# Soft Prompt Tuning for Cross-Lingual Transfer: When Less is More

Fred Philippy$1,2$, Siwen Guo$1$, Shohreh Haddadan$1$,
Cedric Lothritz$2$, Jacques Klein$2$, Tegawendé F. Bissyandé$2$
$1$ Zortify S.A., Luxembourg
$2$ University of Luxembourg, Luxembourg
{fred<EMAIL_ADDRESS><EMAIL_ADDRESS>
{cedric.lothritz, jacques.klein<EMAIL_ADDRESS>
###### Abstract

Soft Prompt Tuning (SPT) is a parameter-efficient method for adapting pre-
trained language models (PLMs) to specific tasks by inserting learnable
embeddings, or soft prompts, at the input layer of the PLM, without modifying
its parameters. This paper investigates the potential of SPT for cross-lingual
transfer. Unlike previous studies on SPT for cross-lingual transfer that often
fine-tune both the soft prompt and the model parameters, we adhere to the
original intent of SPT by keeping the model parameters frozen and only
training the soft prompt. This does not only reduce the computational cost and
storage overhead of full-model fine-tuning, but we also demonstrate that this
very parameter efficiency intrinsic to SPT can enhance cross-lingual transfer
performance to linguistically distant languages. Moreover, we explore how
different factors related to the prompt, such as the length or its
reparameterization, affect cross-lingual transfer performance.

Soft Prompt Tuning for Cross-Lingual Transfer: When Less is More

Fred Philippy$1,2$, Siwen Guo$1$, Shohreh Haddadan$1$, Cedric Lothritz$2$,
Jacques Klein$2$, Tegawendé F. Bissyandé$2$ $1$ Zortify S.A., Luxembourg $2$
University of Luxembourg, Luxembourg {fred<EMAIL_ADDRESS><EMAIL_ADDRESS>{cedric.lothritz, jacques.klein,
<EMAIL_ADDRESS>

##  1 Introduction

Fine-tuning pre-trained language models (PLMs) on task-specific labeled data
requires large amounts of computational resources and may cause catastrophic
forgetting of the pre-trained knowledge (Goodfellow et al., 2015). In
multilingual settings, this may lead to poor cross-lingual transfer
performance (Vu et al., 2022).

To address these challenges, Lester et al. (2021) introduced Soft Prompt
Tuning (SPT), a method that inserts learnable embeddings, or soft prompts, at
the PLM’s input layer. The PLM then makes predictions using the output of its
pre-trained language modeling head. The key advantage of SPT lies in its
ability to leverage the pre-existing knowledge within PLMs while reducing the
reliance on extensive task-specific fine-tuning. SPT has been shown to achieve
remarkable results in various monolingual downstream tasks, especially in few-
shot settings.

Motivated by this success, some recent works have also explored the use of SPT
for cross-lingual transfer, where the goal is to leverage a multilingual
language model (MLLM) to transfer knowledge from a high-resource to a low-
resource language. However, these works have not fully exploited the potential
of SPT. Some have appended a newly initialized classifier to the model (Tu et
al., 2022; Park et al., 2023), hindering the suitability of SPT for few-shot
learning. Others have fine-tuned the entire model along with the prompt (Zhao
and Schütze, 2021; Huang et al., 2022), which reduces the computational
efficiency of SPT.

This is especially problematic given the growing size of state-of-the-art
language models. Therefore, we explore the impact on SPT’s cross-lingual
transfer performance when adhering to the original methodology of Lester et
al. (2021), which involves fine-tuning only the soft prompt while keeping all
model parameters frozen. Specifically, this paper contributes to the field of
cross-lingual SPT by:

  * •

Investigating the impact of model freezing on the cross-lingual transfer
performance of few-shot SPT.

  * •

Demonstrating that by freezing the model, SPT achieves enhanced cross-lingual
transfer, especially to languages linguistically distant from the source
language.

  * •

Exploring further non-linguistic factors that influence the cross-lingual
transfer performance of SPT, in particular prompt length and prompt
reparameterization.

In this study, we conduct experiments on a topic classification dataset in 52
different languages and using 4 different models in few-shot settings. We
believe that our findings can improve the existing methods that aim to enhance
cross-lingual SPT, particularly in the context of current state-of-the-art
models with billions of parameters where parameter efficiency is crucial.

##  2 Related Work

Lester et al. (2021) proposed SPT, a method to leverage a PLM’s pre-trained
language modeling head without appending a new classifier. SPT relies on a
soft prompt, which is a set of learnable embeddings that are concatenated with
the input sequence, and keeps all other model parameters frozen. Since then,
several recent works have explored the use of soft prompts for MLLMs. Zhao and
Schütze (2021) first show that SPT outperforms fine-tuning in few-shot
scenarios for cross-lingual transfer. Huang et al. (2022) introduce a method
to train a language-agnostic soft prompt. However, unlike our study, none of
these works on cross-lingual SPT employ model parameter freezing, leading to a
reduced efficiency in their methods. In contrast, Tu et al. (2022) and Park et
al. (2023) perform model freezing and, in corroboration with Zhao and Schütze
(2021), also show that SPT outperforms fine-tuning for cross-lingual transfer.
However, they append a newly initialized classification head to the model
instead of using the PLM’s pre-trained language modeling head, which diverges
from the original idea of SPT. This setup is unsuitable for few-shot learning,
requiring experiments to be conducted in full-data settings. In addition,
prior studies often focus on smaller ranges of languages, which impedes making
conclusive observations about SPT’s cross-lingual tendencies across different
languages and language families.

##  3 Experimental Setup

Besides adhering to the original setup of SPT, enabling parameter-efficient
and data-efficient training, our study also sets itself apart in its
objectives from the existing literature. Rather than simply demonstrating
superior cross-lingual transfer performance of SPT over fine-tuning, our
research aims to show that the minimal impact on the MLLM’s representation
space not only generally enhances transfer performance but is particularly
effective for linguistically distant languages.

We provide more specific details on our experimental setup in Appendix A.

###  3.1 Soft Prompt

Following Lester et al. (2021), we append a soft prompt to the input sequence
which is passed through an autoregressive language model, generating the
logits for the next token in the input sequence. Each class is linked to a
token from the model’s vocabulary, enabling us to map the token with the
highest logit to the predicted class. Such a mapping is referred to as the
verbalizer (Figure 1).

Figure 1: A simplified illustration of SPT (Lester et al., 2021).
$P_{1},\ldots,P_{n}$ denote the soft prompt tokens, with each token
corresponding to a trainable embedding. Essentially, for a model with an
embedding dimension $d$, a soft prompt of length $n$ forms a $d\times n$
matrix.

###  3.2 Implementation Details

#### Models

With the recent advancement and popularity of autoregressive language models
for various tasks, our research is conducted using two types of MLLMs based on
this architecture: XGLM (Lin et al., 2022) and BLOOM (Scao et al., 2022). For
both models we use 2 different sizes: XGLM-564M and XGLM-1.7B for XGLM, and
BLOOM-560M and BLOOM-1.1B for BLOOM.

#### Data

In our study, we use SIB-200 (Adelani et al., 2023), a topic classification
dataset containing seven distinct topics and covering a diverse range of 200
languages and dialects. We chose this dataset for its broader, more diverse
language range compared to prior studies on cross-lingual SPT, covering almost
all languages our models support, enabling more comprehensive observations.

#### Technical Details

We compare two different settings: tuning the soft prompt with model freezing
(w/ MF) and without model freezing (w/o MF). We perform few-shot fine-tuning
only using English samples. The final cross-lingual transfer performance is
then evaluated on the test sets of all languages supported by the respective
model (30 for XGLM, 38 for BLOOM), using accuracy as the metric. We repeat
each experiment 4 times with different random seeds and report the mean.

##  4 Results

|  | DATA | SYN | GEO | INV | GEN | PHON | FEA
---|---|---|---|---|---|---|---|---
BLOOM-560M | w/o MF | 0,6781 | 0,6457 | 0,2294 | 0,3779 | 0,5081 | 0,4343 | 0,4221
w/ MF | 0,6080 | 0,5742 | 0,2034 | 0,2629 | 0,3676 | 0,4482 | 0,3165
BLOOM-1.1B | w/o MF | 0,6788 | 0,6403 | 0,1693 | 0,4605 | 0,5679 | 0,5272 | -0,4685
w/ MF | 0,4856 | 0,4177 | 0,0290 | 0,2930 | 0,3711 | 0,4283 | 0,3002
XGLM-564M | w/o MF | 0,2672 | 0,6767 | 0,4694 | 0,4016 | 0,3203 | 0,4756 | 0,5949
w/ MF | 0,2453 | 0,6574 | 0,2551 | 0,3410 | 0,2201 | 0,3285 | 0,5185
XGLM-1.7B | w/o MF | 0,2636 | 0,6722 | 0,2566 | 0,3623 | 0,2924 | 0,3213 | 0,5315
w/ MF | 0,2560 | 0,6694 | 0,2949 | 0,3155 | 0,2786 | 0,2779 | 0,4922
Table 1: Pearson correlation between (8-shot) cross-lingual transfer
performance and 6 different linguistic similarity metrics, namely syntactic
(SYN), geographic (GEO), inventory (INV), genetic (GEN), phonological (PHON)
and featural (FEA) distance, as well as the language-specific pre-training
corpus size (DATA).

We provide the full results across all models and languages in Appendix D. The
results reveal that model freezing not only boosts cross-lingual transfer
performance (Figure 2) but additionally is a step towards closing the transfer
gap between linguistically distant and similar languages. This can be seen in
Table 1, which shows that the correlation strength between transfer
performance and language similarity between source and target languages,
measured using 6 different similarity metrics111See Appendix B for more
details. (Littell et al., 2017), decreases when freezing model parameters.
This suggests that the parameter efficiency of SPT mitigates the bias of
cross-lingual transfer towards linguistically similar languages. In other
words, by fine-tuning fewer parameters, cross-lingual transfer, especially to
linguistically distant languages, is enhanced. This improvement over full-
model fine-tuning may be attributed to the reduced impact on the MLLM’s
representation space during fine-tuning (Philippy et al., 2023).

Figure 2: Average cross-lingual transfer performance of SPT with and without
model freezing (MF) for different models across all languages supported by the
respective model.

Figure 3 also shows that, despite the limited number of tunable parameters
when freezing all model parameters, additional training samples further boost
cross-lingual transfer performance.

Figure 3: Average cross-lingual transfer performance of SPT with model
freezing for different number of training samples per class.

#### Parameter efficiency

Besides better cross-lingual transfer performance, model freezing during SPT
also provides parameter efficiency as fine-tuning is restricted to a number of
soft prompt tokens, resulting in only a few thousand parameters in total. This
is less than 0.01% of the parameters fine-tuned in previous studies (Zhao and
Schütze, 2021; Huang et al., 2022).

For illustration, the storage requirement for a copy of the XGLM-1.7B model is
approximately 3.2 GB, whereas a prompt needs less than 100KB. With respect to
training duration, our observations indicate that the time required for
training only the soft prompt is less than half compared to when training all
model parameters. This benefit becomes even more pronounced when considering
the increasing sizes of state-of-the-art models.

##  5 Impact of Prompt Length and Reparameterization

###  5.1 Prompt Length

Using the same configuration as described in Section 3.2, we compare the
transfer performance of prompts with different lengths under the 8-shot
setting. We consider prompt lengths in $\\{1,2,5,10,20,30\\}$ and report the
results for all 4 models. Figure 4 shows that if a soft prompt is too long,
cross-lingual transfer performance degrades.

Figure 4: Average cross-lingual transfer performance, measured as accuracy,
across different prompt lengths for different models.

###  5.2 Reparameterization

Direct fine-tuning of soft prompt embeddings may lead to unstable training and
potentially reduces performance. To address this issue, previous works have
proposed reparameterizing prompt embeddings using different architectures,
such as an LSTM (Liu et al., 2021) or MLP (Li and Liang, 2021), which are
fine-tuned along with the prompt embeddings. Liu et al. (2022) argue that
reparameterization can also have negative effects depending on the task or
dataset.

Motivated by this observation, we investigate the effect of reparameterization
on cross-lingual transfer performance. We adopt the approach proposed by
Razdaibiedina et al. (2023), which uses an MLP with a residual connection and
a "bottleneck" layer for reparameterization. We provide further details on
this method in Appendix C.

Our analysis reveals that BLOOM is significantly more affected by
reparameterization than XGLM (Figure 5 in Appendix C). For both models, the
impact of reparameterization differs across languages — being detrimental for
some and advantageous for others. Notably, for BLOOM, Atlantic-Congo languages
such as Yoruba, Twi, Kinyarwanda, Akan, Fon and Swahili experience the most
significant performance decline due to reparameterization, with drops between
24% to 31%. Conversely, Indo-Aryan languages like Urdu, Hindi, Bengali, and
Nepali, along with Dravidian languages like Malayalam and Tamil see the most
significant improvements, with gains of up to 29%. For XGLM, the outcomes are
more balanced. Nonetheless, we observe that the languages that benefit most
from reparameterization either use Latin script, such as Haitian, German, and
Turkish, or are Dravidian languages such as Telugu and Tamil.

