# Factors Affecting the Performance of Automated Speaker Verification in
Alzheimer’s Disease Clinical Trials

Malikeh Ehghaghi, Marija Stanojevic, Ali Akram, Jekaterina Novikova
Winterlight Labs Inc. Toronto  Canada
{malikeh, marija, aliakram<EMAIL_ADDRESS>

###### Abstract

Detecting duplicate patient participation in clinical trials is a major
challenge because repeated patients can undermine the credibility and accuracy
of the trial’s findings and result in significant health and financial risks.
Developing accurate automated speaker verification (ASV) models is crucial to
verify the identity of enrolled individuals and remove duplicates, but the
size and quality of data influence ASV performance. However, there has been
limited investigation into the factors that can affect ASV capabilities in
clinical environments. In this paper, we bridge the gap by conducting analysis
of how participant demographic characteristics, audio quality criteria, and
severity level of Alzheimer’s disease (AD) impact the performance of ASV
utilizing a dataset of speech recordings from 659 participants with varying
levels of AD, obtained through multiple speech tasks. Our results indicate
that ASV performance: 1) is slightly better on male speakers than on female
speakers; 2) degrades for individuals who are above 70 years old; 3) is
comparatively better for non-native English speakers than for native English
speakers; 4) is negatively affected by clinician interference, noisy
background, and unclear participant speech; 5) tends to decrease with an
increase in the severity level of AD. Our study finds that voice biometrics
raise fairness concerns as certain subgroups exhibit different ASV
performances owing to their inherent voice characteristics. Moreover, the
performance of ASV is influenced by the quality of speech recordings, which
underscores the importance of improving the data collection settings in
clinical trials.

##  1 Introduction

Healthcare systems are increasingly relying on automatic speaker verification
(ASV) models to ensure secure and accurate identification of patients and
healthcare providers, with the aim of preventing fraud, safeguarding patient
privacy, and ensuring the accuracy of medical records (Upadhyay et al., 2022;
Arasteh et al., 2022).

Conducting large-scale clinical trials, involving numerous patients, doctors,
clinics, and even different countries can pose significant challenges in
identifying instances of duplicate participation, which occurs when a single
individual joins the same study more than once, either at different sites or
time points, leading to skewed results and undermining the validity of study
findings (Irum and Salman, 2019). Shiovitz et al. (2013) discovered that as
much as 7.78% of patients involved in a clinical study were duplicated across
different sites.

In some cases, individuals participate in multiple clinical trials
concurrently in order to earn more money. When a trial enrolls an adequate
number of substandard participants, it risks not meeting the primary endpoints
and ultimately causing a multimillion-dollar study to fail. Pinho et al.
(2021) examined the financial effect of duplicate participants on the
pharmaceutical companies conducting a set of short-term study programs across
psychiatric disorders including Schizophrenia, Major Depressive Disorder, and
Bipolar Depression. Based on their results, enrolling ineligible subjects in
the selected studies results in a loss of around $29,680,000 for the sponsor
pharmaceutical company. In addition, duplicate participation results in higher
placebo rates and compromised data integrity. These findings highlight the
importance of addressing the duplicate participant problem and underscore the
need for reliable and accurate ASV methods in healthcare systems to verify
whether an unknown voice belongs to a known enrolled individual (Upadhyay et
al., 2022; Arasteh et al., 2022).

Cognitive impairment has been linked to a decline in vocabulary richness,
syntactic complexity, and speech fluency, according to previous research
(Thomas et al., 2005; Roark et al., 2011; Guinn and Habash, 2012; Meilán et
al., 2012). Therefore, it is important to investigate whether the abnormal
speech patterns exhibited by individuals with cognitive impairment can affect
ASV performance. Despite this concern, there is a paucity of research
examining the relationship between cognitive impairment and ASV in the
existing literature. This research gap motivated us to address this issue by
examining the effect of Alzheimer’s disease (AD) severity level on ASV
performance.

Furthermore, external factors such as participants’ demographic information
(Si et al., 2021), recording environment, or data collection procedure (Woo et
al., 2006; Wan, 2017) may also have an impact on ASV performance, but their
impact is not well-studied in the healthcare industry. An extensive analysis
of these external factors could provide valuable insights into the accuracy
and reliability of ASV models and identify potential sources of bias due to
differences in inherent voice characteristics among subgroups (Si et al.,
2021).

The purpose of this study is to investigate the effectiveness of ASV models in
identifying duplicate patient participation in large-scale clinical trials,
and to explore the factors that influence ASV performance in such settings. To
this end, we utilize a longitudinal clinical dataset of English speech
recordings obtained through multiple speech tasks from 659 participants with
varying levels of AD. We employ the TitaNet model, an end-to-end deep learning
text-independent ASV model pre-trained on a large volume of speech recordings
of English speakers. ASV models can be classified into two groups: text-
dependent (TD) and text-independent (TI). TI ASV models allow for more
flexibility in the enrollment and verification phases without constraints on
the speech content. When pre-trained on extensive audio datasets, TI models
demonstrate a comparable level of accuracy to TD models. We evaluate the
performance of TitaNet on our dataset in a zero-shot setting, achieving a 3.1%
equal error rate (EER). In addition, we analyze the impact of various external
factors on ASV performance, including participant demographic characteristics
(i.e., age, and gender), audio quality criteria (i.e., clinician interference,
background noise, participant accent, and participant clarity), as well as AD
severity level. This study aims to provide valuable insights into the factors
that can affect the performance of ASV models in clinical trial environments,
with the goal of improving the accuracy, fairness, and reliability.

Our findings indicate that voice biometrics may present fairness issues, as
certain subgroups demonstrate differing speaker verification performances due
to their inherent voice characteristics. In addition, the quality of speech
recordings can impact ASV performance, highlighting the importance of
monitoring and enhancing data collection and recording settings during
clinical trials.

##  2 Related Work

Speaker verification technology has been increasingly utilized in various
domains, including healthcare systems. Several studies have analyzed the
feasibility and effectiveness of speaker verification models in healthcare
settings (Hao and Hei, ; Weng et al., 2014). However, the external factors
that can affect the performance of ASV models has not been extensively studied
through research in the healthcare field.

##### Race and Gender Effect:

Si et al. (2021) utilized three state-of-the-art ASV models including the
Xvector-TDNN (Snyder et al., 2018), ECAPA-TDNN (Desplanques et al., 2020), and
DTW (Dutta, 2008) models to explore demographic effects on speaker
verification. For this purpose, they used a subset of the mPower study (Bot et
al., 2016), a Parkinson’s disease mobile dataset, comparing a diverse group of
300 speakers by race and gender. Their results demonstrated that the Latinx
subgroup indicates the worst ASV performance among the four major races in the
dataset (i.e., White, Black, Latinx, and Asian). Based on their findings,
gender represents minor differences in ASV performance between male-only and
female-only subgroups. We did similar gender-level and accent-level analyses
on patients with Alzheimer’s disease to detect the potential sources of bias
in ASV due to the inherent voice characteristics of distinct genders or
English accents.

##### Age Effect:

Kelly and Harte (2011) analyzed the effect of long-term ageing on ASV
performance. They utilized a conventional GMM-UBM system (Irum and Salman,
2019) on a longitudinal voice dataset of a cohort of 13 adult speakers, whose
recordings were collected over a time span of 30-40 years. According to their
results, short-term aging (less than 5 years) does not have a significant
impact on verification performance, compared to normal inter-session
variations. However, for longer periods, aging has a negative effect on
verification accuracy. Moreover, the researchers found that the rate of
verification score decline is more rapid for speakers aged 60 years and above.
However, they only evaluated their models on small cohorts and inter-speaker
differences across different age groups were not further analyzed, while in
the present work, we evaluate the ASV performance across different age groups
over 55 years old and incorporate a larger clinical dataset with 659 speakers
in total, while we controlled for AD effect. Taylor et al. (2020) also
demonstrated that some speech and vocal characteristics (e.g., the spectral
center of gravity, spectral skewness, or spectral kurtosis) undergo
alterations with aging, and these changes can vary between men and women.
These findings suggest that age is an effective factor in speaker’s voice
characteristics and this underscores the importance of assessing age effect on
our ASV model to ensure the fairness of the model across different age groups.

##### Noise Effect:

Wan (2017) applied LibriSpeech corpus (Panayotov et al., 2015) of English
novel reading speech with varying lengths and tested ASV performance across
different types and levels of background noise (e.g., babble, car, office and
airplane noise) with a great mismatch between training and testing speech.
Based on their findings, performance varies across different types of noise
and the number of errors grow with a decrease in the sound-to-noise-ratio
value. However, other metrics of audio quality were not considered in their
study and their models were only trained and tested on healthy speech
recordings. In the present study, we assess the effect of other audio quality
aspects, such as participant clarity, clinician interference, and background
noise, on ASV performance using a dataset of speakers with varying severity
levels of AD, which was collected in clinical environment.

##### Speech Pathology Effect:

Arasteh et al. (2022) have investigated the vulnerability of pathological
speech to re-identification in ASV systems. In a large-scale study, they
explored the effects of different speech pathologies on ASV using a real-world
pathological speech corpus of more than 2,000 test subjects with various
speech and voice disorders. Their results indicated that some types of speech
pathology, particularly dysphonia, regardless of speech intelligibility, are
more vulnerable to a breach of privacy compared to healthy speech. They do not
analyze the effect of AD on ASV performance, even though speech and language
impairment are prevalent issues in moderate to severe stages of AD that may
potentially affect the ASV performance. This motivates us to evaluate ASV
performance across varying severity levels of AD.

##  3 Methods

###  3.1 Datasets

The Alzheimer’s Disease Clinical Trial (ADCT) dataset comprises speech
recordings of English-speaking patients with a clinical diagnosis of mild to
moderate AD who participated in a clinical trial. This is a proprietary
dataset, which was collected every 12 weeks for a 48-week treatment period. It
includes recordings of participants performing a set of self-administered
speech tasks, including picture description (Goodglass et al., 2001; Becker et
al., 1994), phonemic verbal fluency (Borkowski et al., 1967), and semantic
verbal fluency (Tombaugh et al., 1999) tasks.

####  3.1.1 Demographic Information

Demographic data were collected about the participants at the beginning of the
study. This data includes the age, and gender of the individuals upon
consenting. The data collection study was approved by the ethical committee.

####  3.1.2 Transcription and Quality Assessment

All the audio recordings were manually transcribed by 49 trained
transcriptionists based on the CHAT protocol and annotations (MacWhinney,
2014). The transcriptionists utilized an online tool that granted them access
to the recordings and enabled them to transcribe the audio content, segment
the files into utterances, and perform quality assessment. In addition, the
transcriptions manually rated the quality of the recordings according to
different quality criteria. The values range from 0 to 3 for each quality
criterion. Value higher than 0 indicates that the audio recording has minor to
major issues under that quality criterion. The quality criteria consist of
background noise, clinician interference, participant accent, and participant
clarity. The background noise criterion indicates whether there is noise in
the background from the environment. Clinician interference indicates whether
the clinician (or another speaker) interferes with the speech task. The
participant accent criterion indicates whether the participant is a native or
near-native speaker (values of 0) or has a detectable non-native accent (value
higher than 0). Participant clarity indicates whether the participant’s voice
is hard to hear or understand.

