Applications of Fuss-Catalan Numbers to Success Runs of Bernoulli Trials

In a recent paper, the authors derived the exact solution for the probability mass function of the geometric distribution of order k, expressing the roots of the associated auxiliary equation in terms of generating functions for Fuss-Catalan numbers. This paper applies the above formalism for the Fuss-Catalan numbers to treat additional problems pertaining to occurrences of success runs. New exact analytical expressions for the probability mass function and probability generating function and so forth are derived. First, we treat sequences of Bernoulli trials with r ≥ 1 occurrences of success runs of length k with l-overlapping. The case l < 0, where there must be a gap of at least |l| trials between success runs, is also studied. Next we treat the distribution of the waiting time for the rth nonoverlapping appearance of a pair of successes separated by at most k − 2 failures (k ≥ 2).


Introduction
In a recent paper [1], the authors derived the exact analytical solution for the probability mass function of the geometric distribution of order .The roots of the auxiliary equation of the associated recurrence relation were derived in terms of generating functions for Fuss-Catalan numbers.(See the text by Graham et al. [2] for details about Fuss-Catalan numbers.)In this paper, we employ our formalism for the Fuss-Catalan numbers to treat additional problems pertaining to occurrences of success runs in sequences of Bernoulli trials.Throughout our paper, we treat only sequences of independent identically distributed (i.i.d.) Bernoulli trials with constant success probability  (and failure probability  = 1−).The theory of success runs is discussed extensively in the texts by Balakrishnan and Koutras [3] and Johnson et al. [4].Our formalism provides a new perspective to treat problems of success runs in sequences of Bernoulli trials and complements and extends results derived by previous authors (especially Feller [5,).Citations and comparisons to the works of others will be presented in Sections 3 and 4, after we have derived our results.
We treat two main problems in this paper.First, we consider sequences with multiple  ≥ 1 occurrences of success runs of length .The success runs are permitted to overlap, with a maximum of ℓ ≥ 0 overlaps between success runs.This is known as "ℓ-overlapping."The case ℓ < 0 is perhaps surprising at first sight but is also of interest.In this case there must be a gap or "buffer" of at least |ℓ| trials (of arbitrary outcomes) between success runs.We call this "|ℓ|-buffering."We also consider the scenario in which the length of the sequence  is held fixed and the number of success runs  ≥ 0 is allowed to vary.This is the binomial distribution of order  with ℓ-overlapping success runs.An encyclopedia article on binomial distributions of order  has been published by Philippou and Antzoulakos [6].Using Fuss-Catalan numbers, we present new concise expressions for the probability mass functions of these distributions.
In Section 4 we study a different problem.We analyze the distribution of the waiting time for the th nonoverlapping appearance of a pair of successes separated by at most  − 2 failures ( ≥ 2).Our main reference for this problem is the elegant analysis by Koutras [7], who also gives an excellent bibliography on the subject.For  = 1 and  ≥ 2, the problem is a special case of the detection waiting game when a 2-outof- moving (or sliding) window detector is employed.See Koutras [7] for additional details and references.Note that

