In this work, we discuss Bayesian estimation of multinomial probabilities associated with a finite alphabet A under incomplete experimental information. Two types of prior information are considered: (i) number of letters needed to see a particular pattern for the first time, and (ii) the fact that for two fixed words one appeared before the other.Patterns, Stopping times, Incomplete experimental information

Brani Vidakovic

Pepa Ramirez

English

Research Papers in Economics

 
 
Working Paper 08-05 
Statistics and Econometrics Series 03 
February 2008 
Departamento de Estadística 
Universidad Carlos III de Madrid
Calle Madrid, 126
28903 Getafe (Spain)
Fax (34) 91 624-98-49
 
ON BAYESIAN ESTIMATION OF MULTINOMIAL PROBABILITIES UNDER 
INCOMPLETE EXPERIMENTAL INFORMATION
1 
 
Pepa Ramírez and Brani Vidakovic
2 
 
 
 
 
 
Abstract 
 
In this work, we discuss Bayesian estimation of multinomial probabilities associated 
with a finite alphabet A under incomplete experimental information. Two types of prior 
information are considered: (i) number of letters needed to see a particular pattern for 
the first time, and (ii) the fact that for two fixed words one appeared before the other. 
 
 
 
 
Keywords: Patterns, stopping times, incomplete experimental information. 
 
 
 
 
 
                                                           
1 We appreciate the comments by Michael P. Wiper. 
2 Corresponding author: Pepa Ramírez, Universidad Carlos III, Departamento de Estadística, C/ Madrid 
126. 28903 Getafe (Madrid), España, e-mail: jrcobo@est-econ.uc3m.es; Brani Vidakovic, Department of 
Biomedical Engineering, 313 Ferst Drive, Suite 2127, Atlanta, GA 30332-0535 USA, e-mail: 
brani@bme.gatech.edu. 1 Introduction
In this work we take Bayesian point-of-view in the following context. We con-
sider a discrete random variable L that, with probabilities p = (p1,...,pd),
takes the values (l1,...,ld), that is, P(L = lj) = pj,j = 1,...,d. One can
think of the values lj as letters or symbols in an alphabet A. Let An be the
set of all words of length n in A. To generate a particular word ω = ω1ω2 ...ωn,
one selects letters from A sequentially, starting with ω1, and ending with ωn.
Several problems arise from this setting. For instance, Example 7.2.(a) in Ross
(1983), shows how to compute the expected number of symbols N needed to
be observed until some given sequence appears. In addition, the author calcu-
lates the probability that for two given patterns one occurs before the other.
This is done by use of martingales and reformulating the problem in the terms
of fair casino gambling. It can be proven that in this context, N represents a
stopping time related to a certain martingale. Bizley (1972) also considers the
problem of determining the expected number of trials (or symbols) required to
obtain a speciﬁc pattern and gives expressions in terms of probability gener-
ating functions. DasGupta (1993) is also an example in which stopping times
in the Bayesian context are used.
In this note, we estimate the probabilities (p1,...,pd) in Bayesian fashion,
given the stopping time Nω which is deﬁned as the number of observed symbols
until the sequence ω appears for the ﬁrst time.
The organization is as follows. Section 2 provides necessary deﬁnitions and an
important theorem due to Bizley (1972). In Section 3, we develop Bayesian
inference for the Bernoulli model (case for d = 2) and generalize it to the
2multinomial case, providing some examples. Section 4 considers the update of
probabilities with a diﬀerent limited experimental information. A brief con-
clusion is given in Section 5.
2 Theoretical background
We deﬁned A = {l1,l2,...,ld} as the alphabet consisting of d diﬀerent letters
and An as the set of all words of length n in A. We assumed that letter lj is
selected with probability pj, independently of the previously selected letters.
Let Ω = A∞ be the sample space, and F = σ(∪k∈Nσ(Ak)) a σ-algebra on Ω,
where σ(Ak) is the σ-algebra generated by all subsets of Ak. Then, (Ω,F)
is a measurable space. Let Fk = σ(∪i≤kσ(Ai)), k = 1,2,... be a sequence
of nested σ-algebras. In a sequential generation of word ω = ω1ω2 ..., the
statement ρ ∈ Fk means that information wether the pattern ρ appeared up
to index k is available. The set of σ-algebras {Fk}k>1 forms a ﬁltration on
(Ω,F), that is
Fk ⊆ F and k1 6 k2 ⇒ Fk1 ⊆ Fk2.
Deﬁnition 1 Let (Ω,F) be a measurable space and {Fk}k>1 a ﬁltration on it.
A random variable N is said to be a stopping time if, for every n, the event
{N ≤ n} is Fn-measurable.
Let µ be a measure deﬁned of the measurable space (Ω,F). For a word ω =
ω1ω2 ...ωn ∈ An assume that P(ωi = lj) = pj, for i = 1,...,n with
Pd
j=1 pj =
1. A simple measure on Ω can be deﬁned by means of cylindric sets.
Deﬁnition 2 The cylindric set corresponding to the word x = x1x2 ...xn is
3given by Γx = {ω ∈ Ω|ω1 = x1,...,ωn = xn}.
The probability of word x is in fact formally deﬁned as the probability of
Γx ⊆ Ω and then µ(Γx) =
Qn
i=1 P(xi), where P(xi) is the probability of symbol
xi. The measure µ on Ω introduced as above is known as ‘Bernoulli’ measure.
Various other measures are possible. The following deﬁnition can be found in
Bizley (1972).
Deﬁnition 3 The symbol xi in the word x = x1x2 ...xn is said to be critical
if x1x2 ...xi = xn−i+1 ...xn.
Notice that the symbol xn is always critical as x1 ...xn = x1 ...xn For in-
stance, in the word 1010, second and fourth symbols are critical. In the word
1001, ﬁrst, and fourth symbols are critical while in the word 100000, only the
last symbol is critical. Critical symbols in the word ABRACADABRA are
the ﬁrst, fourth and eleventh.
Assume that the word ω contains m critical symbols. Denote by aij the number
of letters li up to and including the jth critical symbol. For example, in the
word 1010, l1 = 0,l2 = 1,a11 = 1,a21 = 1,a12 = 2, and a22 = 2.
Theorem 4 (Bizley, 1972) The number Nω of consecutive symbols needed
to complete a given pattern ω is a stopping time with probability generating
function
Gω(s) =