Hence, we recommend that in cross-lingual settings, the decision to use or
abstain from reparameterization should not be made uniformly. Instead, it
should be tailored based on the specific target languages or language families
in consideration.

##  6 Discussion

Previous works on SPT for cross-lingual transfer in few-shot settings suffers
from two major drawbacks: 1) fine-tuning all model parameters along with the
prompt reduces the computational efficiency of SPT; 2) a bias towards target
languages that are linguistically closer to the source language. Our study
tackles these issues by showing that by simply keeping model parameters frozen
during SPT, we can make progress in addressing both these challenges.

Through our experiments, which covered a wider and more diverse range of
languages than prior work on cross-lingual SPT, we observed intriguing effects
of non-linguistic variables (such as model freezing, prompt length, and
reparameterization) on the transfer performance for individual languages.
Additionally, our results reveal language-specific differences that invite
further inquiry into the possibility of tailoring prompts to the target
language (e.g., applying prompt reparameterization or not depending on the
linguistic distance between the target language family and the source
language) rather than using a single prompt for universal transfer across
languages. We believe that our findings will benefit future work on cross-
lingual SPT and potentially improve the existing techniques (Huang et al.,
2022), becoming more valuable as we adopt larger state-of-the-art models with
billion- and trillion-scale parameters (Lester et al., 2021).

##  7 Conclusion

The objective of our study was to examine the impact of model freezing on the
cross-lingual transfer performance of SPT. Our results demonstrate that SPT, a
method that adjusts less than 0.01% of parameters compared to full-model fine-
tuning, achieves comparable or superior performance for most target languages,
particularly for those that are linguistically more distant. Furthermore, we
found that shorter prompts enhance SPT’s cross-lingual transfer performance,
and that some target language families benefit from reparameterization while
others are adversely affected by it.

## Limitations

Our approach enhances transfer performance for several languages, especially
those that are linguistically more distant. However, we also notice that it
lowers the performance for some languages that are linguistically more
similar. This limitation motivates us to pursue future research that aims to
achieve balanced performance across languages

Another limitation of our approach is the instability of few-shot fine-tuning,
which compromises the robustness of our method’s evaluation. To mitigate this
issue, we ran all experiments four times with different random seeds and
reported the mean and variance of the results. However, we acknowledge that
more research is needed to address the challenges of few-shot fine-tuning.

## Ethics Statement

In this paper, we aim to improve the performance of MLLMs on low-resource
languages, which often suffer from a lack of data and attention in NLP
research. We believe that this is an important and ethical goal, as it enables
NLP advances to benefit a broader range of language communities.

In addition, this paper aims to promote parameter efficiency, which is a
crucial factor for reducing the computational and environmental costs of
training and fine-tuning state-of-the-art language models. We believe that
this aspect will enhance the accessibility and affordability of these models
for researchers and practitioners who face computational constraints.

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##  Appendix A Reproducibility

We provide the code used for our experiments here:
https://github.com/fredxlpy/cross_lingual_prompt_
tuning.

###  A.1 Dataset

Our experiments are based on the SIB-200 dataset (Adelani et al., 2023). The
dataset is based on the FLORES-200 benchmark (NLLB Team et al., 2022), and
consists of 701 training, 99 validation and 204 test samples in each of the
203 languages. The task is to classify each sample into one of the 7 potential
categories: science/technology, travel, politics, sports, health,
entertainment, and geography.

###  A.2 Models

We provide additional information about the models used in our study in Table
2.

Model |  Layers |  Para- meters |  Hidden
size |  Vocab
size
---|---|---|---|---
BLOOM-560M |  24 |  560M |  1.024 |  250.880
BLOOM-1.1B |  |  1.1B |  1.536 |
XGLM-564M |  |  564M |  1.024 |  256.008
XGLM-1.7B |  |  1.7B |  2.048 |
Table 2: Technical details of the models used in our study.

###  A.3 Technical Details

|  |  Batch
size |  Learning
rate |  Prompt
length
---|---|---|---|---
w/ MF |  XGLM
564M |  8 |  0.1 |  10
|  XGLM
1.7B |  |  |
|  BLOOM
560M |  |  |  5
|  BLOOM
1.1B |  |  |  10
w/o MF |  XGLM
564M |  8 |  5e-6 |  10
|  XGLM
1.7B |  |  |
|  BLOOM
560M |  |  1e-6 |  5
|  BLOOM
1.1B |  |  |  10
Table 3: Hyperparameters used in all of our experiments.

We conducted all of our experiments using the Transformers library (Wolf et
al., 2020). In a $k$-shot setting, we fine-tune on $k$ samples per class from
the English train set and use $\frac{k}{4}$ samples per class for validation.
We train all models and prompts for 20 epochs and select the best checkpoint
on the development set. The different hyperparameters used in our experiments
are provided in Table 3.

###  A.4 Soft Prompt

We follow the approach of Lester et al. (2021) and freeze all model parameters
and only fine-tune the soft prompt.

In order to map the tokens predicted by the model to the respective class, we
define a verbalizer $F:T\rightarrow C$, where $T=\\{t_{1},\ldots,t_{K}\\}$ is
a subset of the model’s vocabulary $V$ and $C=\\{1,\ldots,K\\}$ are the
respective classes.

We append a prompt $p=\\{p_{1},\ldots,p_{m}\\}$ to an input sequence
$x=\\{x_{1},\ldots,x_{n}\\}$ and pass
$\\{x_{1},\ldots,x_{n},p_{1},\ldots,p_{m}\\}$ through the autoregressive
language model which outputs the logits for the next token in the input
sequence $\\{l_{1},\ldots,l_{|V|}\\}$.

The predicted token is then $F\left(argmax_{i\in T}l_{i}\right)$

###  A.5 Computing Resources

We conduct all our experiments on 4 A100 40GB GPUs, using 4 different random
seeds, in parallel. All experiments could be run in a few hours.

##  Appendix B Language Distance Metrics

We consider six types of lang2vec222https://github.com/antonisa/lang2vec
(Littell et al., 2017) distances:

  * •

Syntactic Distance (SYN) captures the similarity of syntactic structures
across languages. It is computed as the cosine distance between syntax feature
vectors, which are derived from the World Atlas of Language
Structures333https://wals.info (WALS) (Dryer and Haspelmath, 2013), Syntactic
Structures of World Languages444http://sswl.railsplayground.net/ (SSWL)
(Collins and Kayne, 2011) and Ethnologue555https://www.ethnologue.com/ (Lewis
et al., 2015).

  * •

Geographic Distance (GEO) reflects the spatial proximity of languages. It is
calculated as the shortest distance between two languages on the surface of
the earth’s sphere (i.e., orthodromic distance).

  * •

Inventory Distance (INV) measures the difference in sound inventories across
languages. It is computed as the cosine distance between inventory feature
vectors, which are obtained from the PHOIBLE666https://phoible.org/ database
(Moran et al., 2019).

  * •

Genetic Distance (GEN) indicates the historical relatedness of languages. It
is based on the Glottolog777https://glottolog.org (Hammarström et al., 2015)
tree of language families and is obtained by computing the distance between
two languages in the tree.

  * •

Phonological Distance (PHON) captures the similarity of sound patterns across
languages. It is computed as the cosine distance between phonological feature
vectors, which are sourced from WALS and Ethnologue.

  * •

Featural Distance (FEA) is the cosine distance between feature vectors from a
combination of the 5 above-listed linguistic features.

The values for each distance type range from 0 to 1, where 0 indicates the
minimum distance and 1 indicates the maximum distance.

##  Appendix C Prompt Reparameterization

We follow the residual reparameterization method of Razdaibiedina et al.
(2023) to examine the impact of soft prompt reparameterization. This method
employs a multi-layer perceptron (MLP) architecture for the reparameterization
network, which consists of a down-projection layer and an up-projection layer
with parameter $W_{down}\in\mathbb{R}^{d\times m}$ and
$W_{up}\in\mathbb{R}^{m\times d}$ respectively, where $d$ denotes the model
embedding size and $m$ denotes the hidden representation dimension between
both layers (bottleneck size). A ReLU layer is applied to the hidden
representation, and a normalization layer is applied to the output of the up-
projection layer before summing it with the initial input embedding via a
residual connection. We fine-tune the soft prompt and its reparameterization
network with a bottleneck size of 500 for BLOOM-560M and 200 for XGLM-564M and
report the impact of reparameterization across all target languages in Figure
5. Except for the reparameterization, we adopt the same implementation
settings as described in Section 3.

Figure 5: Impact of reparameterization (expressed in %) on the cross-lingual
transfer performance of BLOOM-560M and XGLM-564M for different target
languages.

##  Appendix D Full Results

The full results discussed in Section 4 are provided in Table 4.

