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NEW PROOF OF A CHARACTERIZATION OF THE GIBBS TYPE PROCESSES

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Abstract

We provide a new proof of the fact that the only Species Sampling Models where the probability of observing a new value does depend just on the sample size n and the number of clusters h are the Gibbs-type Processes.
NEW PROOF OF A CHARACTERIZATION OF THE GIBBS TYPE
PROCESSES
ANDREA AVENI
Abstract. We provide a new proof of the fact that the only Species Sampling Models
where the probability of observing a new value does depend just on the sample size nand
the number of clusters hare the Gibbs-type Processes.
Definition. For any nNand h[n], we denote by n
h:= {cNh:Ph
j=1 cj=n}the
set of all compositions of nof rank (or length) h.
We also write n=n
h=1n
hand ∆ = n1n.
Definition. Given two elements c, d n
hwe say that
cd⇔ ∃σSh: (cj)h
j=1 = (dσ(j))h
j=1
Since Shis a group then is an equivalence relation and we denote define
˜
n
h= ∆n
h/
the set of all number theoretic partitions of the number nof rank h.
Theorem. |n
h|=n1
h1, so that |n|= 2n1
Proof. We classify the compositions cin ∆n
haccording to their first element. If the first
element c1of cis u[n], then (cj)h
j=2 must belong to ∆nu
h1, moreover for any cnu
h1
the partition (u, c) (if exists) is a unique element of ∆n
h. In other words there exists a
bijection between ∆nu
h1and the elements of ∆n
h1starting from u, therefore we must have
|n
h|=
n
X
u=1 |nu
h1|=
n(h1)
X
u=1 |nu
h1|(1)
Now we prove the result by induction on n. The case n= 1 holds since |1
1|=|{(1)}| =
1 = 0
0. Now we assume the formula |n
h|=n1
h1to hold for any (h, n) such that n<N
and we prove it also holds for any (h, N) with h= 1...N.
1
2 ANDREA AVENI
For any such h= 2...N we have by (1) and by inductive hypothesis
|N
h|=
N(h1)
X
u=1 |Nu
h1|
=
N2
X
j=h2|j+1
h1|
=
N2
X
j=h2j
h2=N1
h1
Where the last passage is due to the hockey-stick relation 1. Finally if h= 1 we always
have |n
1|=|{n}| = 1, so, in particular, |N
1|= 1. And this complete the proof.
On the contrary the cardinality of ˜
n
his much more complicated to be analyzed and we
will simply indicate it with ph(n) := |˜
n
h|and p(n) := Pn
h=1 ph(n) = |˜
n|and p(0) := 1
The few known facts about these functions are the following
X
n=0
p(n)zn=
Y
n=1
1
1xn
And this function, defined on |z|<1 has many interesting properties, for instance
gcd(p, q)=1lim
ρ1"(1 ρ) ln
Y
n=1
1
1(ρe2πip/q )n#=π2
6q2.
as I proved in the appendix.
Moreover an asymptotic formula is available
p(n)=Θ(4n3)1exp(πp2n/3)
Now we define the EPPF.
Definition. A EPPF is a function from ˜
to the real numbers (whose restrictions to ˜
n
h
are denoted by pn
h) such that
(PR) For every cn
h, we have
pn
h(c) = pn+1
h+1(c, 1) +
h
X
i=1
pn+1
h(cj+δi
j)h
j=1
(TR) For every cn
h, the ratio pn+1
h+1(c, 1)/pn
h(c)depends only on hand n.
1https ://en.wikipedia.org/wiki/H ockey stick identity
NEW PROOF OF A CHARACTERIZATION OF THE GIBBS TYPE PROCESSES 3
Before tying to characterize EPPFs we would like to consider how may of them we can
reasonably expect. If we denote by ˜
|N:= N
n=1 ˜
n, then the restriction of a EPPF to ˜
|n
is made up by
N
X
n=1 |˜
n|=
N
X
n=1
p(n) =: P(N)
real numbers.
The number of conditions implied by (P R) to the EPPF restricted to ˜
|Nare as many as
the elements in ˜
|N1, so they are P(N1).
On the other hand the condition T R implies the equality of the quantity pn+1
h+1(c, 1)/pn
h(c)
among all the c˜
n
h, so for fixed nand hthe constraints are |˜
n
h|1. Therefore, in this
case, the number of constraints is
N1
X
n=1
n
X
h=1
(ph(n)1) =
N1
X
n=1
p(n)n=P(n1) N(N1)
2.
Finally the degrees of freedom should be
P(N)2P(N1) + N(N1)
2=p(N) + N(N1)
2P(N1)
Unfortunately this quantity is negative for all N > 7. This seems counterintuitive since
there exists some (indeed uncountably many) EPPF. A possible reason is that the set of
constraints we have indicated may be dependent (even though they seem not). Another
possible explanation lies in the non-linearity of the (T R) constraints.
