Table Of ContentA STUDY OF COUNTS OF BERNOULLI STRINGS VIA
CONDITIONAL POISSON PROCESSES
8
FREDW.HUFFER,JAYARAMSETHURAMAN,ANDSUNDERSETHURAMAN
0
0
2
Abstract. A sequence of random variables, each taking values 0 or 1, is
n called a Bernoulli sequence. We say that a string of length d occurs, in a
a Bernoulli sequence, if a success is followed by exactly (d−1) failures before
J the next success. The counts of such d-strings are of interest, and in specific
independent Bernoulli sequences are known to correspond to asymptotic d-
4
cyclecounts inrandompermutations.
1
In this note, we give a new framework, in terms of conditional Poisson
processes,whichallowsforaquickcharacterizationofthejointdistributionof
]
R thecountsofalld-strings,inageneralclassofBernoullisequences,ascertain
mixtures oftheproduct ofPoissonmeasures. Inparticular,this general class
P
includes all Bernoullisequences consideredintheliterature, as wellas ahost
.
h ofnewsequences.
t
a
m
[
1. Introduction
1
v In this note, we study the joint distribution of the counts of certain d-strings
5 of all orders d > 1 arising in Bernoulli sequences. Previous work has used several
1 differentmethods,includingcombinatorial,factorialmoment,andPo´lyaandHoppe
1
urn model methods to identify the joint count distribution with respect to a class
2
of independent Bernoulli sequences. In this context, our main contribution is to
.
1
introduce a new framework, using conditional Poisson processes, which allows for
0
a concise derivation of the joint count distribution as a mixture of the product of
8
0 Poissonmeasures with respect to all Bernoullisequences consideredbefore, as well
: as many others in a general class, including some dependent Bernoulli sequences.
v
A Bernoulli sequence Y = {Y } is a sequence of {0,1}-valued random vari-
i n n≥1
X ables. For d≥1, we say that a d-string occurs if a 1 is followed by exactly (d−1)
r 0’s before the next 1 in the Bernoulli sequence. Specifically, a d-string occurs at
a
time n≥1 if Y =1 where
n,d
Y Y for d=1
Y = n n+1
n,d Y (1−Y )···(1−Y )Y for d≥2,
n n+1 n+d−1 n+d
(cid:26)
that is, if hY ,...,Y i=h1,0,...,0,1i.
n n+d
d−1
LetZ = Y bethecountofalld-strings,ford≥1,andZ=hZ :d≥1i
d n≥1 n,d | {z } d
bethe“countvector”ofstrings. [Ingeneral,Zmayhavedivergentcomponents,but
P
2000 Mathematics Subject Classification. primary60C05; secondary60K99.
Key words and phrases. Bernoulli, cycles, strings, spacings, nonhomogeneous, Poisson pro-
cesses,randompermutations.
Research partially supported by ARO-W911NF-04-1-0333, nsa-h982300510041, and NSF-
DMS-0504193.
1
2 FREDW.HUFFER,JAYARAMSETHURAMAN,ANDSUNDERSETHURAMAN
for the Bernoulli sequences considered in this article it is easily shown (by taking
expectations) that all components Z are finite with probability 1.]
k
In this notation, the general problem is to understand the distribution of Z
and its connection to the underlying sequence Y. Aside from the problem’s basic
interest, d-strings and their counts from specific independent Bernoulli sequences
haveinterpretationswith respectto randompermutations,recordvalues, Bayesian
nonparametrics,and species allocation models through Ewens sampling formula.
d
We will use “=” to signify “equals in distribution,” and L(X) to denote the
law or distribution of the random variable X. Denote also Po(λ) as the Poisson
measure on R with intensity λ, and I(B) as the indicator of a set B.
Example1.1. LetS ={1,2,...,n},andconsidertheFelleralgorithmtogenerate
n
a permutation π :S →S uniformly among the n! choices (cf. Feller (1945)):
n n
1. Drawanelementuniformly fromS , andcallitπ(1). Ifπ(1)=1,a 1-cycle
n
is completed. If π(1)6=1, make another draw uniformly from S \{π(1)},
n
andcallitπ(π(1)). ContinuedrawingfromS \{π(1),π(π(1))},...naming
n
them π(π(π(1))), and so on, until a cycle (of some length) is finished.
