Eﬀective bounds of the variance of statistics on multisets of necklaces

. The variance of a linear statistics on multisets of necklaces is explored. The upper and lower bounds with optimal constants are obtained.


Introduction and results
Let (P, · ) be an initial set of weighted objects and π(j) := {p ∈ P : p = j} < ∞ for every j = 1, 2, . . . . Examine the set G with the extended weight function · of multisets comprised of p ∈ P. Namely, a ∈ G if a = {p 1 , . . . , p r } and a = p 1 + · · · + p r including the empty multiset ∅ of weight 0. Then m(n) := |G n | := {a ∈ G : a = n} = where (k) = 1k 1 + · · · + nk n ifk = (k 1 , . . . , k n ) ∈ N n 0 and n ∈ N 0 := N ∪ {0}. In the present paper, we deal with the multisets for which m(n) = q n , where q 2 is an arbitrary natural number. If q is a prime power, then G may be interpreted as F * q [t], the set of monic polynomials over a finite field F q . Then P is the subset of irreducible polynomials. For an arbitrary such q, there exist combinatorial constructions, called multisets of necklaces satisfying m(n) = q n (see, [1,Example 2.12,p. 43]). For multisets, we have the following relations where in the summations, d runs over natural divisors of n and µ(d) stands for the Möbius function. The equalities are equivalent to the formal power series relation Take an a ∈ G n uniformly at random, that is, sample it with probability ν n ({a}) = q −n , n ∈ N and ν 0 ({∅}) = 1. If k j (a) 0 is the number of elements p i in a ∈ G n of weight j, thenk(a) = k 1 (a), . . . , k n (a) is the structure vector of a ∈ G n satisfying (k(a)) = n. Its distribution is wheres = (s 1 , . . . , s n ) ∈ N n 0 and 1{·} stands for the indicator function. We are interested in the distribution with respect to ν n of the linear statistics The number of components in a is such a function, namely, it equals k 1 (a)+· · ·+k n (a). We refer to [1] for more sophisticated examples. The present paper is devoted to the variance of h(c) which is a sum of dependent random variables (r.vs) as the relation h(j, a) = (k(a)) = n for each a ∈ G n shows. Estimating it, we propose an approach to overcome technical obstacles stemming from dependence.
In the sequel, the expectations and variances with respect to ν n will be denoted by E n and V n while, when the probability space (Ω, F, P ) is not specified, we will respectively use the notation E and V. The summation indexes i, j, l, k, m, m 1 and m 2 will be natural numbers.
Theorem 1 Ifc ∈ R n and n ∈ N 0 , then The sketch of the proof is given at the beginning of Section 2. It is known [1] that, for a fixed j, the r.v. k j (a) converges in distribution to the r.v. γ j distributed according the negative binomial law NB(π(j), q −j ). If {γ 1 , γ 2 , . . . } are mutually independent, define the statistics Y n = c 1 γ 1 + · · · + nγ n . We shall see that the first sum on the right-hand side in (4) is close to VY n ; therefore, estimating V n h(c), we use the following quadratic forms: The inequality becomes an equality for The inequalities are trivial for functions proportional to h(j, a) = n if a ∈ G n , because of V n h(j) = 0 then. A shift ofc eliminates this inconvenience. Observe that either of B n (c − tj) and R n (c − tj) attain their minimums in t ∈ R at Both inequalities are sharp.

Corollary 2
If n 3 andc = αj for every α ∈ R, then The proofs of the last two theorems presented in Section 2 are built upon the ideas and auxiliary results obtained in [4], [2] and [5].

Proof. Actually, this is Lemma 2.2 in [3] stated there for F q [t].
The details remain the same in the more general case.

Lemma 2 For a functional
Proof. Apply Lemma 1 in the double averaging as follows: Proof of Theorem 1. It is straightforward. Applying the last lemma for the relevant Ψ , one can easily find the needed mixed moments of k j (a), 1 j n, and further, the variance of the linear combination h(a).
To prove Theorems 2 and 3, we will apply the following lemmas concerning particular matrices and quadratic forms.
Proof. This is the byproduct of works [4] and [2].
Afterwards, letē r , 1 r n, be the orthogonal basis of R n comprised of the eigenvectors of U andx means the transposed vectorx. Moreover, each bound in (9) and (11) are achieved, respectively, for b m = e rm / √ m, where r = 2, 1, 3 and e rm have been defined in Lemma 3.
Proof. Inequalities (9) are seen from Lemma 3 after the substitution b m = x m / √ m, m n, since the extreme eigenvalues are 1 and −1/2.
After the same substitution, we further examine the quadratic form with the matrix U . Condition (10) reckons the subspace of vectorsx = (x 1 , . . . , x n ) satisfying x 1 + · · · + x j √ j + · · · + x n √ n =x ·ē 1 = 0. This subspace is spanned over the first eigenvector. In other words, under (10), only the form values obtained in the subspace L ⊂ R n spanned over the vectorsē 2 , . . . ,ē n count. Hence Returning to b m , from this we obtain inequality (11).
Proof of Theorem 2. After grouping the summands, expression (4) can be rewritten as follows: Moreover, it becomes an equality if we take c j = c * j satisfying which by the Möbius inversion formula and (1) may be rewritten as (6).
To prove the first assertion of Corollary 1, it suffices to estimate the inner sum in R n (c), namely, Further, using the expression of VY n , we just estimate the remainder: Plugging both estimates into (5), we obtain the first inequality in Corollary 1 with instead of <. In fact, we obtained the strict inequality since Cauchy's inequality applied in the last step is strict ifc is not proportional toj, and in this exceptional case, Vh(c) = 0.
Proof of Theorem 3. Observe that V n h(c) = V n h(c) − tn = V n h(c − tj) for every t ∈ R. Hence the right-hand inequality follows from (5) applied for the shifted statistics.
To get the lower bound of variance, we combine (4) and (11). We start with This and (4) imply the lower bound. Moreover, the latter is sharp since Lemma 4 assures this by a choice of a particular sequenceb m , m n.