| BLOOM-560M | BLOOM-1.1B | XGLM-564M | XGLM-1.7B
---|---|---|---|---
Language | w/o MF | w/ MF | w/o MF | w/ MF | w/o MF | w/ MF | w/o MF | w/ MF
Akan | $\text{22,18}_{\text{9,67}}$ | $\text{34,80}_{\text{6,33}}$ | $\text{19,36}_{\text{6,52}}$ | $\text{35,05}_{\text{0,85}}$ | - | - | - | -
Arabic | $\text{55,51}_{\text{3,56}}$ | $\text{70,22}_{\text{1,62}}$ | $\text{42,03}_{\text{11,9}}$ | $\text{63,48}_{\text{3,50}}$ | $\text{57,60}_{\text{3,34}}$ | $\text{71,69}_{\text{1,89}}$ | $\text{74,75}_{\text{1,67}}$ | $\text{78,68}_{\text{5,49}}$
Assamese | $\text{27,45}_{\text{7,99}}$ | $\text{37,01}_{\text{4,85}}$ | $\text{29,41}_{\text{9,66}}$ | $\text{53,06}_{\text{4,92}}$ | - | - | - | -
Bambara | $\text{16,67}_{\text{3,63}}$ | $\text{26,96}_{\text{8,74}}$ | $\text{17,03}_{\text{5,94}}$ | $\text{29,17}_{\text{3,68}}$ | - | - | - | -
Basque | $\text{43,50}_{\text{12,5}}$ | $\text{61,89}_{\text{1,90}}$ | $\text{38,73}_{\text{7,69}}$ | $\text{63,97}_{\text{13,0}}$ | $\text{67,40}_{\text{1,30}}$ | $\text{71,32}_{\text{2,70}}$ | $\text{71,08}_{\text{3,07}}$ | $\text{72,43}_{\text{6,64}}$
Bengali | $\text{56,62}_{\text{4,48}}$ | $\text{60,78}_{\text{2,41}}$ | $\text{46,69}_{\text{12,2}}$ | $\text{71,81}_{\text{2,90}}$ | $\text{68,14}_{\text{3,51}}$ | $\text{71,45}_{\text{4,24}}$ | $\text{71,57}_{\text{3,05}}$ | $\text{76,23}_{\text{5,22}}$
Bulgarian | - | - | - | - | $\text{72,79}_{\text{5,13}}$ | $\text{77,33}_{\text{2,45}}$ | $\text{78,92}_{\text{2,40}}$ | $\text{81,37}_{\text{4,33}}$
Burmese | - | - | - | - | $\text{63,60}_{\text{6,06}}$ | $\text{71,20}_{\text{3,38}}$ | $\text{72,67}_{\text{3,03}}$ | $\text{73,41}_{\text{7,79}}$
Catalan | $\text{63,48}_{\text{13,1}}$ | $\text{72,30}_{\text{2,28}}$ | $\text{48,77}_{\text{7,93}}$ | $\text{73,77}_{\text{4,03}}$ | $\text{68,50}_{\text{7,52}}$ | $\text{76,35}_{\text{3,03}}$ | $\text{77,33}_{\text{4,01}}$ | $\text{79,04}_{\text{4,13}}$
Chi Shona | $\text{19,98}_{\text{4,79}}$ | $\text{24,88}_{\text{2,67}}$ | $\text{17,89}_{\text{5,69}}$ | $\text{31,00}_{\text{3,95}}$ | - | - | - | -
Chi Tumbuka | $\text{20,34}_{\text{4,54}}$ | $\text{27,70}_{\text{2,55}}$ | $\text{18,14}_{\text{4,95}}$ | $\text{33,70}_{\text{4,62}}$ | - | - | - | -
Chinese | $\text{60,54}_{\text{11,1}}$ | $\text{73,65}_{\text{6,47}}$ | $\text{47,30}_{\text{13,9}}$ | $\text{72,43}_{\text{3,36}}$ | $\text{59,93}_{\text{8,33}}$ | $\text{79,04}_{\text{1,85}}$ | $\text{77,94}_{\text{5,08}}$ | $\text{81,74}_{\text{4,28}}$
English | $\text{75,00}_{\text{5,87}}$ | $\text{74,63}_{\text{2,09}}$ | $\text{69,36}_{\text{2,67}}$ | $\text{75,12}_{\text{2,90}}$ | $\text{78,68}_{\text{1,67}}$ | $\text{79,90}_{\text{2,62}}$ | $\text{80,88}_{\text{2,94}}$ | $\text{82,84}_{\text{5,41}}$
Estonian | - | - | - | - | $\text{72,30}_{\text{3,24}}$ | $\text{75,86}_{\text{3,13}}$ | $\text{76,35}_{\text{1,76}}$ | $\text{81,13}_{\text{5,78}}$
Finnish | - | - | - | - | $\text{76,72}_{\text{1,81}}$ | $\text{79,90}_{\text{1,44}}$ | $\text{79,78}_{\text{1,76}}$ | $\text{82,35}_{\text{5,92}}$
Fon | $\text{19,36}_{\text{10,0}}$ | $\text{25,49}_{\text{7,88}}$ | $\text{13,97}_{\text{3,98}}$ | $\text{26,84}_{\text{5,51}}$ | - | - | - | -
French | $\text{69,61}_{\text{6,52}}$ | $\text{73,16}_{\text{1,89}}$ | $\text{57,23}_{\text{6,29}}$ | $\text{72,92}_{\text{5,51}}$ | $\text{71,94}_{\text{4,26}}$ | $\text{79,29}_{\text{2,98}}$ | $\text{79,04}_{\text{5,48}}$ | $\text{79,90}_{\text{2,80}}$
German | - | - | - | - | $\text{71,57}_{\text{7,19}}$ | $\text{76,23}_{\text{4,67}}$ | $\text{81,62}_{\text{5,04}}$ | $\text{81,62}_{\text{5,79}}$
Greek (modern) | - | - | - | - | $\text{73,90}_{\text{3,47}}$ | $\text{78,19}_{\text{2,93}}$ | $\text{80,27}_{\text{2,70}}$ | $\text{82,97}_{\text{5,11}}$
Gujarati | $\text{41,79}_{\text{7,92}}$ | $\text{37,01}_{\text{9,35}}$ | $\text{27,08}_{\text{7,85}}$ | $\text{54,29}_{\text{10,3}}$ | - | - | - | -
Haitian | - | - | - | - | $\text{65,44}_{\text{1,30}}$ | $\text{68,87}_{\text{2,55}}$ | $\text{74,39}_{\text{1,72}}$ | $\text{74,75}_{\text{6,70}}$
Hindi | $\text{42,52}_{\text{4,28}}$ | $\text{45,59}_{\text{4,47}}$ | $\text{50,12}_{\text{10,0}}$ | $\text{64,95}_{\text{2,85}}$ | $\text{74,14}_{\text{3,28}}$ | $\text{75,37}_{\text{2,41}}$ | $\text{75,74}_{\text{2,95}}$ | $\text{78,19}_{\text{4,88}}$
Igbo | $\text{18,50}_{\text{1,57}}$ | $\text{23,77}_{\text{6,42}}$ | $\text{15,20}_{\text{4,85}}$ | $\text{27,70}_{\text{4,57}}$ | - | - | - | -
Indonesian | $\text{49,26}_{\text{2,55}}$ | $\text{66,91}_{\text{1,86}}$ | $\text{49,14}_{\text{11,9}}$ | $\text{68,75}_{\text{3,38}}$ | $\text{73,90}_{\text{1,29}}$ | $\text{77,57}_{\text{2,45}}$ | $\text{77,21}_{\text{2,48}}$ | $\text{79,90}_{\text{5,34}}$
Isi Zulu | $\text{19,24}_{\text{6,01}}$ | $\text{21,69}_{\text{2,72}}$ | $\text{15,69}_{\text{5,98}}$ | $\text{29,66}_{\text{2,48}}$ | - | - | - | -
Italian | - | - | - | - | $\text{73,41}_{\text{4,82}}$ | $\text{74,75}_{\text{1,52}}$ | $\text{78,43}_{\text{4,95}}$ | $\text{80,02}_{\text{5,52}}$
Japanese | - | - | - | - | $\text{54,29}_{\text{5,98}}$ | $\text{76,84}_{\text{3,89}}$ | $\text{80,64}_{\text{1,47}}$ | $\text{77,94}_{\text{4,65}}$
Kannada | $\text{22,30}_{\text{8,24}}$ | $\text{25,00}_{\text{8,46}}$ | $\text{22,92}_{\text{3,85}}$ | $\text{55,76}_{\text{7,93}}$ | - | - | - | -
Kikuyu | $\text{28,19}_{\text{8,36}}$ | $\text{35,05}_{\text{2,42}}$ | $\text{19,49}_{\text{4,44}}$ | $\text{33,70}_{\text{3,81}}$ | - | - | - | -
Kinyarwanda | $\text{19,00}_{\text{3,21}}$ | $\text{25,74}_{\text{6,26}}$ | $\text{15,69}_{\text{3,80}}$ | $\text{30,39}_{\text{4,33}}$ | - | - | - | -
Korean | - | - | - | - | $\text{73,77}_{\text{1,67}}$ | $\text{74,26}_{\text{2,28}}$ | $\text{74,75}_{\text{4,46}}$ | $\text{77,45}_{\text{5,41}}$
Lingala | $\text{23,90}_{\text{3,85}}$ | $\text{28,19}_{\text{4,74}}$ | $\text{21,69}_{\text{8,43}}$ | $\text{36,15}_{\text{3,29}}$ | - | - | - | -
Malayalam | $\text{23,53}_{\text{11,1}}$ | $\text{21,94}_{\text{7,56}}$ | $\text{30,39}_{\text{9,95}}$ | $\text{59,93}_{\text{4,17}}$ | - | - | - | -
Marathi | $\text{34,68}_{\text{11,1}}$ | $\text{28,31}_{\text{5,83}}$ | $\text{29,78}_{\text{6,21}}$ | $\text{60,05}_{\text{4,41}}$ | - | - | - | -
Nepali | $\text{30,15}_{\text{6,99}}$ | $\text{42,03}_{\text{6,95}}$ | $\text{36,76}_{\text{13,3}}$ | $\text{67,03}_{\text{6,25}}$ | - | - | - | -
Northern Sotho | $\text{20,59}_{\text{6,62}}$ | $\text{28,80}_{\text{0,47}}$ | $\text{18,38}_{\text{4,09}}$ | $\text{33,82}_{\text{2,40}}$ | - | - | - | -
Odia | $\text{34,80}_{\text{7,64}}$ | $\text{31,37}_{\text{6,62}}$ | $\text{25,00}_{\text{5,25}}$ | $\text{47,06}_{\text{9,22}}$ | - | - | - | -
Portuguese | $\text{66,67}_{\text{5,02}}$ | $\text{75,37}_{\text{3,19}}$ | $\text{53,19}_{\text{5,69}}$ | $\text{73,77}_{\text{2,17}}$ | $\text{74,26}_{\text{1,90}}$ | $\text{79,53}_{\text{1,09}}$ | $\text{80,15}_{\text{1,98}}$ | $\text{82,48}_{\text{3,95}}$
Quechua | - | - | - | - | $\text{35,66}_{\text{8,69}}$ | $\text{39,71}_{\text{2,23}}$ | $\text{49,88}_{\text{4,84}}$ | $\text{51,59}_{\text{6,37}}$
Russian | - | - | - | - | $\text{76,96}_{\text{3,23}}$ | $\text{77,21}_{\text{1,98}}$ | $\text{78,19}_{\text{3,43}}$ | $\text{80,27}_{\text{4,30}}$
Spanish | $\text{63,36}_{\text{8,94}}$ | $\text{72,67}_{\text{0,47}}$ | $\text{46,69}_{\text{9,79}}$ | $\text{73,65}_{\text{5,11}}$ | $\text{71,45}_{\text{0,74}}$ | $\text{76,47}_{\text{2,30}}$ | $\text{77,33}_{\text{3,63}}$ | $\text{79,78}_{\text{4,84}}$
Swahili | $\text{35,05}_{\text{7,95}}$ | $\text{49,88}_{\text{6,02}}$ | $\text{25,12}_{\text{6,22}}$ | $\text{49,75}_{\text{7,40}}$ | $\text{61,40}_{\text{8,61}}$ | $\text{69,00}_{\text{2,84}}$ | $\text{73,77}_{\text{2,45}}$ | $\text{72,79}_{\text{7,91}}$
Tamil | $\text{44,85}_{\text{9,58}}$ | $\text{50,74}_{\text{4,09}}$ | $\text{34,44}_{\text{13,1}}$ | $\text{67,40}_{\text{4,71}}$ | $\text{68,75}_{\text{5,68}}$ | $\text{70,59}_{\text{2,12}}$ | $\text{73,90}_{\text{1,01}}$ | $\text{75,86}_{\text{7,91}}$
Telugu | $\text{24,51}_{\text{3,94}}$ | $\text{31,00}_{\text{6,71}}$ | $\text{26,96}_{\text{1,20}}$ | $\text{66,05}_{\text{7,13}}$ | $\text{62,75}_{\text{3,33}}$ | $\text{68,26}_{\text{5,15}}$ | $\text{74,14}_{\text{3,76}}$ | $\text{76,23}_{\text{6,46}}$
Thai | - | - | - | - | $\text{67,77}_{\text{6,42}}$ | $\text{76,35}_{\text{1,16}}$ | $\text{79,53}_{\text{1,72}}$ | $\text{77,33}_{\text{5,02}}$
Turkish | - | - | - | - | $\text{73,16}_{\text{2,84}}$ | $\text{76,96}_{\text{3,18}}$ | $\text{74,63}_{\text{4,30}}$ | $\text{79,17}_{\text{5,89}}$
Twi | $\text{23,41}_{\text{9,5}}$ | $\text{35,29}_{\text{6,64}}$ | $\text{18,75}_{\text{6,83}}$ | $\text{36,52}_{\text{3,32}}$ | - | - | - | -
Urdu | $\text{42,28}_{\text{6,67}}$ | $\text{44,61}_{\text{8,95}}$ | $\text{31,74}_{\text{8,12}}$ | $\text{48,41}_{\text{9,35}}$ | $\text{54,90}_{\text{8,37}}$ | $\text{70,10}_{\text{2,86}}$ | $\text{70,10}_{\text{3,12}}$ | $\text{75,25}_{\text{5,69}}$
Vietnamese | $\text{46,08}_{\text{19,4}}$ | $\text{68,14}_{\text{7,21}}$ | $\text{43,87}_{\text{7,49}}$ | $\text{64,58}_{\text{3,76}}$ | $\text{70,71}_{\text{3,06}}$ | $\text{76,96}_{\text{3,18}}$ | $\text{78,31}_{\text{3,63}}$ | $\text{79,90}_{\text{7,30}}$
Wolof | $\text{25,49}_{\text{7,88}}$ | $\text{34,93}_{\text{4,17}}$ | $\text{21,81}_{\text{9,77}}$ | $\text{41,42}_{\text{4,64}}$ | - | - | - | -
Xhosa | $\text{21,94}_{\text{7,35}}$ | $\text{28,55}_{\text{1,23}}$ | $\text{15,32}_{\text{5,51}}$ | $\text{32,23}_{\text{6,14}}$ | - | - | - | -
Yoruba | $\text{13,36}_{\text{1,62}}$ | $\text{21,94}_{\text{9,49}}$ | $\text{16,30}_{\text{2,28}}$ | $\text{33,21}_{\text{4,76}}$ | - | - | - | -
Table 4: Cross-lingual transfer results, reported as accuracy, along with
standard deviation across 4 runs, after 8-shot soft prompt tuning (SPT) in
English, with and without model freezing (MF).


# On the $\psi$-mixing coefficients of Rényi-type maps

Gabriela Ileana Sebe<EMAIL_ADDRESS>Gheorghe Mihoc-Caius Iacob Institute of
Mathematical Statistics and Applied Mathematics of the Romanian Academy, Calea
13 Sept.13, 050711 Bucharest, Romania  Dan Lascu<EMAIL_ADDRESS>Mircea cel
Batran Naval Academy, 1 Fulgerului, 900218 Constanta, Romania

Abstract

Via dependence with complete connections we investigate the $\psi$-mixing
coefficients of the sequence $(a_{n})_{n\in\mathbb{N}}$ of incomplete
quotients and also of the doubly infinite sequence
$(\overline{a}_{l})_{l\in\mathbb{Z}}$ of extended incomplete quotients of the
Rényi-type continued fraction expansions. A Lévy-type approach allows us to
obtain good upper bounds for these coefficients.
Keywords Rényi-type continued fractions $\cdot$ incomplete quotients $\cdot$
natural extension $\cdot$ $\psi$-mixing coefficients.

Mathematics Subject Classification Primary 11J70 $\cdot$ 11K50; Secondary
60J10

##  1 Introduction

The regular continued fraction (RCF) expansions of real numbers have long been
known as interesting and fruitful because their digits are closely connected
to a dynamical system with nice mixing properties and an explicit invariant
measure.