(Notice that, on the other hand the degreed of freedom of the Gibbs-type solution on ˜
N
are 1 + P(N)P(N1) = p(N)+1>0.)
Despite this pointless discussion on the degreed of freedom we have the following result.
Theorem. The only species sampling models where the probability of observing a value
that has never occurred before depends only on nand his the Gibbs-type processes. In
other words the only non negative weights (((pn
h(c))cn
h)n
h=1)n1such that:
(TR) For every cn
h, the ratio pn+1
h+1(c, 1)/pn
h(c)depends only on hand n.
(PR) For every cn
h, we have
pn
h(c) = pn+1
h+1(c, 1) +
h
X
i=1
pn+1
h(cj+δi
j)h
j=1
(EX) For all ςSh,pn
h((cj)h
j=1) = pn
h((cς(j))h
j=1)
can be written as
pn
h((cj)h
j=1) = Vn
h
h
Y
j=1
(1 σ)cj1
for some σRand for some coefficients ((Vn
h)n
h=1)n1such that
(CL) For every n1,h= 1...n, we have Vn
h=Vn+1
h+1 + (n)Vn+1
h
4 ANDREA AVENI
Proof. First notice that the result trivially holds for p1
1(1), then we proceed by induction
on n; that is, assuming the result holds for all the partitions of any length of any number
up to n=N, then we prove that there must be coefficients {VN+1
h}N
h=1 satisfying condition
(CL) and such that for any h= 1...N + 1 and cN+1
h,
(T H )pN+1
h(c) = VN+1
h
h
Y
j=1
(1 σ)cj1.
Because of assumption (T R) and by inductive hypothesis, we must have that that for any
h∈ {2...N + 1}and for all c= (c1...ch1,1) N+1
h,
pN+1
h(c1...ch1,1)
pN
h1(c1...ch1)=Ch
For some positive constants {Ch}N+1
h=2 . Therefore,
pN+1
h(c1...ch1,1) = VN
h1Ch
h1
Y
j=1
(1 σ)cj1
=VN
h1Ch
h
Y
j=1
(1 σ)cj1.
Finally we define VN+1
h:= VN
h1Chfor any h= 2...N + 1.
So, by (EX) we have that, for any h= 2...N + 1 and for any partition (cj)h
j=1 of N+ 1
with a 1 in it, we must have
cN+1
h: min(c) = 1, pN+1
h((cj)h
j=1) = VN+1
h
h
Y
j=1
(1 σ)cj1.(2)
This is precisely (T H ) for all partitions cof N+1 such that min(c) = 1. Now the problem
is to identify pfor all those partitions that do not have any 1 in them and to verify equation
(CL).
In this respect, we can also define VN+1
1:= (VN
1VN+1
2)/(Nσ) so that we recover (CL)
for h= 1.
Using (PR) and (1), for any partition (c1...ch) of Nwe must have that
pN
h(c1...ch) = pN+1
h+1 (c1...ch,1) +
h
X
i=1
pN+1
h(c1...ci+ 1, ...ch)
h
X
i=1
pN+1
h((ch+δi
j)h
j=1) = hVN
hVN+1
h+1 ih
Y
j=1
(1 σ)cj1
Now we focus our attention to the partitions cN
hthat terminates by 1 (for some
h∈ {2...N}). In this case for each i= 1...h 1, we have that (cj+δi
j)h
j=1 is a partition of
NEW PROOF OF A CHARACTERIZATION OF THE GIBBS TYPE PROCESSES 5
N+ 1 terminating by 1, while (cj+δh
j)h
j=1 is a generic partition of N+ 1 terminating by
2. In this case, thanks to (1), this last equation reads
pN+1
h((cj)h1
j=1 ,2) +
h1
X
i=1
pN+1
h((cj+δi
j)h1
j=1 ,1) = hVN
hVN+1
h+1 ih
Y
j=1
(1 σ)cj1
pN+1
h((cj)h1
j=1 ,2) +
h1
X
i=1
VN+1
h
h1
Y
j=1
(1 σ)cj+δi
j1=hVN
hVN+1
h+1 ih
Y
j=1
(1 σ)cj1
pN+1
h((cj)h1
j=1 ,2) + VN+1
h
h
Y
j=1
(1 σ)cj1
h1
X
i=1
(ciσ) = hVN
hVN+1
h+1 ih
Y
j=1
(1 σ)cj1
pN+1
h((cj)h1
j=1 ,2) = hVN
hVN+1
h+1 VN+1
h(N1(h1)σ)ih
Y
j=1
(1 σ)cj1
Now by exchangeability, we have that if some component of (cj)h1
j=1 is equal to one (notice
that it is always possible to find a partition of rank hof Nwith two elements equal to one
as long as N > 2 and h= 3...N or N= 2 and h= 2) then we must have
VN+1
h(1 σ)
h
Y
j=1
(1 σ)cj1=hVN
hVN+1
h+1 VN+1
h(N1(h1)σ)ih
Y
j=1
(1 σ)cj1
So that we recover that the relation VN
h=VN+1
h+1 + (N)VN+1
hmust hold for every
h3.