1 + (1 − s)
m X
j=1
s
−
P
aij
d Y
i=1
p
−aij
i


−1
, (1)
and expectation
E(Nω) =
m X
j=1
d Y
i=1
p
−aij
i . (2)
4For example, for ω = 010, a11 = 1,a21 = 0,a12 = 2, and a22 = 1. The
probability generating function is
G010(s) =
"
1 + (1 − s)
 
1
(1 − p)s
+
1
(1 − p)2ps3
!#−1
,
and
E(N010) =
1
1 − p
+
1
p(1 − p)2. (3)
3 Bayesian inference
Our goal is to estimate the vector of unknown probabilities p = (p1,...,pd),
pj = P(ωi = lj), in a Bayesian way. The only information we have about the
experiment is the stopping time Nω, the number of symbols observed to obtain
the pattern ω for the ﬁrst time. If the prior distribution for p is π(p), then
using the Bayes formula we can compute the posterior distribution as
π(p|Nω) =
P(Nω|p)π(p)
P(Nω)
. (4)
The estimator of p is the posterior mean. We will illustrate the proposed
Bayesian inference on Binomial and Multinomial cases, i.e, when the alphabet
contains two and more than two symbols.
3.1 Binomial case
Suppose that our alphabet consists only of two letters, say 0,1 correspond-
ing to failure and success. Let p be the probability of success. To be non-
informative, assume that the prior distribution for p is Beta(α,β) with pa-
5rameters α = β = 1, which represents a uniform distribution on (0,1). Notice
that a priori, E(p) = 1/2. Given that we know the stopping time N, we can
estimate p and assess how the available information changes our prior belief.
Several simple examples are provided.
Example 1. Pattern 010 was observed for ﬁrst time after N010 = 10 trials.
Expression (3) is 10 for p = 1/2. The probability P(N010 = 10|p) is com-
puted as
G
(10)
010 (0)/10! = (1 − p)
2p(1 − 6p + 21p
2 − 41p
3 + 52p
4 − 39p
5 + 16p
6 − 3p
7)
which is plotted in Fig 1. The Bayes estimate of probability of success is
equal to
E(p|N010) =
R 1
0 pP(N010|p)π(p)dp
R 1
0 P(N010|p)π(p)dp
= 0.418838.
In order to check sensitivity with respect to the prior we computed the
Bayes estimator for a range of values of α = β. As expected, Fig 2 demon-
strates that the posterior expectation for p tends to 0.5 as α increases.
Example 2. Pattern 011 was observed ﬁrst time after N=10 trials. In this
case, the probability generating function and expected value of N are
G011(s) =
 