Notation and Definitions
We summarize the basic notation and definitions presented in our earlier paper [1].For a sequence of independent identically distributed Bernoulli trials with success probability  (and failure probability  = 1 − ), let   be the waiting time for the first run of  consecutive successes.Then   is said to have the geometric distribution of order .This distribution was studied by Feller in his classic text [5, pp. 322-326].It is also known as the negative binomial distribution of order  with parameter (1, ); see Philippou [8].The probability mass function   of   satisfies the recurrence relation, for  > ,   () =   ( − 1) +   ( − 2) +  2   ( − 3) + ⋅ ⋅ ⋅ +  −1   ( − ) . (1) The initial conditions are   () = 0 for  = 1, . . .,  − 1 and   () =   .We define the auxiliary polynomial The auxiliary equation is A , () = 0. We will drop the subscripts  and  unless necessary.Feller [5] proved that the roots of the auxiliary equation are distinct and also that there is a unique positive real root, and it lies in (0, 1), and the real positive root has a strictly larger magnitude than all the other roots.Additional properties of the roots were derived in [1].We denote the roots by   (, ),  = 0, 1, . . ., −1, where  0 is the positive real root.We call  0 the "principal root" and the other roots "secondary roots."Unless required, we will omit the arguments  and .It is useful to multiply A() by ( − ) to obtain the polynomial Remark 1 (Fuss-Catalan numbers and roots of auxiliary polynomial).Relevant definitions, formulas, and identities for the Fuss-Catalan numbers can be found in the text by Graham et al. [2].The Fuss-Catalan numbers are given by The first form (finite product) is valid in general, while the second form (Gamma functions) is well defined provided ] +  ̸ = 0.The generating function of the Fuss-Catalan numbers is  ] () and [2, p. 363] We will also require the following formula: It was proved in [1] that, for all 0 <  < 1, For /( + 1) <  < 1, the above expression also applies for  0 , while, for 0 <  < /( + 1), For ease of reference, we list several relevant properties of the roots in the following.The proofs of all the results were given in [1], or references cited therein, and are omitted in the following.
Remark 2. All the roots of the auxiliary equation are distinct.Remark 3.For  ∈ (0, 1), the auxiliary equation has a unique positive real root, which lies in (0, 1).We denote the positive real root by  0 , or  0 (, ), as stated above.For any  ∈ (0, 1), exactly one of the three following statements is true: Remark 4. For  ∈ (0, 1), the principal root  0 has a strictly greater magnitude than all the other roots of the auxiliary equation; that is, 0 < |  | <  0 < 1, where   ∈ C \  0 is a root of A().We employ the term "secondary roots" for the set {  ,  = 1, . . .,  − 1}.
Remark 6.For  ∈ (0, 1), let () denote the set of  + 1 roots of the equation In addition to the above properties of the roots, we will also need the following two results, which were not proved in [1], as well as a lemma about sums of series.
Proposition 7 (distinctness of roots for different ).Consider fixed  ∈ (0, 1).Suppose   is a root of the auxiliary equation A() = 0 for  =  1 .Then   is not a root for any other value of .
Lemma 9.For  ≥  ≥ 1 and  ̸ = 1, Next, for  ≥  ≥ 0 and  ̸ = 1, The expressions on the right hand sides of both equations are clearly well defined for all  ̸ = 1.

Multiple 𝑟 ≥ 1 Success Runs in Sequences of Bernoulli Trials
3.1.Probability Generating Function, Mean, and Variance.We now turn to the first problem of interest in this paper, namely, the waiting time to obtain  > 1 success runs of length .
We begin by displaying the following expressions for the case  = 1.They were derived by Feller [5] and will be required in the following.

Journal of Probability and Statistics
We now calculate the probability generating function, mean, and variance for multiple  > 1 overlapping runs.The success runs have length  and there can be at most 0 ≤ ℓ <  overlaps between consecutive success runs.We denote the waiting time by  ,,ℓ .The case ℓ = 0 of nonoverlapping runs was extensively analyzed by Philippou [8], who named the distribution as the negative binomial distribution of order  with vector parameter (, ).Ling (1989) [9] and Hirano et al. [10] derived results for the special case ℓ =  − 1.The text by Balakrishnan and Koutras [3] lists the special cases ℓ = 0 and ℓ =  − 1 as, respectively, Type I and Type III negative binomial distributions of order .
Proof.Define   (=  ,,ℓ ) as the waiting time to complete  success runs.So suppose we have completed  − 1 success runs.Hence by definition the last  trials are all successes.
Then exactly one of the following  − ℓ + 1 mutually exclusive events will occur: (i) The next  − ℓ trials are all successes.This yields the th success run.
(ii) The next  trials are successes, followed by a failure, where  = 0, 1, . . .,  − ℓ − 1.Then we restart the waiting time for the next success run from scratch (conditioned on an initial failure).We denote this additional waiting time by   .Clearly,   has the same distribution as  1 .
Since the events are mutually exclusive, we add the probabilities to obtain Now set  =   and note that   () = E(exp(  )).Hence we obtain the following recurrence relation and solution for   (): Define  as the term in the brackets.After some tedious algebra we obtain In the last line it is necessary to exhibit the dependences on  and ℓ explicitly.Then (24) follows immediately.
Proposition 12 (domain of convergence).The probability generating function for  ≥ 1 success runs  ,,ℓ () converges for Hence the domain of convergence of the probability generating function is the same for all  ≥ 1.
Proposition 13 (mean and variance).The mean  ,,ℓ and variance  2 ,,ℓ for the waiting time for  ≥ 1 success runs are given by Proof.We put  =   and differentiate with respect to  and evaluate at  = 0. We differentiate ln   (  ) to obtain Evaluating at  = 0 and noting that (1, ) = (1, ℓ) = 1 yield This proves (30a).We differentiate again to obtain We again evaluate at  = 0 to obtain This proves (30b).
We now show that various results derived by other authors are special cases of our results above.As stated above, the case ℓ = 0 of nonoverlapping runs was solved by Philippou [8], while Ling (1989) [9] and Hirano et al. [10] treated the case ℓ =  − 1 of overlapping runs.
Remark 15 (Ling (1989)).Ling (1989) Proof.Ling wrote  ()  () and  () 1 () where we have written  ,,ℓ () and (), but the connection between the notations is clear.Setting ℓ =  − 1 in (26a) yields This yields (37a).Next (37b) follows immediately by solving the recurrence relation Similar to Philippou [8], Ling (1989) They also wrote Then they derived the solution They also gave expressions for the mean and variance.The mean is Proof.The connection between the notations is that they write    () and   1 () where we write   () and (), respectively.They employ  as the independent variable, where we use .It is simple to derive that their expression for   1 () in (42) equals that for (, ) in (22).Next, (40) is simply (39) with the changes of notation listed above.Next, setting ℓ =  − 1 and changing the independent variable from  to , This is exactly (42).From (30a), the mean for ℓ =  − 1 is This is exactly (43).Hirano et al. [10] also displayed an expression for the variance.The proof of equivalence with our expression involves merely tedious algebra and is omitted.