####  3.1.3 Clinical Assessment

Participants were assessed on the severity level of AD using the Mini-Mental
State Examination (MMSE) (Henneges et al., 2016) rating scale. MMSE is a brief
cognitive function assessment, which consists of 30 questions that can be
completed in less than 10 minutes. The questions are divided into seven
categories, with each subscore examining a particular aspect of cognition:
Orientation in time (score range 0-5), orientation in place (score range 0-5),
registration (score range 0-3), attention and concentration (score range 0-5),
recall (score range 0-3), language (score range 0-8), and drawing (score range
0-1). MMSE total score ranges from 0 to 30, with higher scores indicating
better cognitive function and lower scores indicating more severe cognitive
impairment. In this study, the participants were categorized into four levels
of AD severity based on MMSE criteria (Wimo et al., 2013): Healthy Control
(HC) (MMSE score > 26 points), Mild AD (MMSE score 21-26 points), Moderate AD
(MMSE score 15-20 points), and Severe AD (MMSE score < 15 points).

####  3.1.4 Dataset Composition

The ADCT data comprises 7084 audio recordings from 659 speakers with
10.7$\pm$7.0 samples on average per each speaker. The average duration of
total audio and speech-only audio are equal to 69.31 and 37.30 seconds,
respectively.

In the dataset, 43.4% of the speakers are male and 56.6% of the speakers are
female. The age range of the subjects spans from 55 to 80 years old. Age
distribution of the subjects is represented in Figure 1, with an average value
equal to 69.7$\pm$6.7.

Figure 1: Age distribution of the speakers in ADCT dataset. Figure 2:
Distribution of AD severity levels in ADCT dataset.

Figure 2 indicates the distribution of MMSE scores in the ADCT dataset,
showing that the majority of the samples consist of mild to severe levels of
AD with scores in the range of 15 to 26 points. The average MMSE score is
equal to 17.3$\pm$4.4. It should also be noted that the severity level of AD
may vary over time for some of the speakers.

###  3.2 Models

In this study, we utilized the TitaNet model (Koluguri et al., 2022), which is
a state-of-the-art end-to-end TI ASV model from the Nvidia NeMo
toolkit111https://github.com/NVIDIA/NeMo that had been pre-trained on an
extensive collection of English speech data, from various publicly-available
resources. The TitaNet model is a neural network model that adopts an encoder-
decoder architecture to extract speaker embeddings from speech. The model
architecture is inspired by ContextNet (Han et al., 2020) model, which
comprises 1D depth-wise separable convolutions followed by squeeze and
excitation (SE) layers combined with channel attention pooling to convert
utterances of varying lengths into a fixed-length embedding. The model
contains 25.3M parameters and it is pre-trained on the VoxCeleb1 Dev (Nagrani
et al., 2017), VoxCeleb2 Dev (Chung et al., 2018), Fisher (Cieri et al.,
2004), Switchboard-Cellular1, Switchboard-Cellular2 (Godfrey and Holliman,
1993), and LibriSpeech (Panayotov et al., 2015) datasets.

We applied the model to the ADCT dataset in a zero-shot setting and used it
for further analysis of the effect of external factors.

###  3.3 Evaluation

To analyze the effect of external factors on ASV performance, we separately
evaluated the performance of the TitaNet model across different subsets of
ADCT data including genders, age groups, audio quality levels, and AD severity
levels. Initially, we produced embeddings for all audio files within each
group. Subsequently, we aggregated these embeddings to create a set of tuples
comprising positive and negative pairs.

Positive tuples refer to pairs of embeddings that belong to the same speaker.
The total number of positive tuples in each group is given by
$\sum_{i=1}^{m}{n_{i}\choose 2}$, which is calculated by summing up all pairs
of $n_{i}$ speech recordings for the ith speaker, where $m$ is the total
number of speakers in the same group.

Negative tuples refer to pairs of embeddings from different speakers within
the same group. The total number of negative pairs is calculated as
$\sum_{i=1}^{m}{n_{i}*(N-n_{i})/2}$, where $N$ is the total number of audio
files in the group, and $n_{i}$ is the number of speech recordings for speaker
$i$. The sum is divided by 2 to avoid counting each pair twice.

After generating all the positive and negative pair tuples, we proceeded to
compute the cosine similarity between the pairs of vector embeddings within
each tuple. Subsequently, we adjusted a threshold value $\theta$ for each
group through manual tuning until the true positive rate equalled the true
negative rate, which enabled us to evaluate the performance of the ASV model
using the equal error rate (EER) metric. If the cosine similarity value
exceeded the threshold, we considered the corresponding tuple as belonging to
the same speaker. Conversely, if the cosine similarity value was below the
threshold, we deemed the two embeddings to represent different speakers.

##  4 Results and Discussion

Gender | Tuned | EER(%) | #Spkrs | #Smpls | Avg #Smpls | Avg Age | Avg MMSE
---|---|---|---|---|---|---|---
Thr. | per Spkr |  | Score
Female | 0.75 | 5.13 | 170 | 2735 | 16.09$\pm$3.94 | 69.53$\pm$6.72 | 17.33$\pm$4.37
Male | 0.75 | 4.98 | 170 | 2671 | 15.72$\pm$4.02 | 69.41$\pm$6.96 | 17.45$\pm$4.45
All | 0.74 | 3.10 | 659 | 7084 | 10.70$\pm$7.00 | 69.55$\pm$6.75 | 17.32$\pm$4.44
Table 1: Benchmarking TitaNet ASV model across different genders on the ADCT English dataset. ‘#Spkrs’ denotes the number of speakers per each gender. ‘#Smpls’ denotes the number of audio recordings per each gender. ‘Tuned Thr.’ denotes tuned threshold. Age Group | Tuned | EER(%) | #Spkrs | #Smpls | Avg #Smpls | Gender | Avg MMSE
---|---|---|---|---|---|---|---
Thr. | per Spkr |  | Score
Age <= 70 | 0.73 | 3.62 | 197 | 3235 | 16.42$\pm$3.86 | Male+Female | 17.09$\pm$4.72
Age > 70 | 0.74 | 4.20 | 195 | 3022 | 15.50$\pm$4.07 | Male+Female | 17.57$\pm$4.11
All | 0.74 | 3.10 | 659 | 7084 | 10.7$\pm$7.0 | Male+Female | 17.32$\pm$4.44
Table 2: Benchmarking TitaNet ASV model across different age groups on the
ADCT English dataset. ‘#Spkrs’ denotes the number of speakers per each age
group. ‘#Smpls’ denotes the number of audio recordings per each age group.
‘Tuned Thr.’ denotes tuned threshold.

In order to have a baseline level of ASV performance, we evaluated the TitaNet
model on all the speech recordings of the ADCT dataset and obtained a 3.1%
EER. To further analyze the impact of participant demographic characteristics,
audio quality and AD severity level, we then recalculated EER and compared the
ASV performance across different subgroups.

Figure 3: Visualization of speaker clusters using TitaNet embeddings of all
audio recordings in ADCT dataset, created using the PaCMAP (Wang et al., 2021)
dimensionality reduction method, where each color represents the gender of the
speaker.

###  4.1 Effect of Participant Demographic Characteristics on ASV Performance

####  4.1.1 Gender Effect

We first analyzed the effect of speaker genders on the performance of ASV. As
shown in Figure 3, two visually distinguishable speaker clusters appeared in
the visualization of speech embeddings of all speakers in our dataset.
Colouring the data points based on their gender indicates that each cluster is
representative of a specific gender. The left cluster mostly comprises female
speakers and the majority of the right cluster consists of male speakers. For
further analysis of the ASV performance, we separately evaluated the model
performance on male and female speech samples within each cluster. To control
for the confounding factors, we randomly downsized the size of the female
subgroup to the number of speakers in the male subgroup and also, matched the
average age and MMSE score between the two subgroups. The results (Table 1)
show that according to the EER metric, the ASV model performs better on the
total dataset comprising diverse genders compared to when it is applied to the
male-only or female-only speakers (Table 1). The ASV performance for the male
subgroup (EER = 4.98%) is slightly better than that for the female subgroup
(5.13% EER), although the difference is not substantial. This is in line with
prior literature (Hanilçi and Ertaş, 2013) demonstrating that male speakers
exhibit higher speaker recognition accuracy compared to female speakers
regardless of the database and classifier used. The results also align with Si
et al. (2021) indicating that there is little difference in gender in terms of
the performance of ASV models in general.

####  4.1.2 Age Effect

We then evaluated the age effect on ASV performance. For this purpose, we
categorized the speakers into two age subgroups with Age <= 70 and Age > 70
according to the speaker ages at the study enrollment date. The threshold was
set to 70 because it is equal to the approximate median and mean value of the
age distribution of the speakers in the ADCT dataset (Section 3.1.4). To
control for the confounding factors, we designed our experiments to ensure
that the age-based subgroups had a comparable number of speakers and average
MMSE scores and included a combination of male and female speakers.

Based on the results indicated in Table 2, EER for participants under 70 is
0.58% lower than the older age group. These results can be explained by Taylor
et al. (2020) revealing that specific attributes of speech and voice
characteristics (e.g., fricative spectral moments, semitone standard
deviation, etc.) vary according to age.

###  4.2 Effect of Audio Quality on ASV Performance

Audio Quality Criterion | Tuned | EER(%) | #Spkrs | #Smpls | Avg #Smpls | Gender | Avg Age | Avg MMSE
---|---|---|---|---|---|---|---|---
Thr. | per Spkr |  |  | Score
Background Noise - No Issue | 0.75 | 2.90 | 125 | 426 | 3.40$\pm$1.45 | M + F | 69.60$\pm$6.72 | 16.94$\pm$5.83
Background Noise - Minor to Major Issue | 0.74 | 3.54 | 125 | 511 | 4.08$\pm$2.08 | M + F | 69.21$\pm$6.45 | 16.78$\pm$5.58
Participant Clarity - No Issue | 0.75 | 2.85 | 112 | 481 | 4.29$\pm$1.70 | M + F | 69.80$\pm$6.38 | 16.81$\pm$5.62
Participant Clarity - Minor to Major Issue | 0.74 | 3.41 | 112 | 432 | 3.85$\pm$1.83 | M + F | 69.23$\pm$6.81 | 16.04$\pm$5.54
Clinician Interference - No Issue | 0.75 | 2.90 | 103 | 659 | 4.30$\pm$2.08 | M + F | 69.40$\pm$6.77 | 17.65$\pm$5.61
Clinician Interf. - Minor to Major Issue | 0.73 | 3.38 | 103 | 399 | 3.87$\pm$1.86 | M + F | 69.43$\pm$6.84 | 14.77$\pm$5.22
Participant Accent - Native | 0.74 | 2.97 | 188 | 901 | 4.79$\pm$2.82 | M + F | 68.63$\pm$6.89 | 17.22$\pm$5.01
Participant Accent - Non-Native | 0.74 | 2.01 | 188 | 594 | 3.16$\pm$1.54 | M + F | 70.45$\pm$6.32 | 17.19$\pm$4.56
All | 0.74 | 3.10 | 659 | 7084 | 10.7$\pm$7.0 | M + F | 69.55$\pm$6.75 | 17.32$\pm$4.44
Table 3: Benchmarking TitaNet speaker verification model across different
levels of audio quality on the ADCT dataset. ‘#Spkrs’ denotes the number of
speakers per each quality subgroup. ‘#Smpls’ denotes the number of audio
recordings per each quality subgroup. ‘Tuned Thr.’ denotes tuned threshold.
For each quality criterion, ‘No Issue’ indicates samples with a rating = 0 and
‘Minor to Major Issue’ indicates samples with a rating > 0 (maximum is 3). ‘M’
denotes male speakers and ‘F’ denotes female speakers. Bold font denotes the
subgroup yielding best ASV performance in each quality criterion.