If, on the other hand, all (cj)h1
j=1 are greater then 1, if we call (c
j)h
j=1 = ((cj)h1
j=1 ,2), then
pN+1
h((cj)h1
j=1 ,2) = VN+1
h(1 σ)
h1
Y
j=1
(1 σ)cj1=VN+1
h
h
Y
j=1
(1 σ)c
j1
And this proves (T H) for all partitions cof N+ 1 such that min(c)=2
Lemma. The EPPF function can always be written as
pn
h((cj)h
j=1) = Vn
h
h
Y
j=1
(1 σ)cj1
for any cN+1
hwith u= min(c)< N + 1.
Proof. Now we proceed by induction on the number uof the last entry of a partition
c= ((cj)h1
j=1 , u) of N. (Notice that (cj)h1
j=1 is a partition of Nuand therefore its length
must be h1∈ {1...N u}so that h∈ {2...N +1 u}. In particular this argument cannot
be carried on when N+ 1 u= 2 or u=N1)
6 ANDREA AVENI
By inductive hypothesis and (?), we have the following
pN+1
h((cj)h
j=1, u + 1) +
h1
X
i=1
pN+1
h((cj+δi
j)h1
j=1 , u) = [VN
hVN+1
h+1 ]
h
Y
j=1
(1 σ)cj1
pN+1
h((cj)h
j=1, u + 1) + VN+1
h
h
Y
j=1
(1 σ)cj1
h1
X
i=1
ciσ= [VN
hVN+1
h+1 ]
h
Y
j=1
(1 σ)cj1
pN+1
h((cj)h
j=1, u + 1) = hVN
hVN+1
h+1 Vn+1
h(Nu(h1)σ)ih
Y
j=1
(1 σ)cj1
Now, if there is some j∈ {1...h 1}such that cju, then, by inductive hypothesis and
(EX ), we find that
VN+1
h(uσ)
h
Y
j=1
(1 σ)cj1=hVN
hVN+1
h+1 Vn+1
h(Nu(h1)σ)ih
Y
j=1
(1 σ)cj1
From which we recover once again the relation VN
h=VN+1
h+1 +VN+1
h(N) for all
h∈ {2...N + 1 u}; if on the contrary for each j= 1..h 1, cj> u then we get
pN+1
h((cj)h
j=1, u + 1) = VN+1
h(uσ)
h
Y
j=1
(1 σ)cj1
And this is proves the lemma.
The only partition that has been left behind is pN+1
1(N+ 1). In this case, because of
(P R) and our definition of VN+1
1, we must have
pN
1(N) = pN+1
1(N+ 1) + p2pN+1
2
pN+1
1= [VN
1VN+1
2](1 σ)N1
pN+1
1=VN+1
1(1 σ)N+11
Now that we have found that (T H) holds for every partition of N+ 1, then it is immediate
to see that (CL) is equivalent to (P R) and therefore (CL) must holds also for h= 2.
Finally it is easy to see that the only Gibbs-type processes where the probability of
observing a value that has never occurred before depends only on nis the Dirichlet Process.
Theorem. The only non-negative weights {{Vn
h}n
h=1}n1such that
V1
1= 1,
For any n1and h∈ {1...n},Vn
h=Vn+1
h+1 +nV n+1
hand
For any n1and h∈ {1...n},Vn+1
h+1 /V n
his constant in h,
NEW PROOF OF A CHARACTERIZATION OF THE GIBBS TYPE PROCESSES 7
can be expressed as
Vn
h=αh1
(1 + α)n1
for some α[0,].
Proof. First of all, we notice that the formula holds for all the couples (n, h) with n1
(there is only one such couple of index, namely (1,1)) then, we proceed by induction, let
us assume that Vn
hhas the specified form for some α[0,) up to n=N, then, if we call
(assuming all weights to be positive) Ω = VN+1
h+1 /V N
h, we have by hypothesis that
VN
N=VN+1
N+1 +NV N+1
N
VN
N= ΩVN
N+NV N+1
N
So we have these two relations
VN+1
N= ΩVN
N1=1
NVN
N.
But this forces Ω = α/(α+N). This, in turn, implies that, for h= 2...N + 1,
VN+1
h= ΩVN,h1=α
α+N
αh2
(1 + α)N1
=αh1
(1 + α)N+11
,
and finally, for h= 1...N,
VN+1
h=1
NVN
h=1
α+N
αh1
(1 + α)N1
=αh1
(1 + α)N+11
.
