#  BEST Impact on Sterile Neutrino Hypothesis

Vladislav Barinov<EMAIL_ADDRESS>Institute for Nuclear Research of the
Russian Academy of Sciences, Moscow 117312, Russia  Physics Department, Moscow
State University, Vorobievy Gory, Moscow 119991, Russia  Dmitry Gorbunov
<EMAIL_ADDRESS>Institute for Nuclear Research of the Russian Academy of
Sciences, Moscow 117312, Russia  Moscow Institute of Physics and Technology,
Dolgoprudny 141700, Russia

###### Abstract

Recently the Baksan Experiment on Sterile Transitions (BEST) has presented
results Barinov  _et al._ (2021) confirming the gallium anomaly—a lack of
electron neutrinos $\nu_{e}$ at calibrations of SAGE Abdurashitov  _et al._
(1999, 2006) and GALLEX Kaether  _et al._ (2010)—at the statistical
significance exceeding 5 $\sigma$. This result is consistent with explanation
of the gallium anomaly as electron neutrino oscillations into sterile
neutrino, $\nu_{s}$. Within this explanation the BEST experiment itself
provides the strongest evidence for the sterile neutrino among all the
previous anomalous results in the neutrino sector. We combine the results of
gallium experiments with searches for sterile neutrinos in reactor
antineutrino experiments (assuming CPT-conservation in the $3+1$ neutrino
sector). While the “gallium” best-fit point in the model parameter space
(sterile neutrino mass squared $m_{\nu_{s}}^{2}\approx 1.25$ eV2, sterile-
electron neutrino mixing $\sin^{2}2\theta\approx 0.34$) is excluded by these
searches, a part of the BEST-favored 2 $\sigma$ region with
$m^{2}_{\nu_{s}}>5$ eV2 is consistent with all of them. Remarkably, the
regions advertised by anomalous results of the NEUTRINO-4 experiment Serebrov
_et al._ (2019) overlap with those of the BEST experiment: the best-fit point
of the joint analysis is $\sin^{2}2\theta\approx 0.38$,
$m_{\nu_{s}}^{2}\approx 7.3$ eV2, the favored region will be explored by the
KATRIN experiment Aker  _et al._ (2021). The sterile neutrino explanation of
the BEST results would suggest not only the extension of the Standard Model of
particle physics, but also either serious modifications of the Standard
Cosmological Model and Solar Model, or a specific modification of the sterile
sector needed to suppress the sterile neutrino production in the early
Universe and in the Sun.

††preprint: INR-TH/2021-017

##  I Introduction

Neutrino oscillations provide us with the only direct evidence for
incompleteness of the Standard Model of particle physics (SM), which fails to
explain them. While the fundamental physics responsible for the oscillations
is still elusive, results of the majority of neutrino experiments can be
phenomenologically described by introducing neutrino mass matrix. The joint
analysis (see e.g. Ref. Esteban  _et al._ (2020) for a recent one) of the
oscillation data determines three mixing angles between the mass and flavor
neutrino eigenstates and two squared-mass differences.

However, results of several experiments Athanassopoulos  _et al._ (1996);
Abdurashitov  _et al._ (1999, 2006); Kaether  _et al._ (2010); Mueller  _et
al._ (2011); Huber (2011); Aguilar-Arevalo  _et al._ (2018); Serebrov  _et
al._ (2019); Aartsen  _et al._ (2020) apparently cannot be described in this
way Kopp  _et al._ (2013); Dentler  _et al._ (2018), and within the
oscillation picture ask for neutrino states additional to those three of the
SM. These results are commonly known as neutrino anomalies and the new states
are known as sterile neutrinos, see e.g. Giunti and Lasserre (2019), since
they mix with the SM (or active) neutrinos but do not participate in the SM
gauge interactions.

All the neutrino anomalies must be thoroughly investigated Abazajian  _et al._
(2012); Böser  _et al._ (2020). Indeed, any unrecognized systematics constrain
the application of neutrino as, e.g. Sramek  _et al._ (2013); Adelberger  _et
al._ (2011) a tool in studying the inner structures of the Earth and Sun. On
the other side, the possible presence of sterile neutrinos in nature will be
the first example of the non-SM physics for which we have been looking for
decades. Moreover, the sterile neutrinos with mixing parameters required to
explain the anomalies Böser  _et al._ (2020) are expected to be produced in
the early Universe and be in equilibrium with the primordial plasma before the
epoch of Big Bang Nucleosynthesis. They would change the standard predictions
for the primordial abundance of the light chemical elements as well as the
subsequent late-time history of our Universe, including the recombination and
formation of the large-scale cosmic structures. Thus, in the present
concordance $\Lambda$CDM cosmological model these sterile neutrinos are
forbidden Aghanim  _et al._ (2020). The situation was quite the opposite about
ten years ago, when both state-of-the-art analyses of the Big Bang
Nucleosynthesis Izotov and Thuan (2010) and cosmic microwave background
anisotropy Komatsu  _et al._ (2011) favored the sterile neutrino hypothesis.
Anyway, with more complicated sterile neutrino sector they can be safe for
cosmology, see e.g. Hannestad  _et al._ (2014); Dasgupta and Kopp (2014); Chu
_et al._ (2015). Moreover, presently we have several tensions in consistent
description of the cosmological data, the most serious are so-called the
Hubble tension and $\sigma_{8}$ tension. So far we have no natural
straightforward resolutions, but some suggestions in literature, see e.g. Zhao
_et al._ (2017); Archidiacono  _et al._ (2020) involve sterile neutrinos as
one of the necessary ingredients.

##  II Results of the BEST experiment

Therefore, both particle physics and cosmology would benefit from settling
this issue with neutrino anomalies, and there are various projects and ongoing
experiments dedicated to clarification of particular anomalous results. One of
these experiments is Baksan Experiment on Sterile Transitions (BEST) aimed at
exploring the gallium anomaly Laveder (2007), which is the lack of electron
neutrino events from compact artificial chromium and argon sources observed by
four (in total) calibrations of SAGE Abdurashitov  _et al._ (1999, 2006) and
GALLEX Kaether  _et al._ (2010) solar neutrino experiments. BEST was placed in
the Baksan Neutrino Observatory of INR RAS and operated in 2019. Closely
following the scheme of the original SAGE calibration its geometry has been
specifically designed to achieve the highest sensitivity to the sterile
neutrino model with parameters tuned to explain the gallium anomaly via
$\nu_{e}\leftrightarrow\nu_{s}$ oscillations.

BEST exploited an artificial chromium source simultaneously irradiating two
volumes filled with gallium, which allowed one to measure the neutrino flux at
two different distances. The BEST experiment has published results Barinov
_et al._ (2021), which confirm the gallium anomaly and are consistent with its
explanation as electron neutrino oscillations into the sterile components. In
the BEST experiment the artificial source was very compact, placed in the
center of the inner spherical volume, which was placed in the center of the
outer cylindrical volume. The neutrino fluxes were measured as
$R_{in}=0.791\pm 0.05$ and $R_{out}=0.766\pm 0.05$ from what are expected,
which implies more than 5 $\sigma$ evidence for the disappearance of electron
neutrinos. Within the sterile neutrino hypothesis this disappearance is
explained as oscillations into sterile neutrino states, so that the electron
neutrino flux at a given distance $r$ from the point-like source is suppressed
by the survival probability

$P\left(E_{\nu},r\right)=1-\sin^{2}2\theta\,\sin^{2}\\!\\!\left(\\!\\!1.27\times\frac{\Delta
m^{2}[\text{\,eV}^{2}]\,r[\text{m}]}{E_{\nu}[\text{\,MeV}]}\\!\right)\\!.$

With the detector size of about a meter and the neutrino energy of about 750
keV (the dominant line of the chromium source) it allows for testing the
oscillations with $\Delta m^{2}\sim 1$ eV2. The geometry details are presented
in Ref. Barinov  _et al._ (2021), and the sterile neutrino parameters (mass
and mixing angle) are defined from the likelihood function given by eqs.
(3)-(5) of Ref. Barinov  _et al._ (2021). Then repeating the joint analysis
for the all gallium experiments one obtains the favorable region in the model
parameter space outlined in Fig. 1;

Figure 1: The regions (in shades of blue) favored by all the gallium
experiments Barinov  _et al._ (2021) for mixing with electron neutrinos.
Assuming the same mixing with electron antineutrinos there are also regions
(in shades of pink) favored by NEUTRINO-4 Serebrov  _et al._ (2021).

the 1-,2- and 3-$\sigma$ contours coincide with those in Fig.3 of Ref. Barinov
_et al._ (2021). We observe that within the sterile neutrino hypothesis the
gallium experiments provides higher than 5 $\sigma$ evidence for the presence
of sterile neutrinos. Actually, one can find that even the single BEST result
exceeds 5 $\sigma$ level.

##  III Joint analysis

The results obtained by the BEST experiment favor rather large mixing angle
and either the region of $\Delta m^{2}\simeq 1$ eV2 or noticeably larger
masses $\Delta m^{2}\gtrsim 3$ eV2. The likelihood contours on the plot of
Fig. 1 show that in the region of large masses the BEST has no sensitivity to
the sterile neutrino masses, which correspond to the oscillation lengths
shorter than the typical meter-scale size of the BEST detector volumes.
Remarkably, parts of this region have been favored by the anomalous results
(2.8 $\sigma$) of one of the reactor neutrino experiments, NEUTRINO-4 Serebrov
_et al._ (2019, 2021). We plot these regions in Fig. 1 for comparison. Note
that the Neutrino-4 likelihood exhibits a local minimum in the region of best
fit point (b.f.p.) of the joint gallium anomaly, $\Delta m^{2}\simeq 1$ eV2.
There are also minima at large masses, and the global minimum at $\Delta
m^{2}\simeq$7 eV2, where 1-$\sigma$ contours of NEUTRINO-4 and gallium
experiments overlap.

The region of masses $\Delta m^{2}\simeq 1$ eV2 has been thoroughly explored
by searches for sterile neutrino in the reactor antineutrino experiments, see
Fig. 2,

Figure 2: The regions favored by all the gallium experiments Barinov  _et al._
(2021) (the same as in Fig. 1) for mixing with electron neutrinos and limits
from searches for sterile neutrinos at reactor antineutrino experiments STEREO
Almazán  _et al._ (2020), PROSPECT Andriamirado  _et al._ (2021) and DANSS
Danilov (2021). The regions of large mixing are excluded by each experiment at
95% C.L.

the strongest constraint obtained by the DANSS experiment Alekseev  _et al._
(2018); Danilov (2021). Note, that while at small masses the favored by
gallium anomaly region is heavily constrained by these searches, at large
masses, $\Delta m^{2}>5$ eV2, there are 2-$\sigma$ and even 1-$\sigma$ patches
favored by gallium experiment and consistent with all these bounds.

Encouraged by the overlap presented on the plot of Fig. 1, we perform a joint
analysis of the all gallium experiments, NEUTRINO-4 and DANSS, which yields
the likelihood contours depicted in Fig. 3.