We examined how different audio quality factors affect the performance of
speaker verification in clinical environments. We divided ADCT samples into
two subgroups based on their quality rating for each criterion: ‘No Issue’ for
samples with a rating of 0 and ‘Minor to Major Issue’ for those with a rating
higher than 0 (maximum is 3). In order to mitigate the influence of possible
confounding factors, we structured our experiments in such a way that the
quality-based subgroups had a similar number of speakers and average MMSE
score, and included both male and female speakers. Table 3 shows the
comparison of EER values between each pair of subgroups per audio quality
criterion. Our results indicate that subgroups of audio samples with no
background noise, and high participant clarity yielded lower EER than
subgroups with varying levels of background noise and poor participant
clarity. Therefore, better control of the data collection setting, along with
the use of high-clarity audio recordings with minimal background noise, would
be recommended in order to improve ASV performance in clinical trials. Our
findings are in line with Eskimez et al. (2018), who demonstrated that
incorporating a DNN-based speech enhancement technique as a front-end noise
reduction module can enhance the ASV performance when applied to noisy speech
data obtained from real customers.

Our results also suggest that clinician interference can negatively impact ASV
performance with 0.48% increase in EER. Therefore, it is recommended that
clinicians refrain from interrupting participants during speech tasks in
recording sessions to prevent any decline in performance.

We also evaluated the speakers’ accents as a quality indicator, and our
results show that ASV performs better on non-native speakers (2.01% EER) than
on native speakers (2.97% EER). This suggests that ASV performance can be
better in trials that involve participants who speak with diverse non-native
English accents as a way to identify unique speech characteristics for each
individual.

Figure 4: Comparison of the performance of the TitaNet ASV model across
different AD severity levels based on EER% metric. Original ADCT refers to the
dataset with the original number of speakers per severity level. Balanced ADCT
refers to the dataset with each group downsized to the number of speakers in
the smallest group, which is the HC group.

###  4.3 Effect of the Severity Level of Alzheimer’s Disease on ASV
Performance

(a) Healthy Control

(b) Severe AD

Figure 5: Speaker cluster visualizations of HC and Severe AD groups based on
TitaNet embeddings for ADCT dataset, created using the PaCMAP (Wang et al.,
2021) dimensionality reduction method, with each colour representing a
distinct speaker.

(a) Severe AD

(b) Moderate AD

(c) Mild AD

(d) Healthy Control

Figure 6: Similarity heatmap visualizations for ADCT dataset using TitaNet
embeddings for different AD severity levels each downsized to the number of
speakers in the smallest group, which is equal to 25. The lighter colours
correspond to a higher level of similarity between the associated speaker
tuples.

We examined how the AD severity level impacts the performance of the ASV
model. For this purpose, we performed a separate recalculation of EER for
subgroups of audio samples consisting of Severe AD (Number of speakers = 218),
Moderate AD (Number of speakers = 436), Mild AD (Number of speakers = 244),
and HC (Number of speakers = 25), while retaining the original number of
speakers. To establish a fair comparison, we then balanced the number of
speakers across the subgroups by downsizing each to the smallest subgroup size
of 25 speakers, which was the size of the HC subgroup. In both scenarios,
higher AD severity levels lowered speaker verification performance, by about
1% to 1.5% of EER (Figure 4). Also, EER was lower within the groups where a
higher number of speakers were included. Figure 5(a) and 5(b) display the
PaCMAP (Wang et al., 2021) visualization of TitaNet speaker embeddings for the
HC and Severe AD groups in the balanced ADCT dataset, where each colour
representing a unique speaker. In both the HC and Severe AD groups, samples
from the same speakers are clustered close to each other, while in the HC
group, the samples from different speakers are more distinguishable compared
to the Severe AD group. Moreover, there is a higher level of similarity
between negative speaker tuples within Moderate and Severe AD subgroups in
comparison to Mild AD and HC subgroups, as indicated in the similarity
heatmaps for different AD severity levels (Figure 6). Overall, our findings
suggest that the unique voice characteristics associated with varying levels
of AD severity (Boschi et al., 2017) can be entangled with the identity of the
speaker and may introduce a potential source of bias in the ASV models.

##  5 Conclusion

Large-scale clinical trials require accurate verification of participants, as
duplicate participation may lead to substandard data quality and significant
financial and health risks. Therefore, developing accurate ASV models for
verifying participant identity is essential in these settings. External
factors such as participant profile or audio quality can cause errors and
biases in ASV performance during the trials, but limited research has been
conducted in this area. In the present work, we utilize a longitudinal speech
dataset of participants with varying levels of AD severity and investigate the
impact of external factors, such as different participant demographic
characteristics, audio quality criteria, and AD severity levels, on the
performance of an end-to-end TI ASV model. Our findings show that variations
in ASV performance can be attributed to the inherent voice characteristics of
different subgroups (e.g., different ages, genders, accents, or AD severity
levels) that are likely to be confused with the identity of the speaker.
Hence, it is critical to reassess this technology to mitigate the risk of
potential biases toward certain subgroups. Based on our results, poor audio
quality with unclear speech, noisy background, and clinician interference also
negatively impacts ASV performance. This highlights the importance of quality
assurance for the speech recordings during the trials. In future work, we aim
to automate the audio quality assessment process by leveraging existing
automated methods such as perceptual evaluation of speech quality (PESQ) (Rix
et al., 2001) or short-time objective intelligibility (STOI) (Taal et al.,
2010), which would reduce the human effort required for this task.

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# Sums of singular series with large sets and the tail of the distribution of
primes

Vivian Kuperberg

###### Abstract.

In 1976, Gallagher showed that the Hardy–Littlewood conjectures on prime
$k$-tuples imply that the distribution of primes in log-size intervals is
Poissonian. He did so by computing average values of the singular series
constants over different sets of a fixed size $k$ contained in an interval
$[1,h]$ as $h\to\infty$, and then using this average to compute moments of the
distribution of primes. In this paper, we study averages where $k$ is
relatively large with respect to $h$. We then apply these averages to the tail
of the distribution. For example, we show, assuming appropriate
Hardy–Littlewood conjectures and in certain ranges of the parameters, the
number of intervals $[n,n+\lambda\log x]$ with $n\leq x$ containing at least
$k$ primes is $\ll x\exp(-k/(\lambda e)).$

The author is supported by NSF GRFP grant DGE-1656518 as well as the NSF
Mathematical Sciences Research Program through the grant DMS-2202128, and
would like to thank Kannan Soundararajan for many helpful comments and
discussions.

##  1\. Introduction

The Hardy–Littlewood prime $k$-tuple conjectures state that if
$\mathcal{H}=\\{h_{1},\dots,h_{k}\\}$ is a set of $k$ distinct integers, then
as $x\to\infty$,

(1) $\sum_{n\leq
x}\prod_{i=1}^{k}\Lambda(n+h_{i})=(\mathfrak{S}(\mathcal{H})+o(1))x,$

where $\mathfrak{S}(\mathcal{H})$ is the singular series

(2) $\mathfrak{S}(\mathcal{H})=\prod_{p\text{
prime}}\frac{1-\nu_{\mathcal{H}}(p)/p}{(1-1/p)^{k}},$

and $\nu_{\mathcal{H}}(p)$ denotes the number of distinct residue classes
modulo $p$ occupied by the elements of $\mathcal{H}$.

Many aspects of the distribution of primes can be understood through the lens
of the Hardy–Littlewood conjectures. For example, in [3], Gallagher showed
that the Hardy–Littlewood conjectures imply that the distribution of primes in
log-size intervals is Poissonian. He did so by showing that for fixed $k$ and
as $h\to\infty$,

(3) $\sum_{\begin{subarray}{c}h_{1},\dots,h_{k}\leq h\\\
\text{distinct}\end{subarray}}\mathfrak{S}(h_{1},\dots,h_{k})\sim\sum_{\begin{subarray}{c}h_{1},\dots,h_{k}\leq
h\\\ \text{distinct}\end{subarray}}1,$

so that singular series for sets of size $k$ have value $1$ on average as
their elements grow large. In [10], Montgomery and Soundararajan computed
second-order terms of this average in order to show that, assuming a version
of the Hardy–Littlewood conjectures with a stronger error term, primes in
somewhat longer intervals obey an appropriate Gaussian distribution. For both
of these analyses, $k$ is fixed throughout.

What about when $k$ is not fixed? Here we study sums of singular series for
sets of size $k$ with elements in $[1,h]$, where $k\to\infty$ as $h\to\infty$.
Put another way, we study the rate at which the average value of
$\mathfrak{S}(\mathcal{H})$ converges to $1$, in order to extend Gallagher’s
proof to larger $k$.

######  Theorem 1.1.

Fix $\delta>0$, and let $h,k\in{\mathbb{N}}$ with $k=O((\log h)^{1-\delta})$.
Let $T_{k}(h)$ be given by

(4) $T_{k}(h):=\sum_{\begin{subarray}{c}h_{1},\dots,h_{k}\leq h\\\
\text{distinct}\end{subarray}}\mathfrak{S}(h_{1},\dots,h_{k}),$

Then there exists a $\beta>0$, dependent only on $\delta>0$, with

$T_{k}(h)=h^{k}+O(h^{k-\beta}).$

In particular, Theorem 1.1 states that (3) holds whenever $k=O((\log
h)^{1-\delta})$ for some $\delta>0$. One might expect the average value of $1$
to extend to still larger $k$; for example, it is reasonable to conjecture
that (3) would hold whenever $k=O((\log h)^{2})$. For arbitrarily large $k$,
Theorem 1.2 provides a bound on the average value of $k$-term singular series
over sets with elements in $[1,h]$.

######  Theorem 1.2.

Let $k,h\in{\mathbb{N}}$, with no conditions on their relative growth rates.
Define $T_{k}(h)$ by (4). Then

(5) $T_{k}(h)\ll h^{k}\prod_{p\leq k^{3}}\frac{1}{(1-1/p)^{k}}\ll h^{k}(3\log
k)^{k}.$

This average value is likely much weaker than the truth, but it has the
advantage of bounding the average value only in terms of $k$. Instead of
taking $p\leq k^{3}$, in our proof we can take $p\leq k^{2+\varepsilon}$ for
any $\varepsilon>0$, which has the effect of replacing the $3^{k}$ in the
final bound with a $(2+\varepsilon)^{k}$. However, the final bound is in any
case $e^{O(k\log\log k)}$. Theorems 1.1 and 1.2 are proven in Section 2.

