4.3 Steiner Distance
4.3.3 Total Steiner k-distance
r∈Z
Ck1,k2H1(2 + χr)s−χr
k1+k∏2−1 i=2
(i + χr) +O(|s|1−k′)
uniformly as |s| → 0 with | arg s| ≤ π − ϵ. Finally, we apply inverse Laplace transform and Proposition 1 of [98] and obtain that, as z → ∞,
C˜[k1,k2](z) = zk1+k2−1Ck1,k2(Ckps+ ϖkps(log2n)) +O(|z|k1+k2−2+ϵ).
In particular,
V˜[k](z) = z2k−1
log 2Ck,k(Ckps+ ϖkps(log2n)) +O(|z|2k−2+ϵ) as z→ ∞.
Remark 22. Note that from the expression of Ck1,k2, we have Ck21,k2 = Ck1,k1Ck2,k2. Thus,
ρ(Pn[k1], Pn[k2]) = Cov(Pn[k1], Pn[k2])
√
Var(Pn[k1])Var(Pn[k2])
∼
√
n2k1+2k2−2Cm,m2 −1(Ckps+ ϖkps(log2n))2
n2k1+2k2−2Cm,mCm−1,m−1(Ckps+ ϖkps(log2n))2 = 1.
Remark 23. Since we already know that Pn[1] satisfies a central limit theo-rem [98], together with the result in the above theo-remark and applying similar argument as of [76], we obtain that
Pn[1]− E(Pn[1])
√
Var(Pn[1])
, . . . ,Pn[k]− E(Pn[k])
√
Var(Pn[k])
−→ (X, . . . , X),d
where X is a standard normal distributed random variable and −→ denotesd weak convergence.
4.3.3 Total Steiner k-distance
Let Sn[k] be the Steiner k-distance. Then, using the same idea as for the k-th total path length, we consider four cases:
1. All k nodes are from one subtree.
SB[k]
n+ Sn[k]−B∗
n.
2. The k nodes are chosen from both subtrees and the root is not chosen.
k−1
3. The root is chosen, the other k
− 1 nodes are all from one subtree.PB[k−1]
4. The root is chosen, the other k
− 1 nodes are from both subtrees.k−2
Note that as for the k-th total path length, here we have a system of recur-rences for the Steiner k-distance. Similar to the analysis of the k-th total path length, we let ˜g[k](z) be the Poisson generating function of the mean of the total Steiner k-distance, ˜W[k1,k2](z) be the Poissonized covariance of the total k1-th Steiner distance and the total k2-th total path length and ˜VS[k](z) be the variance of the k-th Steiner distance. With the help from computer algebra systems, we get the differential-functional equations
˜
and rest of the analysis will be very similar to the one with the k-th total path length, we skip the details and list only the results
E(
Since the leading terms are exactly the same as for the k-th total path length, the same arguments as for Pn[k] gives us the results stated in Theorem 4.3.1.
Chapter 5
A General Framework for Central Limit Theorems
5.1 Framework for m-ary Tries
In this section, we will discuss a general framework for the limiting distribu-tion of additive shape parameters in random digital trees. For m-ary tries and PATRICIA tries, an additive shape parameter is defined as follows: Xn is a sequence of random variables satisfying the distributional recurrence
Xn=d
∑m r=1
X(r)
In(r)
+ Tn, (n≥ n0), (5.1)
where n0 ≥ 0 is an integer, Xn, Xn(1), . . . , Xn(m), (In(1), . . . , In(m)), Tn are inde-pendent and Xn(i) has the same distribution as Xn. The random model we are using is the Bernoulli model which is introduced in Chapter 2. For digital search trees and bucket digital search trees, the distributional recurrence will be
Xn+b=d
∑m r=1
X(r)
In(r)
+ Tn+b, (n ≥ n0), where b≥ 1 is an integer. (5.2) The remaining notations are as in the trie case.
Because of the development of related mathematical techniques, including poissonization, poissonized variance with correction, Mellin transform and contraction method, we have many tools to characterize the asymptotics of additive shape parameters under the Bernoulli model. The authors of [77]
and [75] proposed a systematical way to derive the asymptotics for mean and variance and the limit laws of additive shape parameters of random tries. It
turns out that the same method works for random digital search trees as well.
Definition 5.1.1. If a set P =
{p1, . . . , pm} satisfies that pi ∈ (0, 1) for all 1≤ i ≤ m and ∑ipi = 1, then we say P is a probability family.