Figure 3: The regions (in shades of blue) favored by the joint analysis of the
gallium experiments, DANSS Danilov (2021) and NEUTRINO-4 Serebrov  _et al._
(2021). There are also regions excluded at 95% C.L. from sterile neutrino
searches at reactor antineutrino experiments STEREO Almazán  _et al._ (2020),
PROSPECT Andriamirado  _et al._ (2021). The regions outlined by the green line
is favored at 95% C.L. by the reactor antineutrino anomaly (RAA) Abazajian
_et al._ (2012). The region to the right of the black vertical line is
excluded at 95% C.L. from observations of solar neutrinos Giunti  _et al._
(2021).

We utilize here the original DANSS likelihood 111We thank M. Danilov and N.
Skrobova for sharing the DANSS likelihood distribution. and the NEUTRINO-4
likelihood 222We thank A. Serebrov and R. Samoilov for sharing the Neutrino-4
likelihood distribution. somewhat corrected as advertised in Ref. Giunti  _et
al._ (2021) to be conservative (the NEUTRINO-4 anomaly becomes then at 2.2
$\sigma$ level). The joint anomaly inferred from this analysis is at about 5.7
$\sigma$, and it favors the region of large sterile neutrino masses, $\Delta
m^{2}>5$ eV2. The significance level is dominated by the BEST anomalous
result, while the positions of the local minima of the joint likelihood at
particular values of the sterile neutrino mass are mostly determined by those
of the NEUTRINO-4 experiment.

We also plot in Fig. 3 limits from sterile neutrino searches at STEREO and
PROSPECT experiments (the regions of large mixing are excluded at 95% C.L.),
and the green contour that outlines the region favored at 95% C.L. by the
reactor antineutrino anomaly (RAA), see e.g. Abazajian  _et al._ (2012). One
observes, that the STEREO limits are consistent with the joint anomaly
presented in Fig. 3, while the limits from PROSPECT disfavor 2-$\sigma$
regions except a small part at $\Delta m^{2}\approx 9$ eV2. The reactor
antineutrino anomaly region also has a small spot with 2-$\sigma$ regions of
the joint anomaly.

The analysis above shows that the BEST anomalous result, if explained within
the hypothesis of a single sterile neutrino (dubbed $3+1$ scheme) points at
the region of large mixing and masses, which will be explored at the next
operation stages of the upgraded reactor neutrino experiments and KATRIN
experiment on tritium $\beta$-decay, see Ref. Aker  _et al._ (2021). The
position of the best fit point and the likelihood contours may be refined with
a joint statistical analysis including the likelihoods of other reactor
antineutrino experiments relevant for the task.

Finally, recall that, apart from the direct limits, the typical $3+1$ neutrino
model is strongly constrained from astrophysics and excluded by cosmological
observations within simple extensions of the Standard Cosmological Model.
Indeed, measurements of the solar neutrino flux exclude the models with large
mixing Giunti  _et al._ (2021), the black vertical line in Fig. 3 refers to
the corresponding 95% C.L. bound. Likewise, the large sterile-active neutrino
mixing produces sterile neutrinos in the plasma of the early Universe in the
amount forbidden from analysis of present cosmological data Aghanim  _et al._
(2020). It is tempting to suggest a modification of the simple $3+1$ scheme,
which would circumvent the both indirect constraints.

##  IV Conclusions

To summarize, we investigate how the recent results of BEST experiment can
affect the hypothesis of ’3+1’ scheme in neutrino sector. Note in passing that
the explanation of the observed lack of neutrino events as oscillations to the
sterile neutrino is not the only possibility involving new physics. Moreover,
an overall error in neutrino cross section on gallium or some issues in
extraction efficiency cannot be ruled out, though all the aspects of
experimental procedure have been verified and many have been double-checked
Barinov  _et al._ (2021).

We thank M. Danilov, V. Rubakov and S. Troitsky for stimulating questions and
remarks about the results of the BEST experiment. The work was supported by
the the RSF grant 17-12-01547.

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$\displaystyle={}_{1}F_{1}\left(\frac{k_{i}}{2},\frac{1}{2},\frac{a_{ii}^{2}}{2}\right)+\frac{\sqrt{2}a_{ii}\Gamma\left(\frac{k_{i}+1}{2}\right)}{\Gamma\left(\frac{k_{i}}{2}\right)}{}_{1}F_{1}\left(\frac{k_{i}+1}{2},\frac{3}{2},\frac{a_{ii}^{2}}{2}\right),$

where ${}_{1}F_{1}(a,b,z)$ is the Kummer confluent hypergeometric function.
For $a_{ii}<0$,

$\frac{1}{2^{k_{i}/2}\Gamma\left(\frac{k_{i}}{2}\right)}\int_{0}^{\infty}x^{k_{i}/2-1}\exp\left(-\frac{1}{2}x+a_{ii}\sqrt{x}\right)\mathrm{d}x=\frac{\Gamma\left(k_{i}\right)}{2^{k_{i}-1}\Gamma\left(\frac{k_{i}}{2}\right)}U\left(\frac{k_{i}}{2},\frac{1}{2},\frac{a_{ii}^{2}}{2}\right),$

and $U$ is Kummer’s U function.

##  Appendix B E-Statistics $T^{*}_{N}$ with Variable Stopping Times: the
Filtration Matters

Consider the the t-test as in Example 1.1. Fix some $0<a<b$, and define the
stopping time $N^{*}:=1$ if $|X_{1}|\not\in[a,b]$. $N^{*}=2$ otherwise. Then
clearly $N^{*}$ is not adapted to (hence not a stopping time relative to)
$(M_{n})_{n}$ as defined in that example, since $M_{1}\in\\{-1,1\\}$ coarsens
out all information in $X_{1}$ except its sign. Now let $\delta_{0}:=0$ (so
that ${\cal H}_{0}$ represents the normal distributions with mean $\mu=0$ and
arbitrary variance). Let $T^{*,\delta_{1}}_{n}(X^{n})$ be equal to the GROW
e-statistic $T^{*}_{n}(X^{n})$ as in (14); here we make explicit its
dependence on $\delta_{1}$. For ${\cal H}_{1}$, to simplify computations, we
put a prior $\tilde{\mathbf{\Pi}}^{\delta}_{1}$ on $\Delta_{1}:={\mathbb{R}}$.
We take $\tilde{\mathbf{\Pi}}^{\delta}_{1}$ to be a normal distribution with
mean $0$ and variance $\kappa$. We can now apply Corollary 8.3 (with prior
$\tilde{\mathbf{\Pi}}^{\delta}_{0}$ putting mass $1$ on
$\delta=\delta_{0}=0$), which gives that $\tilde{T}_{n}(X^{n})$ is an
e-statistic, where

$\tilde{T}_{n}(x^{n})=\int\frac{1}{\sqrt{2\pi\kappa^{2}}}\exp\left(-\frac{\delta_{1}^{2}}{2\kappa^{2}}\right)\cdot
T^{*,\delta_{1}}_{n}(x^{n})d\delta_{1}$

coincides with a standard type of Bayes factor used in Bayesian statistics. By
exchanging the integrals in the numerator, this expression can be calculated
analytically. The Bayes factor $\tilde{T}_{1}(x_{1})$ for $x^{1}=x_{1}$ is
found to be equal to $1$ for all $x_{1}\neq 0$, and the Bayes factor for
$(x_{1},x_{2})$ is given by:

$\tilde{T}_{2}(x_{1},x_{2})=\frac{\sqrt{2\kappa^{2}+1}\cdot(x_{1}^{2}+x_{2}^{2})}{\kappa^{2}(x_{1}-x_{2})^{2}+(x_{1}^{2}+x_{2}^{2})}.$

Now we consider the function

$f(x):={\bf E}_{X_{2}\sim N(0,1)}[\tilde{T}_{2}(x,X_{2})].$

$f(x)$ is continuous and even. We want to show that, with $N^{*}$ as above,
$\tilde{T}_{N^{*}}(X^{N^{*}})$ is not an E-variable for some specific choices
of $a,b$ and $\kappa$. Since, for any $\sigma>0$, the null contains the
distribution under which the $X_{i}$ are i.i.d. $N(0,\sigma)$, the data may,
under the null, in particular be sampled from $N(0,1)$. It thus suffices to
show that

${\bf E}_{X_{1},X_{2}\sim N(0,1)}[\tilde{T}_{N^{*}}(X^{N^{*}})]=P_{X_{1}\sim
N(0,1)}(|X_{1}|\not\in[a,b])+{\bf E}_{X_{1}\sim N(0,1)}[{\bf
1}_{|X_{1}|\in[a,b]}f(X_{1})]>1.$

But from numerical integration we find that $f(x)>1$ on $[a,b]$ and $[-b,-a]$
if we take $\kappa=200$, $a\approx 0.44$ and $b\approx 1.70$. Using again
numerical integration, we find that the above expectation is then
approximately equal to $1.19$, which shows that, even though $\tilde{T}_{n}$
is an e-statistic at each $n$ by Corollary 8.3 (it is even a GROW one),
$\tilde{T}_{N^{*}}$ is not an e-statistic (its expectation is $0.19$ too
large), providing the desired counterexample.


n)^{-1}\sum_{i=1}^{n-1}(\hat{\alpha}_{n}\hat{z}_{n}(\hat{\alpha}_{n})+i)^{-1}\to^{p}1$:
Lemma B.5 with
$\hat{\alpha}_{n}\hat{z}_{n}(\hat{\alpha}_{n})=O_{p}(1)=o_{p}(n)$ implies that

$\displaystyle\sum_{i=1}^{n-1}(\hat{\alpha}_{n}\hat{z}_{n}(\hat{\alpha}_{n})+i)^{-1}-\log
n$ $\displaystyle=\psi(n+\hat{\alpha}_{n}\hat{z}_{n}(\hat{\alpha}_{n}))-\log
n-\psi(1+\hat{\alpha}_{n}\hat{z}_{n}(\hat{\alpha}_{n}))$
$\displaystyle\to^{p}0-\psi(1+\alpha
f_{\alpha}^{-1}(\log\operatorname{\it{M}_{\alpha\theta}})),$

so $(\log
n)^{-1}\sum_{i=1}^{n-1}(\hat{\alpha}_{n}\hat{z}_{n}(\hat{\alpha}_{n})+i)^{-1}-1=o_{p}(1)$.
This concludes the proof of (C.12).

###  C.3 Proof of Theorem 3.6

$\hat{\theta}_{n}=\hat{\alpha}_{n}\hat{z}_{n}(\hat{\alpha}_{n})$ and C.4.1
Implies that $\hat{\theta}_{n}\to^{p}\alpha\cdot
f_{\alpha}^{-1}(\log\operatorname{\it{M}_{\alpha\theta}})$, and hence
$\operatorname{(ii)}$ holds. For $\operatorname{(i)}$,
$\sqrt{n^{\alpha}I_{\alpha}}(\hat{\alpha}_{n,\theta}-\alpha)\rightarrow
N/\sqrt{\operatorname{\it{M}_{\alpha\theta}}}$ by 3.8 and
$(\hat{\alpha}_{n}-\hat{\alpha}_{n,\theta})=O_{p}(n^{-\alpha}\log n)$ by
(3.12) leads to

$\displaystyle\sqrt{n^{\alpha}I_{\alpha}}(\hat{\alpha}_{n}-\alpha)$
$\displaystyle=\sqrt{n^{\alpha}I_{\alpha}}(\hat{\alpha}_{n}-\hat{\alpha}_{n,\theta})+\sqrt{n^{\alpha}I_{\alpha}}(\hat{\alpha}_{n,\theta}-\alpha)$
$\displaystyle=O_{p}(n^{-\alpha/2}\log
n)+\sqrt{n^{\alpha}I_{\alpha}}(\hat{\alpha}_{n,\theta}-\alpha)\rightarrow
N/\sqrt{\operatorname{\it{M}_{\alpha\theta}}},$

which concludes the proof.

