Record Values, Poisson Mixtures, and the Joint
Distribution of Counts of Strings in Bernoulli
Sequences
Fred W. Huffer
(joint work with Jayaram Sethuraman and Sunder
Sethuraman)
Talk based on:
Huffer, F.W., Sethuraman, J., and Sethuraman, S. A study of
counts of Bernoulli strings via conditional Poisson processes.
Proceedings of the American Mathematical Society., 137, No. 6,
2125–2134 (2009).
Huffer, F.W., and Sethuraman, J. Joint distributions of counts of
strings in finite Bernoulli sequences. To appear in Journal of
Applied Probability , 49 No. 3 (2012).
Let:
I
U1 , U2 , U3 , . . . be iid continuous random variables.
I
Y1 , Y2 , Y3 , . . . be Bernoulli rv’s which indicate the position of
the record values in U1 , U2 , U3 , . . ., that is,


1 if Uj sets a new record
Yj =
(if Ui < Uj for all i < j),


0 otherwise.
It is well known that:
Y1 , Y2 , Y3 , . . . are independent.
1
I P (Yj = 1) =
for all j.
j
P −1
P
Since
j = ∞, we know
Yj = ∞ almost surely, that is,
there will be infinitely many record values.
I
Example: A sequence of i.i.d. exponentials U1 , U2 , . . . and the
corresponding record indicators Y1 , Y2 , . . ..
The first 30 are shown with records colored red.
1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0
0
5
10
15
20
25
30
Example continued: The same sequence with each record marked
beneath by its duration (how long it stands).
1 10
1 1 4
0
5
145
7
10
15
20
25
30
The duration for a record value Ui is how long the record stands,
that is, the number of trials required to break the record.
Let Zk be the total number of records (over all time) having
duration k.
A Surprising Theorem:
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Zk ∼ Poisson(1/k) for all k, and
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Z1 , Z2 , Z3 , . . . are independent.
This theorem (or various parts of it, and also various extensions of
it) has been proved by various people:
Original Theorem:
I
Goncharov (1944) [in Russian], Zk ∼ Poisson
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Kolchin (1971, Theory Probab. Appl.), Zk independent
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Hahlin (1995) [tech report], Z1
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Diaconis (1996?) [unpublished], Z1
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Emery (1998) [unpublished], Z1
Extensions:
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Arratia-Barbour-Tavare (1992, Ann. Appl. Prob.), Bern(a, 0)
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Joffe-Marchand-Perron-Popadiuk (2004, J. Theoretical Prob.),
Z1 for Bern(1, b).
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Sethuraman and Sethuraman (2004, Rubin Festschrift, IMS),
Bern(1, b)
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Holst (2007, J. Appl. Prob.), Bern(a, b)
The same sequence Y1 , Y2 , Y3 , . . . arises in other contexts:
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The cycles of random permutations of {1, 2, . . . , n}.
Here (Yn , Yn−1 , . . . , Y2 , Y1 ) indicate the endings of cycles in
Feller’s construction of a random permutation.
I
In a special case of the Polya-Blackwell-MacQueen urn
scheme for generating a Dirichlet process.
Here Yi indicates the first appearance of a new value on the
i-th draw from the urn.
Therefore the same theorem holds with a different interpretation.
For example:
(n)
Let Ck be the number of cycles of length k in random
permutation of the integers {1, 2, . . . , n}. Then
(n)
(n)
(n)
d
(C1 , C2 , . . . , Cm ) −→ (Z1 , Z2 , . . . , Zm ) as n → ∞.
How to prove the Theorem?
Various analytic and combinatorial approaches.
One approach is via moments: Show that the moments and joint
moments of (Z1 , . . . , Zm ) coincide with those of independent
Poisson rv’s.
P
P
1
= 1.
Example: Z1 = j Yj Yj+1 so that EZ1 = j 1j · j+1
A simpler approach:
Independent Poisson rv’s arise naturally in the context of
Poisson processes.
So try to embed our problem in a Poisson process.