For many dynamical systems, it is possible to prove the existence of an
invariant measure, however, there are few systems for which this measure is
explicitly known.

We briefly recall some facts about RCFs. Every irrational number $x\in
I:=[0,1]$ can be uniquely expressed as an infinite continued fraction of the
form

$x=\displaystyle\frac{1}{d_{1}(x)+\displaystyle\frac{1}{d_{2}(x)+\displaystyle\frac{1}{d_{3}(x)+\ddots}}}=:[d_{1}(x),d_{2}(x),d_{3}(x),\ldots],$
(1.1)

where the sequence $(d_{n}(x))_{n\geq 1}$ consists of natural integers which
we refer to as RCF digits or incomplete quotients of $x$. Using the Gauss map
$G(x):=1/x\,(\mathrm{mod}\,1)$, $x\in I$, the digits are recursively given by
$d_{1}(x)=\left\lfloor\frac{1}{x}\right\rfloor$, where
$\left\lfloor\cdot\right\rfloor$ denotes the floor function, and
$d_{n+1}(x)=d_{1}\left(G^{n}(x)\right),n\geq 1$. The Gauss map is invariant
under the Gauss measure, $\gamma([0,x])=\log(1+x)/\log 2$. Equipped with the
Gauss measure, we may think of $(I,\mathcal{B}_{I})$ as a probability space
and the sequence of incomplete quotients $d_{i}:I\to\mathbb{N}$ as a sequence
of random variables.

In the case of the RCF, the random variables $(d_{n})$ form a stationary
sequence due to the invariance of the Gauss measure with respect to $G$. The
$d_{n}$s are known to be $\psi$-mixing with respect to the Gauss measure
$\gamma$. This follows from independent work of Kuzmin and Lévy in the late
1920’s, see for example [3] for the proof and a discussion of its history.

The results of Kuzmin and Lévy were then followed by various improvements by
several authors (see [5]).

In 2000, Iosifescu [4] gave a more general result, proving that the sequence
of incomplete quotients of the RCF expansion is $\psi$-mixing under different
probability measures.

In general, just upper bounds for $\psi$-mixing or other dependence
coefficients are derived.

The problem of finding the exact values of $\psi$-mixing coefficients under
Gauss’ measure or upper bounds of them under a particular family of
conditional probability measures of the sequence of RCF digits was solved in
[4].

This paper continues a series of papers [9, 10, 11], dedicated to Rényi-type
continued fraction expansions. These continued fractions are a particular case
of $u$-backward continued fractions studied by Gröchenig and Haas [1].
Starting from the Rényi map $R$ [12], Gröchenig and Haas define a one-
parameter family of interval maps of the form
$T_{u}(x):=\frac{1}{u(1-x)}-\lfloor\frac{1}{u(1-x)}\rfloor$, where $u>0$ and
$x\in[0,1)$. As the parameter $u$ varies in $(0,4)$ there is a viable theory
of $u$-backward continued fractions, which fails when $u\geq 4$. The main
purpose of Gröchenig and Haas was to find an explicit form for the absolutely
continuous invariant measure for $T_{u}$ similar to that of the Gauss measure
${\mathrm{d}x}/{(x+1)}$ for $G$ and Rényi’s measure ${\mathrm{d}x}/{x}$ for
$R:=T_{1}$. They identified in [1] the invariant probability measure for
$T_{u}$ corresponding to the values $u=1/N$, for positive integers $N\geq 2$,
namely ${\mathrm{d}x}/{(x+N-1)}$. So the map $R_{N}:=T_{1/N}$, $N\geq 2$, will
be called a Rényi-type transformation. In [9] we started an approach to the
metrical theory of Rényi-type continued fraction expansions via dependence
with complete connections [6]. Using the natural extensions for Rényi-type
transformation, we obtained an infinite-order-chain representation
$(\overline{a}_{l})_{l\in\mathbb{Z}}$ of the sequence
$({a}_{n})_{n\in\mathbb{N}_{+}}$ of incomplete quotients of these expansions.

Our goal in this paper is to investigate the $\psi$-mixing coefficients
$\left(\psi_{\mu}(n)\right)_{n\geq 1}$, where $\mu\in\mathrm{pr}(I)$, of the
incomplete quotients $({a}_{n})_{n\in\mathbb{N}_{+}}$ under the invariant
probability measure $\rho_{N}$ of $R_{N}$ and under a one-parameter family
$\\{\rho^{t}_{N},t\in I\\}$ of conditional probability measures on
$(I,\mathcal{B}_{I})$ inspired by Doeblin [2]. Here $\mathcal{B}_{I}$ denotes
the $\sigma$-algebra of all Borel subsets of $I$. Since the computation of
these coefficients becomes forbidding for $n\geq 3$, we use a Lévy-type
approach developed in Section 3 (Theorem 3.1) to derive good upper bounds for
them whatever $n\geq 3$. In section 4 we prove that the sequence
$({a}_{n})_{n\in\mathbb{N}_{+}}$ is $\psi$-mixing under $\rho_{N}$ and
$\rho^{t}_{N}$, $t\in I$. Also the doubly infinite sequence
$(\overline{a}_{l})_{l\in\mathbb{Z}}$ of extended incomplete quotients is
$\psi$-mixing under the extended invariant measure $\overline{\rho}_{N}$ and
its $\psi$-mixing coefficients are equal to the corresponding $\psi$-mixing
coefficients under $\rho_{N}$ of $({a}_{n})_{n\in\mathbb{N}_{+}}$.

##  2 Prerequisites

In this section we briefly present known results about Rényi-type continued
fractions (see, e.g., [9]).

###  2.1 Rényi-type continued fraction expansions

Fix an integer $N\geq 2$. Let the Rényi-type continued fraction transformation
$R_{N}:I\rightarrow I$ be given by

$R_{N}(x):=\frac{N}{1-x}-\left\lfloor\frac{N}{1-x}\right\rfloor,x\neq 1;\quad
R_{N}(1):=0.$ (2.1)

For any irrational $x\in I$, $R_{N}$ generates a new continued fraction
expansion of $x$ of the form

$x=1-\displaystyle\frac{N}{1+a_{1}-\displaystyle\frac{N}{1+a_{2}-\displaystyle\frac{N}{1+a_{3}-\ddots}}}=:[a_{1},a_{2},a_{3},\ldots]_{R}.$
(2.2)

Here, $a_{n}$’s are non-negative integers greater than or equal to $N$ defined
by $a_{1}:=a_{1}(x)=\left\lfloor\frac{N}{1-x}\right\rfloor,x\neq
1;a_{1}(1)=\infty$ and $a_{n}:=a_{n}(x)=a_{1}\left(R^{n-1}_{N}(x)\right),n\geq
2,$ with $R_{N}^{0}(x)=x$.

The Rényi-type continued fraction in (2.2) can be viewed as a measure
preserving dynamical system $\left(I,{\mathcal{B}}_{I},R_{N},\rho_{N}\right)$,
and for any integer $N\geq 2$

$\rho_{N}(A):=\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}\int_{A}\frac{\mathrm{d}x}{x+N-1},\quad
A\in{\mathcal{B}}_{I}$ (2.3)

is the absolutely continuous invariant probability measure under $R_{N}$ [1].

###  2.2 Natural extension and extended random variables

Now we recall the natural extension $\overline{R}_{N}$ of $R_{N}$ and its
extended random variables introduced in ([9]). Let
$\left(I,{\mathcal{B}}_{I},R_{N}\right)$ be as in Section 2.1 and define
$(u_{N,i})_{i\geq N}$ by

$u_{N,i}:I\rightarrow I;\quad u_{N,i}(x):=1-\frac{N}{x+i},\quad x\in I.$ (2.4)

For each $i\geq N$, $\left(R_{N}\circ u_{N,i}\right)(x)=x$ for any $x\in I$
and if $a_{1}(x)=i$, then $\left(u_{N,i}\circ R_{N}\right)(x)=x$, $x\in I$.
The natural extension
$\left(I^{2},{\mathcal{B}}^{2}_{I},\overline{R}_{N}\right)$ of
$\left(I,{\mathcal{B}}_{I},R_{N}\right)$ is the transformation
$\overline{R}_{N}$ of the square space
$\left(I^{2},{\mathcal{B}}^{2}_{I}\right):=\left(I,{\mathcal{B}}_{I}\right)\times\left(I,{\mathcal{B}}_{I}\right)$
defined as follows:

$\overline{R}_{N}:I^{2}\rightarrow
I^{2};\quad\overline{R}_{N}(x,y):=\left(R_{N}(x),\,u_{N,a_{1}(x)}(y)\right),\quad(x,y)\in
I^{2}.$ (2.5)

Since $\overline{R}_{N}$ is bijective on $I^{2}$ with the inverse

$(\overline{R}_{N})^{-1}(x,y)=(u_{N,a_{1}(y)}(x),\,R_{N}(y)),\quad(x,y)\in
I^{2}.$ (2.6)

we get the following iterations for each $n\geq 2$:

$\displaystyle\left(\overline{R}_{N}\right)^{n}(x,y)=\left(\,R^{n}_{N}(x),\,[a_{n}(x),a_{n-1}(x),\ldots,a_{2}(x),\,a_{1}(x)+y-1]_{R}\,\right),$
(2.7)
$\displaystyle\left(\overline{R}_{N}\right)^{-n}(x,y)=\left(\,[a_{n}(y),a_{n-1}(y),\ldots,a_{2}(y),\,a_{1}(y)+x-1]_{R},\,R^{n}_{N}(y)\,\right).$
(2.8)

For $\rho_{N}$ in (2.3), we define its extended measure $\overline{\rho}_{N}$
on $\left(I^{2},{\mathcal{B}}^{2}_{I}\right)$ as

$\overline{\rho}_{N}(B):=\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}\int\\!\\!\\!\int_{B}\frac{N\mathrm{d}x\mathrm{d}y}{\left[N-(1-x)(1-y)\right]^{2}},\quad
B\in{\mathcal{B}}^{2}_{I}.$ (2.9)

Then $\overline{\rho}_{N}(A\times I)=\overline{\rho}_{N}(I\times
A)=\rho_{N}(A)$ for any $A\in{\mathcal{B}}_{I}$. The measure
$\overline{\rho}_{N}$ is preserved by $\overline{R}_{N}$, i.e.,
$\overline{\rho}_{N}((\overline{R}_{N})^{-1}(B))=\overline{\rho}_{N}(B)$ for
any $B\in{\mathcal{B}}^{2}_{I}$. With respect to $\overline{R}_{N}$, define
extended incomplete quotients $\overline{a}_{l}(x,y)$,
$l\in\mathbb{Z}:=\\{\ldots,-2,-1,0,1,2,\ldots\\}$ at $(x,y)\in I^{2}$ by

$\overline{a}_{l}(x,y):=\overline{a}_{1}\left(\,(\overline{R}_{N})^{l-1}(x,y)\,\right),\quad
l\in\mathbb{Z},$

with $\overline{a}_{1}(x,y)=a_{1}(x)$, $(x,y)\in I^{2}$. By (2.4) and (2.6) we
have

$\overline{a}_{n}(x,y)=a_{n}(x),\quad\overline{a}_{0}(x,y)=a_{1}(y),\quad\overline{a}_{-n}(x,y)=a_{n+1}(y),$

for any $n\in\mathbb{N}_{+}$ and $(x,y)\in I^{2}$. Since the measure
$\overline{\rho}_{N}$ is preserved by $\overline{R}_{N}$, the doubly infinite
sequence $(\overline{a}_{l}(x,y))_{l\in\mathbb{Z}}$ is strictly stationary
under $\overline{\rho}_{N}$.

Now recall some results obtained in [9].

######  Theorem 2.1.

Fix $(x,y)\in I^{2}$ and let $\overline{a}_{l}:=\overline{a}_{l}(x,y)$ for
$l\in{\mathbb{Z}}$. Set $a:=[\overline{a}_{0},\overline{a}_{-1},\ldots]_{R}$.
For any $x\in I$:

$\overline{\rho}_{N}\left([0,x]\times
I\,|\,\overline{a}_{0},\overline{a}_{-1},\ldots\right)=\frac{Nx}{N-(1-x)(1-a)}\quad\overline{\rho}_{N}\mbox{-}\mathrm{a.s.}$
(2.10)

The stochastic property of $(\overline{a}_{l})_{l\in\mathbb{Z}}$ under
$\overline{\rho}_{N}$ is as follows.