##  Appendix D Omitted proofs

###  D.1 Proof of 3.1

Recall that $p_{\alpha}(j)$ is

$\displaystyle\forall j\in\mathbb{N},\
p_{\alpha}(j):=\frac{\alpha\prod_{i=1}^{j-1}(i-\alpha)}{j!}.$

Then, it is easy to derive $I_{\alpha}$ as

$\displaystyle
I_{\alpha}=-\sum_{j=1}^{\infty}p_{\alpha}(j)\partial_{\alpha}^{2}\log
p_{\alpha}(j)=\sum_{j=1}^{\infty}p_{\alpha}(j)\left(\frac{1}{\alpha^{2}}+\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}\right)$
$\displaystyle=\frac{1}{\alpha^{2}}+\sum_{j=1}^{\infty}p_{\alpha}(j)\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}.$

Hence, we obtained (A). It is important to emphasize that the sum of infinite
series is finite since it is an expectation of a finite function of $j$. Now
Fubini-Tonelli’s implies

$\displaystyle\sum_{j=1}^{\infty}p_{\alpha}(j)\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}$
$\displaystyle=\sum_{i=1}^{\infty}\frac{1}{(i-\alpha)^{2}}\sum_{j=i+1}^{\infty}p_{\alpha}(j).$

Thus, for (B) it remains to show the following:

$\displaystyle\forall i\in\mathbb{N},\
\sum_{j=i+1}^{\infty}p_{\alpha}(j)=\frac{i-\alpha}{\alpha}p_{\alpha}(i)$ (D.1)

We will prove it by induction. Notice that (D.1) holds for $i=1$ since
$\sum_{j=2}^{\infty}p_{\alpha}(j)=1-p_{\alpha}(1)=1-\alpha=(1-\alpha)p_{\alpha}(1)/\alpha$.
Here we assume (D.1) for $i=k\in\mathbb{N}$. Then,

$\displaystyle\sum_{j=(k+1)+1}^{\infty}p_{\alpha}(j)$
$\displaystyle=\sum_{j=k+1}^{\infty}p_{\alpha}(j)-p_{\alpha}(k+1)=\frac{k-\alpha}{\alpha}p_{\alpha}(k)-p_{\alpha}(k+1)$
$\displaystyle=\frac{k-\alpha}{\alpha}\cdot\frac{k+1}{k-\alpha}p_{\alpha}(k+1)-p_{\alpha}(k+1)=\frac{k+1-\alpha}{\alpha}p_{\alpha}(k+1),$

so (D.1) holds for $i=k+1$. Therefore, (D.1) holds for all $i\in\mathbb{N}$.

For the continuity of $I_{\alpha}$, we observe that

$\displaystyle
I_{\alpha}=\frac{1}{\alpha^{2}}+\sum_{j=1}^{\infty}\frac{p_{\alpha}(j)}{\alpha(i-\alpha)}.$

Now we claim that $I_{\alpha}$ converges uniformly on $K$ for any closed
subset $K=[s,t]\in(0,1)$, which clearly concludes the proof of continuity.
Since $p_{\alpha}(j)/\alpha=(\prod_{i=1}^{j-1}(i-\alpha))/j!$ is nonincreasing
on $[s,t]$, we observe that

$\displaystyle\forall j\in\mathbb{N},\ \sup_{\alpha\in
K}\left|p_{\alpha}(j)\frac{1}{\alpha(j-\alpha)}\right|$
$\displaystyle\leq\sup_{\alpha\in
K}\frac{p_{\alpha}(j)}{\alpha}\cdot\sup_{\alpha\in
K}\frac{1}{j-\alpha}=\frac{p_{s}(j)}{s}\frac{1}{j-t}\leq\frac{p_{s}(j)}{s(1-t)},$

and $\sum_{j=1}^{\infty}{p_{s}(j)}/(s(1-t))={1}/({s(1-t)})<+\infty.$ Then, the
Weierstrass M-test implies that $I_{\alpha}$ converges uniformly on $K$, which
concludes the proof.

###  D.2 Proof of Lemma 3.2

Recall that $f_{\alpha}:(-1,\infty)\rightarrow\mathbb{R}$ is defined by

$\displaystyle\forall z\in(-1,\infty),\
f_{\alpha}(z):=\psi(1+z)-\alpha\psi(1+\alpha z),$

where $\psi(z)=\Gamma^{\prime}(z)/\Gamma(z)$ is the digamma function. Then
$\lim_{z\rightarrow 0+}\psi(z)=-\infty$ implies

$\displaystyle\lim_{z\rightarrow-1+}f_{\alpha}(z)=\lim_{z\rightarrow-1+}\psi(1+z)-\alpha\psi(1-\alpha)=-\infty.$

We also observe that $\psi(z)=\log z+o(1)$ for large $z>0$ (see, for example,
[Zwillinger, 2018, Section 6]), which implies that

$\displaystyle f_{\alpha}(z)=\log z-\alpha\log(\alpha z)+o(1)=(1-\alpha)\log
z+O(1)\operatorname{\ \textrm{as}\ }z\rightarrow+\infty$
$\displaystyle\Rightarrow\lim_{z\to\infty}f_{\alpha}(z)=+\infty.$

In contrast, using $\psi^{(1)}(1+z)=\sum_{i=1}^{\infty}(i+z)^{-2}$ for all
$z>-1$, we obtain

$\displaystyle f^{\prime}_{\alpha}(z)$
$\displaystyle=\psi^{(1)}(1+z)-\alpha^{2}\psi^{(1)}(1+\alpha
z)=\sum_{i=1}^{\infty}\left\\{(i+z)^{-2}-\alpha^{2}(i+z\alpha)^{-2}\right\\}$
$\displaystyle=\sum_{i=1}^{\infty}\left\\{(i+z)^{-2}-(i/\alpha+z)^{-2}\right\\}>\sum_{i=1}^{\infty}\left\\{(i+z)^{-2}-(i+z)^{-2}\right\\}=0.$

Thus, $f_{\alpha}$ is strictly increasing. Putting all together, we conclude
that $f_{\alpha}(z)$ is bijective from $(-1,\infty)$ to $\mathbb{R}$. Finally,
we claim $f_{\alpha}^{\prime\prime}(z)<0$. This follows from

$\displaystyle f^{\prime\prime}_{\alpha}(z)$
$\displaystyle=\psi^{(2)}(1+z)-\alpha^{3}\psi^{(2)}(1+\alpha
z)=-2\sum_{i=1}^{\infty}\left\\{(i+z)^{-3}-\alpha^{3}(i+\alpha
z)^{-3}\right\\}$
$\displaystyle=-2\sum_{i=1}^{\infty}\left\\{(i+z)^{-3}-(i/\alpha+z)^{-3}\right\\}<-2\sum_{i=1}^{\infty}\left\\{(i+z)^{-3}-(i+z)^{-3}\right\\}=0,$

where we used $\psi^{(2)}(1+z)=-2\sum_{i=1}^{\infty}(i+z)^{-2}$ and
$0<\alpha<1$.

###  D.3 Proof of 3.3

Recall that Ewens–Pitman partition has the following log-likelihood:

$\displaystyle\ell_{n}(\alpha,\theta)=\sum_{i=1}^{K_{n}-1}\log(\theta+i\alpha)-\sum_{i=1}^{n-1}\log(\theta+i)+\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}\log(i-\alpha).$

First, we derive the convergence of
$\partial_{\theta}^{2}\ell_{n}(\alpha,\theta)$. This is written as

$\partial_{\theta}^{2}\ell_{n}(\alpha,\theta)=-\sum_{i=1}^{K_{n}-1}\frac{1}{(\theta+i\alpha)^{2}}+\sum_{i=1}^{n-1}\frac{1}{(\theta+i)^{2}}.$

Note that $\sum_{i=1}^{K_{n}-1}(\theta+i\alpha)^{-2}$ is a strictly increasing
function of $K_{n}$ and $K_{n}$ is a nondecreasing random variable. Then the
monotone convergence theorem implies that

$\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{(\theta+i\alpha)^{2}}\right]\rightarrow\sum_{i=1}^{\infty}\frac{1}{(\theta+i\alpha)^{2}}<+\infty.$
(D.2)

Using the above displays and $\psi^{(1)}(1+z)=\sum_{i=1}^{\infty}(i+z)^{-2}$,
we obtain

$\displaystyle I_{\theta\theta}^{(n)}$
$\displaystyle=\operatorname{\mathbb{E}}[-\partial_{\theta}^{2}\ell_{n}(\alpha,\theta)]=\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{(\theta+i\alpha)^{2}}\right]-\sum_{i=1}^{n-1}\frac{1}{(\theta+i)^{2}}$
$\displaystyle\rightarrow\sum_{i=1}^{\infty}\frac{1}{(\theta+i\alpha)^{2}}-\sum_{i=1}^{\infty}\frac{1}{(\theta+i)^{2}}$
$\displaystyle=\frac{1}{\alpha^{2}}\left(\sum_{i=1}^{\infty}\frac{1}{(i+\theta/\alpha)^{2}}-\sum_{i=1}^{\infty}\frac{\alpha^{2}}{(i+\theta)^{2}}\right)$
$\displaystyle=\frac{\psi^{(1)}(1+\theta/\alpha)-\alpha^{2}\psi^{(1)}(1+\alpha\cdot\theta/\alpha)}{\alpha^{2}}$
$\displaystyle=\alpha^{-2}f^{\prime}_{\alpha}(\theta/\alpha)<+\infty.$

Next, we derive the leading term for $I_{\alpha\theta}^{(n)}$. Notice that
$\operatorname{\mathbb{E}}\left[\partial_{\theta}\ell_{n}(\alpha,\theta)\right]=0$
holds since $\ell_{n}$ is the log-likelihood. Now from this equation we obtain

$\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{\theta+i\alpha}\right]-\sum_{i=1}^{n-1}\frac{1}{\theta+i}=0.$
(D.3)

In contrast, $\partial_{\theta}\partial_{\alpha}\ell_{n}(\alpha,\theta)$ is
written by

$\displaystyle\partial_{\theta}\partial_{\alpha}\ell_{n}(\alpha,\theta)=-\sum_{i=1}^{K_{n}-1}\frac{i}{(\theta+i\alpha)^{2}}=-\frac{1}{\alpha}\sum_{i=1}^{K_{n}-1}\frac{1}{\theta+i\alpha}+\frac{\theta}{\alpha}\sum_{i=1}^{K_{n}-1}\frac{1}{(\theta+i\alpha)^{2}}.$

Thus, (D.2) and (D.3) result in

$\displaystyle I_{\theta\alpha}^{(n)}$
$\displaystyle=-\operatorname{\mathbb{E}}[\partial_{\theta}\partial_{\alpha}\ell_{n}(\alpha,\theta)]=\frac{1}{\alpha}\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{\theta+i\alpha}\right]-\frac{\theta}{\alpha}\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{(\theta+i\alpha)^{2}}\right]$
$\displaystyle=\frac{1}{\alpha}\sum_{i=1}^{n-1}\frac{1}{\theta+i}+O(1)=\alpha^{-1}\log
n+O(1),$

which completes the proof for $I_{\theta\alpha}^{(n)}$.