Poisson Process (PP) with intensity λ(·)
Let 0 < X1 < X2 < X3 < · · · be arrival times (or points).
Let N(B) be the number of arrivals during B ⊂ R:
P
N(B) = i I (Xi ∈ B)
A PP N(·) with intensity λ(·) satisfies:
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For all intervals B,
Z
N(B) ∼ Poisson(Λ(B))
where Λ(B) ≡
λ(u) du .
B
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For any disjoint intervals B1 , B2 , . . . , Bk ,
N(B1 ), N(B2 ), . . . , N(Bk ) are independent.
Note: λ(t) is the rate of arrivals at time t:
P[arrival in (t, t + dt)] ≈ λ(t) dt
Decomposition of a Poisson Process (a marked PP)
Given: a Poisson process N with intensity λ.
Suppose that each point Xi is independently assigned a positive
integer “mark” Li , according to
P(Li = k | Xi = x) = q(x, k) .
The mark assigned to Xi depends only on the value of Xi , and not
on the other points Xj or mark values Lj (j 6= i).
Let Nk , k ≥ 1, be the process consisting of those points assigned a
mark of k:
P
Nk (B) = i I (Xi ∈ B, Li = k) .
The processes N1 , N2 , N3 , . . . are independent, and, for all k,
Nk is a PP with intensity λk (x) ≡ λ(x)q(x, k).
Thus, for any interval B,
N1 (B), N2 (B), N3 (B), . . . are independent Poisson rv’s with means
Z
ENk (B) =
λk (u) du .
B
Taking B = (0, ∞) counts all the arrivals.
P
Define Zk = Nk ((0, ∞)) = i I (Li = k). We conclude
Z1 , Z2 , Z3 , . . . are independent Poisson rv’s with means
Z ∞
EZk =
λk (u) du .
0
A Marked Poisson Process obtained from the record values of a
sequence of iid exponentials
111
6
46 1
15
1
1
115
1
0
6
1
46
20
40
60
111
80
100
Proof:
By the memoryless property of the exponential distribution, when a
new record is set, it breaks the old record by an amount which is
Exp(1).
Therefore, the spacings between the record values are iid
exponential.
So the record values X1 , X2 , X3 , . . . form a PP with rate 1.
(Remark: If the Ui ’s are iid from some other continuous
distribution, then the record values X1 , X2 , . . . form a
nonhomogeneous Poisson process with intensity function equal to
the hazard function of the distribution.)
Let Lj be the duration of the record Xj .
Given Xj = x, the number of trials Lj required to break this record
is the number of exponential rv’s we must observe until one
exceeds x. Thus Lj has a geometric distribution with pmf
P(Lj = k | Xj = x) = (1 − e −x )k−1 e −x , k ≥ 1
≡ q(x, k) .
This also holds conditional on the earlier records (Xi , Li ), i < j.
We have assigned each Poisson point Xj a mark Lj according to
q(x, k).
Thus, the points marked k form a PP with intensity
λk (x) = 1 · q(x, k).
Since Zk is the number of points marked k, by the result on
decomposing Poisson processes, Z1 , Z2 , Z3 , . . . are independent
Poisson rv’s with
Z ∞
EZk =
(1 − e −x )k−1 e −x dx
0
∞
(1 − e −x )k 1
=
=
k
k
0
QED
By modifications of the Poisson decomposition argument, we can
also prove the various extensions of the theorem.
Unfortunately, in our proofs of the extensions, we lose the simple
“record value” interpretation.
Generalizations of the Theorem (to Bern(a, b) sequences)
Assume a > 0, b ≥ 0.
Let Y1 , Y2 , Y3 , . . . be independent Bernoulli rv’s with
a
a Bern(a, b)
P(Yn = 1) =
for n = 1, 2, 3, . . .
sequence.
a+b+n−1
(The earlier case was a = 1, b = 0.)