######  Corollary 2.2.

For any $i\geq N$, we have

$\overline{\rho}_{N}(\overline{a}_{1}=i|\overline{a}_{0},\overline{a}_{-1},\ldots)=P_{N,i}(a)\quad\overline{\rho}_{N}\mbox{-}\mathrm{a.s.}$
(2.11)

where $a=[\overline{a}_{0},\overline{a}_{-1},\ldots]_{R}$ and

$P_{N,i}(x):=\frac{x+N-1}{(x+i)(x+i-1)}.$ (2.12)

######  Remark 2.3.

The strict stationarity of $\left(\overline{a}_{l}\right)_{l\in\mathbb{Z}}$,
under $\overline{\rho}_{N}$ implies that

$\overline{\rho}_{N}(\overline{a}_{l+1}=i\,|\,\overline{a}_{l},\overline{a}_{l-1},\ldots)=P_{N,i}(a)\quad\overline{\rho}_{N}\mbox{-}\mathrm{a.s.}$
(2.13)

for any $i\geq N$ and $l\in\mathbb{Z}$, where
$a=[\overline{a}_{l},\overline{a}_{l-1},\ldots]_{R}$.

Define extended random variables
$\left(\overline{s}_{l}\right)_{l\in\mathbb{Z}}$ by
$\overline{s}_{l}:=[\overline{a}_{l},\overline{a}_{l-1},\ldots]_{R}$,
$l\in\mathbb{Z}$. Clearly,
$\overline{s}_{l}=\overline{s}_{0}\circ(\overline{R}_{N})^{l}$,
$l\in\mathbb{Z}$. It follows from Corollary 2.2 that
$\left(\overline{s}_{l}\right)_{l\in\mathbb{Z}}$ is a strictly stationary
$[0,1)$-valued Markov process on
$\left(I^{2},{\mathcal{B}}^{2}_{I},\overline{\rho}_{N}\right)$ with the
following transition mechanism. From state $\overline{s}\in I$ the possible
transitions are to any state $1-N/(\overline{s}+i)$ with corresponding
transition probability $P_{N,i}(\overline{s})$, $i\geq N$. Clearly, for any
$l\in\mathbb{Z}$ we have

$\overline{\rho}_{N}(\overline{s}_{l}<x)=\overline{\rho}_{N}(I\times[0,x))=\rho_{N}([0,x)),\quad
x\in I.$ (2.14)

Motivated by Theorem 2.1, we shall consider the one-parameter family
$\\{\rho^{t}_{N}:t\in I\\}$ of (conditional) probability measures on
$\left(I,{\mathcal{B}}_{I}\right)$ defined by their distribution functions

$\rho^{t}_{N}([0,x]):=\frac{Nx}{N-(1-x)(1-t)},\quad x,t\in I.$ (2.15)

Note that $\rho^{1}_{N}=\lambda$. The density of $\rho^{t}_{N}$ is

$h^{t}_{N}(x)=\frac{N(N-1+t)}{[N-(1-x)(1-t)]^{2}},\quad x,t\in I.$ (2.16)

For any $t\in I$ put

$s^{t}_{N,0}:=t,\quad s^{t}_{N,n}:=1-\frac{N}{a_{n}+s^{t}_{N,n-1}},\quad
n\in\mathbb{N}_{+}.$ (2.17)

######  Remark 2.4.

It follows from the properties just described for the process
$(\overline{s}_{l})_{l\in\mathbb{Z}}$ that the sequence
$(s^{t}_{N,n})_{n\in{\mathbb{N}}_{+}}$ is an $I$-valued Markov chain on
$\left(I,{\mathcal{B}}_{I},\rho^{t}_{N}\right)$ which starts at
$s^{t}_{N,0}:=t$ and has the following transition mechanism: from state $s\in
I$ the possible transitions are to any state $1-N/(s+i)$ with corresponding
transition probability $P_{N,i}(s)$, $i\geq N$.

Now, we recall that an $n$-block $(a_{1},a_{2},\ldots,a_{n})$ is said to be
admissible for the expansion in (2.2) if there exists $x\in[0,1)$ such that
$a_{i}(x)=a_{i}$ for all $1\leq i\leq n$. If $(a_{1},a_{2},\ldots,a_{n})$ is
an admissible sequence, we call the set

$I(a_{1},a_{2},\ldots,a_{n}):=\\{x\in
I:a_{1}(x)=a_{1},a_{2}(x)=a_{2},\ldots,a_{n}(x)=a_{n}\\},$ (2.18)

the $n$-th order cylinder. As we mentioned above,
$(a_{1},a_{2},\ldots,a_{n})\in\Lambda^{n}$, with
$\Lambda:=\\{N,N+1,\ldots\\}$.

######  Theorem 2.5 (Generalized Brodén-Borel-Lévy-type formula).

For any $t\in I$ and $n\in\mathbb{N}_{+}$, we have

$\rho^{t}_{N}(R^{n}_{N}<x|a_{1},\ldots,a_{n})=\frac{Nx}{N-(1-x)(1-s^{t}_{N,n})},\quad
x\in I.$ (2.19)

###### Proof.

For any $n\in\mathbb{N}_{+}$ and $x\in I$ consider the conditional probability

$\overline{\rho}_{N}(\left.\overline{R}^{-n}_{N}([0,x]\times
I)\,\right|\,\overline{a}_{n},\ldots,\overline{a}_{1},\overline{a}_{0},\overline{a}_{-1},\ldots).$
(2.20)

Put $t=[\overline{a}_{0},\overline{a}_{-1},\ldots]_{R}$ and note that
$[\overline{a}_{n},\ldots,\overline{a}_{1},\overline{a}_{0},\overline{a}_{-1},\ldots]_{R}=s^{t}_{N,n}$.
On the one hand, it follows from the fact that the measure
$\overline{\rho}_{N}$ is preserved by $\overline{R}_{N}$ and from Theorem 2.1
and Remark 2.3 that the conditional probability (2.20) $\overline{\rho}_{N}$
\- a.s. equals

$\frac{Nx}{N-(1-x)(1-s^{t}_{N,n})}.$

On the other hand, putting

$\overline{\rho}^{t}_{N}(\cdot)=\overline{\rho}_{N}(\left.\cdot\,\right|\,\overline{a}_{0},\overline{a}_{-1},\ldots)$

it is clear that (2.20) $\overline{\rho}_{N}$ \- a.s. equals

$\frac{\overline{\rho}^{t}_{N}\left(\overline{R}^{-n}_{N}([0,x]\times
I)\cap\left(I(a_{1},\ldots,a_{n})\times
I\right)\right)}{\overline{\rho}^{t}_{N}\left(I(a_{1},\ldots,a_{n})\times
I\right)}.$ (2.21)

Since $\overline{R}^{-n}_{N}([0,x]\times I)=\overline{R}^{-n}_{N}([0,x])\times
I$ and $\overline{\rho}^{t}_{N}(A\times I)={\rho}^{t}_{N}(A)$,
$A\in\mathcal{B}_{I}$, the fraction in (2.21) equals

${\rho}^{t}_{N}\left(\left.{R}^{-n}_{N}([0,x])\,\right|\,I(a_{1},\ldots,a_{n})\right)=\rho^{t}_{N}(R^{n}_{N}<x|a_{1},\ldots,a_{n}).$

∎

######  Corollary 2.6.

For any $t\in I$ and $n\in\mathbb{N}_{+}$ we have

$\rho^{t}_{N}(A|a_{1},\ldots,a_{n})=\rho^{s^{t}_{N,n}}_{N}\left(R^{n}_{N}(A)\right)$
(2.22)

whatever the set $A$ belonging to the $\sigma$-algebra generated by the random
variables $a_{n+1},a_{n+2},\ldots$, that is,
$R^{-n}_{N}\left(\mathcal{B}_{I}\right)$.

##  3 A Lévy-type approach

In the sequel we shall obtain estimates for both errors $F^{t}_{N,n}-G_{N}$
and $G^{t}_{N,n}-G_{N}$, $t\in I$, $n\in\mathbb{N}$ where

$\displaystyle F^{t}_{N,n}(x):=\rho^{t}_{N}(R^{n}_{N}<x),x\in
I,n\in\mathbb{N}$ (3.1) $\displaystyle
G^{t}_{N,n}(s):=\rho^{t}_{N}(s^{t}_{N,n}<s),s\in I,n\in\mathbb{N}_{+},$ (3.2)

and

$G^{t}_{N,0}(s):=\left\\{\begin{array}[]{ll}0,&{s\leq t}\\\
1,&{s>t}\end{array}\right.,\quad G_{N}(s):=\rho_{N}([0,s]),\quad s\in I.$
(3.3)

It follows from (2.19) that

$F^{t}_{N,n}(x)=\int_{0}^{1}\frac{Nx}{N-(1-x)(1-s)}\mathrm{d}G^{t}_{N,n}(s)$
(3.4)

for any $x,t\in I$ and $n\in\mathbb{N}$. It is easy to check that

$G_{N}(x)=\int_{0}^{1}\frac{Nx}{N-(1-x)(1-s)}\mathrm{d}G_{N}(s),\quad x\in I$
(3.5)

and

$F^{t}_{N,n}\left(1-\frac{N}{i+1}\right)=G^{t}_{N,n+1}\left(1-\frac{N}{i+1}\right),\quad
n\in\mathbb{N}_{+},t\in I,i\geq N.$ (3.6)

The last equation is still valid for $n=0$ and $t\neq 1$, while

$F^{1}_{N,0}\left(1-\frac{N}{i}\right)=G^{1}_{N,1}\left(1-\frac{N}{i+1}\right),i\geq
N.$ (3.7)

Since $\left(s^{t}_{N,n}\right)_{n\in\mathbb{N}}$ is a Markov chain on
$\left(I,\mathcal{B}_{I},\rho^{t}_{N}\right)$ for any $i\geq N$,
$n\in\mathbb{N}_{+}$, $t\in I$ and $\theta\in[0,1)$ we have

$\displaystyle
G^{t}_{N,n+1}\left(1-\frac{N}{i+\theta}\right)-G^{t}_{N,n+1}\left(1-\frac{N}{i}\right)=\rho^{t}_{N}\left(1-\frac{N}{i}\leq
s^{t}_{N,n+1}<1-\frac{N}{i+\theta}\right)$
$\displaystyle=E\left(\left.\rho^{t}_{N}\left(1-\frac{N}{i}\leq
s^{t}_{N,n+1}<1-\frac{N}{i+\theta}\right)\right|s^{t}_{N,n}\right)=$ (3.8)
$\displaystyle=\int_{0}^{\theta}P_{N,i}(s)\mathrm{d}G^{t}_{N,n}(s),\quad i\geq
N$

while

$\displaystyle
G^{t}_{N,1}\left(1-\frac{N}{i+\theta}\right)-G^{t}_{N,1}\left(1-\frac{N}{i}\right)$
$\displaystyle=$ $\displaystyle\rho^{t}_{N}\left(1-\frac{N}{i}\leq
s^{t}_{N,1}<1-\frac{N}{i+\theta}\right)$ (3.9) $\displaystyle=$
$\displaystyle\int_{0}^{\theta}P_{N,i}(s)\mathrm{d}G^{t}_{N,0}(s),$

that is, (3.8) also holds for $n=0$ if $t\in[0,1)$.

It is easy to check that

$\int_{0}^{\theta}P_{N,i}(s)\mathrm{d}G_{N}(s)=G_{N}\left(1-\frac{N}{i+\theta}\right)-G_{N}\left(1-\frac{N}{i}\right)$
(3.10)

for any $i\geq N$ and $\theta\in[0,1)$.

