Generalized Fibonacci numbers and extreme value laws for the R\'enyi map

In this paper we prove an extreme value law for a stochastic process obtained by iterating the R\'enyi map. Haiman (2018) derived a recursion formula for the Lebesgue measure of threshold exceedance sets. We show how this recursion formula is related to a rescaled version of the $k$-generalized Fibonacci sequence. For the latter sequence we derive a Binet formula which leads to a closed-form expression for the probability of partial maxima of the stochastic process. The proof of the extreme value law is completed by deriving sharp bounds for the dominant root of the characteristic polynomial of our Fibonacci sequence.


Introduction
Extreme value theory for a sequence of i.i.d. random variables (X i ) studies the asymptotic distribution of the partial maximum M n = max(X 0 , . . . , X n−1 ) as n → ∞. Since the distribution of M n has a degenerate limit it is necessary to consider a rescaling. Under appropriate conditions there exist sequences a n > 0 and b n ∈ R for which the limiting distribution of a n (M n −b n ) is nondegenerate. As an elementary example, assume that the variables X i ∼ U(0, 1) are independent. Then with a n = n and b n = 1 it follows for λ > 0 that lim n→∞ P(a n (M n − b n ) ≤ −λ) = lim More generally it can be proven that extreme value distributions for i.i.d. random variables are either a Weibull, Gumbel, or Fréchet distribution [6,7,17]. For extensions of extreme value theory to dependent random variables, see [11].
In the last twenty years, the applicability of extreme value theory has been extended to the setting of deterministic dynamical systems such as iterated maps or differential equations. This development has important applications in the context of climate modelling. Typically, the approach is to show that two mixing conditions are satisfied which guarantee that an extreme value law for a time series generated by a dynamical system can be obtained as if it were an i.i.d. stochastic process. An example of this approach applied to the tent map can be found in [5]. For more details on the subject of extremes in dynamical systems the interested reader is referred to the recent monograph [13] and the extensive list of references therein.
In this paper we consider the Rényi map [16] given by where β > 1. This map is an active topic of study within the field of dynamical systems and ergodic theory. In the special case β = 2 the map f is also known as the doubling map, which is an archetypical example of a chaotic dynamical system of which the properties can be studied in great detail [2]. Other applications involve the study of random number generators [1] or dynamical systems with holes in their state space [10]. From now on we will assume that β ≥ 2 is an integer. In this case the Lebesgue measure is an invariant probability measure of the map f : Proof. For u ∈ [0, 1] we have that P(X ∈ [0, u)) = u. This gives which implies that f (X) ∼ U(0, 1).
For non-integer values of β > 1 the map f can have an invariant measure that is different from the Lebesgue measure; for the case β = ( √ 5 + 1)/2 see [3,16]. Consider the stochastic process (X i ) defined by X i+1 = f (X i ), where X 0 ∼ U(0, 1). Lemma 1.1 implies that the variables X i are identically distributed, but they are no longer independent. Let M n be the partial maximum as defined in (1). Haiman [9] proved the following result: Theorem 1.2. For fixed λ > 0 and the sequence n k = ⌊β k λ⌋ it follows that . Therefore, the result of Theorem 1.2 is in spirit similar to the example in (2), albeit that a subsequence of M n is considered. The fact that the limit is not equal to e −λ has a particular statistical interpretation. The coefficient (β − 1)/β in the exponential is called the extremal index and measures the degree of clustering in extremes arising as a consequence of the dependence between the variables X i ; see [11,13] for more details. The aim of this paper is to give an alternative proof for Theorem 1.2 which relies on asymptotic properties of a rescaled version of the k-generalized Fibonacci numbers.

