k-th Total Path Length

r=1

k!

r!e_k−r+ 2^k− 1

2^k−1− 1, for k ≥ 2.

We state the main result in the form of Theorem 1 because the leading term is the most interesting part and it would be enough for proving the central limit theorem. Also, computing more terms can be extremely complicated.

As we see from the above statements, the difference between the asymptotics of the two shape parameters is _k!(2k−1^nk₋₁₎. This can be explained heuristically.

Let d^[k]n be the difference of the two shape parameters, then d^[k]n ∼ 2d^[k]_n/2+2(ⁿ

2

k

) since the size of both subtrees will be roughly n/2 under the Bernoulli model.

Iterating it, we get d^[k]n = Θ(n^k), which matches the difference above.

Remark 21. Note that the Steiner k-distance is a generalization of the Wiener index, namely, for k = 2 we obtain the Wiener index. Thus, Theorem 4.2.1 is actually a special case of Theorem4.3.1 with k = 2.

4.3.2 k-th Total Path Length

In this section, we start with the recurrence under the Bernoulli model and then use it to get the differential-functional equation of the Poisson model.

The rest of the analysis will focus on the Poisson model, since the depois-sonization is standard with the language of JS-admissible.

Mean of the k-th Total Path Length of DSTs

First, we start with deriving a distributional recurrence relation for the k-th total path length. Recall the notation Pn^[k]for the k-th total path length from the introduction. Moreover, we will use the notation B_n= Binom(n,^{d 1}₂). Let a DST with n + 1 nodes given. Depending on how the k nodes are chosen, there are 4 cases:

1. All k nodes are from one subtree.

The contribution to the k-th total path length will be P_B^[k]

2. The k nodes are chosen from both subtrees and the root is not chosen.

We will have the contribution

k−1

3. The root is chosen, the other k

− 1 nodes are all from one subtree.

It will contribute

4. The root is chosen, the other k

− 1 nodes are from both subtrees.

The contribution will be

k−2

Note that from the above equation, we see that the k-th total path length depends on the 1-st, 2-nd,. . . , (k−1)-th total path length. Thus, we actually have a system of recurrences. The initial conditions are Pn^[0] = 0 for all n and Pn^[k]= 0 for all k > n.

Let ˜f^[k](z) = e^−z∑

n≥0

E(Pn^[k])zⁿ

n! which is the mean in the Poisson model.

Then, from the recurrence relation above, we get

f˜^[k](z) + ˜f^[k]′(z) =2 ˜f^[k]

Note that when k = 1, the above equation will be exactly the same as the one derived in [74] and hence the order of ˜f^[1](z) is known. Thus, by induction and the closure properties of JS-admissibility from [74], we get that

f˜^[k](z) =

{ O(z^k+ϵ), as z → ∞;

O(z^k), as z → 0⁺

uniformly for z with | arg z| ≤ ^π₂ − ϵ, where ϵ > 0 is an arbitrary small con-stant. Applying Laplace transform, we get the differential-functional equa-tion

and divide both sides of above equation by Q(−2s). This yields bound for 1/Q(−2s) obtained in [

74]

1 Q(−2s) =

{ O(s^−b), as s → ∞;

O(1), as s→ 0, where b can be arbitrarily large, we obtain the bounds

L [ ˜¯f^[k]; s] =

Then, by Ritt’s Theorem (Theorem 4.2 of [163]), we derive the bounds

R^[k](s) =

{ O(|s|^−b), as s→ ∞;

O(|s|^−(k+ϵ)), as s→ 0⁺

uniformly for s with | arg z| ≤ π − ϵ. Thus, we may apply the Mellin trans-form:

M [ ¯L^[k]; ω] = 2^2−ω

1− 2^2−ωM [ ¯L^[k−1]; ω]

+ 2^2−ω 1− 2^2−ω

k−1

∑

l=1

∏_l

i=1(ω− i)

l! M [ ¯L^[k−l]; ω− l]

+ 2^2−ω 1− 2^2−ω

k−2

∑

l=1

∏l

i=1(ω− i)

l! M [ ¯L^[k−l−1]; ω− l]

+ 2

1− 2^2−ω

Q(2^ω−k−1)

Q(1) Γ(k− ω)Γ(ω − k + 1)(1 − 2^−k)

+ 2

1− 2^2−ω

Q(2^ω−k)

Q(1) Γ(k + 1− ω)Γ(ω − k)(1 − 2^1−k) +M [R^[k]; ω]

1− 2^2−ω ,

where for convenience, we use the notation M [ ¯L^[k]; ω] for M [ ¯L [ ˜f^[k]; s]; ω].