In the second half of this paper, we discuss one application of sums of
singular series for sets of size $k$ when $k$ varies: namely, the tail of the
distribution of primes. The _maximum_ number of primes in an interval of size
$\lambda\log x$ is closely connected to the study of small gaps between
primes, and has been studied in, among other places, [4], [8], and [11]. In
[4], Granville and Lumley conjecture that if $y\leq\log x$ and $x,y\to\infty$,
the lim sup of the number of primes in intervals $(x,x+y]$ is
$\sim\frac{y}{\log y}$, and if $\log x\leq y=o((\log x)^{2})$, the lim sup
should be given by $\frac{\log x}{\log\left(\frac{(\log x)^{2}}{y}\right)}$.
They also formulate conjectures for larger intervals.

The _expected_ number of primes in such an interval is much smaller; for a
constant $\lambda$, there are on average $\lambda$ primes in an interval
$(x,x+\lambda\log x]$, and Gallagher [3] showed that for fixed $\lambda>0$,
assuming the Hardy–Littlewood conjectures,

$\lim_{x\to\infty}\frac{1}{x}\\#\\{n\leq x:\pi(n+\lambda\log
x)-\pi(n)=k\\}=\frac{\lambda^{k}e^{-\lambda}}{k!}.$

Again, we consider the situation where $k\to\infty$. What bounds can be proven
on the tail of this distribution away from the extreme values? For example,
one can ask how frequently intervals of size $\lambda\log x$ contain at least
$\log\log x$ primes, where the Poisson prediction is that

(6) $\frac{1}{x}\\#\\{n\leq x:\pi(n+\lambda\log x)-\pi(n)\geq\log\log
x\\}\approx\frac{(e\lambda)^{\log\log x}e^{-\lambda}}{(\log\log x)^{\log\log
x}}.$

Since $k=|\mathcal{H}|$ grows with $x$ in our setting, our results rely on a
version of the Hardy–Littlewood conjectures which admits uniformity in the
size of the set $\mathcal{H}$; we state this version here.

######  Conjecture 1.3 (Hardy–Littlewood $k$-tuples conjecture, uniform
version).

There exist two absolute constants $\epsilon>0$ and $C>0$ such that for all
$x$, for all $k\leq(\log\log x)^{3}$, and for all admissible tuples
$\mathcal{H}=\\{h_{1},\dots,h_{k}\\}\subset[0,(\log x)^{2}]$,

(7) $\left|\sum_{n\leq
x}\mathbf{1}_{\mathcal{P}}(n+h_{1})\cdots\mathbf{1}_{\mathcal{P}}(n+h_{k})-\mathfrak{S}(\mathcal{H})\mathrm{li}_{k}(x)\right|\leq
Cx^{1-\varepsilon}.$

Equivalently, for possibly different values of $\varepsilon$ and $C$,

(8) $\left|\sum_{n\leq
x}\Lambda(n+h_{1})\cdots\Lambda(n+h_{k})-\mathfrak{S}(\mathcal{H})x\right|\leq
Cx^{1-\varepsilon}.$

When $k=10$, this conjecture would suggest that for $x=5500$, the error term
above is bounded by $Cx^{1-\varepsilon}$ for some $\varepsilon$ and some $C$.
For several sets of size $10$, computer tests found that bounds of $(\log
x)^{6}x^{1/2}$ held for all $x\leq 5500$; in fact this and other tests for
small values of $k$ suggest that the error term is far smaller, and for
example may be bounded by $Cx^{1/2}(\log x)^{k}$ in this range of $k$ and $h$.
It is also likely possible to extend the range of $k$ and $h$ in this
conjecture; if $k$ is as large as $\log x$, then $\mathrm{li}_{k}(x)$ is small
enough that the conjecture is not so meaningful, but it is difficult to say
when Hardy–Littlewood convergencee should break down.

Gallagher’s proof proceeds by computing moments. One strategy towards
understanding the tail of the distribution is to estimate higher moments of
the distribution, or equivalently, to understand how quickly the $r$th moment
of the distribution of primes converges to the $r$th moment of a Poisson
distribution. Using Conjecture 1.3 as well as Theorems 1.1 and 1.2, we can
prove the following result on moments of the distribution of primes.

######  Theorem 1.4.

Assume Conjecture 1.3. Let $x>0$, and assume that $h=\lambda\log x$ and that
$r\ll(\log h)^{1-\delta}$ for some $\delta>0$. Define the $r$th moment
$m_{r}(x,h)$ of the distribution of primes in intervals of size $h$ by

(9) $m_{r}(x,h)=\frac{1}{x}\sum_{n\leq x}\left(\pi(n+h)-\pi(n)\right)^{r}.$

Then

$m_{r}(x,h)=\left(\sum_{\ell=1}^{r}\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}\lambda^{\ell}\right)(1+o(1)),$

where $\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}$ denotes the Stirling numbers of
the second kind.

######  Remark.

Note that $\lambda$ need not be fixed as $x\to\infty$.

Theorem 1.4 then implies bounds on the tail of the distribution of primes, and
in particular yields the following two corollaries.

######  Corollary 1.5.

Assume Conjecture 1.3. Let $x>0$ and set $h=\lambda\log x$, where $\lambda(x)$
is nondecreasing as $x\to\infty$. Let $k\ll(\log h)^{1-\delta}$ for some
$\delta>0$ and assume that $\frac{k}{\lambda+1}\to\infty$ as $x\to\infty$. Let
$I(x;k,h)$ be given by

(10) $I(x;k,h):=\\#\left\\{n\leq x:\pi(n+h)-\pi(n)\geq k\right\\}.$

If $\lambda\geq 1$, then as $x\to\infty$,

$I(x;k,h)\ll x\mathrm{exp}\left(-\frac{k}{\lambda e}\right).$

Otherwise,

$I(x;k,h)\ll x\mathrm{exp}\left(-\frac{k}{(\lambda+1)e}\right).$

######  Corollary 1.6.

Assume Conjecture 1.3. Let $x>0$, and assume that $h=\lambda\log x$. Let
$I(x;k,h)$ be defined as in (10). Then for any $\delta>0$, as $x\to\infty$,

$I(x;k,h)\ll_{\delta}x\mathrm{exp}\left((\log
h)^{1-\delta}(\log(\lambda+1)+(1-\delta)\log\log h-\log k)\right).$

For example, taking $k=\log h$, Corollary 1.6 says that for all $\delta>0$,
assuming Conjecture 1.3,

$I(x;\log h,h)\ll_{\delta}x\exp\left((\log
h)^{1-\delta}(\log\lambda-\delta\log\log h)\right).$

In [9], Maynard proves lower bounds on the same problem, showing that for any
$x,y\geq 1$ there are $\gg x\exp\left(-\sqrt{\log x}\right)$ integers $n\leq
x$ such that $\pi(n+y)-\pi(n)\gg\log y$, which in this case corresponds to the
condition that there are $\gg\log\log x$ primes in intervals of width
$\lambda\log x$. Both upper and lower bounds are reasonably far from the
Poisson prediction in (6).

It seems reasonable to conjecture that the Poisson prediction should still
hold when $k\sim\log h$ or $k\sim(\log h)^{2}$, and perhaps even larger. At
this point, both upper and lower bounds are far from matching this conjecture.

######  Conjecture 1.7.

Let $x>1$ and let $h=\lambda\log x$, with $\lambda=o((\log x)^{\varepsilon})$
for all $\varepsilon>0$. Let $k\ll(\log h)^{2}$. Define

(11) $\pi_{k}(x;h):=\\#\\{n\leq x:\pi(n+h)-\pi(n)=k\\}.$

Then $\pi_{k}(x;h)\sim x\frac{\lambda^{k}e^{-\lambda}}{k!}$ as $x\to\infty$.

In Section 4, we prove unconditional bounds on the tail of the distribution of
primes. For these arguments we use a Selberg sieve bound instead of applying
the Hardy–Littlewood conjectures. The Selberg sieve bound for prime $k$-tuples
has an extra factor of $2^{k}k!$ from the Hardy–Littlewood prediction. This
factor is larger than our bound on the average of $k$-term singular series in
Theorem 1.2, so we achieve the following unconditional bound for much larger
moments; in particular, for the $r$th moment when $r=o((\log x)^{1/3})$.

######  Theorem 1.8.

Let $x>0$, let $h=\lambda\log x=o(x)$ and let $r=o((\log x)^{1/3})$. Define
the $r$th moment $m_{r}(x,h)$ of the distribution of primes in intervals of
size $h$ as in (9). Then

$m_{r}(x,h)\ll\lambda^{r}r^{2r}e^{O(r\log\log r)}.$

As before, this bound on moments yields the following corollary on intervals
containing many primes.

######  Corollary 1.9.

Let $x>0$, let $h=\lambda\log x$, where $\lambda$ is a nondecreasing function
of $x$. Let $k$ be an integer dependent on $x$ and assume that $k=o((\log
x)^{1/6})$ and that $k/\lambda\to\infty$ as $x\to\infty$. Define $I(x;k,h)$ as
in (10). Then for some constant $C$,

$I(x;k,h)\ll x\exp\left(-\sqrt{\frac{k}{\lambda
e}}2^{C/2}\left(\log\frac{k}{\lambda e}\right)^{-C/2}\right).$

It may also be possible to achieve weaker, yet nontrivial, bounds for larger
$k$, along the lines of the bounds in Corollary 1.5. We predict that the bound
in Corollary 1.5 should hold for any $k\ll(\log h)^{2}$, instead of merely
$k\ll(\log h)^{1-\delta}$.

######  Conjecture 1.10.

Let $x>1$ and let $h=\lambda\log x$, with $\lambda=o((\log x)^{\varepsilon})$
for all $\varepsilon>0$. Let $k\ll(\log h)^{2}$, and define $\pi_{k}(x;h)$ as
in (11). Then $\pi_{k}(x;h)\ll x\mathrm{exp}\left(-\frac{k}{\lambda e}\right)$
as $x\to\infty$.

To put these results in perspective, let us consider the case when
$\lambda=1$, i.e. primes in intervals of width $\log x$. In [3], Gallagher
shows that for fixed $k$, the number of $n\leq x$ such that the interval
$(n,n+\log x]$ contains exactly $k$ primes is asymptotic to the Poisson
prediction $\frac{x}{ek!}$, assuming the Hardy–Littlewood conjectures.
Unconditionally, Gallagher shows in [3] that the number of $n\leq x$ such that
$(n,n+\log x]$ contains exactly $k$ primes is $\lesssim xe^{-Ck}$, for an
absolute constant $C$.

We instead consider the probability that an interval $(n,n+\log x]$ contains
at least $\log\log x$ primes. In this case, the Poisson prediction for the
probability that an interval contains $\log\log x$ primes is
$\frac{(e\lambda)^{\log\log x}e^{-\lambda}}{(\log\log x)^{\log\log x}}$, which
for any $A>0$ is $\ll\frac{1}{(\log x)^{A}}$. In [9], Maynard proves a lower
bound; namely, that at least $\gg x\exp(-\sqrt{\log x})$ intervals $(n,n+\log
x]$, with $n\leq x$, contain $\gg\log\log x$ primes. In Corollary 1.6, we show
that, assuming the uniform Hardy–Littlewood conjectures, for all $\delta>0$,
the number of $n\leq x$ such that $(n,n+\log x]$ contains at least $\log\log
x$ primes is $\ll_{\delta}x\exp\left(-\delta\log\log\log x(\log\log
x)^{1-\delta}\right)$. Unconditionally, we show in Corollary 1.9 that there
exists $C>0$ such that the number of $(n,n+\log x]$ containing at least
$\log\log x$ primes is $\ll x\exp\left(-\sqrt{2^{C}/e}(\log\log
x)^{1/2}(\log\log\log x-1)^{-C/2}\right)$.

