra which
contains
Definition
all
1. The
H
subsets
of
.
of probability measures
finite
family
{ h , h  H } will be said to admit a consistent criterion of
which on these
measures are focused (gathered).
Suppose the contrary, the measures (1) what were disjoint
F1 , F2 and measures  i ( F j )  0, i  j ,
closed sets
have
disjoint
Note 1. If
such that
at
i  j.
mistake is of the i kind, if

criterion we say that the
H i hypothesis is repudiated
when it is true.
measure so that
open set and
assume that  k ( E F )
 i ( Fi U )  0 ,
if
Fi  U is not empty, then we
 1 is true for almost all h , but then
and
Fn , n  1,2,... such increasing sequence of sets that
F  H \  Fn . For each n , we can specify a pair
n
En , En  E
Definition 3. The following probability
i
is called the probability of mistake of the i kind 
criterion.
Note 2. If the family of
probability measures
{ h , h  H ) admits a consistent criterion of hypotheses,
so the probability of mistakes of all kind will be zero.
Let
F1 , F2  H are two closed disjoint sets and
, respectively,
for the measures:
i
closed
sets,
such
that:
n
 h ( E F )  1, h  F ,  h ( E F )  0, h  F .
So, for each closed F  H there exists a closed set E F ,
so that
E and H be a complete metric space,
 : ( E , S )  ( H , B( H )) is continuous consistent
 F ( A)    h ( A) i (dh), i  1,2
of
E F   En . It is clear that F is a closed set and
Let
 1 , 2 measures focused(gathered) on F1 , F2
,
 h ( E n )  1, h  F ;  h ( E n )  1, h  Fn ; E n  E n  
 i ( )   h ( x :  ( x)  hi )
criterion. If
i
by the continuity  k from above. Let F  H a closed set,
hypothesis.
Definition 2. For given
Choosing the
U  H an
y  {x :  ( x)  hk } , then we receive hk
carriers:
E Fi  E , i  1,2;  Fi ( F j )  0
hypotheses if there exist even though one measurable map
 of the space ( E , S ) in ( H , B( H ))
 h ( x :  ( x)  h)  1, h  H .
closed
 h ( E F )   F ( h)
where
 F (h)
,
(2)
the indicator of set F . Use this result, we
investigate the existence of a continuous consistent criteria.
Theorem: In order for a family { h , h  H } of measures
on complete separable metric space ( E , S ) , where H is a
(1)
Fi
then there are two disjoint closed sets E F1 , E F2  E ,
complete metric space, necessary and sufficient condition
for exist once of continuous consistent criterion  is, that
measures (1) for any disjoint closed sets F1 , F2
 H have
disjoint closed supports.
Proof: The necessity of the theorem established above. In
order to prove of sufficient of the condition use (2). Denote
also by
the minimal closed set for which have place
EF
(2). It is obtained from
B  E for whom
that each set F
E F by throw out all balls
 k ( B)  0, h  F
. If we assume,
corresponds set E , then
F
F1  F2
(so far as ). E  E  E
F1
F2
F2
E F1  E F2
and
will be
k
If xi  E hi and
xi  x0 then hi it has no limit
points. Indeed, if
hi  h0 and h0 - an interior point in
U k(mm)1 , then Ehi  U k(mm11) , for sufficiently large i
Ehi  Ek(mm11) , xi  Ek(mm11) , x0  lim xi  Ek(mm11) .
hk( m )  U k( m )
Let
If F  F , then E coincides with the closure of
F
n
E
Fn
F1  F2   and E F  E F   .
1
 ( x)  h0
for
x  Eh \  E
2 n 1 net in H . Than
H  U k(n ) and every point h is inner although bi to one
(n)
that the prototype of U k
. Therefore,
for
 (V ) 
E
1
where
h0 - an
for all
k
and
n
(n )
is E k
every
(n)
k
open
,
set
V
we
 (x)
consequently
have
is
U k( n ) V
measurable. Further 
1
(H \ V ) 
 (E
H
\ E k( n ) ) .
U k( n ) V
k
( n 1)
Suppose also, that for every U k
(n)
there is U k
 U k( n 1)
E F(n ) a closed set in E , corresponding to
(n)
set U k .
Eh be the corresponding to one-point set {h} It is easy,
to see that for each closed set F
have place a
E
E F is a closure for
defined by equality
h
. Let
 (x) is
hF
 ( x)  h , when x  Eh . The map 
defined on a dense set in
E H . We extend it to E H \
E
n
there is
k
if x  E k
(n )
, that
If
H \V  F ,
This means
 1 ( F ) and E F
E
, then
hF
 E F   when U k( n )  V .
 1 ( F )  E F .
Since
h
(n)
then E k
contain a dense subset
E F is dense in  1 ( F ) and hence E F =
 1 ( F ) .
The theorem is proved.
REFERENCE
h
hF
.
If, for some
,
arbitrary fixed value. From the construction  (x) it follows
of the sets U ( n ) .
correspondence:
(1)
k
2
Denote with U ( n ) , K  1,2,... family of closed bolls of
k
Let
point. Assuming
k
n
. We denote by
arbitrary
 ( x0 )  hk( m ) , we define  (x) for all x   E k(1) .
Finally,
radius 2  n , which centers formed
an
k
E F1  F2  E F1  E F2
. Obviously, that
x0  E k(mn ) , x0   E k( m1) .
, then
x   Ek(nn ) , where k n a sequence of natural numbers. Let
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n
us
assume
 ( x)  h , where h  U k(nn ) .This
n
intersection is not empty because
E
(n)
kn
is non-empty
n
and consists for a single point by the completeness of the
space.
Suppose, that for some m ,
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