Notice that this does not only shows the uniqueness of the solution, but it does also show
that our proposed solution is indeed a solution.
Finally, by selecting α=, we get degenerate case where some weights are zero and the
previous argument does not apply.
Definition. The set of all partitions on nletters, denoted by Πnis defined as
Πn:= nπ2[n]:π= [n]∧ ∀πi, πjπ(πi6=πjπiπj=)o
Every partition is thus a set containing mutually disjoint and collectively exhaustive subsets
of [n].
We shall order the elements of each πΠnaccording to their least element. If |π|=h,
then for j= 1 to h, the cardinality of πjwill be indicated cjand its elements will be
denoted as
πj={`i
j}cj
i=1
. Given this notation it is evident that, for each partition πΠn, (cj)h
j=1 is a com-
position in ∆n
h. On the contrary to each composition cn
hthere will correspond (in
general) multiple partitions, for instance, both the (distinct) partitions {{1,2},{3},{4}}
and {{1,3},{2},{4}} correspond to the same composition (2,1,1). It is interesting to find
the cardinality of the set of partitions that correspond to a specific composition cn
h.
8 ANDREA AVENI
Theorem. Given a composition cn
h, the number of partitions πΠncorresponding
to cis
h!
h
Y
j=1
cj!
1
Y
c:j:cj=c|{j:cj=c}|!
1
Proof. The result is evident.
Given this we have that
Bn:= |Πn|=
n
X
h=1
h!X
cn
h
h
Y
j=1
cj!
1
Y
c:j:cj=c|{j:cj=c}|!
1
On the other hand we also have that from any partition πΠn, we can get exactly |π|+ 1
unique partitions of N+ 1, by inserting N+ 1 in one of the elements of πor by adding
{N+ 1}at the end of π. Therefore, we have the recurrence relation
|Πn+1|=X
πΠn|π|+ 1 = |Πn|+X
πΠn|π|
1. Useless Appendix
If we denote by p(n) the number of distinct partitions of the number n, we have that
α(z) =
X
n=0
p(n)zn=
Y
n=1
1
1zn|z|<1
The aim of this appendix is to study the behavior of α(z) on the boundary of its domain.
First of all we can immediately see that if z= exp(2πir) for some rational number rQ,
let us say r=p/q with p, q coprimes, then we have that for every natural number n,
znq = 1 and α(z) will not be defined.
If instead z= exp(2πix) for some xR\Q, then znwill never be equal to one and each
factor in α(z) will be well defined.
Theorem. If gcd(p, q) = 1, then
lim
ρ1"(1 ρ) ln
Y
n=1
1
1(ρe2πip/q )n#=π2
6q2.
NEW PROOF OF A CHARACTERIZATION OF THE GIBBS TYPE PROCESSES 9
Proof.
lim
ρ1"(1 ρ) ln
Y
n=1
1
1(ρe2πip/q )n#=lim
ρ1"(1 ρ)
X
n=1
ln 1(ρe2πip/q )n#
= lim
ρ1"(1 ρ)
X
n=1
X
k=1
(ρe2πip/q )nk
k#
= lim
ρ1"(1 ρ)
X
k=1
1
k
X
n=1
(ρe2πip/q )nk#
= lim
ρ1"(1 ρ)
X
k=1
1
k
1
1(ρe2πip/q )k#
=
X
k=1
1
klim
ρ1
1e2πikp/q
1ρ+e2πikp/q
k1
X
n=0
ρn!1
But, if pand qare coprime, we have that
lim
ρ1"1e2πikp/q
1ρ+e2πikp/q
k1
X
n=0
ρn#=(ke2πik/p ifk/q N
otherwise =(kifk/q N
otherwise
An then, finally
lim
ρ1"(1 ρ) ln
Y
n=1
1
1(ρe2πip/q )n#=
X
k=1
1
k21k/qN
=
X
n=1
1
(qn)2
=π2
6q2
This also implies that limρ1(1 ρ) ln(α(ρexp(2πix)) = 0 if xR\Q.
It would be interesting to see whether α(exp(2πix)) is defined or not for xirrational.
For a normal number x, we could heuristically approximate α(z) in the following way
(zn)n1iid
Unif|z|=1
=1
1zniid
Cauchy(1/4)
<1
1zn= 1/2
10 ANDREA AVENI
Figure 1. A plot of (1 ρ) ln Q
n=1 1
1(ρe2πip/q )nwhen ρ= 1/100.
Where by Cauchy(1/4) we mean the distribution with pdf
1
2π
1
1/4 + t2
But then we have that E(1/(1 zn)) = 1/2 (in the sense of Cauchy principal value). This
argument suggests that α(exp(2πix)) will be zero for any normal x.
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