Probability Mass Function.
We derive an expression for  ,,ℓ (), the probability mass function (p.m.f.) that the th success run of length  with ℓ-overlapping occurs at the th Bernoulli trial, where  ≥ 1 and  ≥ 1. Clearly  ,,ℓ () = 0 for  < ℓ + ( − ℓ) =  − ( − 1)ℓ and  ,,ℓ () =  −(−1)ℓ for  = −(−1)ℓ.An expression for the p.m.f. for the case  = 1 was derived in [1].By definition, the probability generating function is related to the probability mass function via We derived an expression for  ,,ℓ () above and we will use it to derive an expression for  ,,ℓ () in the following.From the second form for (, ) in (22), with  = 1/, Hence for  ≥ 1 success runs, The right hand side is a rational function of two polynomials.From Proposition 7, the auxiliary polynomials A ,ℓ () and A , () have no roots in common.Furthermore, because ℓ < , the numerator polynomial is of a lower degree than the denominator polynomial.We also know that all the roots of the auxiliary polynomials are distinct.Hence we can expand  ,,ℓ () as a sum of partial fractions with repeated roots (of the denominator polynomial) Here the coefficients   are parameters which depend on , , and ℓ but not on .For brevity, we drop the subscripts  and ℓ on  ,,ℓ and also write the roots as   in the following.
The coefficients   can be evaluated explicitly in terms of the roots {  (, ),  = 0, . . .,  − 1} via the standard residues formula Returning to the use of  = 1/, we see that We expand the right hand side using the negative binomial theorem and equate  ,,ℓ () to the coefficient of   .
Proposition 17.The probability mass function  ,,ℓ () for the ℎ success run of length  with ℓ-overlapping is given by Hence  ,,ℓ () is given by a sum of exactly  terms, independently of .Recall from above that   () = 0 for  < −(−1)ℓ so the above formula is only required for  ≥ −(−1)ℓ; hence the binomial coefficients are well defined.
The derivation of the above expression has already been given above, where all notation has been defined and explained.