When $k$ is slightly smaller than $\log\log x$, that is, when $k\ll(\log
h)^{(1-\delta)}$, Corollary 1.5 achieves the bound of $\frac{1}{(\log x)^{A}}$
with $A=\frac{1}{\lambda e}$. Bounding this probability by $\frac{1}{(\log
x)^{A}}$ for any $A>0$ may be within reach even if the Poisson prediction
itself is not. On the other hand, many questions concerning the tail of the
distribution of primes are quite delicate, especially concerning the maximum
and minimum number of primes in an interval of a certain size. For example, in
[7], Maier proved that intervals of size $(\log x)^{A}$ for $A>2$ can contain
surprisingly few or surprisingly many primes. For more information, see [4].
The tail of the distribution of primes is also studied in [2].

##  2\. Averages for large sets

In this section, we prove Theorems 1.1 and 1.2. We begin with Theorem 1.1,
whose proof closely follows Gallagher’s original proof in [3].

###### Proof of Theorem 1.1, following Gallagher..

Let $\nu_{\mathcal{H}}(p):=\\#\mathcal{H}\bmod p$. For a set
$\mathcal{H}=\\{h_{1},\dots,h_{k}\\}$, write
$D_{\mathcal{H}}=\prod_{i<j}(h_{i}-h_{j})$, so that $\nu_{\mathcal{H}}(p)=k$
unless $p|D_{\mathcal{H}}$. Define

$a(p,\nu_{\mathcal{H}}(p)):=\frac{p^{k}-\nu_{\mathcal{H}}(p)p^{k-1}-(p-1)^{k}}{(p-1)^{k}},$

so that the $p$th factor of $\mathfrak{S}(\mathcal{H})$ is given by
$\frac{1-\nu_{\mathcal{H}}(p)/p}{(1-1/p)^{k}}=1+a(p,\nu_{\mathcal{H}}(p))$.

For any prime $p>k$,

$\frac{p^{k}-kp^{k-1}-(p-1)^{k}}{(p-1)^{k}}=\sum_{j=2}^{k}(p-1)^{-j}\binom{k}{j}(1-j)\ll
k^{2}(p-1)^{-2}.$

This and a similar computation for $\nu<k$ shows that for $p>k$,

(12) $|a(p,\nu)|\ll\begin{cases}k^{2}(p-1)^{-2}&\text{ if }\nu=k\\\
k^{2}(p-1)^{-1}&\text{ if }\nu<k.\end{cases}$

For $p\leq k$,

(13) $|a(p,\nu)|=\left|-1+\frac{1-\nu/p}{(1-1/p)^{k}}\right|\ll e^{k/p}.$

For squarefree $q$, write
$a_{\mathcal{H}}(q):=\prod_{p|q}a(p,\nu_{\mathcal{H}}(p))$, so that

$\mathfrak{S}(\mathcal{H})=\prod_{p\leq
k}\frac{1-\nu_{\mathcal{H}}(p)/p}{(1-1/p)^{k}}\sum_{\begin{subarray}{c}q\geq
1\\\ p|q{\Rightarrow}p>k\end{subarray}}a_{\mathcal{H}}(q).$

Using the bounds on $a(p,\nu)$, for any $x$,

$\sum_{\begin{subarray}{c}q>x\\\
p|q{\Rightarrow}p>k\end{subarray}}|a_{\mathcal{H}}(q)|\leq\sum_{\begin{subarray}{c}q>x\\\
p|q{\Rightarrow}p>k\end{subarray}}\frac{\mu^{2}(q)(Ck^{2})^{\omega(q)}}{\phi^{2}(q)}\phi((q,D_{\mathcal{H}})),$

where $\omega(q)$ is the number of prime factors of $q$ and $C$ is an absolute
positive constant.

Writing $q=de$ with $d|D_{\mathcal{H}}$ and $(e,D_{\mathcal{H}})=1$, this is

(14) $\sum_{\begin{subarray}{c}d|D_{\mathcal{H}}\\\
p|d{\Rightarrow}p>k\end{subarray}}\frac{\mu^{2}(d)(Ck^{2})^{\omega(d)}}{\phi(d)}\sum_{\begin{subarray}{c}e>x/d\\\
(e,D_{\mathcal{H}})=1\\\
p|e{\Rightarrow}p>k\end{subarray}}\frac{\mu^{2}(e)(Ck^{2})^{\omega(e)}}{\phi^{2}(e)}.$

Apply Rankin’s trick to bound the inner sum, so that for any $0<\alpha<1$,

$\displaystyle\sum_{\begin{subarray}{c}e>x/d\\\ (e,D_{\mathcal{H}})=1\\\
p|e{\Rightarrow}p>k\end{subarray}}\frac{\mu^{2}(e)(Ck^{2})^{\omega(e)}}{\phi^{2}(e)}$
$\displaystyle\leq\sum_{\begin{subarray}{c}e\geq 1\\\ (e,D_{\mathcal{H}})=1\\\
p|e{\Rightarrow}p>k\end{subarray}}\left(\frac{e}{x/d}\right)^{\alpha}\frac{\mu^{2}(e)(Ck^{2})^{\omega(e)}}{\phi^{2}(e)}$
$\displaystyle=\left(\frac{d}{x}\right)^{\alpha}\prod_{\begin{subarray}{c}p>k\\\
p\nmid
D_{\mathcal{H}}\end{subarray}}\left(1+\frac{Ck^{2}p^{\alpha}}{(p-1)^{2}}\right)$
$\displaystyle\leq\left(\frac{d}{x}\right)^{\alpha}\left(\prod_{\begin{subarray}{c}p>k\end{subarray}}\left(1+\frac{p^{\alpha}}{(p-1)^{2}}\right)\right)^{Ck^{2}}$
$\displaystyle\ll\left(\frac{d}{x}\right)^{\alpha}e^{Ck^{1+\alpha}}.$

Choose $\alpha>0$ to be small enough that $(1+\alpha)(1-\delta)<1$. Then the
sum in (14) is

$\displaystyle\ll\sum_{d|D_{\mathcal{H}}}\frac{\mu^{2}(d)(Ck^{2})^{\omega(d)}}{\phi(d)}\left(\frac{d}{x}\right)^{\alpha}e^{Ck^{1+\alpha}}$
$\displaystyle\ll_{\varepsilon,\delta}(xh)^{\varepsilon}/x^{\alpha},$

for any $k\ll(\log h)^{1-\delta}$. Set $x$ to be $h^{1/2}$, and choose
$\varepsilon>0$ small enough above that
$\frac{3}{2}\varepsilon<\frac{1}{2}\alpha$.

This is true for any set $\mathcal{H}=\\{h_{1},\dots,h_{k}\\}$, so it follows
that

(15) $T_{k}(h)=\sum_{\begin{subarray}{c}q\leq x\\\
p|q{\Rightarrow}p>k\end{subarray}}\sum_{\begin{subarray}{c}r\geq 1\\\
p|r{\Rightarrow}p\leq
k\end{subarray}}\sum_{\begin{subarray}{c}h_{1},\dots,h_{k}\leq h\\\
\text{distinct}\end{subarray}}a_{\mathcal{H}}(qr)+O\left(\frac{(xh)^{\varepsilon}}{x^{\alpha}}\sum_{\begin{subarray}{c}h_{1},\dots,h_{k}\leq
h\\\ \text{distinct}\end{subarray}}\prod_{p\leq
k}\frac{1-\nu_{\mathcal{H}}(p)/p}{(1-1/p)^{k}}\right),$

where we have additionally expanded the terms of the product with $p\leq k$
into the sum over $r$. First consider the error term in (15), which for some
absolute constant $C_{0}$ is

$\displaystyle\ll\frac{(xh)^{\varepsilon}}{x^{\alpha}}h^{k}\prod_{p\leq
k}\left(C_{0}e^{k/p}\right)$
$\displaystyle\ll\frac{(xh)^{\varepsilon}}{x^{\alpha}}h^{k}C_{0}^{\pi(k)}e^{k\sum_{p\leq
k}\tfrac{1}{p}}$
$\displaystyle\ll\frac{(xh)^{\varepsilon}}{x^{\alpha}}h^{k}e^{O(k\log\log k)}$
$\displaystyle\ll\frac{(xh)^{\varepsilon}}{x^{\alpha}}h^{k}e^{(\log
h)^{1-\delta/2}}\ll\frac{(xh)^{\varepsilon}}{x^{\alpha}}h^{k+o_{\delta}(1)}.$

Now consider the main term. The sum over $h_{1},\dots,h_{k}\leq h$ in (15) can
also be written as

$\sum_{\begin{subarray}{c}\vec{\nu}=(\nu_{p})_{p|qr}\\\
\nu_{p}\leq\min\\{p-1,k\\}\end{subarray}}\prod_{p|qr}a(p,\nu_{p})\left(N(\vec{\nu})+O(kh^{k-1})\right),$

where $N(\vec{\nu})$ is the number of $k$-tuples of not necessarily distinct
integers $h_{1},\dots,h_{k}$ with $1\leq h_{1},\dots,h_{k}\leq h$ which occupy
exactly $\nu_{p}$ residue classes mod $p$ for each $p|qr$. We can estimate
$N(\vec{\nu})$ by counting for each $p|qr$ the number of
$h_{1},\dots,h_{k}\bmod p$ that occupy exactly $\nu_{p}$ residue classes and
applying the Chinese Remainder Theorem. Thus

$N(\vec{\nu})=\prod_{p|qr}\binom{p}{\nu_{p}}\sigma(k,\nu_{p})\left(\frac{h}{qr}+O(1)\right)^{k},$

where $\sigma(k,j)$ denotes the number of surjective maps
$[1,k]\twoheadrightarrow[1,j]$; we also have
$\sigma(k,j)=j!\genfrac{\\{}{\\}}{0.0pt}{}{k}{j}$, where
$\genfrac{\\{}{\\}}{0.0pt}{}{k}{j}$ is the Stirling number of the second kind.
Expanding, we get

$N(\vec{\nu})=\prod_{p|qr}\binom{p}{\nu_{p}}\sigma(k,\nu_{p})\left(\left(\frac{h}{qr}\right)^{k}+O\left(\sum_{j=0}^{k-1}\left(\frac{h}{qr}\right)^{j}\binom{k}{j}\right)\right).$

Since $x=h^{1/2}$ and $\prod_{p\leq k}p=e^{O(k)}=e^{O((\log
h)^{1-\delta})}=h^{o(1)}$, for any $q\leq k$ and $r$ with all prime factors
$\leq k$,

$\sum_{j=0}^{k-1}\left(\frac{h}{qr}\right)^{j}\binom{k}{j}\ll
k^{2}\left(\frac{h}{qr}\right)^{k-1}.$

Thus the inner sum in (15) is

$\left(\frac{h}{q}\right)^{k}A(qr)+O\left(k^{2}\left(\frac{h}{qr}\right)^{k-1}B(qr)\right)+O(kh^{k-1}C(qr)),$

where

$\displaystyle A(qr)$
$\displaystyle=\sum_{\begin{subarray}{c}\vec{\nu}=(\nu_{p})_{p|qr}\\\
\nu_{p}\leq\min\\{p-1,k\\}\end{subarray}}\prod_{p|qr}a(p,\nu_{p})\binom{p}{\nu_{p}}\sigma(k,\nu_{p}),$
$\displaystyle B(qr)$
$\displaystyle=\sum_{\begin{subarray}{c}\vec{\nu}=(\nu_{p})_{p|qr}\\\
\nu_{p}\leq\min\\{p-1,k\\}\end{subarray}}\prod_{p|qr}|a(p,\nu_{p})|\binom{p}{\nu_{p}}\sigma(k,\nu_{p}),\text{
and}$ $\displaystyle C(qr)$
$\displaystyle=\sum_{\begin{subarray}{c}\vec{\nu}=(\nu_{p})_{p|qr}\\\
\nu_{p}\leq\min\\{p-1,k\\}\end{subarray}}\prod_{p|qr}|a(p,\nu_{p})|.$

Just as in [3], $A(q)=0$ for $q>1$, and $A(1)=1$.