2. From the elements left in S \{π(1),π(π(1)),...,1} after the first cycle is
n
completed,followtheprocessinstep1withthesmallestremainingnumber
taking the role of “1” to finish a second cycle. Repeat until all elements of
S are exhausted.
n
LetI(n) betheindicatorthatacycleiscompletedatthekthFellerdrawfromS .
k n
A moment’s thought convinces that {I(n)}n are independent Bernoulli random
k k=1
variableswithP(I(n) =1)=1/(n−k+1)as,independent ofthe past,exactly one
k
choiceattime1≤k≤nfromtheremainingn−k+1membersleftinS completes
n
the cycle. Denote C(n) as the number of k-cycles in π,
k
C(n) = I1(n)+ in=−11Ii(n)Ii(+n)1 for k=1
k ( lk=−11(1−Il(n))Ik(n)+ in=−1kIi(n) il=+ik+−11P(1−Il(n))Ii(+n)k for 2≤k ≤n.
Now letQY be the independenPt sequenceQwhere P(Yk = 1) = 1/k for k ≥ 1, so
that Y =d I(n) for 1 ≤ k ≤ n. Then, as Y , and Y n (1−Y ) for
k n−k+1 n n−k+1 l=n−k+2 l
2 ≤ k ≤ n all vanish in probability as n ↑ ∞, we conclude for each k ≥ 1 that
Q
lim C(n) =d Z .
n→∞ k k
Finally, as is well-known, the asymptotic cycle counts {lim C(n)} are dis-
n k k≥1
tributed as independent Poisson random variables with respective means 1/k for
d
k ≥ 1 (cf. Kolchin (1971)). Hence, Z = Po(1/k). [Example 2.1, in section
k≥1
2, gives a derivation in our Poisson process framework. See also Arratia-Barbour-
Q
Tavar´e (1992, 2003) for more discussion with Ewens sampling formula.]
Example 1.2. Consider the standard nonparametric problem of estimating the
unknown distribution function F from independent and identically distributed ob-
servations {X } . A Bayesian may place on F a Dirichlet prior with parameters
i i≥1
aµ where a>0 and µ is a non-atomic probability measure.
Let Y = 1 and for n ≥ 2 define Y = 1 if X is a new observation, that is if
1 n n
X 6∈{X ,...,X }, and Y =0 otherwise. Then, it can be shown that Y is an
n 1 n−1 n
independentBernoullisequencewith P(Y =1)=a/(a+n−1)forn≥1 andthat
n
(logn)−1 n Y → a a.s. The latter result can be interpreted in terms of counts
i=1 i
P
BERNOULLI STRINGS AND POISSON PROCESSES 3
ofstringsinthisBernoullisequence. SeeKorwar-Hollander(1973)formoredetails,
and also Ghosh-Ramamoorthi(2003).
In the literature, to our knowledge, only the count vectors of the following class
of underlying independent Bernoulli sequences have been investigated. Denote the
independent Bernoulli sequence Y where P(Y =1)=a/(a+b+n−1) for n≥1
n
as Y = Bern(a,b). The case a = 1, b = 0 is Example 1.1 (see also Arratia-
Tavar´e (1992)). The case a > 0, b = 0 is Example 1.2. For this case, Arratia-
d
Barbour-Tavar´e (1992) observe that the associated Z = Po(a/k) through
k≥1
connections with Ewens sampling formula. When a = 1, b > 0, Sethuraman-
Q
Sethuraman(2004),employing factorialmoments, show that, given the value x of
0
a Beta(b,1) random variable, Z =d Po((1−xk)/k). Such a distribution will
k≥1 0
be called a “mixture of independent Poisson factors.” When a > 0 and b > 0,
Q
Holst (2007) extends further, using Po´lya and Hoppe urns, and establishes that,
given the value x of a Beta(b,a) random variable, Z =d Po(a(1−xk)/k),
0 k≥1 0
again a mixture of independent Poisson factors. We note also that several inter-
Q
esting studies of 1-strings preceded some of the above work, e.g. an unpublished
manuscript of Diaconis, Chern-Hwang-Yeh (2000), Mo´ri (2001), Joffe-Marchand-
Perron-Popadiuk(2004), and references therein in these and the above papers.
