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Jie Hao and Anant Godbole, ETSU Abstract Poisson, Etc Probability Distributions Motivated by Permutations and Featuring Telescoping Series Jie Hao and Anant Godbole, ETSU GENERALIZED TELESCOPING POISSON: This is a toy example. The probability that a random permutation on THE TELESCOPING GEOMETRIC DISTRIBUTION: The following k Abstract [n] has its first ascent at positions k; k + 1 is given, for 1 ≤ k ≤ n − 1, by (k+1)!. To see this, choose any one of the is a very natural question: in how many permutations, in which the k + 1 elements in positions 1 through k + 1, except for the smallest, to occupy the k + 1st position, and then arrange In this poster, we study the simplest type of infinite telescoping sum, longest increasing subsequence is of length 2, does the first ascent occur the other elements in a monotone decreasing fashion. But, to reformulate this argument via telescoping series, we let namely in positions k; k + 1? Catalan [2] proved the k-fold Catalan convolution 1 1 A be the event that the permutation restricted to the first k integers is monotone decreasing, note that A ⊆ A X X k k+1 k formula pn = (qn − qn+1) = 1; and that the probability f(k) that the first ascent is at k; k + 1 satisfies, for k ≤ n − 1, k ! n=1 n=1 X Y k 2n − k due to our assumptions that 1 = q ≥ q ≥ q ::: ≥ 0 and lim q = 1 1 k Cn;k := Cir−1 = ; 1 2 3 n!1 n f(k) = P (A \ AC ) = − = ; 2n − k n 0. Now the above clearly defines a probability distribution on Z+, but k k+1 k! (k + 1)! (k + 1)! i1+:::+ik=n r=1 1 making unmotivated choices such as q = 4 2 would serve little and in [3] it was proved that there are precisely C permutations on n 3n −2n as before. The chance that the first ascent is at position n is, of course, 1 . Letting n ! 1, we get the discrete n;k purpose, however, so let us state our objective: We wish to propose n! [n] with longest increasing subsequence of size 2 and for which the first distribution on Z+ given by three probability distributions that each x ascent occurs at positions k; k + 1. Thus, for a randomly chosen 123- f(x) = ; x = 1; 2;:::; (1) (i) are similar in form to familiar discrete distributions; (x + 1)! avoiding permutation on [n], the distribution of the location of first which is quite similar to the unit Poisson distribution on 0,1,... with mass function f(x) = e−1=x!. If X is the ascent is given by (ii) arise from the cycle or pattern containment structure of random 1 corresponding first ascent random variable generated, e.g., by a sequence fXng of i.i.d. uniform random variables, C (2n − k − 1)!(n + 1)! permutations; and n=1 f(k) = n;k = k ; k = 1; 2; : : : ; n; then Cn (2n)!(n − k)! (iii) lead to open-ended sets of questions. 1 X x k E(X) = x × which, for small k and large n, may be approximated by f(x) = k+1: Let us go one step further, by asking (x + 1)! 2 x=1 Accordingly, let us define the telescoping geometric-like distribution on = e − 1; (iv) that there be events An; n ≥ 1 in some probability space so that Z+ = 1; 2;::: by q = P (A ) with A ⊆ A so that, as we shall see, p ; n ≥ 1 provides n n n+1 n n and x x + 1 x + 2 a model for first occurrence distributions of certain events. f(x) = x+1 = x − x+1 ; x = 1; 2;:::: 2 2 2 2 2 The viewer of this poster should note that discrete models, other than Var(X) = E(X ) − (E(X)) 2 We see that for a telescoping geometric random variable X, those provided by random permutations, could equally well have been = (e + 1) − (e − 1) used as launching pads for our discrete distributions, and s/he is invited = e(3 − e); 1 x E(X) = X x × to come up with well-motivated first occurrence distributions along the 2x+1 More generally, the m.g.f. of X is given by x=1 lines of the ones in this paper. = 3: 1 X x E(etX) = etx × Thus for a random permutation on a large [n], we expect the first ascent (x + 1)! Poisson, etc. x=1 to be at position e − 1 ≈ 1:718, whereas this value increases to 3 for a ZETA: The zeta distribution with parameter k > 1 has associated prob- −t et t+et = e (1 − e + e ); random 123-avoiding permutation. Moreover, motivated by size-biased ability mass function (pmf) given by compared to the m.g.f. expfet − 1g of the unit Poisson r.v. A θ analog has also been studied by us. permutations, we define a telescoping θ-analog of the above distribution 1 1 defined by f(n) = · ; n = 1; 2;:::; THE TELESCOPING ZETA(2) DISTRIBUTION: There are (n − 1)! permutations on [n] that are unicyclic. Now if ζ(k) nk a permutation on [n + 1] is unicyclic, then its reduction to [n] is also unicyclic. For example, the permutation (θ − 1)2x (θ − 1)x + 1 (θ − 1)(x + 1) + 1 f(x) = = − ; x = 1; 2; : : : ; θ > 1; P1 1 θx+1 θx θx+1 where ζ(k) = n=1 nk is the Riemann zeta function. 123456 POISSON: The Poisson distribution with parameter λ > 0, denoted by 452613; ^ 2 and its properties can be studied, e.g., the MLE of θ θ = 1 + X¯ −1. Po(λ), is the distribution of a random variable X with pmf written in \two line notation," can be reduced, on eliminating the 6 in both the first and second rows and mapping 4 The MOM estimator is the same. Finally consider the null hypothesis H : θ = θ versus the alternative hypothesis H : θ = θ . Assume e−λλx to 3 directly, as the unicyclic 5-permutation 0 0 1 1 f(x) = ; x = 0; 1;:::: 12345 that θ1 > θ0. By the Neyman Pearson theorem, the most powerful test x! ¯ 45231; rejects H0 for small values of X, where the critical value is determined and the so-called moment generating function (m.g.f.) of X is by the level of significance used. a general definition of reduction can now be easily formulated. Let n ≥ 1 and consider π 2 Sn+1. On setting An tX t to be the event that the reduced permutation on [n] is unicyclic, we have that A ⊆ A , and so P (A \ AC ) = MX(t) = E e = expfλ(e − 1)g: n+1 n n n+1 P (An) − P (An+1). It follows that References (n − 1)! n! 1 GEOMETRIC: The geometric random variable counts the number of − = [1] M. B´ona(2004). Combinatorics of Permutations, Chapman and independent Bernoulli trials with success probability p that need to be n! (n + 1)! n(n + 1) Hall, Boca Raton. conducted in order to get the first success. We have, for x ≥ 1, is the probability that π 2 Sn+1 has a unicyclic reduction but we have π(n+1) = n+1, so that π itself is not unicyclic, and has an n cycle on [n] followed by a 1-cycle on fn + 1g. Given a random permutation in S , we let X = 1; 2; : : : ; n [2] E. Catalan (1887). Sur les nombres de Segner, Rend. Circ. Mat. 1 1 − p n f(x) = (1 − p)x−1 · p; E(X) = ; Var(X) = : be the largest i ≤ n such that the reduction of π to [i] is unicyclic. We then have Palermo 1, 190{201. p p2 1 1 1 [3] S. Connolly, Z. Gabor, and A. Godbole (2014+). The location of pi = P (X = i) = − = ; i = 1; 2; : : : ; n − 1; (2) the first ascent in a 123-avoiding permutation, submitted, available i i + 1 i(i + 1) at http://arxiv.org/pdf/1401.2691.pdf with 1 P (X = n) = : n.
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