For a probability family P ={p1, . . . , pm}, if there exists a constant a ∈ R and a sequence {ki}mi=1, ki ∈ N for all 1 ≤ i ≤ m such that pi = aki for all i, then we say P is periodic. Otherwise, P is said to be aperiodic.
For a probability family P ={p1, . . . , pm}, we define a function Λ(s) = 1− p−s1 − · · · − p−sm .
We let Z be the set of roots of Λ(s) = 0 and define the following notations Z<α=Z ∩ {ℜ(z) < α} and Z=α =Z ∩ {ℜ(z) = α}.
Then from [55] and [67], we have the following properties
Theorem 5.1.2. Depending on the real part of the solutions of Λ(s), we
have three cases:(i) If ℜ(s) < −1, then Λ(s) has no solutions. In other words, Z<−1 =∅.
(ii) If ℜ(s) = −1, then Z=−1 ={−1} ∪ S where
S =
{−1 + χk|χk = 2kπi/ log a, k∈ Z \ {0}} , Pis periodic;
∅,
P is aperiodic.
(iii) If ℜ(s) > −1, then there exists a positive constant η such that for any solutions ω1, ω2, we have |ω1− ω2| > η.
Lemma 5.1.3. Let ˜
f (z) and ˜h(z) be entire functions satisfying a functional equation of the formf (z) =˜
∑m r=1
f (p˜ rz) + ˜h(z) (5.3)
where{p1, . . . , pm} forms a probability family. We denote by h = −
∑m r=1
prlog pr. If ˜h(z)∈ J Sα,γ with 0≤ α < 1 and ˜f (0) = ˜f′(0) = 0, then
f (z) =˜ 1 h
∑
ωk∈Z<−α−ϵ
G(ωk)z−ωk+O(zα+ϵ), where the sum expression is infinitely differentiable and
G(ω) =
∫ ∞
0
zω−1˜h(z)dz =M [˜h; ω].
Proof. Since ˜h(z) ∈ J Sα,γ with 0 ≤ α < 1, by a similar proof as of
By the converse mapping theorem, Theorem 3.3.10 and Theorem 5.1.2, we get the desired result.
Now, we consider the moment generating function of Xn Mn(y) :=E( generating function, we get that
µn= Mn′(0) = ∑ n0 = 2. For more general cases, our method will also apply with slight modifications.
Now, we utilize the idea of Poissonization which was already used in previous sections. We let
f˜1(z) = e−z∑
then (5.4) yields that
Next, we utilize the idea of Poissonized variance with correction and let V˜X(z) = ˜f2(z)− ˜f1(z)2− z ˜f1′(z)2,
V˜T(z) = ˜h2(z)− ˜h1(z)2− z˜h′1(z)2. From (5.5), we derive that
V˜X(z) =
Before we go on to derive asymptotic expressions, we introduce the Hadamard product of Poisson generating functions.
Definition 5.1.4. Given two Poisson generating functions
F˜1(z) = e−z∑ we define the Hadamard product of these two functions asF˜3(z) := ˜F1(z)⊙ ˜F2(z) = e−z∑
n≥0
anbn n! zn.
Note that the definition is different from the usual one since we consider the exponential generating function.
Subsequently, we will use Hadamard products to handle the function ˜ϕ1 in (5.6). For this, we will need the following theorem which shows that JS-admissibility is closed under the Hadamard product.
Theorem 5.1.5. If ˜
F1 ∈ J Sα1,β1 and ˜F2 ∈ J Sα2,β2, then ˜F3 ∈ J Sα1+α2,β1+β2. More precisely, we haveF˜3(z) = ˜F1(z) ˜F2(z) + z ˜F1′(z) ˜F2′(z) +O(
|z|α1+α2−2(log+|z|)β1+β2) , uniformly as |z| → ∞ and | arg(z)| ≤ θ, where 0 < θ < π/2.
Proof. See the proof of Proposition 3.5 of [75].
We now can state the result on asymptotic expressions of mean and vari-ance. (This result was first obtained by Fuchs et al. in [75].)
Proposition 5.1.6. If ˜h
1(z)∈ J Sα1,γ1 with 0≤ α1 < 1, then E (Xn) = 1h
∑
ωk∈Z<−α1−ϵ
GE(ωk)n−ωk +O(nα1+ϵ),
where the sum expression is infinitely differentiable and
GE(ω) =M [˜h1; ω] =
∫ ∞
0
˜h(z)zω−1dz.