Let Zk be the number of occurrences of pairs of successes which
are k apart, that is, pairs of successes separated by exactly k − 1
failures:
P
Z1 =
I {(Yi , Yi+1 ) = (1, 1)}
Pi
Z2 =
I {(Yi , Yi+1 , Yi+2 ) = (1, 0, 1)}
Pi
Z3 =
i I {(Yi , Yi+1 , Yi+2 , Yi+3 ) = (1, 0, 0, 1)}
etc.
• Bern(a, 0) sequences arise in connection with “biased
permutations” studied by Arratia-Barbour-Tavare (which relate to
the Ewen’s sampling formula).
• Bern(a, b) sequences arise in Polya-Blackwell-MacQueen urn
scheme.
Generalizations of Theorem
• If b = 0, then Z1 , Z2 , Z3 , . . . are independent Poisson rv’s with
EZk =
a
.
k
• (Holst, 2007) If b > 0, then the joint distribution of
(Z1 , Z2 , Z3 , . . .) can be expressed as a mixture of independent
Poissons:
One can construct a probability space with a random
variable W ∼ Beta(b, a) such that conditional on
W = w , the rv’s Z1 , Z2 , Z3 , . . . are independent Poisson
rv’s with Zk having (conditional) mean λk given by
λk =
a(1 − w k )
.
k
This result is an instance of the
Mixture of Independent Poissons (MIP) property: there is
a random variable (or vector) W conditioned on which
the count variables Z1 , Z2 , . . . are independent Poisson.
Finite Bernoulli Sequences
Holst (2007, 2008) studied finite Bernoulli sequences.
Let Bern(a, b, n) denote the distribution of the finite initial segment
(Y1 , . . . , Yn ) of a Bern(a, b). Holst obtained expressions for:
• joint factorial moments of (Z1 , Z2 , . . . , Zk ) from Bern(a, 0, n),
• factorial moments of Z1 from Bern(a, b, n).
Note: Now Z1 ≡
n−1
X
I {(Yi , Yi+1 ) = (1, 1)}, etc.
i=1
In this case, the Zi ’s are no longer exactly Poisson or exactly
independent or expressible as a MIP.
In our latest work, we:
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Consider initial segments (Y1 , Y2 , . . . , Yζ ) of Bern(a, b) which
are finite but have a random length ζ.
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Describe a general class of random variables ζ for which the
counts Z1 , Z2 , . . . retain the MIP property.
I
For this class of ζ, give a general expression for the joint
factorial moments of (Z1 , Z2 , . . . , Zk ).
Method of Proof: extension of the Poisson process embedding
argument.
Simplest case of our general class of ζ:
If ζ is independent of Y1 , Y2 , . . . and has a geometric distribution
(allowing ζ = 0), then the MIP property holds.
(We obtain the mixture explicitly. See next slide.)
Suppose P(ζ = k) = (1 − ξ)ξ k for k ≥ 0 where 0 < ξ < 1.
By specializing our results to this case and using a power series
expansion in ξ, we deduce the conditional moments of Z1 , Z2 , . . .
given {ζ = n}.
This yields the results of Holst for Bern(a, b, n) and Bern(a, 0, n),
and also yields a general expression for the joint factorial moments
for Bern(a, b, n).
Explicit formulas for mixture when b = 0
• Conditional on {T =t}, the counts Z1 , Z2 , . . . are independent
Poisson random variables with means given by
E (Zk | T =t) =
aξ k t k
k
for k = 1, 2, 3, . . .
• The rv T satisfies P(0 ≤ T < 1) = 1 and
ξ (1 − t)a
P(T > t) =
(1 − ξ t)a
for 0 ≤ t < 1.
(When b > 0 the formulas are messier.)
Another member of our general class of ζ
Observe the sequence Y1 , Y2 , . . . and define
ζ = first time we observe m consecutive zeros
= inf{j : Yj−m+1 = Yj−m+2 = · · · = Yj−1 = Yj = 0} .
When the Yj ’s indicate record values in an iid sequence (i.e.,
a = 1, b = 0), this ζ corresponds to stopping the competition when
a record has stood for m trials.