Now consider $C(qr)$, which can be estimated using the bounds (12) and (13)
for $a(p,\nu)$. Write

$C(qr)=\prod_{p|qr}\left(\sum_{\nu=1}^{p}|a(p,\nu)|\right).$

If $p>k$, the $p$th factor is $\ll\frac{pk^{2}}{p-1}$, whereas if $p\leq k$,
this factor is $\ll pe^{k/p}.$ Thus

(16)
$C(qr)\leq(C_{1}k^{2})^{\omega(q)}\frac{q}{\phi(q)}C_{1}^{\omega(r)}e^{k\sum_{p|r}1/p},$

for some absolute constant $C_{1}$, where without loss of generality
$C_{1}\geq 1$.

After summing (16) over all $q$, the contribution to $T_{k}(h)$ from the
factors corresponding to $C(qr)$ is

$\displaystyle\ll kh^{k-1}\sum_{\begin{subarray}{c}q\leq x\\\
p|q{\Rightarrow}p>k\end{subarray}}\sum_{\begin{subarray}{c}r\geq 1\\\
p|r{\Rightarrow}p\leq
k\end{subarray}}(C_{1}k^{2})^{\omega(q)}\frac{q}{\phi(q)}C_{1}^{\omega(r)}e^{k\sum_{p|r}1/p}$
$\displaystyle\ll kh^{k-1}\prod_{p\leq
k}\left(1+C_{1}e^{k/p}\right)\sum_{\begin{subarray}{c}q\leq x\\\
p|q{\Rightarrow}p>k\end{subarray}}(C_{1}k^{2})^{\omega(q)}\frac{q}{\phi(q)}$
$\displaystyle\ll kh^{k-1}(2C_{1})^{\pi(k)}e^{k\sum_{p\leq
k}\tfrac{1}{p}}\sum_{\begin{subarray}{c}q\leq x\\\
p|q{\Rightarrow}p>k\end{subarray}}(C_{1}k^{2})^{\omega(q)}\frac{q}{\phi(q)}$
$\displaystyle\ll kh^{k-1}h^{o_{\delta}(1)}\ll h^{k-1+o_{\delta}(1)},$

since $k\ll(\log h)^{1-\delta}$.

Finally consider $B(qr)$. Just as with $C(qr)$, $B(qr)$ is multiplicative, and
the $p$th factor of $B(qr)$ is given by

$\sum_{\nu=1}^{p}|a(p,\nu)|\binom{p}{\nu}\sigma(k,\nu),$

which by (13) and (12) is

$\ll\begin{cases}\frac{k^{2}p^{k}}{p-1}&\text{ if }p>k\\\ e^{k/p}p^{k}&\text{
if }p\leq k.\end{cases}$

Thus for some absolute constant $C_{2}\geq 1$, and summing over all $q$, the
contribution to $T_{k}(h)$ from the $B(qr)$ factors is

$\displaystyle\ll\sum_{\begin{subarray}{c}q\leq x\\\
p|q{\Rightarrow}p>k\end{subarray}}\sum_{\begin{subarray}{c}r\geq 1\\\
p|r{\Rightarrow}p\leq
k\end{subarray}}k^{2}\left(\frac{h}{qr}\right)^{k-1}\frac{(C_{2}k^{2})^{\omega(q)}q^{k}}{\phi(q)}C_{2}^{\omega(r)}r^{k}e^{k\sum_{p|r}\tfrac{1}{p}}$
$\displaystyle\ll k^{2}h^{k-1}\sum_{\begin{subarray}{c}q\leq x\\\
p|q{\Rightarrow}p>k\end{subarray}}\frac{(C_{2}k^{2})^{\omega(q)}q}{\phi(q)}\prod_{p\leq
k}\left(1+C_{2}e^{k/p}p\right).$

For $k\ll(\log h)^{1-\delta}$, the product over $p\leq k$ is $\ll e^{(\log
h)^{1-\delta/2}}=h^{o_{\delta}(1)}$, so the overall sum is

$\displaystyle\ll k^{2}h^{k-1+o_{\delta}(1)}\sum_{\begin{subarray}{c}q\leq
x\\\
p|q{\Rightarrow}p>k\end{subarray}}\frac{(C_{2}k^{2})^{\omega(q)}q}{\phi(q)}$
$\displaystyle\ll h^{k-1+o_{\delta}(1)}.$

Putting everything together,

$T_{k}(h)=h^{k}+O\left(h^{k-1+o_{\delta}(1)}+\frac{(xh)^{\varepsilon}}{x^{\alpha}}h^{k+o_{\delta}(1)}\right),$

which for $x=h^{1/2}$ is $h^{k}+O(h^{k-\beta})$ for
$\beta<\frac{1}{2}\alpha-\frac{3}{2}\varepsilon,$ as desired. ∎

The techniques relying on the Chinese Remainder Theorem break down for larger
$k$, where they do not give a bound with the correct power of $h$. We will now
turn to Theorem 1.2, which bounds $T_{k}(h)$ for any $k$ where the dependence
on $h$ is $h^{k}$, which is approximately the number of terms in the sum. In
other words, Theorem 1.2 provides a bound that is uniform in $h$ on the
average value of $k$-term singular series for sets with elements that are at
most $h$. The proof of Theorem 1.2 relies on Lemma 2.1, which is a uniform
bound on $\mathfrak{S}(\mathcal{H})$ for $\mathcal{H}\subset[1,h]$ satisfying
the conditions of Theorem 1.2. The proof is then completed via an application
of Perron’s formula.

######  Lemma 2.1.

Let $\mathcal{H}=\\{h_{1},\dots,h_{k}\\}$ be a set of distinct integers.
Define $\mathfrak{S}(\mathcal{H})$ as in (2). Then

(17) $\mathfrak{S}(\mathcal{H})\ll\prod_{p\leq
k^{3}}\frac{1}{(1-1/p)^{k}}\prod_{p>k^{3}}\frac{1-k/p}{(1-1/p)^{k}}\binom{k}{2}^{-1}\sum_{1\leq
i<j<\leq k}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}.$

###### Proof.

By definition,

$\displaystyle\mathfrak{S}(\mathcal{H})$ $\displaystyle=\prod_{p\text{
prime}}\frac{1-\nu_{\mathcal{H}}(p)/p}{(1-1/p)^{k}}$
$\displaystyle\leq\prod_{\begin{subarray}{c}p\text{ prime}\\\ p\leq
k^{3}\end{subarray}}\frac{1}{(1-1/p)^{k}}\prod_{\begin{subarray}{c}p\text{
prime}\\\ p>k^{3}\end{subarray}}\frac{1-\nu_{\mathcal{H}}(p)/p}{(1-1/p)^{k}}$
(18) $\displaystyle=\prod_{p\leq
k^{3}}\frac{1}{(1-1/p)^{k}}\prod_{p>k^{3}}\frac{1-k/p}{(1-1/p)^{k}}\prod_{\begin{subarray}{c}p|\Delta(\mathcal{H})\\\
p>k^{3}\end{subarray}}\frac{p-\nu_{\mathcal{H}}(p)}{p-k}.$

Rewrite the product inside via

$\displaystyle\prod_{\begin{subarray}{c}p|\Delta(\mathcal{H})\\\
p>k^{3}\end{subarray}}\frac{p-\nu_{\mathcal{H}}(p)}{p-k}$
$\displaystyle\leq\exp\Big{(}\sum_{\begin{subarray}{c}p|\Delta(\mathcal{H})\\\
p>k^{3}\end{subarray}}\frac{r-\nu_{\mathcal{H}}(p)}{p-k}\Big{)}$
$\displaystyle\ll\exp\Big{(}2\sum_{1\leq i<j\leq
k}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}$
$\displaystyle\leq\binom{k}{2}^{-1}\sum_{1\leq i<j<\leq
k}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)},$

where the last step comes from Jensen’s inequality; plugging this into (18)
yields the result. ∎

With Lemma 2.1 in hand, we now turn to the proof of Theorem 1.2.

###### Proof of Theorem 1.2.

By Lemma 2.1, for each set $\mathcal{H}$ under consideration,

$\mathfrak{S}(\mathcal{H})\ll\prod_{p\leq
k^{3}}\frac{1}{(1-1/p)^{k}}\prod_{p>k^{3}}\frac{1-k/p}{(1-1/p)^{k}}\binom{k}{2}^{-1}\sum_{1\leq
i<j<\leq k}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}.$

Sum over $\mathcal{H}$ to get

$T_{k}(h)\ll\prod_{p\leq
k^{3}}\frac{1}{(1-1/p)^{k}}\prod_{p>k^{3}}\frac{1-k/p}{(1-1/p)^{k}}\sum_{\mathcal{H}\subset[1,h]}\binom{k}{2}^{-1}\sum_{1\leq
i<j<\leq k}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}.$

Let $S_{k}(h)$ refer to the sums above, so that

$S_{k}(h):=\sum_{\mathcal{H}\subset[1,h]}\binom{k}{2}^{-1}\sum_{1\leq i<j<\leq
k}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}.$

Then

$\displaystyle S_{k}(h)$ $\displaystyle\ll\binom{k}{2}^{-1}\sum_{1\leq i<j\leq
k}\sum_{\begin{subarray}{c}\mathcal{H}\subset[1,h]\\\
|\mathcal{H}|=k\end{subarray}}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}$
$\displaystyle\ll\binom{k}{2}^{-1}\sum_{1\leq i<j\leq
k}\sum_{\begin{subarray}{c}h_{i},h_{j}\leq h\\\
\text{distinct}\end{subarray}}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|(h_{i}-h_{j})\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}h^{k-2}$
$\displaystyle=h^{k-2}\sum_{\begin{subarray}{c}\ell\leq
h\end{subarray}}(h-\ell)\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|\ell\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}$ $\displaystyle\leq
h^{k-1}\sum_{\begin{subarray}{c}\ell\leq
h\end{subarray}}\exp\Big{(}2\binom{k}{2}\sum_{\begin{subarray}{c}p|\ell\\\
p>k^{3}\end{subarray}}\frac{1}{p}\Big{)}.$