Finally, we derive the leading term for $I_{\alpha\alpha}^{(n)}$. Note

$\partial_{\alpha}^{2}\ell_{n}(\alpha,\theta)=-\sum_{i=1}^{K_{n}-1}\frac{i^{2}}{(\theta+i\alpha)^{2}}-\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}},$

where

$\displaystyle\frac{i^{2}}{(\theta+i\alpha)^{2}}=\frac{1}{\alpha^{2}}-\frac{2\theta}{\alpha^{2}}\frac{1}{\theta+i\alpha}+\frac{\theta^{2}}{\alpha^{2}(\theta+i\alpha)^{2}}.$

Then, putting together the above displays and (D.2) and (D.3), one obtains

$\displaystyle I^{(n)}_{\alpha\alpha}$
$\displaystyle=-\operatorname{\mathbb{E}}[\partial_{\alpha}^{2}\ell_{n}(\alpha,\theta)]=\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{i^{2}}{(\theta+i\alpha)^{2}}\right]+\sum_{j=2}^{n}\operatorname{\mathbb{E}}[S_{n,j}]\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}$
$\displaystyle=\frac{\operatorname{\mathbb{E}}[K_{n}]-1}{\alpha^{2}}-\frac{2\theta}{\alpha^{2}}\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{\theta+i\alpha}\right]+\frac{\theta^{2}}{\alpha^{2}}\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{(\theta+i\alpha)^{2}}\right]+\sum_{j=2}^{n}\operatorname{\mathbb{E}}[S_{n,j}]\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}$
$\displaystyle=\frac{1}{\alpha^{2}}\operatorname{\mathbb{E}}[K_{n}]+\sum_{j=2}^{n}\operatorname{\mathbb{E}}[S_{n,j}]\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}-\frac{2\theta}{\alpha^{2}}\sum_{i=1}^{n-1}\frac{1}{\theta+i}+\frac{\theta^{2}}{\alpha^{2}}\operatorname{\mathbb{E}}\left[\sum_{i=1}^{K_{n}-1}\frac{1}{(\theta+i\alpha)^{2}}\right]-\frac{1}{\alpha^{2}}$
$\displaystyle=\frac{\operatorname{\mathbb{E}}[K_{n}]}{\alpha^{2}}+\sum_{j=2}^{n}\operatorname{\mathbb{E}}[S_{n,j}]\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}+O(\log
n).$

Here we take
$X_{n}:=\alpha^{-2}K_{n}+\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}{(i-\alpha)^{-2}}$.
Then, one obtains

$\displaystyle I_{\alpha\alpha}^{(n)}=\operatorname{\mathbb{E}}[X_{n}]+O(\log
n),$

Thus, it remains to show
$\operatorname{\mathbb{E}}[X_{n}]/n^{\alpha}\to\operatorname{\mathbb{E}}[\operatorname{\it{M}_{\alpha\theta}}]I_{\alpha}$.
We will show it by the dominated convergence theorem. First, we observe that
$X_{n}$ is bounded by $K_{n}$ up to a constant:

$\displaystyle 0\leq
X_{n}=\sum_{j=1}^{n}S_{n,j}\left(\frac{1}{\alpha^{2}}+\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}\right)=\sum_{j=1}^{n}S_{n,j}\sum_{i=0}^{j-1}\frac{1}{(i-\alpha)^{2}}$
$\displaystyle\leq\sum_{j=1}^{n}S_{n,j}\cdot\underbrace{\sum_{i=0}^{\infty}\frac{1}{(i-\alpha)^{2}}}_{=:C_{\alpha}}=\sum_{j=1}^{n}S_{n,j}\cdot
C_{\alpha}=K_{n}C_{\alpha},$

Next A.1.1 applied with $g(j)=\sum_{i=0}^{j-1}(i-\alpha)^{-2}$ and
$K_{n}/n^{\alpha}\rightarrow\operatorname{\it{M}_{\alpha\theta}}$ (a.s.)
result in

$\displaystyle\frac{X_{n}}{n^{\alpha}}$
$\displaystyle=\frac{K_{n}}{n^{\alpha}}\sum_{j=1}^{\infty}\frac{S_{n,j}}{K_{n}}\sum_{i=0}^{j-1}\frac{1}{(i-\alpha)^{2}}\rightarrow\operatorname{\it{M}_{\alpha\theta}}\sum_{j=1}^{\infty}p_{\alpha}(j)\sum_{i=0}^{j-1}\frac{1}{(i-\alpha)^{2}}$
$\displaystyle=\operatorname{\it{M}_{\alpha\theta}}\left(\frac{1}{\alpha^{2}}+\sum_{j=2}^{\infty}p_{\alpha}(j)\sum_{i=1}^{j-1}\frac{1}{(i-\alpha)^{2}}\right)=\operatorname{\it{M}_{\alpha\theta}}I_{\alpha}\
\text{(a.s.)}$

where we used (A) of 3.1 for the last equality. Conversely, putting together
the above displays and
$K_{n}/n^{\alpha}\rightarrow\operatorname{\it{M}_{\alpha\theta}}$ in mean by
(A) of Lemma 2.2, the dominated convergence implies that

$\displaystyle
n^{-\alpha}\operatorname{\mathbb{E}}[X_{n}]=\operatorname{\mathbb{E}}[n^{-\alpha}X_{n}]\rightarrow\operatorname{\mathbb{E}}[\operatorname{\it{M}_{\alpha\theta}}I_{\alpha}]=\operatorname{\mathbb{E}}[\operatorname{\it{M}_{\alpha\theta}}]I_{\alpha}.$

This concludes the proof.

###  D.4 Derivation of the condition Equation B.2

Recall that $\partial_{x}\ell_{n}(\cdot,\theta^{\star})$ is written by

$\displaystyle\forall x\in(-\theta^{\star}\vee 0,1),\
\partial_{x}\ell_{n}(x,\theta^{\star})=\sum_{i=1}^{K_{n}-1}\frac{i}{\theta+ix}-\sum_{j=2}^{n}{S_{n,j}}\sum_{i=1}^{j-1}\frac{1}{i-x}$

Here we prove that $\partial_{x}\ell_{n}(\cdot,\theta^{\star})=0$ has a unique
solution on $(-\theta^{\star}\vee 0,1)$ if and only if

$\displaystyle 1<K_{n}<n,\
\Theta_{n}:=\frac{K_{n}(K_{n}-1)}{2\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}i^{-1}}>\theta.$
(D.4)

Here, it is important to emphasize that
$\partial_{x}\ell_{n}(x,\theta^{\star})$ is decreasing in $x$:

$\displaystyle\partial_{x}^{2}\ell_{n}(x,\theta^{\star})$
$\displaystyle:=-\sum_{i=1}^{K_{n}-1}\frac{i^{2}}{(\theta+ix)^{2}}-\sum_{j=2}^{n}{S_{n,j}}\sum_{i=1}^{j-1}\frac{1}{(i-x)^{2}}\leq
0$

First, we prove the sufficiency of (D.4). Since
$\partial_{x}\ell_{n}(x,\theta^{\star})$ is decreasing, it suffices to show
$\lim_{x\rightarrow 1-}\partial_{x}\ell_{n}(x,\theta^{\star})(x)<0$ and
$\lim_{x\rightarrow(-\theta\vee
0,1)+}\partial_{x}\ell_{n}(x,\theta^{\star})>0$. Notice that $K_{n}<n$ is
equivalent to $\sum_{j\geq 2}S_{n,j}>0$. Then we can rewrite
$\partial_{x}\ell_{n}(x,\theta^{\star})$ by

$\displaystyle\partial_{x}\ell_{n}(x,\theta^{\star})=-\frac{1}{1-x}\sum_{j=2}^{n}{S_{n,j}}+\sum_{i=1}^{K_{n}-1}\frac{i}{\theta+ix}-\sum_{j=3}^{n}{S_{n,j}}\sum_{i=2}^{j-1}\frac{1}{i-x},$

where, the first term goes to $-\infty$ as $x\rightarrow 1-$ while the others
converge to a finite value. Thus, $\lim_{x\rightarrow
1-}\partial_{x}\ell_{n}(x,\theta^{\star})(x)<0$ holds.

For $\lim_{x\rightarrow(-\theta\vee
0,1)+}\partial_{x}\ell_{n}(x,\theta^{\star})>0$, we consider the two cases
$\\{\theta>0\\}$ and $\\{-\alpha<\theta\leq 0\\}$ separately. In the former,
observe that

$\displaystyle\lim_{x\rightarrow(-\theta\vee
0)+}\partial_{x}\ell_{n}(x,\theta^{\star})=\partial_{x}\ell_{n}(0,\theta^{\star})$
$\displaystyle=\sum_{i=1}^{K_{n}-1}\frac{i}{\theta}-\sum_{j=2}^{n}{S_{n,j}}\sum_{i=1}^{j-1}\frac{1}{i}=\frac{K_{n}(K_{n}-1)}{2\theta}-\sum_{j=2}^{n}{S_{n,j}}\sum_{i=1}^{j-1}\frac{1}{i}$
$\displaystyle=\frac{\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}i^{-1}}{\theta}\left(\frac{K_{n}(K_{n}-1)}{2\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}i^{-1}}-\theta\right)$
$\displaystyle=\frac{\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}i^{-1}}{\theta}\left(\Theta_{n}-\theta\right)>0,$

where we used
$\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}i^{-1}>\sum_{j=2}^{n}S_{n,j}>0$ and the
assumption $\Theta_{n}>\theta$. In the latter case, i.e., when
$-\alpha<\theta\leq 0$, $-\theta\vee 0=-\theta$ holds. If we rewrite as

$\displaystyle\partial_{x}\ell_{n}(x,\theta^{\star})=\frac{1}{\theta+x}+\sum_{i=2}^{K_{n}-1}\frac{i}{\theta+ix}-\sum_{j=2}^{n}{S_{n,j}}\sum_{i=1}^{j-1}\frac{1}{i-x},$

it is clear that the first term goes to $+\infty$ while the other terms
converge to finite values as $x\rightarrow(-\theta)+$. This concludes the
proof of $\lim_{x\rightarrow(-\theta\vee
0)+}\partial_{x}\ell_{n}(x,\theta^{\star})>0$.