With this background,ourmainidea isthatitiseasiertostudy Zstartingfrom
an extrinsic “conditional marked Poisson process model” (CMPP) rather than di-
rectly from the Bernoulli sequence. Namely, we prove that when the underlying
Bernoulli sequence Y is generated through a CMPP model, the count vector Z is
distributed as a mixture of independent Poissonfactors in terms of model parame-
ters (Theorem 2.2). As remarked earlier, the Poisson process techniques used here
are different from previous methods and allow quick derivations. Perhaps inter-
estingly, the sequences Y found in our model include many dependent Bernoulli
sequences (some explicit examples are in section 5). However,the most generalse-
quencestudied till now,the independent sequenceBern(a,b)witha>0 andb≥0,
canalso be realizedin our framework(Proposition3.1), yielding a new proofof its
count vector distribution.
Our conditional marked Poisson process model also yields a new class of inde-
pendent Bernoulli sequences which we call Bern (a,b). Denote the independent
1
Bernoullisequence Y where P(Y =1)=1, and P(Y =1)=a/(a+b+n−2) for
1 n
n≥ 2 as Y = Bern (a,b). The Bern (a,b) sequence appends a 1 to the Bern(a,b)
1 1
sequenceandpicksuponemored-stringcontributedbyanyleading0’sinBern(a,b).
We showthat the distribution ofthe countvectorZ forBern (a,b)for a>0,b≥1
1
is a mixture of independent Poisson factors (Proposition 4.1). This result fails for
0 ≤ b < 1, and in this case even the distribution of Z , the count of 1-strings in
1
Bern (a,b), is not a mixture of Poisson distributions (Proposition 4.5). However,
1
the distribution of Z in Bern (a,b) can be expressed through a recurrence relation
1
for all values of b including 0≤b<1 (Proposition 4.3).
The planof the article is to discuss the CMPP model, and provethe main theo-
rem in section 2. In sections 3 and 4, the main theorem is applied to independent
sequences Bern(a,b) and Bern (a,b) respectively. Last, in section 5, two explicit
1
dependent Bernoulli sequences, arising from the CMPP model, are given.
4 FREDW.HUFFER,JAYARAMSETHURAMAN,ANDSUNDERSETHURAMAN
2. CMPP models
The following “Poissonprocess” derivation of the distribution of Z with respect
to Bern(1,0) (cf. Example 1.1) motivates subsequent development.
Example 2.1. Consider the following standard way to generate a Bern(1,0) se-
quence. Let{β } be independent, identicallydistributed(iid) Uniform[0,1]ran-
i i≥1
dom variables, and define Y = I(β is a record), n ≥ 1. R`enyi’s theorem shows
n n
that {Y } are independent and P(Y = 1) = 1/n for n ≥ 1, that is Y =
n n≥1 n
Bern(1,0). Let{X } be the recordvalues among{β } . Notice that the point
i i≥1 i i≥1
process N on [0,1] defined by N(A) = δ (A) is a nonhomogeneous Poisson
i≥1 Xi
processon[0,1]withintensity1/(1−x)(cf. Resnick(1994)). ForeachpointX ,we
i
P
canassociateaGeometric(1−X )variableL (a“mark”)correspondingtothenum-
i i
ber of uniform random variables in {β } to the next record. Then, by thinning
i i≥1
decompositions,Z = I(L =k)= δ ([0,1])I(L =k)fork ≥1arein-
k i≥1 i i≥1 Xi i
dependentPoissonvariableswithrespectivemeans 1(1−x)−1xk−1(1−x)dx=1/k
P P 0
for k ≥1.
R
In a sense, the thrust of the following CMPP model and our main result (The-
orem 2.2) below is to reverse the procedure in Example 2.1. By beginning with a
givenPoissonprocessandspacingvariables,whichthemselvesdetermine the count
vector Z, we then see what associated Bernoulli sequence Y arises.
Considerasequenceofrandomvariables(X,L)={(X ,L )} onR×Nwhere
i i i≥0
N = {1,2,...}, and the point process N on R given by N(A)= δ (A). Let
i≥1 Xi
also g : R → [0,∞) be a probability density function (pdf), and for each x ∈ R
r(x,·),q(x,·):N→[0,1] be probability mass functions, and λ :PR→[0,∞) be an
x
intensity function.