The relation with generalized Fibonacci numbers
In this section we fix the numbers k ∈ N and u = β −k . For any integer i ≥ 0 we define the set where the dependence on k is suppressed in the notation for convenience. Then where Leb denotes the Lebesgue measure. Based on self-similarity arguments Haiman [9] derived the following recursion formula: The same idea was used earlier by Haiman to study extreme value laws for the tent map [8].
For n ∈ Z we define the following numbers:  Figure 1: Illustration of the sets E 0 , . . . , E 5 for β = 2 and k = 2. Each set E n is a union of β n intervals. Intervals which are disjoint from (resp. contained in) the intervals comprising E 0 , . . . , E n−1 are drawn in blue (resp. red). For n ≥ 2 the number F n equals the number of subintervals of the set E n−1 which need to be added to E 0 ∪ · · · ∪ E n−2 in order to obtain E 0 ∪ · · · ∪ E n−1 . The figure clearly shows that F 2 = 1, F 3 = 2, F 4 = 3, F 5 = 5, and F 6 = 8 which are the starting numbers of the Fibonacci sequence.
These numbers have the following geometric meaning. Note that the sets E i can be written as a union of β i intervals: For n ≥ 2 the number F n equals the number of subintervals of the set E n−1 which need to be added to E 0 ∪ · · · ∪ E n−2 in order to obtain E 0 ∪ · · · ∪ E n−1 . Figure 1 illustrates this for the special case β = 2 and k = 2.
Lemma 2.1. For any k, n ∈ N it follows that Proof. For n ≥ k + 2 equation (4) gives or, equivalently, The proof is completed by substituting n for n − k − 1.
The following result provides the connection between the sequence (B n ) and generalizations of the Fibonacci numbers. In particular, for β = 2 the sequence (F n ) is the well-known k-generalized Fibonacci sequence. (ii) The sequence (F n ) defined in (5) satisfies In particular, F n = (β − 1)β n−2 for 2 ≤ n ≤ k + 1.
Proof. Assume that statement (i) holds. By definition F 1 = 1 and for 2 ≤ n ≤ k + 1 equation We proceed with induction on n. For any n ≥ k + 1 equation (4) gives In particular, for n = k + 1 we have Assume that for some n ≥ k + 1 it follows that First using equation (5) and then equation (4) twice gives where the last equality follows from (7). Finally, the induction hypothesis implies that Hence, statement (ii) follows. Conversely, assume that statement (ii) holds. In particular, F n = (β − 1)β n−2 for 2 ≤ n ≤ k + 1 so that by equation (5) it follows that Equation (3) now follows by recalling that B 1 = u.
We proceed by strong induction on n. We have Recalling that B 1 = u = β −k , equation (5) implies that which shows that equation (4) holds for n = k + 1. Assume that there exists m ∈ N such that (4) holds for all k + 1 ≤ n ≤ m. Observe that Therefore, The induction hypothesis gives Hence, statement (i) follows.

The Binet formula
Let the sequence (F n ) be as defined in (6), where β ≥ 2 is assumed to be an integer. In this section we will derive a closed-form expression for F n as a function of n along the lines of Spickerman and Joyner [18] and Dresden and Du [4]. Levesque [12] derived a closed-form expression for sequences of the form (6) in which each term is multiplied with a different factor. Another interesting paper by Wolfram [19] considers explicit formulas for the kgeneralized Fibonacci sequence with arbitrary starting values, but we will not pursue those ideas here. The characteristic polynomial of the sequence (F n ) is given by The following result concerns properties of the roots of this polynomial. The proof closely follows Miller [15]. For alternative proofs for the special case β = 2, see [14,19].
Lemma 3.1. Let k ≥ 2 and β ≥ 2 be integers. Then (i) the polynomial p k has a real root 1 < r k,1 < β; (ii) the remaining roots r k,2 , . . . , r k,k of p k lie within the unit circle of the complex plane; (iii) the roots of p k are simple.
Note that p k has no root r such that |r| > r k,1 . Indeed, if such a root exists, then p(r) = 0 so that the triangle inequality implies that p(|r|) ≤ 0, which contradicts observation (O1).
Finally, p k has no root r with either |r| = 1 or |r| = r k,1 but r = r k,1 . Indeed, if such a root exists, then q k (r) = (r − 1)p k (r) = 0, which implies βr k = r k+1 + β − 1 and If the inequality in (9) is strict, then q k (|r|) > 0. Since q k (1) = 0 and q k (r k,1 ) = 0 it then follows that |r| = 1 and |r| = r k,1 . If the inequality in (9) is an equality, then r k+1 must be real. Since q k (r) = 0, it follows that r k = ((β − 1) + r k+1 )/β is real as well and hence r itself is real. An application of Descartes' rule of signs to q k implies that when k is even p k has one negative root, and when k is odd p k has no negative root. If k is even, then p k (0) = −(β − 1) and p k (−1) = 1. By the Intermediate Value Theorem it follows that −1 < r < 0. We conclude that no root of p k , except r k,1 itself, has absolute value 1 or r k,1 .
(iii) If p k has a multiple root, then so has q k . In that case, there exists r such that q k (r) = q ′ k (r) = 0. Note that q ′ k (r) = 0 implies that r = 0 or r = βk/(k + 1). Clearly, r = 0 is not a root of q k . By the Rational Root Theorem it follows that the only rational roots of q k can be integers that divide β − 1. Hence, r = βk/(k + 1) is not a root of q k either. We conclude that q k , and thus p k , cannot have multiple roots.
The proof of the following result closely follows the method of Spickerman and Joyner [18] and then uses a rewriting step as in Dresden and Du [4].
Lemma 3.2. The sequence (F n ) as defined in (6) is given by the following Binet formula: where r k,1 , . . . , r k,k are the roots of the polynomial defined in (8).
Proof. The generating function of the sequence (F n ) is given by The equation Finally, using that F 1 = 1 and F n = (β − 1)β n−2 for 2 ≤ n ≤ k − 1 implies that Note that 1/r is a root of the denominator of G if and only if r is a root of the characteristic polynomial p k . By Lemma 3.1 part (iii) we can expand the generating function in terms of partial fractions as follows: where the coefficients are given by Observe that This results in .
Finally, we have that Substituting the values for the coefficients completes the proof.
For the special case β = 2 Dresden and Du [4] go one step further and derive the following simplified Binet formula: We expect that this formula can be proven for all integers β > 1 for n sufficiently large, where the lower bound on n may depend on both β and k. However, we will not pursue this question in this paper.