Binomial Distribution of Order 𝑘 with ℓ-Overlapping.
Consider a sequence of Bernoulli trials of fixed length  > 0 and let  ,,ℓ denote the number of success runs of length  with a maximum of ℓ overlaps between success runs.This is the binomial distribution of order  with ℓ-overlapping success runs and has been reviewed in the encyclopedia article by Philippou and Antzoulakos [6].Good overviews have also been given by Makri and Philippou [11] and Makri et al. [12]; see the bibliographies in both references.Ling (1988) [13] introduced the case of ℓ =  − 1 and called it the "Type II binomial distribution of order ." The case  = 0 is the probability that the longest success run in the first  trials has length less than .It is also known as the probability that the waiting time to attain the first success run of length  exceeds  trials.This scenario has been solved by many authors.For example, Feller [5] presented an asymptotic solution in terms of the principal root.In our paper [1], we extended Feller's solution to include all the roots.Solutions have also been derived by Burr and Cane [14], Godbole [15], Philippou and Makri [16], and Muselli [17], all of whom expressed their results using (possibly nested) binomial or multinomial sums.
Proof.We solve the problem as follows.Suppose the th success run is completed on the th trial.Then by definition the outcomes of the last  trials are all successes.There are now two mutually exclusive and exhaustive possibilities, according as the ( − 1)th success run is contiguous with (and possibly overlaps) the th success run, or there is at least one failure between the runs.
(i) In the former case, the (−1)th success run terminates at the trial  −  + ℓ. (This event is null if  = 1.)The outcome of the ( − )th trial is a success.
(ii) In the latter case, the outcome of the ( − )th trial is a failure.The first  −  − 1 trials contain exactly  − 1 success runs.

Success
Runs with ℓ < 0. The case ℓ < 0 is not without interest.In this scenario, there must be a gap or buffer of at least |ℓ| trials (of arbitrary outcomes) between success runs.We call this scenario "|ℓ|-buffering."First, we derive the probability mass function, probability generating function, mean, and variance of the negative binomial distribution of order  for  ≥ 1 success runs of length  with ℓ-overlapping.Next, we treat sequences with a fixed total length  and study the binomial distribution of order  with buffer |ℓ|.The value of  of the number of success runs spans the interval 0 ≤  ≤ ⌊( − ℓ)/( − ℓ)⌋.This is the same formula as for ℓ ≥ 0. We derive an expression for the probability mass function for the above distribution.
Most of the published literature for the case ℓ < 0 has treated sequences of fixed length .Inoue and Aki [21, Section 4.2] published results for sequences of Markov trials.They derived an expression for the p.g.f. as a nested sum of multinomial terms [21,Proposition 4].Han and Aki [22] treated sequences of i.i.d.Bernoulli trials.They derived a recurrence relation for the p.g.f.[22,Theorem 4.1].
Proof.We omit the indices  and ℓ in the following.Consider   (), where  success runs have taken place, ending at trial .Hence the last  outcomes are all successes.We then have the following mutually exclusive and exhaustive possibilities.(Note that if ℓ = −1, this set is empty.)The probability of this event is The events are mutually exclusive and exhaustive (noting that there can be at most one success run completed from trials  −  + ℓ + 1 through  −  − 1 because of the buffering requirement); hence we add the probabilities to obtain Rearranging terms and replacing  by  +  − ℓ and  by  + 1 yield (62).For ℓ = −1, the last sum is absent and the above expression is the same as (53).