Then, we say (X,L) is the conditional marked Poisson process M(g,r,λ,q) if
the following hold:
1. X has pdf g,
0
2. conditional on X = x , N is a nonhomogeneous Poisson process with
0 0
intensity function λ (·),
x0
3. P(L =k|X)=r(X ,k) for k ≥1, and
0 0
4. P(L =k|X,L ,L ,...,L )=q(X ,k) for k,n≥1.
n 0 1 n−1 n
LetL∗ =L ,andL∗ =L∗ +L forr ≥1. WenowdefineaBernoullisequence
0 0 r r−1 r
Y based on (X,L) as follows: Y = 1 if n is of the form L∗ for some r ≥ 0, and
n r
Y =0 otherwise. Another way to say this is
n
0 when n<L∗, or L∗ <n<L∗ for r ≥0
Y = 0 r r+1 (2.1)
n 1 when n=L∗ for r≥0.
(cid:26) r
Then, the count vector Z is given by
Z = I(L =k), for k ≥1. (2.2)
k n
n≥1
X
We note the zeroth mark L is not included in the above summation since any Y
0 i
with i < L is part of an initial segment of zeros of the sequence not preceded by
0
a 1, and so does not contribute to any d-string, for d≥1.
Theorem 2.2. Suppose λ (x)q(x,k)dx < ∞ for all w ∈ R and k ≥ 1. Then,
w
the count vector Z associated with sequence Y, defined through CMPP (X,L) =
R
BERNOULLI STRINGS AND POISSON PROCESSES 5
M(g,r,λ,q), is distributed as follows. Given the value X =x ,
0 0
d
Z = Po λ (x)q(x,k)dx .
x0
k≥1 (cid:18)Z (cid:19)
Y
Remark 2.3. The distribution of Z does not depend on the transition function r,
consistent with the discussion of L before the theorem.
0
Also, for a given k ≥ 1, Z is infinite with positive probability exactly when
k
there is a set B such that P(X ∈B)>0 and λ (x)q(x,k)dx=∞ for w ∈B.
0 w
Proof of Theorem 2.2. Recall the count vRector representation (2.2). Condi-
tional on X = x , the point process M on R × N given by M(A × {k}) =
0 0
δ (A)I(L = k) is a Poisson process on R × N with intensity function
i≥1 Xi i
λ (x)q(x,k) (cf. Proposition 4.10.1 (b) Resnick (1994)). Hence, it follows that,
gPixv0enX =x ,thevariablesM(R×{k})= I(L =k)=Z areindependent
0 0 n≥1 n k
Poissonvariables with respective means λ (x)q(x,k)dx, for k ≥1. (cid:4)
Px0
3. The sequeRnce Bern(a,b)
We now derive the count vector distribution for the sequence Bern(a,b) using a
CMPP model. Denote, as usual, for α,β >0, the Beta function
Γ(α)Γ(β)
B(α,β) = , (3.1)
Γ(α+β)
and let
1. g¯(x)=xb−1(1−x)a−1/B(b,a) on 0<x<1, the Beta(b,a) pdf,
2. r¯(x,k)=xk−1(1−x) for k≥1,
3. λ¯ (x)=[a/(1−x)]I(w <x<1), and
w
4. q¯(x,k)=xk−1(1−x) for k ≥1.
Proposition 3.1. The model (X,L) = M(g¯,r¯,λ¯,q¯) produces an independent
d
Bernoulli sequence Y = Bern(a,b) for a > 0 and b > 0 whose count vector
Z, conditional on the value x of a Beta(b,a) random variable, is distributed as
0
Po(a(1−xk)/k).
k≥1 0
Remark 3.2. As acorollary,by takingb↓0,we recoverthe countvectordistribu-
Q
d
tionforBern(a,0)alreadyconsideredintheliteratureassimplyZ= Po(a/k).
k≥1
Note that (X ,L )→(0,1) in distribution as b↓0.
0 0
The Poisson process in the above CMPP model with intensityQλ¯ (·) can be
w
generated in the following way. First, the point process formed by the record
values from an iid sequence of Beta(1,a) random variables is a Poisson process
with intensity a/(1−x), the Beta(1,a)failure rate (cf. Resnick (1994)Proposition
d
4.11.1(b)). Next,wethinthisprocessasfollows. LetX =Beta(b,a),and{X }
0 i i≥1
be the record values from an iid sequence of Beta(1,a) random variables, subject
to X > X for i ≥ 1. Then, conditional on X =x , the point process N¯ defined
i 0 0 0
by N¯(A) = δ (A) is the desired Poisson process with intensity function
i≥1 Xi
λ¯ (x)=[a/(1−x)]I(x <x<1).