The fundamental strip of the above expression will be the half planeℜ(ω) >

k + 1. To apply the inverse Mellin transform, we need to figure out all the singularities of the above expression. Since the case k = 1 is already solved in [74] and the general case k will be determined by 1, . . . , k− 1, we get that for k≥ 2 the expression can be simplified as

M [ ¯L^[k]; ω] = 2^2−ω 1− 2^2−ω

k−1

∑

r=1

∏_r

i=1(ω− i)

r! M [ ¯L^[k−r]; ω− r]

+ 1

1− 2^2−ω

Q(2^ω−k−1)

Q(1) Γ(k− ω)Γ(ω − k + 1)(2 − 2^1−k) + ¯g_k(ω) where ¯g_k(ω) is the sum of all the remaining terms in the expression. From the bound we derived for R^[k](s) and L [ ˜¯f^[k]; s] and the properties of the Mellin transform [62], we get that if α is a singularity of ¯g_k(ω), then ℜ(α) ≤ k.

From [74], we have that

M [ ¯L^[1]; ω] = G₁(ω) 1− 2^2−ω, where

G₁(ω) = Q(2^ω−2)

Q(1) Γ(ω)Γ(1− ω).

Plugging this into the recurrence and iterating, we get that for k≥ 2 M [ ¯L^[k]; ω] =

∏k−1

i=1(ω− i)

1− 2^k+1−ω G₁(ω− k + 1)Ak(ω) + T_k(ω)G₁(ω− k + 1) + gk(ω) where g_k(ω) is defined recursively by g₁(ω) = 0, g₂(ω) = ¯g₂(ω) and

g_k(ω) = 2^2−ω 1− 2^2−ω

k−1

∑

r=1

∏r

i=1(ω− i)

r! g_k−r(ω− r) + ¯gk(ω).

Again, by similar argument as above, we have that if α is a singularity of g_k(ω), then ℜ(α) ≤ k. The function Ak(ω) is defined recursively as A₁(ω) = 1, A2(ω) = 1

2^ω−2− 1 and

A_k(ω) = 2^2−ω 1− 2^2−ω

k−1

∑

r=1

Ak−r(ω− r)

r! .

Also, T_k(ω) is defined recursively as T₁(ω) = 0, T₂(ω) = ₄₍₁₋₂⁶_2−ω) and

T_k(ω) = 2^2−ω 1− 2^2−ω

k−1

∑

r=1

∏k−1

i=1(ω− i)

r! T_k−r(ω− r) + 2(1− 2^−k) 1− 2^2−ω . Note that one can easily prove that

A_k(k + 1 + χ_m) = A_k(k + 1) = 1 (k− 1)!

for χm = 2iπm

log 2 , m∈ Z by induction. Moreover, the Laurent series of Ak(ω) at ω = k + 1 + χr is given as

A_k(ω) = 1

(k− 1)! + d_k(ω− k − 1) + O((ω − k − 1)²), where {dk}_k≥1 is a sequence which is defined recursively as d₁ = 0 and

d_k= 1 2^k−1− 1

k−1

∑

r=1

d_k−r

r! − 2^k−1 2^k−1− 1

log 2 (k− 1)!.

Because we have the explicit form of G₁(ω), we rewrite the expression as M [ ¯L^[k]; ω] = Q(2^ω−k−1)

(1− 2^k+1−ω)Q(1)Γ(ω)Γ(k− ω)Ak(ω) + g_k(ω).

Finally, applying the inverse Mellin transform and collecting residues, we get

Finally, we apply Proposition 1 of [74] and obtain that, as z → ∞, f˜^[k](z) =z^klog z

Variance and Covariance of the k-th Total Path Length

Next, let us consider the variance. Here we introduce the poissonized variance and covariance as For detailed explanation of why we choose them this way, see [74]. Note that when k₁ = k₂ = k, ˜V^[k](z) = ˜C^[k1,k2](z). Thus, we will consider only C˜^[k1,k2](z) in this section.

