NOTE ON THE EXPONENTIAL RECURSIVE k -ARY TREES

. In the present paper, we consider exponential recursive trees with no node of outdegree greater than k , called exponential recursive k -ary trees ( k ≥ 2). At each step of growing of these trees, every external node (insertion position) is changed into a leaf with probability p , or fails to do so with probability 1 − p . We investigate limiting behavior of fundamental parameters such as size, leaves and distances in exponential recursive k -ary trees.

. The steps of evolution of a recursive tree of size 5. changed into a leaf with probability p ∈ (0, 1), or not to be changed with probability q = 1 − p. Figure 3 shows all exponential recursive ternary trees of ages 0, 1 and 2 and their probabilities. In the figure, external nodes and internal nodes are represented by squares and circles, respectively. The probability of getting each tree from the previous one appears above the arrow leading to it. For example, the leftmost tree in the third row occurs with probability p 2 q 2 .

Size
Let S n be the size of a random exponential recursive k-ary tree of index p and age n (k ≥ 2), and let Y n be the number of external nodes in the extension. Thus, we have Y n = (k − 1)S n + 1. (2.1) Let the notation Bin(m, p) stands for a binomial random variable that counts the number of successes in m independent identically distributed trials with success probability p in each trial. Each external node that succeeds to be converted into a leaf adds k − 1 external nodes; we have Y n = Y n−1 + (k − 1)Bin(Y n−1 , p).
Let F n be the sigma field generated by the first n steps of evolution. Then, E[Y n |F n−1 ] = Y n−1 + (k − 1)pY n−1 = ((k − 1)p + 1)Y n−1 .  The probability of getting each tree from the previous one appears above the arrow leading to it.
It follows that Y n /((k − 1)p + 1) n is a martingale with bounded moments That is, E[Y n ] = ((k − 1)p + 1) n and, by (2.1), According to the martingale convergence theorem (Thm. 6.6.9 in [2]), we have where Y * is an integrable limit random variable. Since for a limiting random variable S * . We now characterize the limiting distribution of S n /((k − 1)p + 1) n by inductively-constructed moments. The calculation techniques were taken from [1].
Theorem 2.1. Let S n be the number of nodes in an exponential recursive k-ary tree of age n and index p (k ≥ 2). As n → ∞, we have almost sure convergence where the limiting random variable S * has moments a m := E[S m * ] defined inductively by Proof. After one step of evolution, the initial external node either succeeds to recruit a leaf (event R) or not. Let I be the indicator of event R, that assumes the value 1, if R occurs, and is 0 otherwise. If R occurs, the extended exponential recursive k-ary tree has an internal node with k external nodes. By time n, these k external nodes produce k independent exponential recursive k-ary subtrees of sizes S n−1 , . . ., S (k) n−1 . These k random variables are independent and each is distributed like S n−1 . Alternatively, if R does not occur, the size of the exponential recursive k-ary tree at step n is the same as S n−1 . Hence, the size satisfies the distributional equation n−1 + · · · + S (k) Take expectations both sides of the equation (2.7). By the independence relations and identical distribution in the k subtrees, we get Scale the above recurrence by ((k − 1)p + 1) nm to get Taking limits, as n → ∞, we get Isolating the cases i j = m, for j = 1, 2, . . . , k, and rearranging, we get the claimed recurrence. Now, by induction on m, we show that a m /m! ≤ 1/(k − 1). This assertion is true for m = 1. For m ≥ 2 and for all complex z, with |z| < 1. Therefore, by Theorem 30.1 in [3], we conclude the uniqueness of the distribution with the moments a m , m ≥ 1. This verifies the convergence in distribution S n /((k − 1)p + 1) n D −→ S, for some random variable S with the moments a m = E[S m ], m ≥ 1. Since S n /((k − 1)p + 1) n a.s. −→ S * , so it converges in distribution to S * . That is S d = S * . Thus, S * has the moments given by the stated recurrence.