Now, by (3.4) and (3.5) we have

$\displaystyle F^{t}_{N,n}(x)-G_{N}(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{Nx}{N-(1-x)(1-s)}\mathrm{d}\left(G^{t}_{N,n}(s)-G_{N}(s)\right)$
$\displaystyle=$
$\displaystyle-\int_{0}^{1}\left(G^{t}_{N,n}(s)-G_{N}(s)\right)\frac{\partial}{\partial
s}\left(\frac{Nx}{N-(1-x)(1-s)}\right)\mathrm{d}s$

for any $x,t\in I$ and $n\in\mathbb{N}$. Setting

$\alpha^{t}_{N,n}:=\sup_{s\in I}\left\lvert
G^{t}_{N,n}(s)-G_{N}(s)\right\rvert,\quad t\in I,n\in\mathbb{N},$ (3.11)

we obtain

$\left\lvert
F^{t}_{N,n}(x)-G_{N}(x)\right\rvert\leq\alpha^{t}_{N,n}\int_{0}^{1}\frac{Nx(1-x)}{[N-(1-x)(1-s)]^{2}}\mathrm{d}s=\alpha^{t}_{N,n}\frac{x(1-x)}{N-1+x},$

hence

$\left\lvert F^{t}_{N,n}(x)-G_{N}(x)\right\rvert\leq\left(\sup_{x\in
I}\frac{x(1-x)}{N-1+x}\right)\alpha^{t}_{N,n}=\beta_{N}\cdot\alpha^{t}_{N,n},$
(3.12)

$t,x\in I$ and $n\in\mathbb{N}$, where

$\beta_{N}:=\frac{(1-N+\sqrt{N^{2}-N})(N-\sqrt{N^{2}-N})}{\sqrt{N^{2}-N}}=2N-1-2\sqrt{N(N-1)}.$
(3.13)

Let us note that

$\alpha^{t}_{N,0}=\max\left(G_{N}(t),1-G_{N}(t)\right),\quad t\in I.$

######  Theorem 3.1.

For any $n\in\mathbb{N}_{+}$ and $t\in I$ we have

$\sup_{x\in I}\left\lvert
F^{t}_{N,n}(x)-G_{N}(x)\right\rvert\leq\left\\{\begin{array}[]{ll}(0.171)(0.251)(0.348)^{n-1},&{N=2}\\\
\beta_{N}\cdot\delta_{N}\cdot c^{n-1}_{N},&{N\geq 3},\end{array}\right.$
$\sup_{x\in I}\left\lvert
G^{t}_{N,n}(x)-G_{N}(x)\right\rvert\leq\left\\{\begin{array}[]{ll}(0.251)(0.348)^{n-1},&{N=2}\\\
\delta_{N}\cdot c^{n-1}_{N},&{N\geq 3},\end{array}\right.$

where $\beta_{N}$ is as in(3.13) and

$\displaystyle\delta_{N}$ $\displaystyle:=$
$\displaystyle\frac{2}{N+1}-\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}\log\left(\frac{N^{2}}{N^{2}-1}\right),$
$\displaystyle c_{N}$ $\displaystyle:=$ $\displaystyle
2N-1-2\sqrt{N(N-1)}+\frac{N-1}{N(N+1)}.$

###### Proof.

For any $i\geq N$, $n\in\mathbb{N}_{+}$, $t\in I$ and $\theta\in[0,1)$ we have

$\displaystyle
G^{t}_{N,n+1}\left(1-\frac{N}{i+1+\theta}\right)-G_{N}\left(1-\frac{N}{i+1+\theta}\right)$
$\displaystyle=G^{t}_{N,n+1}\left(1-\frac{N}{i+1}\right)-G_{N}\left(1-\frac{N}{i+1}\right)$
$\displaystyle+G^{t}_{N,n+1}\left(1-\frac{N}{i+1+\theta}\right)-G^{t}_{N,n+1}\left(1-\frac{N}{i+1}\right)$
$\displaystyle+G_{N}\left(1-\frac{N}{i+1}\right)-G_{N}\left(1-\frac{N}{i+1+\theta}\right)$

and by (3.6), (3.8),(3.9), (3.10) and (3.12) we obtain

$\displaystyle\left\lvert
G^{t}_{N,n+1}\left(1-\frac{N}{i+1+\theta}\right)-G_{N}\left(1-\frac{N}{i+1+\theta}\right)\right\rvert$
$\displaystyle\leq\left\lvert
F^{t}_{N,n}\left(1-\frac{N}{i+1}\right)-G_{N}\left(1-\frac{N}{i+1}\right)\right\rvert$
$\displaystyle+\left\lvert\int_{0}^{\theta}P_{N,i+1}(s)\mathrm{d}\left(G^{t}_{N,n}(s)-G_{N}(s)\right)\right\rvert\leq\beta_{N}\alpha^{t}_{N,n}$
$\displaystyle+\left\lvert\int_{0}^{\theta}\left(G_{N}(s)-G^{t}_{N,n}(s)\right)\mathrm{d}P_{N,i+1}(s)+P_{N,i+1}(\theta)\left(G^{t}_{N,n}(\theta)-G_{N}(\theta)\right)\right\rvert$
$\displaystyle\leq\left(\beta_{N}+\beta_{N}(i,\theta)\right)\alpha^{t}_{N,n}$

where

$\beta_{N}(i,\theta)=\int_{0}^{\theta}\left\lvert\frac{\mathrm{d}P_{N,i+1}(s)}{\mathrm{d}s}\right\rvert\mathrm{d}s+P_{N,i+1}(\theta).$
(3.14)

Since

$\displaystyle\left\lvert\frac{\mathrm{d}P_{N,i+1}(s)}{\mathrm{d}s}\right\rvert$
$\displaystyle=$
$\displaystyle\left\lvert\frac{i-N+1}{(s+i)^{2}}-\frac{i-N+2}{(s+i+1)^{2}}\right\rvert$
$\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{(s+2)^{2}}-\frac{2}{(s+3)^{2}},&{i=N=2\mbox{
and }s\in[0,\sqrt{2}-1]}\\\
\displaystyle-\frac{1}{(s+2)^{2}}+\frac{2}{(s+3)^{2}},&{i=N=2\mbox{ and
}s\in(\sqrt{2}-1,1]}\\\
\displaystyle-\frac{1}{(s+N)^{2}}+\frac{2}{(s+N+1)^{2}},&{i=N\geq 3\mbox{ and
}s\in[0,1]}\\\
\displaystyle\frac{i-N+1}{(s+i)^{2}}-\frac{i-N+2}{(s+i+1)^{2}},&{i\geq
N+1,N\geq 2\mbox{ and }s\in[0,1]}\end{array}\right.$

we get for any $\theta\in[0,1)$

$\displaystyle\int_{0}^{\theta}\left\lvert\frac{\mathrm{d}P_{N,i+1}(s)}{\mathrm{d}s}\right\lvert\mathrm{d}s$
$\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle\int_{0}^{\theta}\left[\frac{1}{(s+2)^{2}}-\frac{2}{(s+3)^{2}}\right]\mathrm{d}s,\quad{i=N=2\mbox{
and }\theta\in[0,\sqrt{2}-1]}\\\ \\\
\displaystyle\int_{0}^{\sqrt{2}-1}\left[\frac{1}{(s+2)^{2}}-\frac{2}{(s+3)^{2}}\right]\mathrm{d}s+\int_{\sqrt{2}-1}^{\theta}\left[-\frac{1}{(s+2)^{2}}+\frac{2}{(s+3)^{2}}\right]\mathrm{d}s,&i=N=2\\\
&\mbox{ and }\theta\in(\sqrt{2}-1,1]\\\
\displaystyle\int_{0}^{\theta}\left[-\frac{1}{(s+N)^{2}}+\frac{2}{(s+N+1)^{2}}\right]\mathrm{d}s,\quad{i=N\geq
3\mbox{ and }\theta\in[0,1)}\\\ \\\
\displaystyle\int_{0}^{\theta}\left[\frac{i-N+1}{(s+i)^{2}}-\frac{i-N+2}{(s+i+1)^{2}}\right]\mathrm{d}s,\quad{i\geq
N+1,N\geq 2\mbox{ and }\theta\in[0,1)}.\end{array}\right.$

Actually,

$\beta_{N}(i,\theta)=\left\\{\begin{array}[]{ll}\displaystyle
2P_{2,3}(\theta)-\frac{1}{6},&{i=N=2\mbox{ and }\theta\in[0,\sqrt{2}-1]}\\\
\\\ \displaystyle 6-4\sqrt{2}-\frac{1}{6},&{i=N=2\mbox{ and
}\theta\in(\sqrt{2}-1,1)}\\\ \\\ \displaystyle\frac{N-1}{N(N+1)},&{i=N\geq
3\mbox{ and }\theta\in[0,1)}\\\ \\\ \displaystyle
2P_{N,i+1}(\theta)-\frac{N-1}{i(i+1)},&{i\geq N+1,N\geq 2\mbox{ and
}\theta\in[0,1)}.\end{array}\right.$ (3.17)

Now, if $i\geq N+1$, with $N\geq 2$, $\beta^{\prime}_{N}(i,\theta^{*})=0$ for

$\theta^{*}=1-N+\sqrt{(i+1-N)(i+2-N)}.$

For any real $\theta$, $\beta_{N}(i,\theta)$ is increasing if
$\theta\leq\theta^{*}$ and decreasing if $\theta\geq\theta^{*}$

Since

$\theta^{*}\in\left\\{\begin{array}[]{ll}(1,+\infty),&{i\geq 3\mbox{ and
}N=2}\\\ (0,1),&{i=4\mbox{ and }N=3}\\\ (1,+\infty),&{i\geq 5\mbox{ and
}N=3}\\\ (-\infty,0),&{i=5\mbox{ and }N=4}\\\ (0,1),&{i=6\mbox{ and }N=4}\\\
(1,+\infty),&{i\geq 7\mbox{ and }N=4}\\\ (-\infty,0),&{i=6\mbox{ and }N=5}\\\
(-\infty,0),&{i=7\mbox{ and }N=5}\\\ (0,1),&{i=8\mbox{ and }N=5}\\\
(1,+\infty),&{i\geq 9\mbox{ and }N=5}\\\ \vdots\end{array}\right.$

and $\theta\in[0,1)$ we get

$\displaystyle\sup_{\begin{array}[]{c}i\geq N+1,\\\ N\geq
2,\theta\in[0,1)\end{array}}\beta_{N}(i,\theta)$
$\displaystyle=\left\\{\begin{array}[]{ll}\displaystyle\sup_{i\geq
3}\beta_{2}(i,1)=\beta_{2}(3,1)=0.11667,&{N=2}\\\ \displaystyle\sup_{i\geq
5}\left\\{\beta_{3}(4,\theta^{*}),\beta_{3}(i,1)\right\\}=\beta_{3}(4,\theta^{*})=0.10204,&{N=3}\\\
\displaystyle\sup_{i\geq
7}\left\\{\beta_{4}(5,0),\beta_{4}(6,\theta^{*}),\beta_{4}(i,1)\right\\}=\beta_{4}(5,0)=0.1,&{N=4}\\\
\displaystyle\sup_{i\geq
9}\left\\{\beta_{5}(6,0),\beta_{5}(7,0),\beta_{5}(8,\theta^{*}),\beta_{5}(i,1),\right\\}=\beta_{5}(6,0)=0.07143,&{N=5}\\\
\displaystyle\beta_{N}(N+1,0)&{N\geq 6}.\end{array}\right.$

Finally, it is easy to check that for any $N\geq 2$

$\displaystyle\sup_{\begin{array}[]{c}i\geq N,\\\
\theta\in[0,1)\end{array}}\beta_{N}(i,\theta)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle
6-4\sqrt{2}-\frac{1}{6}=0.17647,&{N=2}\\\
\displaystyle\frac{1}{6}=0.16666,&{N=3}\\\
\displaystyle\frac{3}{20}=0.15,&{N=4}\\\ \vdots\end{array}\right.$
$\displaystyle=$ $\displaystyle\left\\{\begin{array}[]{ll}\displaystyle
0.17647,&{N=2}\\\ \displaystyle\frac{N-1}{N(N+1)},&{N\geq
3}.\end{array}\right.$

Hence

$\displaystyle\alpha^{t}_{N,n+1}=\sup_{\begin{array}[]{c}i\geq N,\\\
\theta\in[0,1)\end{array}}\left\lvert
G^{t}_{N,n+1}\left(1-\frac{N}{i+1+\theta}\right)-G_{N}\left(1-\frac{N}{i+1+\theta}\right)\right\rvert$
(3.23)
$\displaystyle\leq\left(2N-1-2\sqrt{N(N-1)}+\sup_{\begin{array}[]{c}i\geq
N,\\\ \theta\in[0,1)\end{array}}\beta_{N}(i,\theta)\right)\alpha^{t}_{N,n}$
(3.26)
$\displaystyle=\left\\{\begin{array}[]{ll}\displaystyle(9-\frac{1}{6}-6\sqrt{2})\alpha^{t}_{N,n}=0.348\ldots\alpha^{t}_{N,n},&{N=2}\\\
\displaystyle\left(2N-1-2\sqrt{N(N-1)}+\frac{N-1}{N(N+1)}\right)\alpha^{t}_{N,n}=:c_{N}\cdot\alpha^{t}_{N,n},&{N\geq
3}\end{array}\right.$ (3.29)

for any $t\in I$ and $n\in\mathbb{N}_{+}$.

Finally, by (3.6) and (3.9) we have

$G^{1}_{N,1}\left(1-\frac{N}{i+1+\theta}\right)=F^{1}_{N,0}\left(1-\frac{N}{i+1}\right)$

and

$\displaystyle G^{t}_{N,1}\left(1-\frac{N}{i+1+\theta}\right)$
$\displaystyle=$ $\displaystyle
G^{t}_{N,1}\left(1-\frac{N}{i+1}\right)+\int_{0}^{\theta}P_{N,i+1}(s)\mathrm{d}G^{t}_{N,0}(s)$
$\displaystyle=$ $\displaystyle\left\\{\begin{array}[]{ll}\displaystyle
F^{t}_{N,0}\left(1-\frac{N}{i+1}\right),&{0\leq\theta\leq t}\\\ \displaystyle
F^{t}_{N,0}\left(1-\frac{N}{i+1}\right)+P_{N,i+1}(t),&{\theta>t}\end{array}\right.$

for any $t\in[0,1)$, $\theta\in[0,1)$ and $i\geq N$, $N\geq 2$. It is easy to
see that

$\displaystyle\alpha^{t}_{N,1}$ $\displaystyle=$
$\displaystyle\sup_{\begin{array}[]{c}i\geq N,\\\
\theta\in[0,1)\end{array}}\left\lvert
G^{t}_{N,1}\left(1-\frac{N}{i+1+\theta}\right)-G_{N}\left(1-\frac{N}{i+1+\theta}\right)\right\rvert$
(3.33) $\displaystyle\leq$
$\displaystyle\frac{2}{N+1}-\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}\log\left(\frac{N^{2}}{N^{2}-1}\right)=:\delta_{N},\quad
t\in I.$ (3.34)

It follows from (3.29) and (3.34) that

$\alpha^{t}_{N,n}\leq\left\\{\begin{array}[]{ll}\displaystyle(0.251)(0.348)^{n-1},&{N=2}\\\
\displaystyle\delta_{N}\cdot c^{n-1}_{N},&{N\geq 3}\end{array}\right.$

for any $t\in I$ and $n\in\mathbb{N}_{+}$. By (3.12) the proof is complete. ∎

##  4 $\psi$-mixing coefficients of $\left(a_{n}\right)_{n\in\mathbb{N}_{+}}$

A mixing property of a stationary stochastic process
$(\ldots,X_{-1},X_{0},X_{1},\ldots)$ reflects a decay of the statistical
dependence between the past $\sigma$-algebra $\sigma\left(\\{X_{k}:k\leq
0\\}\right)$ and the asymptotic future $\sigma$-algebra
$\sigma\left(\\{X_{k}:k\geq n\\}\right)$ as $n\rightarrow\infty$. The various
mixing properties are described by corresponding measures of dependence
between $\sigma$-algebras.