From the given definition, we derive that

C˜^[k1,k2](z) + ˜C^[k1,k2]′(z) = ˜f₂^[k1,k2](z) + ˜f₂^[k1,k2]′(z)− ˜f^[k1](z) ˜f^[k2](z)

− z ˜f^[k1]′(z) ˜f^[k2]′(z)− ˜f^[k1]′(z) ˜f^[k2](z)

− ˜f^[k1](z) ˜f^[k2]′(z)− ˜f^[k1]′(z) ˜f^[k2]′(z)

− z ˜f^[k1]′′(z) ˜f^[k2]′(z)− z ˜f^[k1]′(z) ˜f^[k2]′′(z).

From the recurrence of P_n+1^[k] , we derive the differential-functional equations of ˜f₂^[k] and ˜f₂^[k1,k2] and plug them into the above equation. Thus, by the same argument we used in the mean case, we find the bounds

C˜^[k1,k2](z) =

{ O(z^k1+k2−1+ϵ), as z → ∞;

O(z^max{k1,k2}), as z→ 0⁺

uniformly for z with | arg z| ≤ ^π₂ − ϵ. With the help of computer algebra systems, we get that

C˜^[k1,k2](z) + ˜C^[k1,k2]′(z) =2 list the whole expression here. For the later computation, we only need the property that ˜g₂^[k1,k2](z) = O(z^k1+k2−2) as z → ∞. Similar to our analysis of the mean, we apply Laplace transform to the differential-functional equations and divide both sides by Q(−2s). Let k^′ = k1+ k2, then with the function L(s) defined as

L(s) =L^{(k′−r1−r2−k)}[ ˜C^[r1,r2]; 2s].

Before we proceed to apply the Mellin transform, we derived similar bounds as in the analysis of the mean:

L [ ˜¯C^[k1,k2]; s] =

{ O(|s|^−b), as s→ ∞;

O(|s|^−(k′+ϵ)), as s→ 0⁺,

where b is a constant which can be arbitrarily large. Note that the bounds hold uniformly for | arg s| ≤ π − ϵ. Now, we apply the Mellin transform on both sides of the above equalities.

Again, we use the simplified notationM [ ¯L^[k1,k2]; ω] =M [ ¯L [ ˜C^[k1,k2]]; ω].

Then, the equation becomes M [ ¯L^[k1,k2]; ω] =2^2−ω where Ar1,r2(ω) satisfies the recurrence

A_k1,k2(ω) = 2^2−ω singulari-ties with real part larger than k^′ − 1. From above recurrence, we can easily prove that for all k∈ Z

by induction. For convenience, we set C_k1,k2 = A_k1,k2(k₁+ k₂). Applying the inverse Mellin transform and collecting residues, we get

L [ ˜¯C^[k1,k2]; ω] = s^−k′ log 2

∑

r∈Z

C_k1,k2H₁(2 + χ_r)s^−χr

k1+k∏2−1 i=2

(i + χ_r) +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) = z^k1+k2−1C_k1,k2(C_kps+ ϖ_kps(log₂n)) +O(|z|^k1+k2−2+ϵ).

In particular,

V˜^[k](z) = z^2k−1

log 2C_k,k(C_kps+ ϖ_kps(log₂n)) +O(|z|^2k−2+ϵ) as z→ ∞.

Remark 22. Note that from the expression of C_k1,k2, we have C_k²_1,k2 = C_k1,k1C_k2,k2. Thus,

ρ(P_n^[k1], P_n^[k2]) = Cov(Pn^[k1], Pn^[k2])

√

Var(Pn^[k1])Var(Pn^[k2])

∼

√

n^2k1+2k2−2C_m,m² ₋₁(C_kps+ ϖ_kps(log₂n))²

n^2k1+2k2−2C_m,mC_m−1,m−1(C_kps+ ϖ_kps(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



P^n[1]− E(P^n[1])

√

Var(Pn^[1])

, . . . ,Pn^[k]− E(P^n[k])

√

Var(Pn^[k])



−→ (X, . . . , X),^d

where X is a standard normal distributed random variable and −→ denotes^d weak convergence.

在文檔中 隨機數位樹上加法性參數之機率分析 (頁 103-112)