x0 P 0
ProofofProposition3.1. Thesecondpartonthecountvectordistributionfollows
from Theorem 2.2, noting for k ≥1, that
1 1 a(1−xk)
λ¯ (x)q¯(x,k)dx = axk−1dx = 0 . (3.2)
x0 k
Z0 Zx0
6 FREDW.HUFFER,JAYARAMSETHURAMAN,ANDSUNDERSETHURAMAN
Forthefirstpart,weobservethatthedistributionof{Y } giventhrough(2.1)
i i≥1
is uniquely determined by the probabilities of cylinder sets of the form
E(k ,...,k )=(L =k ,L =k ,...,L =k ) (3.3)
0 n 0 0 1 1 n n
= Y =1 for t∈{K ,K ,...,K },and Y =0 otherwise for 1≤t≤K
t 0 1 n t n
where k(cid:16),k ,...,k are positive integers and K = k ,K = K +k ,...,K(cid:17)=
0 1 n 0 0 1 0 1 n
def
K + k are their partial sums. If the probability of sets of the form E =
n−1 n
E(k ,...,k ) is a product of appropriate marginal probabilities then {Y ,n ≥ 1}
0 n n
will be the Bernoulli sequence Bern(a,b). We will proceed to establish this.
LetA ={0<x <x <···<x <1}. UsingtheBetavariablesrepresentation
n 0 1 n
in Remark 3.2, write
n
P(E) = g¯(x )r¯(x ,k ) P(X ∈dx |X >x )q¯(x ,k ) dx .
0 0 0 i i i i−1 i i 0
ZAn iY=1h i
SinceP(X ∈dx |X >x )=a(1−x )a−1/(1−x )adx for1≤i≤n,wehave
i i i i−1 i i−1 i
further that the last line equals
an n
xb+k0−2 xki−1(1−x )adx ...dx (3.4)
B(b,a) 0 i n 0 n
ZAn i=1
Y
B(b+K −1,a+1) an
n
= ·
B(b,a) n−1(b+K −1)
s=0 s
and, noting (3.1) and αΓ(α)=Γ(α+1), that (3.4) becomes
Q
a Kn−2(b+r) an Kn b+i−1 n a
r=0 · =
rKQ=n0−1(a+b+r) sn=−01(b+Ks−1) i=1a+b+i−1r=0b+Kr−1
Y Y
whQich is exactly Ki=n1PQ(Yi = 0) nr=0[P(YKr = 1)/P(YKr = 0)] with Y specified
as Bern(a,b). (cid:4)
Q Q
4. The sequence Bern (a,b)
1
We will derive the count vector distribution for the sequence Bern (a,b), and
1
show a dichotomy depending on whether b≥1 or b<1. We first consider the case
where a>0 and b>1. Define
1. g∗(x)=xb−2(1−x)a/B(b−1,a+1)on0<x<1,the Beta(b−1,a+1) pdf,
2. r∗(x,1)=1,
3. λ∗(x)=[a/(1−x)]I(w <x<1), and
w
4. q∗(x,k)=xk−1(1−x) for k ≥1.
Proposition 4.1. The CMPP model (X,L)=M(g∗,r∗,λ∗,q∗) produces an inde-
d
pendent Bernoulli sequence Y =Bern (a,b) for a >0 and b >1, and, conditional
1
on a Beta(b−1,a+1) variable X = x , the distribution of its count vector Z is
0 0
Po(a(1−xk)/k).
k≥1 0
QRemark 4.2. Asa corollary,by takingb↓1,wefind the countvectordistribution
d
forBern (a,1)tobesimplyZ= Po(a/k). [Infact,Bern (a,1)coincideswith
1 k≥1 1
the sequence Bern(a,0) mentioned earlier in Remark 3.2.]
Q
Also, we note the Poisson process in the above CMPP model with intensity
λ∗ can be generated, as in Proposition 3.1, by taking X =d Beta(b −1,a+ 1),
0
BERNOULLI STRINGS AND POISSON PROCESSES 7
and {X } as the sequence of records from an iid sequence of Beta(1,a) random
i i≥1
variables, subject to the condition X >X .
