Optimality

Optimal sampling strategies 283
Claim (2): G (n) ∈ ∆n(M1, . . . , MP) and G (n) satisfies (7.1).
(Initial Condition) The claim is trivial for n = 0.
(Induction Step) Clearly from (3.4) and (3.5)
(7.6)
�
k
g (n+1)
k
= 1 + �
k
g (n)
k
= n + 1,
and 0≤g (n+1)
k ≤ Mk. Thus G (n+1) ∈ ∆n+1(M1, . . . , MP). We now prove that
G (n+1) satisfies property (7.1). We need to consider pairs i, j as in (7.1) for which
either i = m or j = m because all other cases directly follow from the fact that
G (n) satisfies (7.1).
Case (i) j = m, where m is defined as in (3.5). Assuming that g (n+1)
m < Mm, for
all i�= m such that g (n+1)
i > 0 we have
�
ψi g (n+1)
� �
i − ψi g (n+1)
� �
i − 1 = ψi g (n)
� �
i − ψi g (n)
�
i − 1
�
≥ ψm g (n)
� �
m + 1 − ψm g (n)
�
m
(7.7)
�
≥ ψm g (n)
� �
m + 2 − ψm g (n)
�
m + 1
�
= ψm g (n+1)
� �
m + 1 − ψm g (n+1)
�
m .
that
(7.8)
Case (ii) i = m. Consider j�= m such that g (n+1)
j
ψm
�
g (n+1)
m
�
− ψm
Thus Claim (2) is proved.
�
g (n+1)
� �
m − 1 = ψm g (n)
�
m + 1
�
≥ ψj g (n)
�
j + 1
�
= ψj g (n+1)
�
j + 1
< Mj. We have from (3.5)
− ψm
− ψj
�
− ψj
�
g (n)
m
g (n)
j
�
�
�
g (n+1)
j
It only remains to prove the next claim.
Claim (3): h(n), or equivalently � (n)
k ψk(g k ), is non-decreasing and discreteconcave.
Since ψk is non-decreasing for all k, from (3.4) we have that � (n)
k ψk(g k ) is a
non-decreasing function of n. We have from (3.5)
h(n + 1)−h(n) = � �
ψk(g
k
(n+1)
k )−ψk(g (n)
k )
�
(7.9)
�
ψk(g (n)
(n)
k + 1)−ψk(g k )
�
.
= max
k:g (n)
k