1 + (1 − s)
1
(1 − p)p2s3
!−1
and E(N011) =
1
(1 − p)p2.
P(N011 = 10|p) = p
2 − p
3 − 5p
4 + 10p
5 − 2p
6 − 9p
7 + 9p
8 − 3p
9,
thus, E(p|N011) = 0.573.
FIGURES 1 AND 2 ABOUT HERE
63.2 Multinomial case
If we consider the general case where the alphabet consists of d letters, l1,...,ld
with vector of probabilities p = (p1,...,pd), the previous calculations can be
generalized in a straightforward way. The natural prior distribution for p is
the Dirichlet distribution with a density
π(p) = π(p|α1,...,αd) =
Γ(
Pd
i=1 αi)
Qd
i=1 Γ(αi)
d Y
i=1
p
αi−1
i , αi > 0,
d X
i=1
pi = 1.
Suppose that Nω, the number of consecutive symbols to complete a given
pattern ω is known to be n. Then, from (1), P(Nω = n|p) = G(n)
ω (0)/n!. By
Bayes formula (4), we obtain the posterior distribution:
π(p|Nω = n) =
P(Nω = n|p)π(p)
R
P(Nω = n|p)π(p)dp
. (5)
To ﬁnd the expected posterior probability for symbol li we compute
E(pi|Nω = n) =
1 Z
0
piπ(pi|Nω = n)dpi, (6)
where the marginal posterior π(pi|Nω = n) is obtained from (5).
Example 3. As an illustration, we take the case where d = 3, and assume
that the letters are 0,1 and 2. We are interested in Bayesian estimation of
p1,p2,p3 = 1 − p1 − p2, given that the word ω = 012 appeared for the ﬁrst
time after N012 = 10 trials. This word has only one critical symbol (the last
one), so a11 = a12 = a13 = 1 and thus, the probability generating function
and expected value are
G012(s) =
s3p1p2p3
1 + s3p1p2p3 − s
, E(N012) =
1
p1p2p3
.
7The likelihood P(N012 = 10|p1,p2,p3) = p1p2p3(1 − 5p1p2p3 + 3p2
1p2
2p2
3), for
some values of p is depicted in Fig 3.
FIGURE 3 ABOUT HERE
If prior paramaters are, for instance α1 = α3 = 3 and α2 = 5, then the
prior is π(p) ∝ p2
1p4
2p2
3 and the expected a priori probabilities are E(p1) =
E(p3) = .2727 and E(p3) = .4545, that is, we favored symbol 1 against
0 and 2, before observing Nω = 10. Applying expression (6) we ﬁnd that
a posteriori, the expected values change to E(p1|Nω = 10) = E(p3|Nω =
10) = .2843 and E(p2|Nω = 10) = .4312. Thus, having observed the pattern
012 for the ﬁrst time after 10 trials, our beliefs about p1,p2 and p3 are
modiﬁed: the expected values of p1 and p3 increase and that of p2 decreases.
4 Updating p if the word ω appeared before the word ρ
In this Section we consider a diﬀerent type of limited experimental infor-
mation. The only information available is the fact that a particular word ω
appeared before another ﬁxed word ρ (we will denote this as ω ≺ ρ). Let Nω
and Nρ be as before and Nρ|ω be the number of additional observations neces-
sary to obtain the word ρ when the word ω is just observed. The deﬁnition of
Nω|ρ is analogous. To apply Bayes formula and update p, the following result
can be utilized.
Theorem 5 P(ω ≺ ρ|p) =
E(Nρ) + E(Nω|ρ) − E(Nω)
E(Nω|ρ) + E(Nρ|ω)
. (7)
Proof: The proof is straightforward. Denote by a∧b the minimum of a and b.
If Nρ is represented as Nρ = (Nρ ∧Nω)+Nρ −(Nρ ∧Nω), and Nρ −(Nρ ∧Nω)