1 0
ProofofProposition4.1. WeneedonlyestablishthedistributionofY,asthelast
statement follows from Theorem 2.2 and the computation (3.2). The calculations
are similar to the proof of Proposition 3.1. Let k = 1,k ,k ,...,k be positive
0 1 2 n
integers, and K = k = 1,K = K +k ,...,K = K +k be their partial
0 0 1 0 1 n n−1 n
sums. Recall the cylinder set defined in (3.3) and let
def
E = E(1,k ,...,k ) = (L =1,L =k ,...,L =k ),
1 1 n 0 1 1 n n
and set A = {0 < x < x < ··· < x < 1}. Write, using the construction in
n 0 1 n
Remark 4.2, that
1
P(E ) = xb−2(1−x )a ·1
1 B(b−1,a+1) 0 0
ZAn
n (cid:2) (cid:3)
× a(1−x )a−1/(1−x )a xki−1(1−x ) dx ...dx
i i−1 i i 0 n
i=1
Y(cid:2) (cid:3)(cid:2) (cid:3)
an n
= xb−2 xki−1(1−x )adx ...dx .
B(b−1,a+1) 0 i n 0 n
ZAn i=1
Y
Then, with (3.1) and αΓ(α)=Γ(α+1), the last line equals
B(b+K −2,a+1) an
n
·
B(b−1,a+1) (b−1) n−1(b+K −2)
s=1 s
= rK=n0−2(b−1+rQ) · an
Kn−2(a+b+r) (b−1) n−1(b+K −2)
Qr=0 s=1 s
KQn−1 b+i−1 n aQ
=
a+b+i−1 b+K −2
r
i=1 r=1
Y Y
which is exactly P(Y =1) Kn P(Y =0) n [P(Y =1)/P(Y =0)] with Y
1 i=2 i r=1 Kr Kr
specified as Bern (a,b). (cid:4)
1
Q Q
We now give the distribution of the count vector under Bern (a,b) for all a>0
1
andb≥0byconditioningonthelocationofthesecond1inthesequenceY. Denote
Z(a,b)asthecountvectorwithrespecttoBern1(a,b)fora>0andb≥0. LetWn
be the sequence whose nth co-ordinate is 1 and all the other co-ordinates are zero,
for n≥1. Let also
a for n=2
p = a+b
n a n−3 b+r for n≥3
(cid:26) a+b+n−2 r=0 a+b+r
be the probability that the second 1 inQBern1(a,b) occurs at time n≥2, and note
p =1.
n≥2 n
PProposition 4.3. For a>0 and b≥0, we have
L(Z(a,b)) = p L Z(a,b+n−1)+W , (4.1)
n n−1
nX≥2 (cid:16) (cid:17)
and Z(a,b+n−1), conditional on the value x of a Beta(b+n−2,a+1) random
0
variable, is distributed as Po(a(1−xk)/k), for b>0 and n≥2.
k≥1 0
Q
8 FREDW.HUFFER,JAYARAMSETHURAMAN,ANDSUNDERSETHURAMAN
Remark 4.4. The special case b = 0 is interesting. The sequence Bern (a,0) is
1
the independent sequence where Y = Y = 1 and P(Y = 1) = a/(a+n−2) for
1 2 n
n≥3. That is, starting from time n=2, the sequence is Bern (a,1)=Bern(a,0).
1
Hence,byProposition3.1(seeRemark3.2),Z(a,0)isdistributedasZˆ+W where
1
Zˆ =d Po(a/k) is the count vector for Bern(a,0). This agrees with (4.1), since
k≥1
p =1 (when b=0) and Z(a,1)=Zˆ.
2 Q
Proof of Proposition 4.3. The distribution of Z(a,b) follows by conditioning on
the first time that Y = 1 for n ≥ 2. The distributions of Z(a,b + n −1) are
n
completely specified by Proposition 4.1 and Remark 4.2, since b+n−1 ≥ 1 for
n≥2. (cid:4)
From (4.1), it is not clear whether the distribution of Z(a,b) is a mixture of
product Poisson factors or not for 0 ≤ b < 1. We show now that even the first
component Z (a,b) is not a mixture of Poissons when 0≤b<1.
