In this final section, we will briefly sketch some further examples. It should by now be clear that our approach essentially rests on the transfer theorem. Once such a result is established, the remaining proof is rather automatic.
Number of Key Comparisons for Insertion and Depth in Binary Search Trees. These examples are similar but more easier than the examples discussed in the previous section. For instance, let Xn denote the number of key comparisons when inserting a random node in a random binary search tree build from n records (this quantity is also called “unsuccessful search”; see Chapter 2 in [11] for background). Then, for n ≥ 1,
Xn| (In = j)=d
(Xj + 1, with probability (j + 1)/(n + 1), Xn−1−j+ 1, with probability (n − 1 − j)/(n + 1)
with P (In = j) = 1/n, 0 ≤ j < n and X0 = 0. From this, a straightforward computation reveals that the underlying recurrence (with a scaling factor n + 1 as in the previous section) is given by
an= 2 n
n−1
X
j=0
aj+ bn, (n ≥ 1) (10)
with a0 = 0. A transfer theorem for this recurrence of similar type as in the previous section is easily derived and can be found in [8].
Proposition 7. Consider (10).
(i) Let bn= O (n1−) with > 0 suitable small. Then,
an = cn + O n1− , where c is a suitable constant.
(ii) Let bn= n logαn with α ∈ {0, 1, . . .}. Then, an = 2n logα+1n
α + 1 + nPolα(log n) + O n1− , where > 0 is suitable small.
(iii) Let bn= O (n logαn) with α ∈ {0, 1, . . .}. Then, an= O n logα+1n.
(iv) Item (iii) holds with O replaced by o as well.
Hence, our approach applies as in the last section (the technical details being easier) and we obtain the following theorem.
Theorem 4. As n → ∞, we have
Xn− 2 log n
√2 log n
−→ N (0, 1).d
Similarly, the depth of a random node satisfies almost the same distributional recurrence (again see Chapter 2 in [11] for background). Hence, again a central limit theorem follows from the above transfer theorem by applying our approach.
Depth of Variants of Binary Search Trees. The previous example of the depth can be extended to several extensions of binary search trees. Here, we are going to discuss three of them, namely, median-of-(2t + 1) binary search trees (see [3]), m-ary search trees (see [2]), and quadtrees (see [1]). Subsequently, let Xn denote the depth of a randomly chosen record in the random tree build from n records. Moreover, the underlying recurrence will be satisfied by all centered and non-centered moments multiplied by n.
First, for median-of-(2t+1) binary search trees, Xnsatisfies the distributional recurrence for n ≥ 2t+1
Xn| (In= j)=d
Xj + 1, with probability j/n,
Xn−1−j + 1, with probability (n − 1 − j)/n, 0, with probability 1/n
with P (In = j) = jt n−1−j
t / 2t+1n , 0 ≤ j < n and suitable initial conditions. Hence, the underlying recurrence is given by
an = 2
n 2t+1
n−1
X
j=0
j t
n − 1 − j t
aj + bn, (n ≥ 2t + 1) (11)
with suitable initial conditions. This recurrence was extensively studied in [3]. In particular, the following transfer theorem can be proved with the tools of the latter paper.
Proposition 8. Consider (11).
(i) Let bn= O (n1−) with > 0 suitable small. Then,
an = cn + O n1− , where c is a suitable constant.
(ii) Let bn= n logαn with α ∈ {0, 1, . . .}. Then, an = n logα+1n
(H2t+2− Ht+1)(α + 1) + nPolα(log n) + O n1− , where > 0 is suitable small and Hn=Pn
j=11/j denotes the n-th harmonic number.
(iii) Let bn= O (n logαn) with α ∈ {0, 1, . . .}. Then, an= O n logα+1n.
(iv) Item (iii) holds with O replaced by o as well.
Hence, our approach applies and yields the following result (see [4] for a different approach).
Theorem 5. As n → ∞, we have
Xn− log n/(H2t+2− Ht+1) q
(H2t+2(2) − Ht+1(2)) log n/(H2t+2− Ht+1)3
−→ N (0, 1),d
where Hn(2) =Pn
j=11/j2.
Next, for the m-ary search tree, we have for n ≥ m − 1
Xn| In[1] = j1, . . . , In[m] = jm d
=
Xj1 + 1, with probability j1/n, ...
Xjm+ 1, with probability jm/n, 0, with probability (m − 1)/n
with P (In[1] = j1, . . . , In[m] = jm) = 1/ m−1n , j1, . . . , jm ≥ 0, j1+ . . . + jm = n − m + 1 and X0 = · · · = Xm−2 = 0. The underlying recurrence is given by
an= m
n m−1
n−m+1
X
j=0
n − 1 − j m − 2
aj + bn, (n ≥ m − 1) (12)
with a0 = · · · = am−2 = 0. Also, this recurrence was already investigated before and transfer theorems can be found in [2] and [5].
Proposition 9. Consider (12).
(i) Let bn= O (n1−) with > 0 suitable small. Then,
an = cn + O n1− , where c is a suitable constant.