Moreover, if ˜VT(z)∈ J Sα2,γ2 with 0≤ α2 < 1 and ˜h2(z)∈ J S , then
Var (Xn)∼ 1 h
∑
ωk∈Z=−1
GV(ωk)n−ωk,
where the sum expression is infinitely differentiable and
GV(ω) =M [ ˜VT + ˜ϕ1+ ˜ϕ2; ω] =
∫ ∞
0
(V˜T(z) + ˜ϕ1(z) + ˜ϕ2(z) )
zω−1dz.
Proof. The expression of the mean follows directly from (5.5), Lemma5.1.3 and depoissonization.
For the variance, we start from (5.6). We apply Theorem 5.1.5 to ˜g(z),
Now, we turn to ˜ϕ2(z). First, by applying the Mellin transform to (5.5), we get that for −2 < ℜ(ω) < −1,
M [ ˜f1; ω] = GE(ω) Λ(ω) . Thus, from inverse Mellin transform,
f˜1′(z) = d
where the latter follows from the fact that the integral has no poles atℜ(ω) =
−1. As a result, ˜ϕ2(z) = o(|z|) as z → ∞ which in turn shows that M [˜ϕ2; ω]
has no poles at ℜ(ω) = −1. Now, the converse mapping theorem proves the claimed expansion for ˜V (z). Moreover, by JS-admissibility, the expansion holds for Var(Xn), as well.
Now, we can state the general central limit theorem.
Theorem 5.1.7. Suppose that ˜h
1(z)∈ J Sα1,γ1 with 0≤ α1 < 1/2, ˜h2(z)∈ J S and ˜VT(z) ∈ J Sα2,γ2 with 0 ≤ α2 < 1. Moreover, we assume that∥Tn∥s = o(√
n) with 2 < s ≤ 3 and V(Xn)≥ cn for all n large enough and some c > 0. Then, as n→ ∞,
Xn− E(Xn)
√V(Xn)
−→ N (0, 1).d
Proof. From Proposition5.1.6, we get that E(Xn) =1
h
∑
ωk∈Z<−α1−ϵ
GE(ωk)n−ωk+O(nα1+ϵ), V(Xn)∼1
h
∑
ωk∈Z=−1
GV(ωk)n−ωk.
From the assumption, we can choose ϵ such that α1+ ϵ < 1/2. Next, we set ϖ1(x) = ∑
ωk∈Z<−α1−ϵ
G1(ωk) h x−ωk, ϖ2(x) = ∑
ωk∈Z=−1
G2(ωk)
h x−ωk−1.
To apply the contraction method, we need to verify the following conditions:
(a) (
In(r)ϖ2(In(r)) nϖ2(n)
)1/2 Ls
−→ Ar,
∑m r=1
A2r = 1 and P(∃r : Ar = 1) < 1.
(b)
(nϖ2(n))−1/2 (
Tn− ϖ1(n) +
∑m r=1
ϖ1(In(r)) )
Ls
−→ 0.
We begin with the verification of (a). By the strong law of large number and the dominating converge theorem,
In(r)−→ pLp r, 1≤ r ≤ m. (5.7)
Moreover, by the definition of ϖ2(x), we have that
ϖ2(prn) = ϖ2(n) for all 1≤ r ≤ m and ϖ′2(n) =O(n−1).
By the Taylor series expansion of ϖ2:
ϖ2(In(r)) = ϖ2(n) +O(
Combining (5.7) and (5.8), we get ( and hence the condition (a) is verified.
Now, we turn to the verification of condition (b). Note that from the assumption on ∥Tn∥s and V(Xn), the term Tn can be dropped from (b).
Therefore, we only need to check that
(nϖ2(n))−1/2 Chernoff’s bound yields that
P(Acn) = O
Again, we compute the Taylor expansion of ϖ1 (on An):
We estimate the terms in (5.10) individually. First, we consider
ϖ1′(prn)−ϖ′1(psn) = ∑
ωk∈Z<−α1−ϵ
G1(ωk)
h (−ωk)n−ωk−1(p−ωr k−1−p−ωs k−1) = o(n).
Together with the assumption on V(Xn), we get
Similarly, we also have Substituting (5.11) and (5.12) back into (5.9) shows that (b) holds.