1
Proposition 4.5. The distribution of Z ≡ Z (a,b), the count of 1-strings in the
1 1
Bern (a,b) sequence, is not a mixture of Poissons when 0≤b<1, that is, there is
1
no measure µ on [0,∞) such that
E exp{tZ } = ev(et−1)dµ(v). (4.2)
1
h i Z[0,∞)
Proof. It is well known that when (4.2) holds, the variable Z is over-dispersed,
1
def
that is O(Z ) = Var(Z )−E(Z ) ≥ 0. The proof now follows by the expression
1 1 1
for O(Z ) in (4.4) below. Let Y =Bern (a,b). Then,
1 1
Z = Y +Zˆ =Y +Y Y +Z+ (4.3)
1 2 1 2 2 3 1
where Zˆ = Y Y and Z+ = Y Y , and the latter is independent
1 i≥2 i i+1 1 i≥3 i i+1
of Y . Furthermore Zˆ , Z+ are the counts of strings of order 1 from Bern(a,b),
2 P 1 1 P
Bern(a,b+1),respectively,andtheirdistributionsareknownfromProposition3.1.
Hence, by easy calculations
E(Zˆ )= a2 , E(Z+)= a2 , E(Zˆ 2)= a3(a+1) + a2 .
1 (a+b) 1 (a+b+1) 1 (a+b)(a+b+1) (a+b)
From the identities in (4.3), we have
a(a+1) a(a+1) a2(a+1)(a+2)
E(Z )= , E(Z2)= + .
1 (a+b) 1 (a+b) (a+b)(a+b+1)
This leads to
a2(a+1)(b−1)
O(Z ) = (4.4)
1 (a+b)2(a+b+1)
which is negative for b<1, and positive for b>1. (cid:4)
5. Some dependent Bernoulli sequences
Two examples of dependent Bernoulli sequences, arising in CMPP models with
simple structures, whose count vector distributions are mixtures of independent
Poissonfactors are given.
First Sequence. For a > 0 and b > 0, denote P as the probability distri-
a,b
bution of the CMPP M(g¯,r¯,λ¯,q¯) described in Proposition 3.1 which gives rise to
BERNOULLI STRINGS AND POISSON PROCESSES 9
the Bernoulli sequence Bern(a,b). Let now r+(x,k) = kxk−1(1−x)2 for k ≥ 1.
Consider the associated CMPP model M(g¯,r+,λ¯,q¯) with g¯,λ¯,q¯ the same as in
Proposition 3.1. Denote the probability measure under this model as P+ =P+ .
a,b
Note that r+(x,k)=k[r¯(x,k)−r¯(x,k+1)] where r¯(x,k)=xk−1(1−x). Recall
def
the cylinder set E = E(k ,...,k ) from (3.3) where k ,k ,...,k are positive
0 n 0 1 n
integers, and K ,K ,...,K their partial sums. It is easy to see that
0 1 n
P+(E) = k P E(k ,...,k ) −P E(k +1,k ,...,k ) .
0 a,b 0 n a,b 0 1 n
h (cid:16) (cid:17) (cid:16) (cid:17)i
From this expression, the distribution of Y can be recovered,and shown to be not
that of independent Bernoulli variables. For instance,
a(a+1)
P+(Y =1) = P (Y =1)−P (Y =0,Y =1) = ,
1 a,b 1 a,b 1 2
(a+b)(a+b+1)
and analogously
a2(a+2)+2ba(a+1)
P+(Y =1) = .
2
(a+b)(a+b+1)(a+b+2)
Thus
a2(a+1)(a2+2a+2ba+2b)
P+(Y =1)P+(Y =1) = ,
1 2 (a+b)2(a+b+1)2(a+b+2)
which does not match
a2(a+2)
P+(Y =1,Y =1) =
1 2
(a+b)(a+b+1)(a+b+2)
for a,b>0.
Finally, by Remark 2.3, we note the count vectors under P and P+ have
a,b
the same distribution, and by Proposition 3.1 conditional on the value of x of a
0
Beta(b,a) variable, the count vectors are distributed as Po(a(1−xk)/k).
k≥1 0
Second Sequence. Consider P1,0, the measure for thQe CMPP model discussed
in Example 2.1 and Remark 3.2, with respect to Bernoulli sequence Bern(1,0),
where(X ,L )≡(0,1),{X } aretherecordsfromaniidUniform[0,1]sequence,
0 0 i i≥1
and L are Geometric(1−X ) for i≥1.