Exponentially growing sequences
In preparation to the proof of Theorem 1.2 we will prove two facts on sequences that exhibit exponential growth. The first result is a variation on a well-known limit: is a sequence such that lim k→∞ ka k = c, then Proof. Let ε > 0 be arbitrary. Then there exists N ∈ N such that |ka k − c| ≤ ε, or, equivalently, Since ε > 0 is arbitrary, the result follows.
The next result provides sufficient conditions under which the difference of two exponentially increasing sequences grows at a linear rate: Proof. The algebraic identity It suffices to show that lim k→∞ S k = 1. To that end, note that the assumption implies that lim k→∞ b k = 0 so that −1 < −b k /a < 0 for k sufficiently large. Bernoulli's inequality gives which implies that for k sufficiently large. Moreover, the assumption implies that lim k→∞ kb k = 0. An application of the Squeeze Theorem completes the proof.

Proof of the extreme value law
Let λ > 0 and define n k = ⌊β k λ⌋. Combining Lemma 2.1 and 3.2 gives where r k,i are the roots of p k . Recall that r k,1 is the unique root in the interval (1, β), and that |r k,i | < 1 for i = 2, . . . , k. In the remainder of this section Theorem 1.2 will be proven by a careful analysis of the asymptotic behaviour of the dominant root r k,1 .
We define the following numbers: The number r k,max is obtained by applying a single iteration of Newton's method to p k using the starting point x = β. The number r k,min is a correction of r k,max with an exponentially decreasing factor.
Proof. (i) For x = 1 we have In particular, for k ≥ 2 it follows that It suffices to show that the expression between brackets is positive for k sufficiently large. Lemma 4.2 gives Hence, for k sufficiently large it follows that and the right-hand side is positive for k > 2β/(β − 1).
(ii) Similar to the proof of part (i) it follows that It suffices to show that the expression between brackets is positive for k sufficiently large. Lemma 4.2 gives Hence, for k sufficiently large it follows that This gives and the right-hand side is positive for k sufficiently large.
(iii) By the Intermediate Value Theorem there exists a point c ∈ (r k,min , r k,max ) such that p k (c) = 0. Note that c > 1 for k sufficiently large. Since r k,1 is the only zero of p k which lies outside the unit circle it follows that c = r k,1 .
The inequality β k λ − 1 ≤ n k ≤ β k λ combined with (10) implies that A similar result holds for the sequence (a k ). Hence, (11) together with the Squeeze Theorem applied to (12) completes the proof.
This completes the proof. whereby Theorem 1.2 has been proven.