Leaves
Let L n be the number of leaves in an exponential k-ary tree of age n and index p. After one step of evolution, if the initial external node succeeds to recruit a leaf, then the extended exponential recursive k-ary tree has an internal node with k external nodes. In this case, for i = 1, . . . , k, let J (i) n−1 be an indicator of the event that the ith external node does not progress into a tree of age n − 1. By the nth step, these k external nodes produce k independent exponential recursive k-ary subtrees of age n − 1. Let the number of leaves arising from these k subtrees be respectively L n−1 . These k random variables are independent and each is distributed like L n−1 . After n steps of evolution, the root of the tree is a leaf if and only if J Alternatively, if the initial external node does not succeed to recruit a leaf, then the number of leaves in the exponential recursive k-ary tree at step n is the same as L n−1 . Hence, the number of leaves satisfies the distributional equation Note that, for each n ≥ 1, the vectors L From (3.2), in particular, we can calculate the exact mean, and find its asymptotic equivalent.
Proposition 3.1. Let L n be the number of leaves in an exponential k-ary tree of age n and index p. Then we have as n → ∞.
Proof. Take the expectation of (3.1), to write The solution to the recurrence of the mean is The sum is a geometric series, yielding This implies the assertion.
The calculation techniques were taken from [1].
Theorem 3.2. Let L n be the number of leaves in an exponential recursive k-ary tree of age n and index p. We have the convergence in distribution Proof. Scale the recurrence (3.2) by (k − 1)p + 1 nm to get Thus, inductively the sequence of the limit b m satisfies the recurrence Using this recurrence and b 1 < 1/(k − 1), by (2.8), we can conclude that bm m! < 1/(k − 1) for m ≥ 1. So, for |z| < 1, the series ∞ m=0 bm m! z m converges. Therefore, by Theorem 30.1 in [3], L n /((k − 1)p + 1) n converges in distribution to a unique limit L * .

External path length
Let W n be the external path length, which is the sum of the root-to-external node distances (the sum of the depth of external nodes). We use T n for an exponential recursive k-ary tree of age n and index p. Assume, we number the external nodes of T n arbitrarily with the numbers 1, . . . , Y n . Let D i be the depth of external node i, for i = 1, . . . , Y n . Then the external path length of T n is Similar to the equation (2.6), for n ≥ 1, W n and Y n satisfy the distributional equations Here W

The expectation and variance
The pair of equations (4.1) and (4.2) is sufficient to determine the means and the quadratic order moments exactly.
Theorem 4.1. Let W n be the external path length in an exponential k-ary tree of age n and index p. We have Proof. Raise both sides of (4.2) to the second power. So we get Take the expectation of (4.4) and observe the independence Y (i) with initial condition E[Y 2 0 ] = 1. Iterating this formula, we find Taking the expectation of (4.1), we get with initial condition E[W 0 ] = 0. By iterating, the mean value follows. Then multiplying (4.1) and (4.2), we deduce that, We next take the expectation and get the following recurrence for E[W n Y n ], Now, we take the square of the equation (4.1). Then, by taking expectation, we have Again, by iterating, we obtain This implies the claim for the variance of W n .

Characterizing the limiting distribution
Here, we use the contraction method by Roesler and Neininger to state the limiting distribution ofŴ n := W n /kpn ((k − 1)p + 1) n . Recently, this method has been applied to characterize the limiting distribution of the protected node path length in [6]. By equation (4.1), we have All the conditions of the contraction method are satisfied (Thm. 3 of [9], page 8), then Theorem 4.2. The scaled external path lengthŴ n := W n /kpn ((k − 1)p + 1) n converges in distribution to some random variableŴ with distribution, the unique solution of the following equation whereŴ (i) , i = 0, 1, . . . , k, are independent copies ofŴ and independent of I.
Proof. The proof is based on the Lemma 3.1 page 502 of the paper [8]. Let M 2 be the space of all distributions with finite absolute second moment. Consider the transformation T : M 2 → M 2 defined by whereŴ (i) , i = 0, 1, . . . , k and I are independent andŴ (i) have µ as distribution. At a first step we have to prove that with respect to the metric L 2 2 defined on M 2 by , where X, Y have µ and ν as distribution, respectively , the transformation have a unique fixed point. In fact let µ and ν be two measures of M 2 Since 1/ (k − 1)p + 1 < 1, then T is a contracted transform and it has only one fixed point. Now, ifŴ n converges in distribution, the limit will be the unique fixed point of the transformation T . Namely, we have to prove that lim n→∞ L 2 2 Ŵ ,Ŵ n = 0, to conclude thatŴ n converges in distribution and in L 2 toŴ . We have

Almost sure convergence
In this subsection, the calculation techniques were taken from [1].
Theorem 4.3. Let W n be the external path length in an exponential k-ary tree of age n and index p. As n → ∞, we have the almost sure and L 2 convergence, Proof. Let B k be a sequence of independent Bernoulli (p) random variables, defined on the same probability space as the trees. Consider B k ∞ k=1 to be independent of the structure of the tree at any age, too. If the ith external node recruits a leaf, it is converted into an internal node (a leaf node) at its own level and it adds k external nodes at the next level (as the children of the new leaf). The contribution of the ith external node of T n−1 into W n , is D i + (k(D i + 1) − D i )B i , and we write Taking an expectation conditioned of F n−1 , we obtain (4.10) Let M n := α n W n + β n Y n , for specified factors α n and β n that transformed M n a martingale. So, by (2.2), (4.10) and a martingalization procedure, we have E[M n |F n−1 ] = α n (k − 1)p + 1 W n−1 + kpα n Y n−1 + β n (k − 1)p + 1 Y n−1 = M n−1 = α n−1 W n−1 + β n−1 Y n−1 .