i i
LetP′ standforthemeasureunderthe“switched”CMPPmodelwhere(X ,L )
1 1
and (X ,L ) are interchanged. The probabilities of Y on cylinder sets (cf. (3.3),
2 2
under P′, is given by
P′ E(1,k ,...,k ) = P′(L =k ,...,L =k )
1 n 1 1 n n
(cid:16) (cid:17) = P (L =k ,L =k , and L =k for 3≤i≤n)
1,0 2 1 1 2 i i
for positive integers k = 1,k ,...,k , with K = 1,K = K + k ,...,K =
0 1 n 0 1 0 1 n
K +k as their partial sums. Under both models P and P′, as only two
n−1 n 1,0
terms (L ,L ) exchange places, the associated count vectors are the same, and by
1 2
Proposition 3.1 distributed as Po(1/k).
k≥1
We now show that {Y } is not an independent sequence under P′. From the
i i≥1
Q
calculation in (3.4) with (X ,L )≡(0,1), Y ≡1 and r¯(x,1)=1 (take b↓0), and
0 0 1
10 FREDW.HUFFER,JAYARAMSETHURAMAN,ANDSUNDERSETHURAMAN
a=1, we can write
P′(Y =1) = P (L =1) = P (L =k,L =1)
2 1,0 2 1,0 1 2
k≥1
X
= xk−1(1−x )dx dx = 1/4.
1 2 1 2
k≥1Z0<x1<x2<1
X
Also,
P′(Y =1,Y =1) = P (L =1,L =1) = P (Y =1,Y =1) = 1/6,
2 3 1,0 1 2 1,0 2 3
P′(Y =0,Y =1) = P (L =2)
2 3 1,0 2
= xk−1x (1−x )dx dx = 5/36,
1 2 2 1 2
k≥1Z0<x1<x2<1
X
which give P′(Y = 1) = 11/36. However, P′(Y = 1)P′(Y = 1) = 11/144 6=
3 2 3
1/6=P′(Y =1,Y =1).
2 3
References
[1] Arratia, R., Barbour, A.D. and Tavar´e, S. (1992) Poisson process approximations for the
Ewenssamplingformula.Ann.Appl.Probab.2519-535.
[2] Arratia, R., Barbour, A.D. and Tavar´e, S. (2003) Logarithmic Combinatorial Structures: A
ProbabilisticApproach.EuropeanMathematical Society, Zu¨rich.
[3] Arratia,R.,andTavar´e,S.(1992)Thecyclestructureofrandompermutations.Ann.Probab.
201567-1591.
[4] Chern,H.-H.,Hwang,H.-K.andYeh,Y.-N.(2000)Distributionofthenumberofconsecutive
records.RandomStructures andAlgorithms17169-196.
[5] Feller,W.(1945)Thefundamentallimittheoremsinprobability.Bull.Amer.Math.Soc.51
800-832.
[6] Ghosh,J.K.,andRamamoorthi,R.V.(2003)BayesianNonparametrics,SpringerVerlag,New
York.
[7] Holst, Lars (2007) Counts of failure strings in certain Bernoulli sequences. to appear in J.
Appl.Probab.
[8] Joffe,A.,Marchand,E.,Perron,F.andPopadiuk,P.(2004)OnsumsofproductsofBernoulli
variablesandrandompermutations. JournalofTheoreticalProbability17285-292.
[9] Kolchin, V.F. (1971) A problem of the allocation of particles in cells and cycles of random
permutations.TheoryProbab.Appl.1674-90.
[10] Korwar,R.M.,and Hollander,M.(1973) Contributions tothe theory ofDirichletprocesses.
Ann.Probab.1705-711.
[11] Mo´ri, T. F. (2001) On the distribution of sums of overlapping products. Acta Scientarium
Mathematica(Szeged) 67833-841.
[12] Resnick,S.I.(1994)Adventures inStochastic Processes.Second Ed.Birkha¨user,Boston.
[13] Sethuraman, Jayaram and Sethuraman, Sunder (2004) On counts of Bernoulli strings and
connections to rank orders and random permutations. In A festschrift for Herman Rubin.
IMSLectureNotes MonographSeries45140-152.
Departmentof Statistics, Florida StateUniversity,Tallahassee, FL 32306.
e-mail: [email protected]
Departmentof Statistics, Florida StateUniversity,Tallahassee, FL 32306.
e-mail: [email protected]
Department of Mathematics, 396 Carver Hall, Iowa State University, Ames, IA
50011,USA.
e-mail: [email protected]