The sum over $\ell$ is a sum over a multiplicative function $f_{k}(\ell)$,
with $f_{k}(p^{j})=0$ if $p\leq k^{3}$ and $f_{k}(p^{j})=\exp(k(k-1)/p)$ if
$p>k^{3}$, regardless of $j$. The function $f_{k}(\ell)$ satisfies

$f_{k}(\ell)=\sum_{d|\ell}g_{k}(d),$

where $g_{k}$ is a multiplicative function given by $g_{k}(p^{j})=0$ if $p\leq
k^{3}$ or $j\geq 2$ and $g_{k}(p)=\exp(k(k-1)/p)-1$ for $p>k^{3}$ prime. Then

$\sum_{\ell\leq h}f_{k}(\ell)=\sum_{\ell\leq
h}\sum_{d|\ell}g_{k}(d)=\sum_{d\leq
h}g_{k}(d)\left\lfloor\frac{h}{d}\right\rfloor\leq
h\sum_{d=1}^{\infty}\frac{g_{k}(d)}{d}.$

The sum over $d$ can be rewritten as

$\displaystyle\sum_{\begin{subarray}{c}p\text{ prime}\\\
p>k^{3}\end{subarray}}\frac{1}{p}\left(\exp(k(k-1)/p)-1\right)$
$\displaystyle=\sum_{\begin{subarray}{c}p\text{ prime}\\\
p>k^{3}\end{subarray}}\sum_{j\geq
1}\frac{1}{p}\left(\frac{k(k-1)}{p}\right)^{j}$
$\displaystyle<\sum_{\begin{subarray}{c}p\text{ prime}\\\
p>k^{3}\end{subarray}}\sum_{j\geq 1}\frac{1}{p^{1+j/3}}.$

This sum is bounded by a constant independent of $k$, so that $S_{k}(h)\ll
h^{k}$.

Finally, return to the contribution from the small primes and $T_{k}(h)$,
which is bounded by

$T_{k}(h)\ll h^{k}\prod_{p\leq k^{3}}\frac{1}{(1-1/p)^{k}}\ll h^{k}(3\log
k)^{k},$

as desired. ∎

##  3\. Proof of Theorem 1.4 and its corollaries

Throughout, consider an interval of size $h=\lambda\log x$. Fix $\delta>0$ and
assume that $r\ll(\log h)^{1-\delta}$. We begin with the proof of Theorem 1.4.

The $r$th moment $m_{r}(x,h)$, defined in (9), is given by

$\displaystyle m_{r}(x,h)$ $\displaystyle=\sum_{\begin{subarray}{c}1\leq
h_{1},\dots,h_{r}\leq
h\end{subarray}}\frac{1}{x}\sum_{\begin{subarray}{c}n\leq
x\end{subarray}}\mathbf{1}_{\mathcal{P}}(n+h_{1})\cdots\mathbf{1}_{\mathcal{P}}(n+h_{r})$
$\displaystyle=\sum_{\ell=1}^{r}\frac{\sigma(r,\ell)}{\ell!}\sum_{\begin{subarray}{c}1\leq
h_{1},\dots,h_{\ell}\leq h\\\
\text{distinct}\end{subarray}}\frac{1}{x}\sum_{n\leq
x}\mathbf{1}_{\mathcal{P}}(n+h_{1})\cdots\mathbf{1}_{\mathcal{P}}(n+h_{\ell}),$

where $\sigma(r,\ell)$ is the number of surjective maps
$[1,r]\twoheadrightarrow[1,\ell]$, and
$\frac{\sigma(r,\ell)}{\ell!}=\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}$, the
Stirling number of the second kind.

Apply Conjecture 1.3 to replace the sum over correlations of primes with a sum
over singular series, yielding

$m_{r}(x,h)=\sum_{\ell=1}^{r}\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}\frac{1}{(\log
x)^{\ell}}\sum_{\begin{subarray}{c}1\leq h_{1},\cdots,h_{\ell}\leq h\\\
\text{distinct}\end{subarray}}(\mathfrak{S}(\\{h_{1},\dots,h_{\ell}\\})+o(1)).$

Estimating this moment now depends on the average of the singular series
constants, and in particular how quickly this average converges to $1$. We
apply our results from Section 2 bounding sums of singular series for large
sets, and in particular Theorem 1.1, which requires our assumption that
$r\ll(\log h)^{1-\delta}$. For larger $r$, one could also apply the weaker
result in Theorem 1.2 to yield a weaker moment bound.

By Theorem 1.1, for some $\beta>0$ dependent only on $\delta>0$,

$\sum_{\begin{subarray}{c}1\leq h_{1},\cdots,h_{k}\leq h\\\
\text{distinct}\end{subarray}}\mathfrak{S}(h_{1},\dots,h_{k})=h^{k}+O(h^{k-\beta}).$

Then for any $r\leq k$,

$\displaystyle m_{r}(x,h)$
$\displaystyle=\sum_{\ell=1}^{r}\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}\frac{1}{(\log
x)^{\ell}}\left(h^{\ell}+O(h^{k-\beta})\right)+o\left(\sum_{\ell=1}^{r}\frac{1}{(\log
x)^{\ell}}\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}h^{\ell}\right)$
$\displaystyle=\left(\sum_{\ell=1}^{r}\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}\lambda^{\ell}\right)(1+o(1)),$

where the error term is uniform in $r$. This completes the proof of Theorem
1.4. We now proceed to prove Corollary 1.5.

###### Proof of Corollary 1.5.

Applying a Markov bound to the $r$th moment $m_{r}(x,h)$, we get that

$I(x;k,h)\leq\frac{x}{k^{r}}m_{r}(x,h)\ll\frac{x}{k^{r}}\sum_{\ell=1}^{r}\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}\lambda^{\ell}.$

The Stirling numbers of the second kind are bounded above by
$\genfrac{\\{}{\\}}{0.0pt}{}{r}{\ell}\leq\frac{1}{2}\binom{r}{\ell}\ell^{r-\ell},$
so

$I(x;k,h)\ll\frac{x}{k^{r}}\sum_{\ell=1}^{r}\binom{r}{\ell}\ell^{r-\ell}\lambda^{\ell}\ll\frac{x}{k^{r}}(\lambda+r)^{r}.$

If $\lambda\geq 1$, then as $x\to\infty$ eventually
$\frac{k}{\lambda}\geq\frac{\lambda e}{\lambda-1}$, which in turn implies that
we can choose $r=\frac{k}{\lambda e}$ and get $(\lambda+r)^{r}\leq(\lambda
r)^{r}$. With this choice of $r$, we thus get $I(x;k,h)\ll xe^{-k/\lambda e}$,
as desired.

Meanwhile if $\lambda<1$, we can choose $r=\frac{k}{(\lambda+1)e}$ to get the
desired result. ∎

Corollary 1.6 follows via the same argument as the proof of Corollary 1.5, but
where $r$ is taken to be $(\log h)^{1-\delta}$.

##  4\. Unconditional bounds

We can also achieve weaker unconditional bounds on the moments $m_{r}(x,h)$
and the tail of the distribution via replacing the use of the Hardy–Littlewood
conjectures by an application of the Selberg sieve. More precisely, we will
make use of the following theorem, which is proven in Section 4.1.

######  Theorem 4.1.

Let $x\geq 2$, let $k=o((\log x)^{1/3})$, and let
$\mathcal{H}=\\{h_{1},\dots,h_{k}\\}$ be a set of $k$ distinct natural
numbers. For any $\varepsilon>0$,

$\displaystyle\left|\left\\{n\leq x:n+h_{i}\text{ prime for all
}i\right\\}\right|$
$\displaystyle\leq(2+\varepsilon)^{k}k!\mathfrak{S}(\mathcal{H})\frac{x}{\log^{k}x}\left(1+O\left(\frac{\log\log
3x+k^{3}+k\log\log 3|D_{\mathcal{H}}|}{\log x}\right)\right),$

where $D_{\mathcal{H}}:=\prod_{i<j}(h_{i}-h_{j})$.

Theorem 4.1 extends the work of Klimov in [6], who shows an analogous bound
for $k$ fixed as $x\to\infty$.

We now turn to the proofs of Theorem 1.8, as well as that of Corollary 1.9.

###### Proof of Theorem 1.8.

As in the proof of Theorem 1.4, we have

$m_{r}(x,h)=\sum_{\ell=1}^{r}\sigma(r,\ell)\sum_{\begin{subarray}{c}1\leq
h_{1},\cdots,h_{\ell}\leq h\\\
\text{distinct}\end{subarray}}\frac{1}{x}\sum_{n\leq
x}\mathbf{1}_{\mathcal{P}}(n+h_{1})\cdots\mathbf{1}_{\mathcal{P}}(n+h_{\ell}).$

For our choice of $h$ and $r$, the error term in Theorem 4.1 is $O(1)$.
Applying Theorem 4.1, the $r$th moment is then bounded by

$\displaystyle m_{r}(x,h)$
$\displaystyle\ll\sum_{\ell=1}^{r}\frac{\sigma(r,\ell)}{(\log
x)^{\ell}}\sum_{\begin{subarray}{c}1\leq h_{1},\cdots,h_{\ell}\leq h\\\
\text{distinct}\end{subarray}}(2+\varepsilon)^{\ell}\ell!\mathfrak{S}(\mathcal{H})$
$\displaystyle\ll\sum_{\ell=1}^{r}\frac{\sigma(r,\ell)}{(\log
x)^{\ell}}(2+\varepsilon)^{\ell}\ell!h^{\ell}e^{O(\ell\log\log\ell)},$

where the last step follows by applying Theorem 1.2. By a similar calculation
as in the proof of Theorem 1.4, this quantity is

$\ll\lambda^{r}r^{r+1}r!(2+\varepsilon)^{r}e^{O(r\log\log r)},$

which gives the result. ∎

###### Proof of Corollary 1.9.

The proof of this corollary proceeds along the same lines as the proofs of
Corollaries 1.5 and 1.6. In this case, we know unconditionally from Theorem
1.8 that

$I(x;k,h)\leq\frac{x}{k^{r}}m_{r}(x,h)\ll\frac{x}{k^{r}}\lambda^{r}r^{2r}e^{O(r\log\log
r)}.$

There exists some constant $C>0$ with

$I(x;k,h)\ll\frac{x}{k^{r}}\lambda^{r}r^{2r}e^{Cr\log\log r},$

for large enough $r$. Choosing $r=\left(\frac{k}{\lambda
e}\right)^{1/2}2^{C/2}\left(\log\frac{k}{\lambda e}\right)^{-C/2}$ gives the
desired result. ∎

###  4.1. Selberg’s Sieve: Proof of Theorem 4.1

Selberg’s sieve has previously been used to bound the frequency of prime
$k$-tuples; see for example [5], which we will refer to throughout this
section, as well as [1] and [6]. In [5], Halberstam and Richert proceed along
a very similar calculation, with the only material difference being that we
are not taking $k$ to be a constant in terms of the other parameters, and thus
we keep track of the dependence on $k$ throughout. We proceed along the lines
of Theorem 5.7 from [5]. To do so, there are several lemmas that we will want
to adapt to this setting.