8is conditioned on the event {ω ≺ ρ}, it follows
E(Nρ)=E(Nρ ∧ Nω) + E(Nρ − (Nρ ∧ Nω)|ω ≺ ρ)P(ω ≺ ρ|p)
=E(Nρ ∧ Nω) + E(Nρ|ω)P(ω ≺ ρ|p). (8)
Equivalently, E(Nω) = E(Nρ ∧ Nω) + E(Nω|ρ)P(ρ ≺ ω|p).
Since P(ρ ≺ ω|p) = 1 − P(ω ≺ ρ|p), by solving (8) and the counterpart for
E(Nω), we get (7). 2
4.1 Calculating Nρ|ω
In order to calculate Expression (7) in Theorem 5 we need a eﬃcient way of cal-
culating E(Nω|ρ) and E(Nρ|ω). Suppose that ω = ω1ω2 ...ωn1 is observed and
assume that the expected number of trials to obtain the word ρ = ρ1ρ2 ...ρn2
is of interest. Let j be such that ωn1−j+1 ...ωn1 = ρ1 ...ρj. We assume j > 0,
since if j = 0, ρ does not overlap ω, and E(Nρ|ω) = E(Nρ). In addition, we also
assume that j < min{n1,n2}. If n2 ≤ n1 and j = n2, then we already observed
ρ in ω and Nρ|ω = 0. If n1 ≤ n2 and j = n1, then E(Nρ|ω) = E(Nρj+1...ρn2).
Theorem 6
E(Nρ|ω)=(n2 − j) + (1 − P(ρj+1 ...ρn2))(E(Nρ) − (n2 − j)) + (9)
+
n2−j X
i=1
iP(ρj+1ρj+2 ...ρj+i−1)Q(ρj+i).
Proof: Denote by P(ωiωi+1 ...ωk) the probability of the pattern ωiωi+1 ...ωk,
and Q(·) = 1 − P(·). Then,
E(Nρ|ω)=Q(ρj+1)(1 + E(Nρ)) + P(ρj+1)Q(ρj+2)(2 + E(Nρ)) + ...
+P(ρj+1ρj+2 ...ρn2−1)Q(ρn2)(n2 − j + E(Nρ)) (10)
+P(ρj+1ρj+2 ...ρn2−1ρn2). 2
9Example 4. As an illustration, consider the case where ω = 1001, and ρ =
010. The information we have about the experiment is that ω appeared
before ρ. If the probability of the symbol ‘1’ is p, then
E(N1001) =
1
p
+
1
p2q2, and E(N010) =
1
q
+
1
pq2.
Also, E(N010|1001) = 1+pE(N010) and E(N1001|010) = p(1+E(N1001))+q2(2+
E(N1001))+2pq. Thus, P(1001 ≺ 010|p) = p2(3−3p+p2)/(1+2p2−2p3+p4)
(see Fig.4). Under the uniform prior on p, E(p|ω ≺ ρ) = .666884.
FIGURE 4 ABOUT HERE
5 Summary and future directions
In this work we took Bayesian point of view to infer the probabilities of let-
ters in an alphabet A. First, we considered the case where the information
about the experiment is the value of a stopping time variable, representing
the number of trials necessary to obtain a ﬁxed pattern. Several examples for
the Bernouilli and multinomial models are provided. Second, we update the
probabilities of letters if we know that a particular pattern appeared before the
other ﬁxed pattern. Our results help in inference of binomial and multinomial
probabilities when the information from the experiment is limited.
10References
Bizley, M. T. L., Patterns in repeated trials., J. Inst. Actu., 1972, London, 88,
360-366.
DasGupta, A., Rinott, Y., & Vidakovic, B., Stopping times related to diag-
nostics and outliers. Presented at the Bayesian BOAT conference, Basel
1993.
Ross, S., Stochastic Processes, 1983, John Wiley & Sons.
11

On Bayesian estimation of multinomial probabilities under incomplete experimental information

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On Bayesian estimation of multinomial probabilities under incomplete experimental information

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