The truncated geometric election algorithm: Duration of the election

1. Motivation This is a sequel paper to Kalpathy and Ward (2014), which also appears in Statistics and Probability Letters. The earlier paper introduces a randomized leader election algorithm, in which a truncated geometric number of participants survive from round to round. (Louchard and Prodinger (2009), constitutes a good starting point for readers who are unfamiliar with randomized leader election algorithms; they also have extensive references, to provide a broader context.) Kalpathy and Ward (2014) precisely analyze the number of rounds that a particular contestant survives in the election. That analysis, however, had three key shortcomings. (1) It also focused on only the duration of a particular contestant; it did not analyze the length of the entire election (which should be more interesting and more useful in practice). (2) It only analyzed the mean and variation of the duration of a particular contestant (who was not necessarily the winner), but it did not discuss the asymptotic distribution of the duration (the analysis contained in the present investigation is more informative and useful). (3) The method of election in the earlier paper did not guarantee a unique winner. The present paper addresses all three of these issueswith the earlier paper.We focus on the duration of the entire election, not just of one participant. We go beyond the mean and variance, and analyze also the asymptotic distribution of the entire election. We also analyze two variants of the election, namely, the version from the original paper, and also a new style of election in which a unique winner is guaranteed to appear at the end of the election process. ∗ Corresponding author. E-mail addresses: [email protected] (G. Louchard), [email protected] (M.D. Ward). http://dx.doi.org/10.1016/j.spl.2015.02.018 0167-7152/© 2015 Elsevier B.V. All rights reserved. G. Louchard, M.D. Ward / Statistics and Probability Letters 101 (2015) 40–48 41 As another point of motivation for this sequel paper, the authors discover (in the proofs of Theorems 3.1 and 3.4) that the total length of each of these styles of election, when starting with n participants, can be decomposed into a sum of n− 1 independent random variables that do not have identical distributions. This surprising point about the decomposition of the length of the whole election is enlightened in the analysis by using moment generating functions. More precisely (in the proof of Theorem 3.1), from the representation of φn(t) := E(etXn) for n ≥ 2, shown in Eq. (1), it is surprising to see a decomposition of Xn in a sum of n − 1 independent random variables. Indeed, we see that Xn has the same distribution as Z2 + Z3 + · · · + Zn, where the Zj’s are independent, non-negative random variables, and Z2 has moment generating function φ2(t) = etp(1+q) 1−q2(q+etp) , while Zj has moment generating function 1−qj 1−qj(q+etp) (for 3 ≤ j ≤ n). An analogous decomposition holds in the proof of Theorem 3.4, because as we see in Eq. (5), we have a decomposition of Yn into a sum of n − 1 independent, non-negative random variables that do not have identical distributions. 2. Definitions We consider elections for which, if n contestants are present in a round, then Kn contestants proceed to the next round, where Kn is a truncated geometric random variablewith parameters p and q := 1 − p. So the mass of Kn is P(Kn = l) = pq 1 − qn+1 , for l = 0, 1, . . . , n. We study the number of rounds needed for the election is two distinct situations. The difference occurs in how the one of the base conditions is handled, namely, what happens when Kn = 0. 1. Setup #1. As in Kalpathy and Ward (2014), if Kn = 0 in one of the rounds, the election stops, and the remaining n participants can all be treated as winners, or (alternatively) all considered as losers, but the main point is that no additional rounds of the election take place. (It is easy to show that the Kalpathy–Ward election, starting with n participants, will end without a unique winner with probability 1 1+q n j=3 1−qj (1−pqj−qj+1) .) 2. Setup #2. In the literature, it is very common that, when all of the current participants fail to advance to the next round, all of them are given another chance to participate in one more (renewal) round. We define Xn and Yn as the number of rounds, when starting with n participants, for the elections given in setups #1 and #2, respectively. For example, consider an election with 20 initial participants. If 8 survive in round 1 (and 12 are eliminated), and 5 survive in round 2 (and 3 are eliminated), and 0 survive in round 3 (all 5 are eliminated), then X20 = 3 because 3 rounds were needed for the election. On the other hand, in such a situation, setup #2mandates that the 5 participants from round 3 are all resurrected and get to continue for subsequent rounds. Suppose that these 5 participants are resurrected, and exactly 1 of the 5 survives in round 4. Then Y20 = 4 because 4 rounds were needed for the election. 3. Main results First we give some results about the Kalpathy–Ward model from Kalpathy and Ward (2014) (Setup #1). Theorem 3.1. The expected number of rounds E(Xn) in an election (in setup #1), for n ≥ 2, is E(Xn) = p + ∞ 