Pairs of Successes Separated by At Most 𝑘−2 Failures
In this section we study a different problem.We treat the distribution of waiting time for the th nonoverlapping appearance of a pair of successes separated by at most  − 2 failures ( ≥ 2).Our main reference is the elegant analysis by Koutras [7], who also gives an excellent bibliography on the subject.To avoid cluttering the notation in this paper with too many symbols, we will reuse some of the symbols such as  , () for the probability mass function, and so forth.It should be understood that we are treating a new problem, and the following notation is self-contained.We begin with  = 1.Koutras [7] gave a recurrence relation for the probability mass function  , ().We will suppress the indices  and  unless required.We derive the exact solutions for the roots of the auxiliary polynomial associated with the recurrence relation, in terms of Fuss-Catalan numbers.We also derive various pertinent properties of the roots.We then solve a Vandermonde matrix system of equations to derive an expression for the p.m.f. as a sum over powers of the roots.We also derive an expression for the probability of the waiting time to exceed  trials.Let us denote the waiting time by  , .Note that Koutras [7] writes  , , but we write  , to maintain consistency with the notation in the earlier parts of our paper.We begin with the case  = 1 and drop the subscripts.
The initial conditions are [7, eq.3.2] The auxiliary polynomial associated with the above recurrence relation is The auxiliary equation is A() = 0.
Proposition 23 (properties of roots).For fixed 0 <  < 1, the roots of the auxiliary polynomial have the following properties: (a) There are no repeated roots.
(b) There is a unique positive real root.
(d) If  is odd, there are no other real roots.If  is even, there is exactly one negative real root.
(e) The magnitude of the positive real root exceeds that of all the other roots.
Proof.Both A() and A  () must vanish simultaneously at a repeated root.Next Hence A  () vanishes at  = 0 (not a root of A()) or  = ( − 1)/.Note that 0 < ( − 1)/ < .Now for 0 <  < , Hence for 0 <  < , Hence A() ̸ = 0 for  = ( − 1)/.Hence A() has no repeated roots.Next note that A(0) = − −1 , A() = − −1 , and A(1) = (1 −  −1 ).Hence A() has an odd number of positive real roots for  ∈ (, 1).Now from (67), A  () > 0 for  > (1 − 1/), so A  () > 0 for  > .It follows that A() has exactly one positive real root, and it lies in the interval  ∈ (, 1).Also if  is odd then A() < 0 for  < 0 and there are no negative real roots.If  is even then A() increases as  decreases through negative values; hence for even , A() has exactly one negative real root.Next, if   is a root, by the triangle inequality, Hence The inequality is strict unless   is real and positive (so that both    and  −1  are real and positive) and we have shown that there is only one real positive root.Hence the real positive root has a larger magnitude than all the other roots.
We will call the positive real root the "principal root" and refer to all the other roots as "secondary roots."We will denote the roots by   (, ),  = 0, . . .,  − 1, where the principal root is  0 .Although our the following analysis is for 0 <  < 1, it is helpful to note the following limiting cases for  = 0 and  = 1.
As already stated earlier in this paper, details about the Fuss-Catalan numbers can be found in the text by Graham et al. [2].See the expressions above, in (4), (5a), and (5b), which will be essential in the following.We require the following lemma for the domain of convergence of the generating functions of the Fuss-Catalan numbers.
Lemma 26 (domain of convergence).The generating function of the Fuss-Catalan numbers  ] () converges for The proof was derived in our earlier paper [1,Corollary 22 and Proposition 23] and is omitted.The proof was actually derived for  +1 () but note (this is important) that  does not have to be an integer; hence we write ] above.It was also proved in [1] that the domain of convergence includes the circle of convergence.

(76b)
For  * ≤  ≤ 1, the principal and secondary roots are given by For  =  * , either set of solutions may be employed.
We now derive an expression for the probability mass function in terms of a sum over the roots.

Proposition 28. The probability mass function 𝑓(𝑛) is given by
Next we treat the case of  > 1 nonoverlapping appearances of a pair of successes separated by at most  − 2 failures.Following Koutras [7], the probability mass function and probability generating function are defined via (also following Koutras we drop the subscript  and write   =  , , etc.) Following the procedure in the earlier part of our paper, we will derive an expression for the p.m.f.  () via the p.g.f.  () and a partial fraction decomposition.Since the enumeration is nonoverlapping and the trials are i.i.d.random variables, the p.g.f. is given by [7, Theorem Hence   () is given by a sum of exactly  terms, independently of .Note that   () = 0 for  < 2, because of the nonoverlapping enumeration.An alternative method to derive the probability mass function   () for  > 1 was given by Koutras, where the p.m.f. is obtained via a recurrence relation (unnumbered equations after Theorem 4.1 in [7]).Our expression in (110) is essentially the solution of that recurrence.
Next consider a fixed number of trials .Following Koutras [7], let  , denote the number of occurrences of a strand of  (at most) trials containing two successes in the first  outcomes.Then [7, eq.(112) We now use ( 17) and ( 18) to evaluate the sum over  (note that  ≤  so all the binomial coefficients are nonzero, also Because each   is itself a sum of  terms, the number of summands on the right hand side is ( 2 ).However, the overall computational complexity is independent of .

( i )
The outcome of trial  −  is a success.Then the ( − 1)th success run must end at trial  −  + ℓ.The |ℓ| trials in the sequence from  −  + ℓ + 1 through  −  constitute the buffer between the two success runs.The probability of this event is  +1  −1 (−+ℓ).(This event is null if  = 1.Note that  0,,ℓ ( −  + ℓ) = 0.) (ii) The outcome of trial  −  is a failure.Then we must attain  − 1 success runs by trial  −  + ℓ.However, we must subtract the possibility that the th success run ends at one of the trials  −  + ℓ + 1 through  −  − 1.
sum of exactly  terms, independently of .Note also that   ̸ = 1 for 0 ≤  ≤  − 1; hence the right hand side is well defined for all the roots.Substituting in (112) yields  ( , ≥ ) =  .