Explicit formula for mixture when b = 0
• Conditional on {T =t}, the counts Z1 , Z2 , . . . are independent
Poisson random variables with means given by
E (Zk | T =t) =
at k
I (k ≤ m)
k
• The rv T satisfies P(0 < T < 1) = 1 and
Z t
ax m
P(T > t) = exp −
dx
2
0 (1 − x)
for 0 ≤ t < 1.
The general class of ζ (preserving the MIP)
Let:
Y1 , Y2 , Y3 , . . . be a Bern(a, b) sequence.
L0 , L1 , L2 , . . . be the waiting times (number of trials) between
successes (1’s) in this sequence.
Example:
00010000001 1 001000000000001···
L0
L1
L2 L3
L4
L0 = 4, L1 = 7, L2 = 1, L3 = 3, L4 = 12
Let φ = (φ1 , φ2 , . . .) and ψ = (ψ1 , ψ2 , . . .) be arbitrary sequences
of constants in [0, 1].
Suppose we observe the values Y1 , Y2 , Y3 . . . one by one.
Every time a ‘1’ is observed, a random decision is made whether or
not to halt (stop observing) the sequence at that point, with the
halting probability depending only on the waiting time since the
previous ‘1’.
Two Cases:
I
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Halt at L0 with probability 1 − φL .
0
For n ≥ 1, halt at
n
X
Li with conditional probability 1 − ψL .
n
i=0
Let τ be the halting point.
Now back up 1: define ζ = τ − 1.
Theorem: The sequence Y1 , Y2 , . . . , Yζ has counts Z1 , Z2 , . . .
satisfying the MIP property.
(We give an explicit mixture.)
In our two examples
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ζ ∼ Geometric: equivalent to taking φk = ψk = ξ k for all k.
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Stop if m consecutive zeros: equivalent to taking
φk = ψk = 1 for k ≤ m and φk = ψk = 0 for k > m.
Idea of Proof: (in Bern(a, 0) case with L0 = 1)
Consider a marked Poisson process
0 < X1 < X2 < X3 < · · ·
L1
L2
L3
···
(points)
(marks)
with intensity λ∗ (·) and mark distribution q ∗ (x, k).
Define a Bernoulli sequence Y1 , Y2 , . . . with 1’s in positions
K0 = L0 , K1 = L0 + L1 , K2 = L0 + L1 + L2 , . . .
P
The n-th ‘1’ is in position Kn = ni=0 Li for n = 0, 1, 2, . . .
Let T > 0 be a rv independent of the marked PP.
Suppose we observe only the initial part of the marked PP: just the
points Xi less than T .
This leads to a finite Bernoulli sequence Y1 , Y2 , . . . , Yζ consisting
of the initial part of Y1 , Y2 , . . ..
Here ζ = τ − 1 where τ = inf{Ki : Xi ≥ T } .
For this finite sequence, the counts are:
X
Count points marked k
Zk =
I (Li = k)
occurring before T .
i : i≥1,
Xi <T
Since
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The points marked k form a PP with intensity λ∗ (x)q ∗ (x, k)
(independent for each k), and
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T is independent of the marked PP,
we obtain the MIP property:
Conditional on T ,
I
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Z1 , Z2 , Z3 , . . . are independent Poisson rv’s
RT
E (Zk |T ) = 0 λ∗ (x)q ∗ (x, k) dx
If λ∗ , q ∗ , and T are chosen in just the right way, the finite
sequence Y1 , . . . , Yζ coincides with that obtained in the sequential
halting process (based on the φ and ψ) in our Theorem.
Choosing in “just the right way” requires the “death mark”
construction in our forthcoming JAP paper.
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counts of Bernoulli strings via conditional Poisson processes. Proc. Amer.
Math. Soc. 137, 2125-2134.
Huffer, F.W., and Sethuraman, J. (2012) Joint distributions of counts of
strings in finite Bernoulli sequences. To appear in Journal of Applied
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