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and thus
|
( Π k V ) i
( Π k V ) i |≤
c.
(9.40)
As this holds for any i ,wehave
Π k V
V .
Π k V
V
(9.41)
which completes the proof.
Thus, averaging classifiers themselves are compatible with value iteration and
policy iteration. Does this, however, still hold for a set of classifiers that provides
its prediction by (9.21)? Let us first consider the special case when the mixing
functions are constant:
Lemma 9.2. Let the global model V be given by V = k G k Π k V ,where Π k
is the approximation operator of averaging classifiers, and the G k 's are constant
diagonal mixing matrices with non-negative elements such that k G k = I holds.
Then, this model forms a non-expansion with respect to the maximum norm.
Proof. Themethodtoprooftheabovestatement is similar to the one used to
prove Lemma 9.1: firstly, we show that for any two vectors V , V such that
V
ΠV and
Π ( V + b e )=( ΠV )+ b e ,where Π = k G k Π k . Then, non-expansion is shown
by applying these properties to the maximum absolute difference between two
arbitrary vectors.
ΠV
V , an arbitrary scalar b ,and e =(1 ,..., 1) T
we have ΠV
ΠV
follows from observing that for the vector a = V
V with
non-negative elements, all elements of k G k Π k a are non-negative due to the
non-negativity of the elements of G k ,and Π k a
0 , and thus
ΠV =
k
G k Π k V
ΠV .
G k Π k a
(9.42)
k
Also, as Π k ( V + b e )=( Π k V )+ b e and k G k ( V k + b e )= b e + k G k V k for
any K vectors
{ V k }
,wehave
Π ( V + b e )= ΠV + b e .
(9.43)
Let V , V now be to arbitrary vectors, not bound to V
V ,andlet c =
V =max i |
V i
V i |
V
be their maximum absolute distance. Thus, for
any i ,
V i
V i
c
V i + c.
(9.44)
Given the properties of Π it follows that
( ΠV ) i
( ΠV ) i
c
( ΠV ) i + c,
(9.45)
and therefore
( ΠV ) i |≤
|
( ΠV ) i
c,
(9.46)
from which follows that
ΠV
V ,
ΠV
V
(9.47)
which completes the proof.
 
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