We study the $\psi$-mixing coefficients of
$\left(a_{n}\right)_{n\in\mathbb{N}_{+}}$ under either $\rho^{t}_{N}$, $t\in
I$, or $\rho_{N}$.

For any $k\in\mathbb{N}_{+}$ let
$\mathcal{B}^{k}_{1}=\sigma(a_{1},\ldots,a_{k})$ and
$\mathcal{B}^{\infty}_{k}=\sigma(a_{k},a_{k+1},\ldots)$ denote the
$\sigma$-algebras generated by the random variables $a_{1},\ldots,a_{k}$,
respectively, $a_{k},a_{k+1},\ldots$. Clearly, $\mathcal{B}^{k}_{1}$ is the
$\sigma$-algebra generated by the closures of the fundamental intervals of
rank $k$ while $\mathcal{B}^{\infty}_{k}=R^{-k+1}_{N}(\mathcal{B}_{I})$,
$k\in\mathbb{N}_{+}$.

For any $\mu\in\mathrm{pr}(\mathcal{B}_{I})$ consider the $\psi$-mixing
coefficients

$\psi_{\mu}(n)=\sup\left\lvert\frac{\mu(A\cap
B)}{\mu(A)\mu(B)}-1\right\rvert,\quad n\in\mathbb{N}_{+},$ (4.1)

where the supremum is taken over all $A\in\mathcal{B}^{k}_{1}$ and
$B\in\mathcal{B}^{\infty}_{k+n}$ such that $\mu(A)\mu(B)\neq 0$ and
$k\in\mathbb{N}_{+}$. Define

$\varepsilon_{N,n}=\sup\left\lvert\frac{\rho^{t}_{N}(B)}{\rho_{N}(B)}-1\right\rvert,\quad
n\in\mathbb{N}_{+},$ (4.2)

where the supremum is taken over all $t\in I$ and
$B\in\mathcal{B}^{\infty}_{n}$ with $\rho_{N}(B)>0$. Note that the sequence
$\left(\varepsilon_{N,n}\right)_{n\in\mathbb{N}_{+}}$ is non-increasing since
$\mathcal{B}^{\infty}_{n+1}\subset\mathcal{B}^{\infty}_{n}$ for any
$n\in\mathbb{N}_{+}$. We shall show that $\varepsilon_{N,n}$ can be expressed
in terms of $F^{t}_{N,n-1}$, $t\in I$ and $G_{N}$, namely
$\varepsilon_{N,n}=\varepsilon^{\prime}_{N,n}$ with

$\varepsilon^{\prime}_{N,n}=\sup_{t,x\in
I}\left\lvert\frac{{\mathrm{d}F^{t}_{N,n-1}(x)}/{\mathrm{d}x}}{\nu_{N}(x)}-1\right\rvert,\quad
n\in\mathbb{N}_{+},$ (4.3)

where $\nu_{N}$ is the density measure of $\rho_{N}$, i.e.,

$\nu_{N}(x)=\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}\frac{1}{x+N-1}.$
(4.4)

Indeed, by the very definition of $\varepsilon^{\prime}_{N,n}$, for any
$t,x\in I$ we have

$\varepsilon^{\prime}_{N,n}\nu(x)\geq\left\lvert\frac{\mathrm{d}F^{t}_{N,n-1}(x)}{\mathrm{d}x}-\nu_{N}(x)\right\rvert.$
(4.5)

By integrating the above inequality over $B\in\mathcal{B}^{\infty}_{n}$ we
obtain

$\displaystyle\rho_{N}(B)\varepsilon^{\prime}_{N,n}$ $\displaystyle\geq$
$\displaystyle\int_{B}\left\lvert\frac{\mathrm{d}F^{t}_{N,n-1}(x)}{\mathrm{d}x}-\nu_{N}(x)\right\rvert{\mathrm{d}x}$
$\displaystyle\geq$
$\displaystyle\left\lvert\int_{B}\mathrm{d}F^{t}_{N,n-1}(x)-\int_{B}\nu_{N}(x){\mathrm{d}x}\right\rvert=\left\lvert\rho^{t}_{N}(B)-\rho_{N}(B)\right\rvert$

for any $B\in\mathcal{B}^{\infty}_{n}$, $n\in\mathbb{N}_{+}$ and $t\in I$.
Hence $\varepsilon^{\prime}_{N,n}\geq\varepsilon_{N,n}$, $n\in\mathbb{N}_{+}$.
On the other hand, for any arbitrarily given $n\in\mathbb{N}_{+}$ let
$B^{+}_{x,k}:=\left(x\leq R^{n-1}_{N}<x+k\right)\in\mathcal{B}^{\infty}_{n}$,
with $x\in[0,1)$, $k>0$, $x+k\in I$, and $B^{-}_{x,k}:=\left(x-k\leq
R^{n-1}_{N}<x\right)\in\mathcal{B}^{\infty}_{n}$, with $x\in(0,1]$, $k>0$,
$x-k\in I$. Clearly,

$\varepsilon_{N,n}\geq\max\left(\left\lvert\frac{\rho^{t}_{N}\left(B^{+}_{x,k}\right)}{\rho_{N}\left(B^{+}_{x,k}\right)}-1\right\rvert,\left\lvert\frac{\rho^{t}_{N}\left(B^{-}_{x,k}\right)}{\rho_{N}\left(B^{-}_{x,k}\right)}-1\right\rvert\right)$

for any $t\in I$ and suitable $x\in I$ and $k>0$. Letting $k\rightarrow 0$ we
get $\varepsilon_{N,n}\geq\varepsilon^{\prime}_{N,n}$, $n\in\mathbb{N}_{+}$.
Therefore $\varepsilon_{N,n}=\varepsilon^{\prime}_{N,n}$,
$n\in\mathbb{N}_{+}$.

It is easy to compute $\varepsilon^{\prime}_{N,1}=\varepsilon_{N,1}$ and
$\varepsilon^{\prime}_{N,2}=\varepsilon_{N,2}$. Since
$F^{t}_{N,0}(x)=\rho^{t}_{N}([0,x])$, $t,x\in I$, we have

$\displaystyle\varepsilon_{N,1}$ $\displaystyle=$
$\displaystyle\displaystyle\sup_{t,x\in
I}\left\lvert\frac{{\mathrm{d}F^{t}_{N,0}(x)}/{\mathrm{d}x}}{\nu_{N}(x)}-1\right\rvert=\sup_{t,x\in
I}\left\lvert\frac{\left.\left(\frac{Nx}{N-(1-x)(1-t)}\right)^{\prime}\right|_{x}}{\frac{\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}}{x+N-1}}-1\right\rvert$
$\displaystyle=$ $\displaystyle\sup_{t,x\in
I}\left\lvert\frac{N(N-1+t)(N-1+x)}{[N-(1-x)(1-t)]^{2}}\cdot\log\left(\frac{N}{N-1}\right)-1\right\rvert.$

As

$N-1\leq\frac{N(N-1+t)(N-1+x)}{[N-(1-x)(1-t)]^{2}}\leq N,\quad x,t\in I,$

it follows that

$\varepsilon_{N,1}=N\log\left(\frac{N}{N-1}\right)-1.$

Next, as

$F^{t}_{N,1}(x)=\rho^{t}_{N}\left(R_{N}<x\right)=\rho^{t}_{N}\left(R^{-1}_{N}(0,x)\right)$

and since

$R^{-1}_{N}(0,x)=\bigcup_{i\geq N}\left(1-\frac{N}{i},1-\frac{N}{x+i}\right]$

it follows that

$\displaystyle F^{t}_{N,1}(x)$ $\displaystyle=$ $\displaystyle\sum_{i\geq
N}\rho^{t}_{N}\left(1-\frac{N}{i},1-\frac{N}{x+i}\right]=\sum_{i\geq
N}\frac{x(t+N-1)}{(x+i+t-1)(i+t-1)}$ $\displaystyle=$
$\displaystyle(t+N-1)\sum_{i\geq
N}\left(\frac{1}{i+t-1}-\frac{1}{x+i+t-1}\right),\quad x,t\in I.$

Then

$\displaystyle\varepsilon^{\prime}_{N,2}$ $\displaystyle=$
$\displaystyle\displaystyle\sup_{t,x\in
I}\left\lvert\frac{{\mathrm{d}F^{t}_{N,1}(x)}/{\mathrm{d}x}}{\nu_{N}(x)}-1\right\rvert$
$\displaystyle=$ $\displaystyle\sup_{t,x\in
I}\left\lvert\left(\log\left(\frac{N}{N-1}\right)\right)(x+N-1)(t+N-1)\sum_{i\geq
N}\frac{1}{(x+i+t-1)^{2}}-1\right\rvert.$

A simple computation yields

$1+N^{2}\zeta(2,N+1)-N\zeta(2,N)\leq(x+N-1)(t+N-1)\sum_{i\geq
N}\frac{1}{(x+i+t-1)^{2}}\leq 1+(N-1)^{2}\zeta(2,N),$

Hence

$\displaystyle\varepsilon^{\prime}_{N,2}=\max\left\\{\begin{array}[]{ll}\left\lvert\left(1+(N-1)^{2}\zeta(2,N)\right)\log\left(\frac{N}{N-1}\right)-1\right\rvert,\\\
\\\
\left\lvert\left(1+N^{2}\zeta(2,N+1)-N\zeta(2,N)\right)\log\left(\frac{N}{N-1}\right)-1\right\rvert.\end{array}\right.$

For $N=2$, we get

$\displaystyle\varepsilon_{2,2}=\varepsilon^{\prime}_{2,2}$ $\displaystyle=$
$\displaystyle\max\\{\left\lvert[1+4\zeta(2,3)-2\zeta(2,2)]\log
2-1\right\rvert,\left\lvert[1+\zeta(2,2)]\log 2-1\right\rvert\\}$
$\displaystyle=$ $\displaystyle\max\\{\left\lvert[2(\zeta(2)-1)]\log
2-1\right\rvert,\left\lvert\zeta(2)\log 2-1\right\rvert\\}$ $\displaystyle=$
$\displaystyle\max\\{1-[2(\zeta(2)-1)]\log 2,\zeta(2)\log 2-1\\}=\zeta(2)\log
2-1=0.14018\ldots$

Clearly, for $n\geq 3$ the computation of $\varepsilon_{N,n}$ becomes very
difficult. Instead, Theorem 3.1 can be used to derive good upper bounds for
$\varepsilon_{N,n}$ whatever $n\geq 3$.

######  Proposition 4.1.

We have $\varepsilon_{N,1}\leq\left(\log\left(\frac{N}{N-1}\right)\right)\cdot
K_{N}$ and

$\varepsilon_{N,n}\leq\left\\{\begin{array}[]{ll}(\log
2)(7.55)(0.251)(0.348)^{n-2},&{N=2}\\\
\left(\log\left(\frac{N}{N-1}\right)\right)K_{N}\cdot\delta_{N}\cdot
c^{n-1}_{N},&{N\geq 3}\end{array}\right.,\quad n\geq 2,$ (4.7)

where
$K_{N}:=N+\frac{N^{3}}{(N-1)^{2}}-(N-1)\left[\frac{(2N-1)N}{(N-1)^{2}+N^{2}}+\frac{2N+1}{2N}\right]$
and $\delta_{N}$ and $c_{N}$ are as in Theorem 3.1.