Finally, we show that the condition (D.4) is necessary. Suppose $K_{n}=1$,
which is equivalent to $S_{n.n}=1$. Then,
$\partial_{y}\ell_{n}(x,\theta^{\star})=-n\sum_{i=1}^{n-1}(i-x)^{-1}<0$ for
all $x\in(-\theta\vee 0,1)$, so it has no roots on $(-\theta\vee 0,1)$. If
$K_{n}=n$, i.e., $S_{n,1}=n$, we observe
$\partial_{x}\ell_{n}(x,\theta^{\star})=n^{-1}\sum_{i=1}^{n-1}i/(\theta+ix)>0$
for all $x\in(-\theta\vee 0,1)$, so again it has no roots. Finally suppose
that $1<K_{n}<n$ but $\theta\geq\Theta_{n}$. In this case, $\theta$ must be
positive since

$\displaystyle\theta\geq\Theta_{n}=\frac{K_{n}(K_{n}-1)}{2\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}i^{-1}}>0.$

Thus, $(-\theta\vee 0,1)=(0,1)$ holds. Since
$\partial_{x}\ell_{n}(x,\theta^{\star})$ is decreasing in $x$, we have

$\displaystyle\forall x\in(0,1),\
\partial_{x}\ell_{n}(x,\theta^{\star})<\partial_{x}\ell_{n}(0,\theta^{\star})=\frac{\sum_{j=2}^{n}S_{n,j}\sum_{i=1}^{j-1}i^{-1}}{\theta}(\Theta_{n}-\theta)\leq
0,$

so again, it has no roots.

###  D.5 Proof of Lemma A.1

We define
$\mathcal{N}_{j}:=\left\\{\lim_{n\rightarrow\infty}\mathbb{P}_{n}(j)\neq\mathbb{P}(j)\right\\}$
and $\mathcal{N}:=\cup_{j=1}^{\infty}\mathcal{N}_{j}$.
$\Pr(\mathcal{N}_{j})=0$ by (B) of Lemma 2.2 implies
$\Pr(\mathcal{N})=\Pr(\cup_{j=1}^{\infty}\mathcal{N}_{j})\leq\sum_{j=1}^{\infty}\Pr(\mathcal{N}_{j})=0$,
i.e., $\mathcal{N}$ is null set. Thus, we can exclude the event $\mathcal{N}$.
Observe

$\displaystyle\left|\mathbb{P}_{n}(j)-\mathbb{P}(j)\right|\leq\mathbb{P}_{n}(j)+\mathbb{P}(j),\
\sum_{j=1}^{\infty}\left\\{\mathbb{P}_{n}(j)+\mathbb{P}(j)\right\\}=1+1=2.$

Then, Fatou’s lemma implies that

$\displaystyle
2-\limsup_{n\rightarrow\infty}\sum_{j=1}^{\infty}\left|\mathbb{P}_{n}(j)-\mathbb{P}(j)\right|=\liminf_{n\rightarrow\infty}\left(2-\sum_{j=1}^{\infty}\left|\mathbb{P}_{n}(j)-\mathbb{P}(j)\right|\right)$
$\displaystyle=\liminf_{n\rightarrow\infty}\left\\{\sum_{j=1}^{\infty}\left(\mathbb{P}_{n}(j)+\mathbb{P}(j)-\left|\mathbb{P}_{n}(j)-\mathbb{P}(j)\right|\right)\right\\}$
$\displaystyle\geq\sum_{j=1}^{\infty}\left\\{\liminf_{n\rightarrow\infty}\left(\mathbb{P}_{n}(j)+\mathbb{P}(j)-\left|\mathbb{P}_{n}(j)-\mathbb{P}(j)\right|\right)\right\\}=\sum_{j=1}^{\infty}(2\mathbb{P}(j)+0)=2.$

Subtracting $2$ from both sides, we obtain
$\limsup_{n\rightarrow\infty}\sum_{j=1}^{\infty}\left|\mathbb{P}_{n}(j)-\mathbb{P}(j)\right|=0$,
which is equivalent to
$\lim_{n\rightarrow\infty}\sum_{j=1}^{\infty}\left|\mathbb{P}_{n}(j)-\mathbb{P}(j)\right|=0$.

###  D.6 Proof of Lemma A.3

Recall that $\Psi:(0,1)\rightarrow\mathbb{R}$ is defined by

$\displaystyle\forall x\in(0,1),\ \Psi(x):=\frac{1}{x}-\mathbb{P}g_{x},$ (D.5)

where $\mathbb{P}g_{x}=\sum_{j=1}^{\infty}p_{\alpha}(j)g_{x}(j)$ and
$g_{x}(j)=\sum_{i=1}^{j-1}(i-x)^{-1}$. We claim that $\Psi(x)$ converges for
each $x\in(0,1)$ by the followings:

  * •

$p_{\alpha}(j)=O(j^{-(1+\alpha)})$ (see Section 2.1)

  * •

$\sum_{i=1}^{j-1}(i-x)^{-1}\sim\log{j}$ as $j\to\infty$.

  * •

$\int_{1}^{\infty}\log{x}\cdot
x^{-(1+\alpha)}\mathop{}\\!\mathrm{d}x=\alpha^{-2}<+\infty$.

First, we prove (A): $\Psi$ is of class $C^{1}$ on $(0,1)$. For any closed
subset $K=[s,t]\subset(0,1)$, observe

$\displaystyle\forall j\in\mathbb{N},\ \sup_{x\in
K}\left|\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}x}g_{x}(j)\right|=\sup_{x\in
K}\sum_{i=1}^{j-1}\frac{1}{(i-x)^{2}}<\sum_{i=1}^{\infty}\frac{1}{(i-t)^{2}}:=C.$

Since $\Psi(x)$ takes a form of expectation of $g_{x}$ with respect to
$\mathbb{P}$ (see (D.5)), Weierstrass’s M-test implies that $\Psi^{\prime}(x)$
converges uniformly on $K$. As $K$ is arbitrary, it means that
$\Psi^{\prime}(x)$ converges compactly on $(0,1)$. Thus, we conclude that
$\Psi$ is class $C^{1}$ on $(0,1)$.

Now (B) immediately follows: The compact convergence that we have shown
implies that we can change the summation and differentiation, which leads to

$\displaystyle\Psi^{\prime}(x)$
$\displaystyle=\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}x}\left(\frac{1}{x}-\mathbb{P}g_{x}\right)=-\frac{1}{x^{2}}-\mathbb{P}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}x}g_{x}$
$\displaystyle=-\frac{1}{x^{2}}-\sum_{j=1}^{\infty}p_{\alpha}(j)\sum_{i=1}^{j-1}\frac{1}{(i-x)^{2}}\overset{(\star)}{=}-I_{\alpha}<0,$

where $(\star)$ follows from (A) of 3.1.

For (C), the differentiation of both side of
$\sum_{j=1}^{\infty}p_{\alpha}(j)=1$ by $\alpha$ implies that
$\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\alpha}\sum_{n=1}^{\infty}p_{\alpha}(j)=0$.
Now we claim that we can interchange the differential and summation on the
right-hand side. Then, the definition
$p_{\alpha}(j):=\alpha\prod_{i=1}^{j-1}(i-\alpha)/(j!)$ implies that

$\displaystyle 0$
$\displaystyle=\sum_{j=1}^{\infty}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\alpha}p_{\alpha}(j)=\sum_{j=1}^{\infty}p_{\alpha}(j)\left(\frac{1}{\alpha}-\sum_{i=1}^{j-1}\frac{1}{i-\alpha}\right)$
$\displaystyle=\frac{1}{\alpha}-\sum_{j=2}^{\infty}p_{\alpha}(j)\sum_{i=1}^{j-1}\frac{1}{i-\alpha}\overset{(\star
2)}{=}\Psi(\alpha),$

where we used (B) of 3.1. This concludes the proof of $(C)$. It remains to
prove
$\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\alpha}\sum_{n=1}^{\infty}p_{\alpha}(j)=\sum_{n=1}^{\infty}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\alpha}p_{\alpha}(j)$.
We prove this by the compact convergence of
$\sum_{j=1}^{\infty}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\alpha}p_{\alpha}(j)$
on $(0,1)$, i.e. the uniform convergence of
$\sum_{j=1}^{\infty}\frac{d}{d\alpha}p_{\alpha}(j)$ on $K$ for any closed
interval $K=[s,t]\subset(0,1)$. Note that
$\alpha^{-1}{p_{\alpha}(j)}=\prod_{i=1}^{j-1}(i-\alpha)/{j!}$ is nonincreasing
function on $(0,1)$ for all $j\in\mathbb{N}$. Then, we obtain

$\displaystyle\sup_{\alpha\in[s,t]}\left|\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\alpha}p_{\alpha}(j)\right|=\sup_{\alpha\in[s,t]}\left|p_{\alpha}(j)\left(\frac{1}{\alpha}-\sum_{i=1}^{j-1}\frac{1}{i-\alpha}\right)\right|$
$\displaystyle\leq\sup_{\alpha\in[s,t]}\left\\{\frac{p_{\alpha}(j)}{\alpha}\left(1+\sum_{i=1}^{j-1}\frac{\alpha}{i-\alpha}\right)\right\\}\leq\sup_{\alpha\in[s,t]}\left(\frac{p_{\alpha}(j)}{\alpha}\right)\times\left(1+\sup_{\alpha\in[s,t]}\sum_{i=1}^{j-1}\frac{\alpha}{i-\alpha}\right)$
$\displaystyle=\frac{p_{s}(j)}{s}\left(1+\sum_{i=1}^{j-1}\frac{t}{i-t}\right)=:M_{j}^{s,t},$

and we observe

$\displaystyle\sum_{j=1}^{\infty}M_{j}^{s,t}=\sum_{j=1}^{\infty}\frac{p_{s}(j)}{s}\left(1+t\sum_{i=1}^{j-1}\frac{1}{i-t}\right)=\frac{1}{s}+\frac{t}{s}\sum_{j=1}^{\infty}p_{s}(j)\sum_{i=1}^{j-1}\frac{1}{i-t}<+\infty.$

Here the right-hand side converges based on the same argument at the beginning
of this proof. Thus, Weierstrass’s M-test implies that
$\sum_{j=1}^{\infty}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}\alpha}p_{\alpha}(j)$
converges uniformly on $[s,t]$.

###  D.7 Proof of Lemma B.1

We observe

$\displaystyle\hat{\Psi}_{n,\theta^{\star}}(x)-\Psi(x)=-\frac{1}{xK_{n}}-\frac{\theta^{\star}}{xK_{n}}\sum_{i=1}^{K_{n}-1}\frac{1}{\theta^{\star}+ix}-(\mathbb{P}_{n}g_{x}-\mathbb{P}g_{x}),$
(D.6)

where $g_{x}(j)=\sum_{i=1}^{j-1}(i-x)^{-1}$. Then, $K_{n}\to\infty$ (a.s.) and
$(\mathbb{P}_{n}g_{x}-\mathbb{P}g_{x})\to^{p}0$ by Lemma A.2 implies
$\hat{\Psi}_{n,\theta^{\star}}(x)-\Psi(x)=o_{p}(1)$, which concludes the proof
of (A).