(ii) Let bn= n logαn with α ∈ {0, 1, . . .}. Then, an= n logα+1n
(Hm− 1)(α + 1) + nPolα(log n) + O n1− , where > 0 is suitable small and Hn=Pn
j=11/j denotes the n-th harmonic number.
(iii) Let bn= O (n logαn) with α ∈ {0, 1, . . .}. Then, an= O n logα+1n.
(iv) Item (iii) holds with O replaced by o as well.
Then, again by our approach, the following result can be proved (see also [4] and [12] for different approaches).
Theorem 6. As n → ∞, we have
Xn− log n/(Hm− 1) q
(Hm(2)− 1) log n/(Hm− 1)3
−→ N (0, 1),d
where Hn(2) =Pn
j=11/j2.
Finally, for d-dimensional quadtrees, we have for n ≥ 1
Xn|
In[1] = j1, . . . , In[2d] = j2d
d
=
Xj1 + 1, with probability j1/n, ...
Xj
2d + 1, with probability j2d/n, 0, with probability 1/n with X0 = 0 and
P (In[1] = j1, . . . , In[2d] = j2d) =
n − 1 j1, . . . , j2d
Z
[0,1]d
q1(x)j1· · · q2d(x)jd2dx, where j1, . . . , j2d ≥ 0, j1+ · · · + j2d = n − 1, x = (x1, . . . , xd) and
qh(x) =
d
Y
i=1
((1 − bi)xi+ bixi) , (1 ≤ h ≤ 2d)
with (b1, . . . , bd)2the binary representation of h − 1. From this, we obtain for the underlying recurrence
an= 2d
n−1
X
j=0
πn,jaj+ bn, (n ≥ 1) (13)
with a0 = 0 and
πn,j =n − 1 j
Z
[0,1]d
(x1· · · xd)j(1 − x1· · · xd)n−1−jdx.
This recurrence was studied in [1]. The following transfer theorem can be proved with tools from the latter paper.
Proposition 10. Consider (13).
(i) Let bn= O (n1−) with > 0 suitable small. Then,
an = cn + O n1− , where c is a suitable constant.
(ii) Let bn= n logαn with α ∈ {0, 1, . . .}. Then, an = 2n logα+1n
d(α + 1) + nPolα(log n) + O n1− , where > 0 is suitable small and Hn=Pn
j=11/j denotes the n-th harmonic number.
(iii) Let bn= O (n logαn) with α ∈ {0, 1, . . .}. Then, an= O n logα+1n.
(iv) Item (iii) holds with O replaced by o as well.
Using our approach then gives the following result (see also [4] and [6] for different approaches).
Theorem 7. As n → ∞, we have
Xn− 2 log n/d p2 log n/d2
−→ N (0, 1).d
Number of Collisions in the β(2, b)-Coalescent. This is an example from coalescent theory (see [10]
for background). Let Xnbe a sequence of random variables satisfying Xn = Xd n−In+ 1. (n ≥ 2) with X1 = 0 and (In)n≥1independent of (Xn)n≥1with distribution
πn,j = P (In= j) = Γ(n − j + b − 1)Γ(n + 1)
(j + 1)Γ(n − j)Γ(n + b)H(n, b) (1 ≤ j ≤ n − 1), where b > 0 and
H(n, b) = b
b + n − 1 + Ψ(b + n − 1) − Ψ(b) − 1.
The authors of [10] asked for a proof of their main result (a central limit theorem for Xn suitable cen-tralized and normalized) directly from the above recurrence. Indeed, our approach is able to solve this problem once a suitable transfer theorem for the underlying recurrence is proved. Therefore, note that the underlying recurrence (without a scaling factor) is given by
an=
n−1
X
j=1
πn,jan−j+ bn, (n ≥ 2), (14)
where a1 = 0. Unfortunately, due to the more complicated nature of πn,j this recurrence is more involved.
In particular, we have not been able to prove an analogous result to part (i) of the transfer results above.
However, we strongly conjecture that the following claim holds true.
Conjecture 1. Consider (14). Let bn= O (1/n) with > 0 suitable small. Then, an = c + O (1/n) ,
where c is a suitable constant.
As before, apart from this property, we need a couple of other transfer properties. However, once this conjecture is established, the other properties can be deduced from it.
Proposition 11. Assume that the above conjecture holds.
(i) Let bn= logαn with α ∈ {−1, 0, 1, . . .}. Then, an= logα+2n
(α + 2)m1
+ nPolα+1(log n) + O (1/n) ,
where > 0 is suitable small and m1 = ζ(2, b) with ζ(z, b) the Hurwitz zeta function.
(ii) Let bn= O (logαn) with α ∈ {−1, 0, 1, . . .}. Then, an= O logα+2n.
(iii) Item (ii) holds with O replaced by o as well.
Proof. All these properties follow from the conjecture by using similar ideas as in [10].
Finally, by applying our approach, we obtain the following result.
Theorem 8. As n → ∞, we have
Xn− log2n/(2m1) q
m2log3n/(3m31)
−→ N (0, 1),d
where m2 = 2ζ(3, b).
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