THE ASYMPTOTIC BEHAVIOR OF ELEMENTARY SYMMETRIC FUNCTIONS ON A PROBABILITY DISTRIBUTION

The problem on asymptotic of the value &#960;(m,n)=m!&#963;m(p(1,n),p(2,n),&#8230;,p(n,n)) is considered, where &#963;m(x1,x2,&#8230;,xn) is the mth elementary symmetric function of n variables. The result is interpreted in the context of nonequiprobable random mappings theory.

The following relations are valid for n → ∞ π(m, n) → e − 1 2 ( π(m, n) → e − 1 2 α 2 x 2 as m n(log n) −1 → x, γ = 1 2 ; (5) π(m, n) → L (γ, αx) as m n γ → x, 0 < γ < 1 2 (6) where L(γ, z) is the entire analytic function defined by the infinite product For 0 < γ < 1 2 and for the values |x| < γ α the following representation is also valid. Note a specification of the theorem above in the spirit of random mappings theory [1]. Denote Ξ(n) = {1, 2, . . . , n} for a positive integer n. Let also the set Ξ(n) be endowed with a probability measure µ n,q given by equalities µ n,q (k) = p(k, n) where the function q and the numbers p(k, n) are as in (1). Consider now the random mapping F defined by Generally speaking, this mean that the "appeal" of a point n is proportional to its weight p(k, n). If q ≡ 1, then F q,n is a completely random mapping. We emphasize here that the completely random mappings is essentially a purely combinatorial object. The theory of completely random mappings is quite well developed, using specifically combinatorial methods. See [1] and bibliography therein.
For a mapping f : Ξ(n) → Ξ(n) and for an element k ∈ Ξ(n) we denote by Q(k, f ) the first recurrence time, that is Q(k, f ) = min{i : f i (k) = f j (k), for some j < i}. Designate further by Q(x, f ) the scaled distribution function where #S denotes the cardinality of the finite set S. The function Q is a random function if f is constructed as realization of the random mapping F . Thus, we can consider the mathematical expectation Q γ,n (x). The theorem above implies the following corollary immediately.

Corollary 1
The following relations are valid for n → ∞ Random mapping with similar asymptotic of weights of elements arises naturally, for instance, in the analysis of discretizations of random mappings where the box counting dimension of the invariant measure differs from its correlation dimension [3].
To conclude this section we note that the equality (5) can be obtained also as a corollary of Theorem 1 [2].

Proof
The proof of the theorem is based on the following two lemmas.
Lemma 1 Let f (z) and f n (z), n = 1, 2, . . ., be entire analytic functions of z ∈ C such that f n (z) → f (z) as n → ∞ uniformly with respect to z from any bounded circle |z| ≤ R. Let µ(n) be integer-valued function and ν(n) be real-valued one such that Then (1 + p(k, n)z) .
Then uniformly with respect to z from any bounded circle |z| ≤ R the following relations are valid where L(γ, z) is the entire analytic function defined by (7).
Proof of Theorem 1. Since the proof is identical for the cases 0 < γ < 1 2 , γ = 1 2 and 1 2 < γ ≤ 1 it will be presented only for the last case.
Remark that the value of 1 m! π(m, n) by the Viète theorem coincides with the coefficient at the term z m in the Taylor expansion of the function f (n, z) defined by (9). Then by the theorem on residue of an analytic function From this and from Lemmas 1 and 2 the statement of Theorem for the case 1 2 < γ ≤ 1 immediately follows.
Although Lemma 1 is in line with statements from the theory of integral's asymptotics proven with the help of the saddle point method (see. e.g. [4,5]), we failed to find the appropriate reference to the exact formulation. Because of this and for the sake of completeness of presentation, below is given the full proof of Lemma 1.
Proof of Lemma 1. By the theorem on residue of an analytic function for any value of ρ > 0. Hence, choosing ρ = µ(n) ν(n) , the value of F (n) may be represented as where Estimate the value of |ξ(n)|. Fulfil the substitution z = µ(n) ν(n) e iϕ in the above integral. Then Now, using the Stirling's formula we may write that with an appropriate constant c 0 where In view of (10) and (11) it remains to show that To do it introduce auxiliary constants Clearly χ > 0, c 1 < ∞ and ε(n) → 0 as n → ∞ in view of uniform convergence of the sequence {f n (z)} to f (z) on any bounded circle. Hence and, by substitution µ(n)χϕ = ψ, we obtain that where By definition of the value ε(n), we have for |ψ| ≤ (µ(n)) 1/4 and thus At the same time, by definition of the constant c 1 we have From (13), (14) it follows (12) which completes the proof of the lemma.
In the proof of Lemma 2 properties of the numbers p(k, n) play an important role. So, establish these properties prior to pass to the proof of Lemma 2. From the theorem on mean value of integral it follows that Therefore where and thus, in view of continuity of the function β(t), From the continuity of the function β(t) it follows that the product of the first two multipliers in (15) is uniformly (with respect to n and k ∈ [1, n]) bounded. At the same time, for a given γ ∈ (0, 1] the numbers k 1−γ (k γ − (k − 1) γ ) are also uniformly bounded. Then by (16) such a constant p * = p(α, γ) can be chosen that and simultaneously Proof of Lemma 2. Case 1 2 < γ ≤ 1. Set g(n, z) = f (n, (here the latter equality follows from (1) and (2)). By (19), 0 ≤ p(k, n) √ n ≤ p * n 1 2 −γ and so p(k, n) √ n → 0 as n → ∞ uniformly with respect to k. Hence for any R < ∞ such a value n(R) can be chosen that |p(k, n) √ nz| ≤ 1 2 for n ≥ n(R), |z| ≤ R uniformly with respect to k. Then, for such n and z, the representation By substituting (21) in (20) we obtain that g(n, z) = e − n 2 ( P n k=1 p 2 (k,n))z 2 + e θ(n,z) where θ(n, z) = n 3 2 n k=1 θ(k, n, z)p 3 (k, n) z 3 .
From (1) it is seen that p(k, n) = 1 n q(t k,n ) with an appropriate t k,n ∈ k−1 n , k n . Thus, given arbitrary ε > 0, we can write

From (19) it follows that
with an appropriate constant c 0 , and thus On the other hand, in view of summability of the function q 2 (t) on the interval [ε, 1], S 2 (ε, n) → 1 2  |θ(k, n, z)|p 3 (k, n) z 3 ≤ with an appropriate constant c 1 . Thus From (23), (26), (27) and from the definition of the function g(n, z) the statement of the lemma for the case 1 2 < γ ≤ 1 follows.
The lemma is completely proved and so is the theorem.