For (B), differentiation of (D.6) leads to

$\displaystyle\hat{\Psi}^{\prime}_{n,\theta^{\star}}(x)-\Psi^{\prime}(x)$
$\displaystyle=\frac{1}{x^{2}K_{n}}+\frac{\theta^{\star}}{x^{2}K_{n}}\sum_{i=1}^{K_{n}-1}\frac{1}{\theta^{\star}+ix}-\frac{\theta^{\star}}{xK_{n}}\sum_{i=1}^{K_{n}-1}\frac{i}{(\theta^{\star}+ix)^{2}}$
$\displaystyle-(\mathbb{P}_{n}h_{x}-\mathbb{P}h_{x}),$

where $h_{x}$ is the function defined by
$h_{x}(j)=\partial_{x}g_{x}(j)=\sum_{i=1}^{j-1}(i-x)^{-2}$. Thus, we can bound
$\sup_{x\in I}|\hat{\Psi}^{\prime}_{n,\theta^{\star}}(x)-\Psi^{\prime}(x)|$ as
follows:

$\displaystyle\sup_{x\in
I}\left|\hat{\Psi}^{\prime}_{n,\theta^{\star}}(x)-\Psi^{\prime}(x)\right|$
$\displaystyle\leq\sup_{x\in I}\frac{1}{K_{n}x^{2}}+\sup_{x\in
I}\frac{|\theta^{\star}|}{x^{2}K_{n}}\sum_{i=1}^{K_{n}-1}\frac{1}{\theta^{\star}+ix}$
$\displaystyle+\sup_{x\in
I}\frac{\theta^{\star}}{xK_{n}}\sum_{i=1}^{K_{n}-1}\frac{i}{(\theta^{\star}+ix)^{2}}+\sup_{x\in
I}\left|\mathbb{P}_{n}h_{x}-\mathbb{P}h_{x}\right|.$

Here we write $I=[s,t]\subset(-\theta^{\star}\vee 0,1)$. Then each term is
bounded:

$\displaystyle 0\leq\sup_{x\in I}\frac{1}{K_{n}x^{2}}\leq\frac{1}{K_{n}s^{2}}$
$\displaystyle 0\leq\sup_{x\in
I}\frac{|\theta^{\star}|}{x^{2}K_{n}}\sum_{i=1}^{K_{n}-1}\frac{1}{\theta^{\star}+ix}\leq\frac{|\theta^{\star}|}{s^{2}}\frac{1}{K_{n}}\sum_{i=1}^{K_{n}-1}\frac{1}{\theta^{\star}+is}=O(K_{n}^{-1}\log
K_{n})$ $\displaystyle 0\leq\sup_{x\in
I}\frac{|\theta^{\star}|}{xK_{n}}\sum_{i=1}^{K_{n}-1}\frac{i}{(\theta^{\star}+ix)^{2}}\leq\frac{|\theta^{\star}|}{sK_{n}}\sum_{i=1}^{K_{n}-1}\frac{i}{(\theta^{\star}+is)^{2}}=O(K_{n}^{-1}\log
K_{n})$ $\displaystyle 0\leq\sup_{x\in
I}|\mathbb{P}_{n}h_{x}-\mathbb{P}h_{x}|\leq\sum_{j=1}^{\infty}|\mathbb{P}_{n}(j)-\mathbb{P}(j)|\cdot\sup_{x\in
I}\|h_{x}\|_{\infty}$
$\displaystyle=\sum_{j=1}^{\infty}|\mathbb{P}_{n}(j)-\mathbb{P}(j)|\cdot\sum_{i=1}^{\infty}\frac{1}{(i-t)^{2}}=O(1)\sum_{j=1}^{\infty}|\mathbb{P}_{n}(j)-\mathbb{P}(j)|.$

Then $K_{n}\to\infty$ (a.s.) and
$\sum_{j=1}^{\infty}|\mathbb{P}_{n}(j)-\mathbb{P}(j)|\rightarrow 0$ (a.s.) by
Lemma A.1 implies that they converge to $0$ (a.s.), which concludes the proof.

###  D.8 Proof of Lemma C.1

The following proof is basically the same as [Carlton, 1999, Lemma 5.3], but
we write it to make our proof self-contained. We fix $x\in(0,1)$ and define
$\hat{\Phi}_{n,x}:(-x,\infty)\rightarrow\mathbb{R}$ by

$\displaystyle\forall y>-x,\
\hat{\Phi}_{n,x}(y):=\partial_{y}\ell_{n}(x,y)=\sum_{i=1}^{K_{n}-1}\frac{1}{y+ix}-\sum_{i=1}^{n-1}\frac{1}{y+i}.$
(D.7)

Then, $1<K_{n}$ implies

$\displaystyle\lim_{y\rightarrow-x+}\hat{\Phi}_{n,x}(y)=\lim_{y\rightarrow-x+}\frac{1}{y+x}+\sum_{i=2}^{K_{n}-1}\frac{1}{(i-1)x}-\sum_{i=1}^{n-1}\frac{1}{-x+i}=+\infty.$

Now we claim $\hat{\Phi}_{n,x}(y)<0$ for sufficiently large $y$: we observe

$\displaystyle\hat{\Phi}_{n,x}(y)$
$\displaystyle=\sum_{i=1}^{K_{n}-1}\left(\frac{1}{y+ix}-\frac{1}{y+i}\right)-\sum_{K_{n}}^{n-1}\frac{1}{y+i}=\sum_{i=1}^{K_{n}-1}\frac{(1-x)i}{(y+ix)(y+i)}-\sum_{i=K_{n}}^{n-1}\frac{1}{y+i}$
(D.8)
$\displaystyle\leq\frac{K_{n}^{2}(1-x)}{(y+x)(y+1)}-\frac{1}{y+K_{n}}=\frac{K_{n}^{2}(1-x)(y+K_{n})-y^{2}-(1+x)y-x}{(y+x)(y+1)(y+K_{n})},$

where the numerator in the last term is quadratic, and its coefficient is
$-1$. Thus, the numerator takes a negative value for sufficiently large $y$
while the denominator remains positive. Putting together the above two
displays and the intermediate value theorem, we see that $\hat{\Psi}_{n,x}$
has a root in $(-x,\infty)$.

Next, we prove the uniqueness. Observe

$\displaystyle\hat{\Phi}^{\prime}_{n,x}(y)=-\sum_{i=1}^{K_{n}-1}\frac{1}{(y+ix)^{2}}+\sum_{i=1}^{n-1}\frac{1}{(y+i)^{2}}$
$\displaystyle=-\sum_{i=1}^{K_{n}-1}\left(\frac{1}{(y+ix)^{2}}-\frac{1}{(y+i)^{2}}\right)+\sum_{i=K_{n}}^{n-1}\frac{1}{(y+i)^{2}}$
$\displaystyle=-\sum_{i=1}^{K_{n}-1}\frac{i(1-x)(y+ix+y+i)}{(y+ix)^{2}(y+i)^{2}}+\sum_{i=K_{n}}^{n-1}\frac{1}{(y+i)^{2}}$
$\displaystyle<-\sum_{i=1}^{K_{n}-1}\frac{i(1-x)(y+ix+y+ix)}{(y+ix)^{2}(y+i)^{2}}+\sum_{i=K_{n}}^{n-1}\frac{1}{(y+i)^{2}}$
$\displaystyle=-2\sum_{i=1}^{K_{n}-1}\frac{(1-x)i}{(y+ix)(y+i)^{2}}+\sum_{i=K_{n}}^{n-1}\frac{1}{(y+i)^{2}}$
$\displaystyle\leq-\frac{2}{y+K_{n}-1}\underbrace{\sum_{i=1}^{K_{n}-1}\frac{(1-x)i}{(y+ix)(y+i)}}_{:=Q_{n}}+\sum_{i=K_{n}}^{n-1}\frac{1}{(y+i)^{2}}.$

If $\hat{\Phi}^{\prime}_{n,x}(y)\geq 0$, then
$Q_{n}\leq(({y+K_{n}-1})/{2})\cdot\sum_{i=K_{n}}^{n-1}(y+i)^{-2}$. This and
(D.8) result in

$\displaystyle\hat{\Phi}_{n,x}(y)$
$\displaystyle=Q_{n}-\sum_{i=K_{n}}^{n-1}\frac{1}{y+i}\leq\frac{y+K_{n}-1}{2}\sum_{i=K_{n}}^{n-1}\frac{1}{(y+i)^{2}}-\sum_{i=K_{n}}^{n-1}\frac{1}{y+i}$
$\displaystyle=\sum_{i=K_{n}}^{n-1}\frac{y+K_{n}-1-2(y+i)}{2(y+i)^{2}}$
$\displaystyle=\sum_{i=K_{n}}^{n-1}\frac{K_{n}-1-2i-y}{2(y+i)^{2}}<\sum_{i=K_{n}}^{n-1}\frac{K_{n}-2i}{2(y+i)^{2}}<-\frac{K_{n}}{2(y+K_{n})^{2}}<0,$

for all $y>-x(>-1)$. In summary, we get

$\displaystyle\forall y\in(-x,\infty),\ \hat{\Phi}^{\prime}_{n,x}(y)\geq
0\Rightarrow\hat{\Phi}_{n,x}(y)<0,$ (D.9)

which implies the uniqueness of the solution of $\hat{\Phi}_{n,x}(y)=0$ on
$(-x,\infty)$. Let $\hat{y}_{n}(x)$ be the unique solution. Then,
$\hat{\Phi}_{n,x}^{\prime}(\hat{y}_{n}(x))<0$ is obvious from (D.9) and
$\hat{\Phi}_{n,x}(\hat{y}_{n}(x))=0$.

###  D.9 Proof of Lemma C.3

Section D.8 implies that $\hat{y}_{n}(x)$ is the unique root of
$\hat{\Phi}_{n,x}$ on $(-x,\infty)$, which is defined by (D.7), and hence,
$\hat{y}_{n}(x)$ satisfies the following equality:

$\displaystyle\forall x\in(0,1),\
\sum_{i=1}^{K_{n}-1}\frac{1}{\hat{y}_{n}(x)+ix}=\sum_{i=1}^{n-1}\frac{1}{\hat{y}_{n}(x)+i}\text{
and }\hat{y}_{n}(x)>-x.$ (D.10)

Then, we have

$\displaystyle\hat{\Psi}_{n}(x)=\Psi_{n}(x,\hat{y}_{n}(x))$
$\displaystyle=\frac{1}{K_{n}}\sum_{i=1}^{K_{n}-1}\frac{i}{\hat{y}_{n}(x)+ix}-\sum_{j=2}^{n}\frac{S_{n,j}}{K_{n}}\sum_{i=1}^{j-1}\frac{1}{i-x}$
$\displaystyle=\frac{1}{x}-\frac{1}{K_{n}x}-\frac{1}{K_{n}x}\sum_{i=1}^{K_{n}-1}\frac{\hat{y}_{n}(x)}{\hat{y}_{n}(x)+ix}-\mathbb{P}_{n}g_{x}$
$\displaystyle\overset{(\star
1)}{=}\frac{1}{x}-\frac{1}{K_{n}x}-\frac{1}{K_{n}x}\sum_{i=1}^{n-1}\frac{\hat{y}_{n}(x)}{\hat{y}_{n}(x)+i}-\mathbb{P}_{n}g_{x},$
$\displaystyle\overset{(\star
2)}{=}\frac{1}{x}-\frac{1}{K_{n}x}-\frac{1}{K_{n}}\sum_{i=1}^{n-1}\frac{\hat{z}_{n}(x)}{x\hat{z}_{n}(x)+i}-\mathbb{P}_{n}g_{x},$

where $(\star 1)$ follows from (D.10) and we used
$\hat{z}_{n}(x)=\hat{y}_{n}(x)/x$ for $(\star 2)$. This concludes the proof.