We begin by defining notation. Let $\mathcal{P}$ be the set of all primes, and
for $z>0$, let $P(z):=\prod_{p\leq z}p$. Let
$\mathcal{H}=\\{h_{1},\dots,h_{k}\\}$ be a set of $k$ distinct natural
numbers, so that $D_{\mathcal{H}}\neq 0$. Define

$\mathcal{A}:=\left\\{\prod_{i=1}^{k}(n+h_{i}):n\leq x\right\\},$

and define $A_{p}$ to be the number of elements of $\mathcal{A}$ that are
divisible by a prime $p$, with
$A_{p}=\frac{\nu_{\mathcal{H}}(p)}{p}x+O(\nu_{\mathcal{H}}(p)).$ Let $A_{d}$
be the number of elements of $\mathcal{A}$ that are relatively prime to $d$,
so that $A_{d}=\frac{\nu_{\mathcal{H}}(d)}{d}x+O(\nu_{\mathcal{H}}(d)),$ where
$\nu_{\mathcal{H}}(d)=\prod_{p|d}\nu_{\mathcal{H}}(p)$. Let
$R_{d}=A_{d}-\frac{\nu_{\mathcal{H}}(d)}{d}x$, so that
$|R_{d}|\leq\nu_{\mathcal{H}}(d)$. Our goal is to estimate the quantity

(19) $S(\mathcal{A};\mathcal{P},z):=|\\{a:a\in\mathcal{A},(a,P(z))=1\\}|.$

Halberstam and Richert define three conditions on a sieve problem in order to
apply the Selberg sieve, which they denote ($R$), ($\Omega_{1}$), and
$(\Omega_{2}(\kappa,L))$. These are:

($R$) $\displaystyle|R_{d}|$ $\displaystyle\leq\nu_{\mathcal{H}}(d)\text{ if
}\mu(d)\neq 0;$ ($\Omega_{1}$) $\displaystyle 0$
$\displaystyle\leq\frac{\nu_{\mathcal{H}}(p)}{p}\leq
1-\frac{1}{\alpha_{1}}\text{ for some constant $\alpha_{1}\geq 1$};$
($\Omega_{2}(\kappa,L)$) $\displaystyle-L$ $\displaystyle\leq\sum_{w\leq
p<z}\frac{\nu_{\mathcal{H}}(p)\log
p}{p}-\kappa\log\frac{z}{w}\leq\alpha_{2}\text{ if }2\leq w\leq z.$

For the final condition, $\alpha_{2}$ and $L$ are constants, each $\geq 1$,
which are independent of $z$ and $w$. For our purposes, we will need to keep
track of the values $\alpha_{1}$, $\alpha_{2}$, and $L$, and in particular
their dependence on $k$. The parameter $\kappa$ is the dimension of the sieve,
which in this case is equal to $k$.

######  Lemma 4.2.

For a set $\mathcal{H}=\\{h_{1},\dots,h_{k}\\}$ and $E$, $\mathcal{A}$,
$A_{d}$, and $R_{d}$ defined as above, the conditions ($R$), ($\Omega_{1}$),
and ($\Omega_{2}(\kappa,L)$) are satisfied with $\alpha_{1}=k+1$,
$\alpha_{2}=O(k)$, $\kappa=k$, and $L=k\log\log 3|D_{\mathcal{H}}|$.

###### Proof.

We first see that condition ($R$) is satisfied, since
$|R_{d}|\leq\nu_{\mathcal{H}}(d)$. We also have
$\frac{\nu(p)}{p}\leq\frac{\min\\{k,p-1\\}}{p}\leq 1-\frac{1}{k+1},$ so that
($\Omega_{1}$) is satisfied with $\alpha_{1}=k+1$.

For any $z$ and any $w<z$,

$\sum_{w\leq p<z}\frac{\nu(p)\log p}{p}=k\sum_{w\leq p<z}\frac{\log
p}{p}-\sum_{w\leq p<z}\frac{k-\rho(p)}{p}\log p,$

so that

$\sum_{w\leq p<z}\frac{\nu(p)\log p}{p}\leq k\log\frac{z}{w}+O(k),$

and

$\sum_{w\leq p<z}\frac{\nu(p)\log p}{p}\geq
k\log\frac{z}{w}-O(k)-\sum_{p|E}\frac{k}{p}\log p\geq
k\log\frac{z}{w}-O(k)-O(k\log\log 3|D_{\mathcal{H}}|),$

so ($\Omega_{2}(\kappa,L)$) is satisfied with $\kappa=k$, $\alpha_{2}=O(k)$
and $L=O(k\log\log 3|D_{\mathcal{H}}|)$. ∎

For $d$ squarefree, let $g(d)=\frac{\nu(d)}{d\prod_{p|d}(1-\nu(p)/p)}$, let
$G(z)=\sum_{d<z}\mu(d)^{2}g(d),$ and more generally for any $x$ define
$G(x,z)=\sum_{\begin{subarray}{c}d<x\\\ d|P(z)\end{subarray}}\mu(d)^{2}g(d)$,
so that $G(z)=G(z,z)$. Let $W(z)=\prod_{p<z}\left(1-\frac{\nu(p)}{p}\right).$

######  Lemma 4.3 (After Lemma 5.4 from [5]).

Let $G,W$ be defined as above, and let $z>0$ be such that for sufficiently
large constants $B_{L}$ and $B_{k}$, $L\leq\frac{1}{B_{L}}\log z$ and
$k^{2}\leq\frac{1}{B_{k}}\log z$. Then

$\frac{1}{G(z)}=W(z)e^{\gamma k}\Gamma(k+1)\left(1+O\left(\frac{L+k^{3}}{\log
z}\right)\right).$

###### Proof.

This proof closely follows the proof of Lemma 5.4 from [5], so here we simply
highlight the differences, which arise only in that here we keep track of the
dependence on k in the form of the constants $\alpha_{1},\alpha_{2}$,
$\kappa$, and $L$.

In [5], Halberstam and Richert show that for $0<z\leq x$,

$\displaystyle\sum_{\begin{subarray}{c}d<x\\\ d|P(z)\end{subarray}}g(d)\log
d=$ $\displaystyle\sum_{\begin{subarray}{c}d<x\\\
d|P(z)\end{subarray}}g(d)\sum_{p<\min\\{x/d,z\\}}\frac{\nu_{\mathcal{H}}(p)}{p}\log
p$ $\displaystyle+\sum_{\begin{subarray}{c}xz-2\leq d<x\\\
d|P(z)\end{subarray}}g(d)\sum_{\begin{subarray}{c}\sqrt{x/d}\leq
p<\min\\{x/d,z\\}\\\ p\nmid
d\end{subarray}}\frac{g(p)\nu_{\mathcal{H}}(p)}{p}\log p.$

For the first inner sum, they use ($\Omega_{2}(k,L)$) to bound it above. For
the second, by applying ($\Omega_{1}$), ($\Omega_{2}(k,L)$), and the fact that
$\frac{g(p)\nu_{\mathcal{H}}(p)}{p}=g(p)-\frac{\nu_{\mathcal{H}}(p)}{p}$, we
get that

$\sum_{\begin{subarray}{c}\sqrt{x/d}\leq p<\min\\{x/d,z\\}\\\ p\nmid
d\end{subarray}}\frac{g(p)\nu_{\mathcal{H}}(p)}{p}\log
p\leq(\alpha_{1}-1)(\alpha_{2}+O(k))=O(k^{2}).$

Combining these estimates, we get

$\displaystyle\sum_{\begin{subarray}{c}d<x\\\ d|P(z)\end{subarray}}g(d)\log
d=$ $\displaystyle\sum_{\begin{subarray}{c}x/z\leq d<x\\\
d|P(z)\end{subarray}}g(d)\left(k\log\frac{x}{d}+O(L)\right)$
$\displaystyle+\sum_{\begin{subarray}{c}d<x\\\ d|P(z)\end{subarray}}g(d)(k\log
z+O(L))+O(k^{2}G(x,z))$ $\displaystyle=k\sum_{\begin{subarray}{c}d<x\\\
d|P(z)\end{subarray}}g(d)\log\frac{x}{d}-k\sum_{\begin{subarray}{c}d<x/z\\\
d|P(z)\end{subarray}}g(d)\log\frac{x/z}{d}+O((L+k^{2})G(x,z)).$

We then follow the remainder of the proof of Lemma 5.4 in [5], to get that

$G(z)=c_{3}(\log z)^{k}\left(1+O\left(\frac{L+k^{2}}{\log z}\right)\right),$

where

$c_{3}=\frac{1}{\Gamma(k+1)}\prod_{p}\left(1-\frac{\nu_{\mathcal{H}}(p)}{p}\right)^{-1}\left(1-\frac{1}{p}\right)^{k}.$

By equation (2.3.12) in [5],

(20) $\frac{1}{W(z)}\ll e^{O(k^{3})}(\log z)^{k}.$

By tracing through the constants in Lemmas 5.2 and 5.3 of [5], while using
(20), we get that

(21) $W(z)=\mathfrak{S}(\mathcal{H})\frac{e^{-\gamma k}}{(\log
z)^{k}}\left(1+O\left(\frac{L+k^{3}}{\log z}\right)\right).$

∎

We also make use of Theorem 3.1 from [5], which we cite without modification.

######  Theorem 4.4 (Theorem 3.1, Halberstam–Richert, [5]).

Using the notation of this section, with $S(\mathcal{A};\mathcal{P},z)$
defined in (19), we have

$S(\mathcal{A};\mathcal{P},z)\leq\frac{x}{G(z)}+\frac{z^{2}}{W^{3}(z)}.$

We are now ready to prove Theorem 4.1. Fix $z<x$ to be chosen later; we
estimate $S(\mathcal{A};\mathcal{P},z)$, which is an upper bound for our
desired quantity. By Theorem 4.4,

$S(\mathcal{A};\mathcal{P},z)\leq\frac{x}{G(z)}+\frac{z^{2}}{W^{3}(z)}.$

Applying Lemma 4.3 yields

$S(\mathcal{A};\mathcal{P},z)\leq xW(z)e^{\gamma
k}\Gamma(k+1)\left(1+O\left(\frac{L+k^{3}}{\log
z}\right)\right)+xW(z)\frac{z^{2}}{xW^{4}(z)}.$

By (20), we have

$\frac{z^{2}}{W^{3}(z)}=xW(z)\frac{z^{2}}{xW^{4}(z)}=xW(z)O\left(\frac{z^{2}(\log
z)^{4k}e^{O(k^{3})}}{x}\right),$

so that

$S(\mathcal{A};\mathcal{P},z)\leq xW(z)\left(e^{\gamma
k}\Gamma(k+1)\left(1+O\left(\frac{L+k^{3}}{\log
z}\right)\right)+O\left(\frac{z^{2}(\log
z)^{4k}e^{O(k^{3})}}{x}\right)\right).$

Plugging in (21), we get

$S(\mathcal{A};\mathcal{P},z)\leq\Gamma(k+1)\mathfrak{S}(\mathcal{H})\frac{x}{(\log
z)^{k}}\left(1+O\left(\frac{L+k^{3}}{\log z}\right)+O\left(\frac{z^{2}(\log
z)^{4k}e^{O(k^{3})}}{x}\right)\right).$

Set $z=x^{1/(2+\varepsilon)}$ to complete the proof, keeping in mind that
$k=o((\log x)^{1/3})$.

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