Discrete Probabilities on Natural Numbers: Existence Theorems

On Certain Discrete Probabilities
Theorem 1. For every real number s >1there exists a discrete probability Pon ¡N,2N¢such that
P(kN)=k−s
for all k ∈N.
Proof. Putting P¡{n}¢:=ζ(s)−1n−sfor all singletons {n}∈2N, where
ζ(s):=X
n≥1
n−s
is Riemann’s zeta function, we get a well defined descrete measure P:2N→R+which satisfies
P(kN)=X
n≥1
P¡{kn}¢=ζ(s)−1k−sX
n≥1
n−s=k−s
as desired, and in particular it holds P(N)=1, so that Pis a probability measure. ä
Theorem 2. Let s ∈]0,1] be a real number. Then there is no probability measure Pon ¡N,2N¢such that
P(kN)=k−s
for all k ∈N.
Proof. Let Pdenote the set of all primes and put Pn:=©p∈P¯
¯n≤pªas well as Pn,m:=©p∈P¯ ¯n≤p≤mª
for all n,m∈N. It holds limm→∞ Pn,m=Pnand limn→∞ Pn= ;. We apply a reasoning ad absurdum and
assume that there exists a probability measure Psuch that P(kN)=k−s. Then we have
1−Pµ[
p∈Pn,m
pN¶=X
Q⊆Pn,m
(−1)card QPµ\
p∈Q
pN¶=X
Q⊆Pn,m
(−1)card QP
µY
p∈Q
p¶N
=X
Q⊆Pn,m
(−1)card QY
p∈Q
p−s=Y
p∈Pn,m
¡1−p−s¢.
Now recall that, given a sequence (am)m≥m0of real numbers am∈]0,1[ , we have the equivalence
X
m≥m0
am= ∞ ⇐⇒ Y
m≥m0
(1 −am)=0.
But it is well known (and easy to prove) that
X
p∈Pn
p−s= ∞ ,
hence
Y
p∈Pn
¡1−p−s¢=0,
and it follows
P³©k∈N¯
¯k<nª´≤1−Pµ[
p∈Pn
pN¶=1−lim
m→∞
Pµ[
p∈Pn,m
pN¶
=lim
m→∞ Y
p∈Pn,m
¡1−p−s¢=Y
p∈Pn
¡1−p−s¢=0
and finally
1=P(N)=lim
n→∞
P³©k∈N¯
¯k<nª´=0,
which is a contradiction. This proves the theorem. ä