###  D.10 Proof of Lemma C.4

#### Proof of (A)

Suppose $1<K_{n}<n$. Then, for all $x\in(0,1)$, Section D.8 implies that
$\hat{y}_{n}(x)$ is the unique solution of

$\displaystyle\Phi_{n}(x,y)=\partial_{y}\ell_{n}(x,y)=\sum_{i=1}^{K_{n}-1}\frac{1}{y+ix}-\sum_{i=1}^{n-1}\frac{1}{y+i}=0$

and $\partial_{y}\Phi_{n}(x,\hat{y}_{n}(x))<0$. Therefore, the implicit
function theorem implies

$\displaystyle\hat{y}_{n}^{\prime}(x)=-\partial_{x}\Phi_{n}(x,y)/\partial_{y}\Phi_{n}(x,y).$

Then, the derivative of $\hat{z}_{n}(x)=\hat{y}_{n}(x)/x$ can be written as

$\displaystyle\hat{z}_{n}^{\prime}(x)=\frac{-\hat{y}_{n}(x)+x\hat{y}_{n}^{\prime}(x)}{x^{2}}=\frac{-\hat{y}_{n}(x)\partial_{y}\Phi_{n}(x,\hat{y}_{n}(x))-x\partial_{x}\Phi_{n}(x,\hat{y}_{n}(x))}{x^{2}\partial_{y}\Phi_{n}(x,\hat{y}_{n}(x))}.$
(D.11)

Note that the numerator of (D.11) is calculated as

$\displaystyle-\hat{y}_{n}(x)\partial_{y}\Phi_{n}(x,\hat{y}_{n}(x))-x\partial_{x}\Phi_{n}(x,\hat{y}_{n}(x))$
$\displaystyle=\sum_{i=1}^{K_{n}-1}\frac{\hat{y}_{n}(x)}{(\hat{y}_{n}(x)+ix)^{2}}-\sum_{i=1}^{n-1}\frac{\hat{y}_{n}(x)}{(\hat{y}_{n}(x)+i)^{2}}+\sum_{i=1}^{K_{n}-1}\frac{ix}{(\hat{y}_{n}(x)+ix)^{2}}$
$\displaystyle=\sum_{i=1}^{K_{n}-1}\frac{1}{\hat{y}_{n}(x)+ix}-\sum_{i=1}^{n-1}\frac{\hat{y}_{n}(x)}{(\hat{y}_{n}(x)+i)^{2}}$
$\displaystyle=\sum_{i=1}^{n-1}\frac{1}{\hat{y}_{n}(x)+i}-\sum_{i=1}^{n-1}\frac{\hat{y}_{n}(x)}{(\hat{y}_{n}(x)+i)^{2}}=\sum_{i=1}^{n-1}\frac{i}{(\hat{y}_{n}(x)+i)^{2}}=\sum_{i=1}^{n-1}\frac{i}{(x\hat{z}_{n}(x)+i)^{2}}>0,$

where we use (D.10) for the third equality. In contrast, the denominator of
(D.11) is negative owing to $\partial_{y}\Phi_{n}(x,\hat{y}_{n}(x))<0$ and is
written as

$\displaystyle 0>x^{2}\partial_{y}\Phi_{n}(x,\hat{y}_{n}(x))$
$\displaystyle=-\sum_{i=1}^{K_{n}-1}\frac{x^{2}}{(\hat{y}_{n}(x)+ix)^{2}}+\sum_{i=1}^{n-1}\frac{x^{2}}{(\hat{y}_{n}(x)+i)^{2}}$
$\displaystyle=-\sum_{i=1}^{K_{n}-1}\frac{1}{(\hat{z}_{n}(x)+i)^{2}}+\sum_{i=1}^{n-1}\frac{x^{2}}{(x\hat{z}_{n}(x)+i)^{2}}.$

Therefore, we have

$\displaystyle
0>\hat{z}^{\prime}_{n}(x)=\frac{\sum_{i=1}^{n-1}{i}/{(x\hat{z}_{n}(x)+i)^{2}}}{-\sum_{i=1}^{K_{n}-1}(\hat{z}_{n}(x)+i)^{-2}+\sum_{i=1}^{n-1}{x^{2}}/{(x\hat{z}_{n}(x)+i)^{2}}},$

thereby completing the proof of (A).

Toward the proof of (B) and (C), we derive the following nonasymptotic
inequality.

######  Lemma D.1.

Suppose $1<K_{n}<n$. Then, for all $x\in(0,1)$, it holds that

$\displaystyle|\hat{z}_{n}(x)|\leq 1\vee\left(A_{n}(x)\wedge B_{n}(x)\wedge
C_{n}(x;\alpha)\right),$

where $A_{n}(x),B_{n}(x)$ and $C_{n}(x;\alpha)$ are defined by

$\displaystyle A_{n}(x)$ $\displaystyle:=\frac{nK_{n}}{x(n-K_{n})},$
$\displaystyle B_{n}(x)$
$\displaystyle:=K_{n}\left\\{\left(\frac{n+\hat{z}_{n}(x)}{1+\hat{z}_{n}(x)}\right)^{x}-1\right\\}^{-1},$
$\displaystyle C_{n}(x;\alpha)$
$\displaystyle:=n^{\frac{\alpha-x}{1-x}}\left(2\frac{K_{n}+\hat{z}_{n}(x)}{n^{\alpha}}\right)^{\frac{1}{1-x}}.$

###### Proof.

We fix $x\in(0,1)$. Suppose $\hat{z}_{n}(x)<1$. Then the assertion is obvious
from $1>\hat{z}_{n}(x)=\hat{y}_{n}(x)/x>-x/x=-1$ so that $|\hat{z}_{n}(x)|<1$.

Now we assume $\hat{z}_{n}(x)\geq 1$. Note that
$x\cdot\Phi_{n}(x,x\hat{z}_{n}(x))=x\cdot 0=0$ is equivalent to

$\displaystyle\sum_{i=1}^{K_{n}-1}\frac{x}{x\hat{z}_{n}(x)+ix}-\sum_{i=1}^{n-1}\frac{x}{x\hat{z}_{n}(x)+i}=0\Leftrightarrow\sum_{i=1}^{n-1}\frac{x}{x\hat{z}_{n}(x)+i}=\sum_{i=1}^{K_{n}-1}\frac{1}{\hat{z}_{n}(x)+i},$

which implies

$\displaystyle\frac{x(n-1)}{x\hat{z}_{n}(x)+n-1}\leq\sum_{i=1}^{n-1}\frac{x}{x\hat{z}_{n}(x)+i}=\sum_{i=1}^{K_{n}-1}\frac{1}{\hat{z}_{n}(x)+i}\leq\frac{K_{n}-1}{\hat{z}_{n}(x)+1}$
$\displaystyle\Rightarrow\frac{\hat{z}_{n}(x)}{K_{n}-1}+\frac{1}{K_{n}-1}\leq\frac{\hat{z}_{n}(x)}{n-1}+\frac{1}{x}$
$\displaystyle\Rightarrow\hat{z}_{n}(x)\leq\frac{(n-1)(K_{n}-1-x)}{x(n-K_{n})}<\frac{nK_{n}}{x(n-K_{n})}=A_{n}(x).$
(D.12)

In contrast, elementary calculus implies

$\displaystyle\forall m\in\mathbb{N}_{\geq 2},\ \forall\delta>0,\
\log\left(\frac{\delta+m}{\delta+1}\right)<\sum_{i=1}^{m-1}\frac{1}{\delta+i}$
$\displaystyle<\log\left(\frac{\delta+m-1}{\delta}\right)<\log\left(\frac{\delta+m}{\delta}\right).$

Then, this and the assumption $\hat{z}_{n}(x)\geq 1>0$ result in

$\displaystyle
x\log\left(\frac{\hat{z}_{n}(x)+n}{\hat{z}_{n}(x)+1}\right)<\sum_{i=1}^{n-1}\frac{x}{\hat{z}_{n}(x)+i}\leq\sum_{i=1}^{n-1}\frac{x}{x\hat{z}_{n}(x)+i}$
$\displaystyle=\sum_{i=1}^{K_{n}-1}\frac{1}{\hat{z}_{n}(x)+i}<\log\left(\frac{\hat{z}_{n}(x)+K_{n}}{\hat{z}_{n}(x)}\right)$
$\displaystyle\Rightarrow\left(\frac{\hat{z}_{n}(x)+n}{\hat{z}_{n}(x)+1}\right)^{x}<\frac{\hat{z}_{n}(x)+K_{n}}{\hat{z}_{n}(x)}$
(D.13)
$\displaystyle\Leftrightarrow\frac{K_{n}}{\hat{z}_{n}(x)}>\left(\frac{\hat{z}_{n}(x)+n}{\hat{z}_{n}(x)+1}\right)^{x}-1$
$\displaystyle\Leftrightarrow\hat{z}_{n}(x)<K_{n}\left\\{\left(\frac{\hat{z}_{n}(x)+n}{\hat{z}_{n}(x)+1}\right)^{x}-1\right\\}^{-1}=B_{n}(x).$
(D.14)

Furthermore, (D.13) and the assumption $\hat{z}_{n}(x)\geq 1>0$ yield

$\displaystyle\frac{\hat{z}_{n}(x)}{(\hat{z}_{n}(x)+1)^{x}}<\frac{\hat{z}_{n}(x)+K_{n}}{(\hat{z}_{n}(x)+n)^{x}}<\frac{\hat{z}_{n}(x)+K_{n}}{(0+n)^{x}}=\frac{\hat{z}_{n}(x)+K_{n}}{n^{x}}$
$\displaystyle\Rightarrow\hat{z}_{n}(x)^{1-x}<\left(\frac{1+\hat{z}_{n}(x)}{\hat{z}_{n}(x)}\right)^{x}\frac{\hat{z}_{n}(x)+K_{n}}{n^{x}}$
$\displaystyle\leq\left(\frac{1+1}{1}\right)^{x}\frac{\hat{z}_{n}(x)+K_{n}}{n^{x}}<n^{\alpha-x}\cdot
2\frac{\hat{z}_{n}(x)+K_{n}}{n^{\alpha}}$
$\displaystyle\Rightarrow\hat{z}_{n}(x)<n^{\frac{\alpha-x}{1-x}}\left(2\frac{\hat{z}_{n}(x)+K_{n}}{n^{\alpha}}\right)^{\frac{1}{1-x}}=C_{n}(x;\alpha).$
(D.15)

Combining (D.12), (D.14), and (D.15), we get

$\displaystyle\hat{z}_{n}(x)\geq 1\Rightarrow|\hat{z}_{n}(x)|<A_{n}(x)\wedge
B_{n}(x)\wedge C_{n}(x;\alpha),$

thereby completing the proof. ∎

#### Proof of (B)

We fix $x\in(0,1)$. For $A_{n}(x)$ in Lemma D.1,
$K_{n}=O_{p}(n^{\alpha})=o_{p}(n)$ implies

$\displaystyle\frac{A_{n}(x)}{n}=\frac{K_{n}}{x(n-K_{n})}=o_{p}(1),$

and hence $|\hat{z}_{n}(x)|\leq 1\vee A_{n}(x)=o_{p}(n)$. Considering this,
for $B_{n}(x)$ in Lemma D.1, it holds that

$\displaystyle\frac{B_{n}(x)}{n^{\alpha}}=\frac{K_{n}}{n^{\alpha}}\cdot\left\\{\left(\frac{n+\hat{z}_{n}(x)}{1+\hat{z}_{n}(x)}\right)^{x}-1\right\\}^{-1}\rightarrow\operatorname{\it{M}_{\alpha\theta}}\cdot\
0=0\text{ in probability},$

which implies $|\hat{z}_{n}(x)|\leq 1\vee B_{n}(x)=o_{p}(n^{\alpha})$. Then,
for $C_{n}(x;\alpha)$ in Lemma D.1, we get

$\displaystyle n^{-\frac{\alpha-x}{1-x}}\cdot
C_{n}(x;\alpha)=\left(2\frac{K_{n}+\hat{z}_{n}(x)}{n^{\alpha}}\right)^{\frac{1}{1-x}}\rightarrow\left(2\operatorname{\it{M}_{\alpha\theta}}+0\right)^{\frac{1}{1-x}}=(2\operatorname{\it{M}_{\alpha\theta}})^{\frac{1}{1-x}}\text{
in probability},$

where we used
$K_{n}/n^{\alpha}\rightarrow\operatorname{\it{M}_{\alpha\theta}}$ (a.s.).
Therefore,

$\displaystyle|\hat{z}_{n}(x)|\leq 1\vee C_{n}(x;\alpha)=1\vee
O_{p}(n^{\frac{\alpha-x}{1-x}})=\left\\{\begin{array}[]{ll}O_{p}(n^{\frac{\alpha-x}{1-x}})&0<x<\alpha\\\
O_{p}(1)&\alpha\leq x<1\end{array}\right.,$

thereby completing the proof.

#### Proof of (C)

Step 1. $\sup_{x\in[\alpha\pm\delta_{n}]}|\hat{z}_{n}(x)|=O_{p}(1)$: Note that
we can take $(s,t)$ satisfying $0<s<\alpha<t<1$ and
$[\alpha\pm\delta_{n}]\subset[s,t]\subset(0,1)$ for sufficiently large $n$ by
$\delta_{n}=o(1/\log n)=o(1)$. Considering $\hat{z}_{n}(x)$ is monotone on
$(0,1)$ by (A), (B) implies

$\displaystyle\sup_{x\in[\alpha\pm\delta_{n}]}|\hat{z}_{n}(x)|\leq\sup_{x\in[s,t]}|\hat{z}_{n}(x)|\leq|\hat{z}_{n}(s)|+|\hat{z}_{n}(t)|=O_{p}(n^{\frac{\alpha-s}{1-s}})+O_{p}(1)=o_{p}(n^{\alpha}).$

Then, for $C_{n}(x;\alpha)$ in Lemma D.1, it holds that

$\displaystyle\log C_{n}(x=\alpha\pm\delta_{n};\alpha)$
$\displaystyle=\frac{\mp\delta_{n}}{1-\alpha\mp\delta_{n}}\log
n+\frac{1}{1-\alpha\mp\delta_{n}}\log\left(2\frac{K_{n}+\hat{z}_{n}(\alpha\pm\delta_{n})}{n^{\alpha}}\right)$
$\displaystyle\rightarrow
0+\frac{1}{1-\alpha}\log(2\operatorname{\it{M}_{\alpha\theta}}+0)=O_{p}(1),$

and hence $C_{n}(\alpha\pm\delta_{n};\alpha)=O_{p}(1)$. Therefore, the
monotonicity of $\hat{z}_{n}(x)$ on $(0,1)$ and Lemma D.1 imply

$\displaystyle\sup_{x\in\alpha\pm\delta_{n}}|\hat{z}_{n}(x)|\leq|\hat{z}_{n}(\alpha\pm\delta_{n})|\leq
1\vee C_{n}(\alpha\pm\delta_{n};\alpha)=O_{p}(1),$

which completes the proof.

Step 2.
$\sup_{x\in[\alpha\pm\delta_{n}]}|f_{\alpha}(\hat{z}_{n}(x))-\log(K_{n}/n^{\alpha})|=o_{p}(1)$:
Denote $\hat{z}_{n}(\alpha\pm\delta_{n})$ by $\hat{z}_{n}^{\pm}$ and
$\alpha\pm\delta_{n}$ by $\alpha_{n}^{\pm}$. Note that
$\hat{z}^{\prime}_{n}(x)<0$ by (A) and $f^{\prime}_{\alpha}(z)>0$ by Lemma 3.2
imply $(f_{\alpha}\circ\hat{z}_{n})^{\prime}(x)<0$ for all $x\in(0,1)$, and
hence, $f_{\alpha}\circ\hat{z}_{n}$ is monotone on $(0,1)$. Therefore, it
suffices to show

$\displaystyle f_{\alpha}(\hat{z}_{n}^{\pm})-\log(K_{n}/n^{\alpha})=o_{p}(1).$

Now we claim
$\alpha_{n}^{\pm}\cdot\Phi_{n}(\alpha_{n}^{\pm},\alpha_{n}^{\pm}\cdot\hat{z}_{n}^{\pm})=\alpha_{n}^{\pm}\cdot\
0=0$: This is because
$\alpha^{\pm}_{n}\hat{z}_{n}^{\pm}=\hat{y}_{n}(\alpha_{n}^{\pm})$ by
definition and $\hat{y}_{n}$ is the function s.t.
$\Phi_{n}(x,\hat{y}_{n}(x))=0$ for all $x\in(0,1)$ (see (C.1)). Then, we have

$\displaystyle 0$
$\displaystyle=\sum_{i=1}^{K_{n}-1}\frac{\alpha_{n}^{\pm}}{\alpha_{n}^{\pm}\cdot\hat{z}_{n}^{\pm}+i\alpha_{n}^{\pm}}-\sum_{i=1}^{n-1}\frac{\alpha_{n}^{\pm}}{\alpha_{n}^{\pm}\cdot\hat{z}_{n}^{\pm}+i}=\sum_{i=1}^{K_{n}-1}\frac{1}{\hat{z}_{n}^{\pm}+i}-\sum_{i=1}^{n-1}\frac{\alpha_{n}^{\pm}}{\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i}$
$\displaystyle=\sum_{i=1}^{K_{n}-1}\frac{1}{\hat{z}_{n}^{\pm}+i}-\sum_{i=1}^{n-1}\frac{\alpha_{n}^{\pm}-\alpha}{\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i}-\alpha\sum_{i=1}^{n-1}\left(\frac{1}{\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i}-\frac{1}{\alpha\hat{z}_{n}^{\pm}+i}\right)-\sum_{i=1}^{n-1}\frac{\alpha}{\alpha\hat{z}_{n}^{\pm}+i}$
$\displaystyle=\psi(K_{n}+\hat{z}_{n}^{\pm})-\psi(1+\hat{z}_{n}^{\pm})-\sum_{i=1}^{n-1}\frac{\alpha_{n}^{\pm}-\alpha}{\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i}-\alpha\sum_{i=1}^{n-1}\frac{(\alpha-\alpha_{n}^{\pm})\hat{z}_{n}^{\pm}}{(\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i)(\alpha\hat{z}_{n}^{\pm}+i)}$
$\displaystyle-\alpha\psi(n+\alpha\hat{z}_{n}^{\pm})+\alpha\psi(1+\alpha\hat{z}_{n}^{\pm}),$

where $\psi$ is the digamma function. Using
$f_{\alpha}(z)=\psi(1+z)-\alpha\psi(1+\alpha z)$, the above equality is
equivalent to

$\displaystyle f_{\alpha}(\hat{z}_{n}^{\pm})-\log(K_{n}/n^{\alpha})$
$\displaystyle=(\psi(K_{n}+\hat{z}_{n}^{\pm})-\log
K_{n})-\alpha(\psi(n+\alpha_{n}^{\pm}\hat{z}_{n}^{\pm})-\log n)$
$\displaystyle-\sum_{i=1}^{n-1}\frac{\alpha_{n}^{\pm}-\alpha}{\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i}-\sum_{i=1}^{n-1}\frac{\alpha(\alpha-\alpha_{n}^{\pm})\hat{z}_{n}^{\pm}}{(\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i)(\alpha\hat{z}_{n}^{\pm}+i)}.$

Note that $\hat{z}_{n}^{\pm}=O_{p}(1)$ by Step 1. Then, the first and second
terms are $o_{p}(1)$ by Lemma B.5 applied with
$\hat{z}_{n}^{\pm}/K_{n}=O_{p}(1)/K_{n}=o_{p}(1)$ and
$\hat{z}_{n}^{\pm}/n=O_{p}(1)/n=o_{p}(1)$. As for the third and fourth terms,
$\alpha_{n}^{\pm}\in[s,t]\subset(0,1)$ for sufficiently large $n$ and
$\hat{z}_{n}^{\pm}=O_{p}(1)$ by Step 1 result in

$\displaystyle\left|\sum_{i=1}^{n-1}\frac{\alpha_{n}^{\pm}-\alpha}{\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i}\right|$
$\displaystyle\leq\delta_{n}\sum_{i=1}^{n-1}\frac{1}{i-\alpha_{n}^{\pm}}\leq\delta_{n}\sum_{i=1}^{n-1}\frac{1}{i-t}$
$\displaystyle=o(1/\log n)O_{p}(\log n)=o_{p}(1),$
$\displaystyle\left|\sum_{i=1}^{n-1}\frac{\alpha(\alpha-\alpha_{n}^{\pm})\hat{z}_{n}^{\pm}}{(\alpha_{n}^{\pm}\hat{z}_{n}^{\pm}+i)(\alpha\hat{z}_{n}^{\pm}+i)}\right|$
$\displaystyle\leq\alpha\delta_{n}|\hat{z}_{n}^{\pm}|\sum_{i=1}^{n-1}\frac{1}{(-t+i)(-\alpha+i)}$
$\displaystyle=o(1/\log n)O_{p}(1)O(1)=o_{p}(1).$

Therefore, we have
$f_{\alpha}(\hat{z}_{n}^{\pm})-\log(K_{n}/n^{\alpha})=o_{p}(1)$, thereby
completing the proof.