###### Proof.

It follows from (3.4) and (3.5) that

$\frac{\mathrm{d}F^{t}_{N,n}(x)}{\mathrm{d}x}=\int_{0}^{1}\frac{N(N-1+s)}{[N-(1-x)(1-s)]^{2}}\mathrm{d}G^{t}_{N,n}(s)$

and

$\nu(x)=\int_{0}^{1}\frac{N(N-1+s)}{[N-(1-x)(1-s)]^{2}}\mathrm{d}G_{N}(s)$

for any $x,t\in I$ and $n\in\mathbb{N}$. Using the last two equations,
integration by parts yields

$\displaystyle\left\lvert\frac{\mathrm{d}F^{t}_{N,n}(x)}{\mathrm{d}x}-\nu(x)\right\rvert$
$\displaystyle=$
$\displaystyle\left\lvert\int_{0}^{1}\frac{N(N-1+s)}{[N-(1-x)(1-s)]^{2}}\mathrm{d}\left(G^{t}_{N,n}(s)-G_{N}(s)\right)\right\rvert$
$\displaystyle=$
$\displaystyle\left\lvert\int_{0}^{1}\left(G^{t}_{N,n}(s)-G_{N}(s)\right)\frac{\partial}{\partial
s}\left(\frac{N(N-1+s)}{[N-(1-x)(1-s)]^{2}}\right)\mathrm{d}s\right\rvert$
$\displaystyle\leq$ $\displaystyle\sup_{s\in I}\left\lvert
G^{t}_{N,n}(s)-G_{N}(s)\right\rvert\cdot\int_{0}^{1}N\left\lvert\frac{N-(1-x)(2N-1+s)}{[N-(1-x)(1-s)]^{3}}\right\rvert\mathrm{d}s.$

But

$\displaystyle\int_{0}^{1}N\left\lvert\frac{N-(1-x)(2N-1+s)}{[N-(1-x)(1-s)]^{3}}\right\rvert\mathrm{d}s$
$\displaystyle=\left\\{\begin{array}[]{ll}N\displaystyle\int_{0}^{1}\frac{-N+(1-x)(2N-1+s)}{[N-(1-x)(1-s)]^{3}}\mathrm{d}s,&{x\in\left[0,\frac{N-1}{2N}\right)}\\\
\\\
N\int_{0}^{\frac{(2N-1)x-N+1}{1-x}}\frac{N-(1-x)(2N-1+s)}{[N-(1-x)(1-s)]^{3}}\mathrm{d}s+N\int^{1}_{\frac{(2N-1)x-N+1}{1-x}}\frac{-N+(1-x)(2N-1+s)}{[N-(1-x)(1-s)]^{3}}\mathrm{d}s,&{x\in\left[\frac{N-1}{2N},\frac{N}{2N-1}\right]}\\\
\\\
N\displaystyle\int_{0}^{1}\frac{N-(1-x)(2N-1+s)}{[N-(1-x)(1-s)]^{3}}\mathrm{d}s,&{x\in\left(\frac{N}{2N-1},1\right]}\end{array}\right.$
$\displaystyle=\left\\{\begin{array}[]{ll}\displaystyle\frac{N(N-1)}{(x+N-1)^{2}}-1,&{x\in\left[0,\frac{N-1}{2N}\right)}\\\
\\\
\displaystyle\frac{-N(N-1)}{(x+N-1)^{2}}+\frac{1}{2x(1-x)}-1,&{x\in\left[\frac{N-1}{2N},\frac{N}{2N-1}\right]}\\\
\\\ \displaystyle
1-\frac{N(N-1)}{(x+N-1)^{2}},&{x\in\left(\frac{N}{2N-1},1\right]}\end{array}\right.$

and

$\displaystyle(x+N-1)\int_{0}^{1}N\left\lvert\frac{N-(1-x)(2N-1+s)}{[N-(1-x)(1-s)]^{3}}\right\rvert\mathrm{d}s=$
$\displaystyle=\left\\{\begin{array}[]{ll}\displaystyle\frac{N(N-1)}{x+N-1}-(x+N-1)\leq
1,&{x\in\left[0,\frac{N-1}{2N}\right)}\\\ \\\
\displaystyle\frac{-N(N-1)}{x+N-1}+\frac{x+N-1}{2x(1-x)}-(x+N-1),&{x\in\left[\frac{N-1}{2N}\leq
K_{N},\frac{N}{2N-1}\right]}\\\ \\\ \displaystyle
x+N-1-\frac{N(N-1)}{x+N-1}\leq
1,&{x\in\left(\frac{N}{2N-1},1\right]}\end{array}\right.$

where

$K_{N}=N+\frac{N^{3}}{(N-1)^{2}}-(N-1)\left[\frac{(2N-1)N}{(N-1)^{2}+N^{2}}+\frac{2N+1}{2N}\right].$

Therefore

$\sup_{x,t\in
I}\left\lvert\frac{\mathrm{d}F^{t}_{N,n}(x)}{\mathrm{d}x}-\nu(x)\right\rvert\leq\left(log\left(\frac{N}{N-1}\right)\right)\cdot
K_{N}\sup_{s,t\in I}\left\lvert G^{t}_{N,n}(s)-G_{N}(s)\right\rvert,\quad
n\in\mathbb{N}.$

Then by Theorem 3.1 the proof is complete. ∎

######  Theorem 4.2.

For any $t\in I$ we have

$\psi_{\rho^{t}_{N}}(n)\leq\frac{\varepsilon_{N,n}+\varepsilon_{N,n+1}}{1-\varepsilon_{N,n+1}},\quad
n\in\mathbb{N}_{+}.$ (4.11)

Also,

$\psi_{\rho_{N}}(n)=\varepsilon_{N,n},\quad n\in\mathbb{N}_{+}.$ (4.12)

###### Proof.

Let $(I\left(i^{(k)}\right)$ denote the cylinder $I(i_{1},\ldots,i_{k})$ for
$k\in\mathbb{N}$. It follows from (2.22) that for any $t\in I$ we have

$\varepsilon_{N,n}=\sup\left\lvert\frac{\rho^{t}_{N}\left(B|I\left(I\left(i^{(k)}\right)\right)\right)}{\rho_{N}(B)}-1\right\rvert$
(4.13)

where the supremum is taken over all $B\in\mathcal{B}^{\infty}_{k+n}$ with
$\rho_{N}(B)>0$, $i^{(k)}\in\Lambda^{k}$, and $k\in\mathbb{N}$. For
arbitrarily given $k,l,n\in\mathbb{N}_{+}$, $i^{(k)}\in\Lambda^{k}$ and
$j^{(l)}\in\Lambda^{l}$ put

$A=I\left(i^{(k)}\right),\quad
B=\left((a_{k+n},\ldots,a_{k+n+l-1})=j^{(l)}\right)$

and note that $\rho^{t}_{N}(A)\rho^{t}_{N}(B)\neq 0$ for any $t\in I$. By
(4.12) we have

$\left\lvert\rho^{t}_{N}(B|A)-\rho_{N}(B)\right\rvert\leq\varepsilon_{N,n}\rho_{N}(B)$
(4.14)

and

$\left\lvert\rho^{t}_{N}(B)-\rho_{N}(B)\right\rvert\leq\varepsilon_{N,n+k}\rho_{N}(B).$
(4.15)

It follows from (4.13) and (4.14) that

$\left\lvert\rho^{t}_{N}(B|A)-\rho^{t}_{N}(B)\right\rvert\leq(\varepsilon_{N,n}+\varepsilon_{N,n+k})\rho_{N}(B)$

whence

$\left\lvert\rho^{t}_{N}(A\cap
B)-\rho^{t}_{N}(A)\rho^{t}_{N}(B)\right\rvert\leq(\varepsilon_{N,1}+\varepsilon_{N,n+k})\rho^{t}_{N}(A)\rho_{N}(B).$

It follows from (4.14) that

$\rho_{N}(B)\leq\frac{\rho^{t}_{N}(B)}{1+\varepsilon_{N,n+k}}.$

Since the sequence $\left(\varepsilon_{N,n}\right)_{n\in\mathbb{N}_{+}}$ is
non-increasing, we have

$\frac{\varepsilon_{N,n}+\varepsilon_{N,n+k}}{1-\varepsilon_{N,n+k}}\leq\frac{\varepsilon_{N,n}+\varepsilon_{N,n+1}}{1-\varepsilon_{N,n+1}},n\in\mathbb{N}_{+},$

which completes the proof of (4.11).

To prove (4.12) we first note that putting $A\in I\left(i^{(k)}\right)$ for
any given $k\in\mathbb{N}_{+}$ and $i^{(k)}\in\Lambda^{k}$ by (4.13) we have

$\left\lvert\rho^{t}_{N}(A\cap
B)-\rho^{t}_{N}(A)\rho_{N}(B)\right\rvert\leq\varepsilon_{N,n}\rho^{t}_{N}(A)\rho_{N}(B)$

for any $t\in I$, $B\in\mathcal{B}^{\infty}_{k+n}$, and $n\in\mathbb{N}_{+}$.
By integrating the above inequality over $t\in I$ with respect to $\rho_{N}$
and taking into account that

$\int_{I}\rho^{t}_{N}(E)\rho_{N}(\mathrm{d}t)=\rho_{N}(E),\quad
E\in\mathcal{B}_{I}$

we get

$\psi_{\rho_{N}}(n)\leq\varepsilon_{N,n},\quad n\in\mathbb{N}_{+}.$

To prove the converse inequality, remark that the $\psi$-mixing coefficients
under the extended measure $\overline{\rho}_{N}$ of the doubly infinite
sequence $(\overline{a}_{l})_{l\in\mathbb{Z}}$ of extended incomplete
quotients, are equal to the corresponding $\psi$-mixing coefficients under
$\rho_{N}$ of $({a}_{n})_{n\in\mathbb{N}_{+}}$. This is obvious by the very
definitions of $(\overline{a}_{l})_{l\in\mathbb{Z}}$ and $\psi$-mixing
coefficients. As $(\overline{a}_{l})_{l\in\mathbb{Z}}$ is strict1y stationary
under $\overline{\rho}_{N}$, we have

$\psi_{\rho_{N}}(n)=\psi_{\overline{\rho}_{N}}(n)=\sup\left\lvert\frac{\overline{\rho}_{N}\left(\overline{A}\cap\overline{B}\right)}{\overline{\rho}_{N}\left(\overline{A}\right)\overline{\rho}_{N}\left(\overline{B}\right)}-1\right\rvert,\quad
n\in\mathbb{N}_{+},$

where the upper bound is taken over all
$\overline{A}=\sigma(\overline{a}_{n},\overline{a}_{n+1},\ldots)$ and
$\overline{B}=\sigma(\overline{a}_{0},\overline{a}_{-1},\ldots)$ for which
$\overline{\rho}_{N}\left(\overline{A}\right)\overline{\rho}_{N}\left(\overline{B}\right)\neq
0$. Clearly, $\overline{A}=A\times I$ and $\overline{B}=I\times B$, with
$A\in\mathcal{B}^{\infty}_{n}=R^{-n+1}_{N}\left(\mathcal{B}_{I}\right)$ and
$B\in\mathcal{B}_{I}$. Then

$\psi_{\rho_{N}}(n)=\sup_{\begin{array}[]{c}A\in
R^{-n+1}_{N}\left(\mathcal{B}_{I}\right)\\\ B\in\mathcal{B}_{I}\\\
\rho_{N}(A)\rho_{N}(B)\neq
0\end{array}}\left\lvert\frac{\overline{\rho}_{N}(A\times
B)}{\rho_{N}(A)\rho_{N}(B)}-1\right\rvert,\quad n\in\mathbb{N}_{+}.$ (4.16)

Now, it is easy to check that

$\overline{\rho}_{N}(A\times
B)=\int_{A}\rho_{N}(\mathrm{d}t)\rho^{t}_{N}(B)=\int_{B}\rho_{N}(\mathrm{d}u)\rho^{u}_{N}(A)$

for any $A,B\in\mathcal{B}_{I}$. It then follows from (4.16) and the very
definition of $\varepsilon_{N,n}$ that

$\psi_{\rho_{N}}(n)\geq\sup_{\begin{array}[]{c}A\in
R^{-n+1}_{N}\left(\mathcal{B}_{I}\right)\\\ u\in I,\rho_{N}(A)\neq
0\end{array}}\left\lvert\frac{{\rho}^{u}_{N}(A)}{\rho_{N}(A)}-1\right\rvert=\varepsilon_{N,n},\quad
n\in\mathbb{N}_{+}.$

This completes the proof of (4.12). ∎

######  Corollary 4.3.

The sequence $\left(a_{n}\right)_{n\in\mathbb{N}+}$ is $\psi$-mixing under
$\rho_{N}$ and $\rho^{t}_{N}$, $t\in I$. For any $t\in I$ we have

$\psi_{\rho^{t}_{N}}(1)=\frac{\varepsilon_{N,1}+\varepsilon_{N,2}}{1-\varepsilon_{N,2}}$

and

$\psi_{\rho^{t}_{N}}(n)=\frac{\left(\log\left(\frac{N}{N-1}\right)\right)K_{N}\cdot\delta_{N}\cdot
c^{n-2}_{N}(1+c_{N})}{1-\left(\log\left(\frac{N}{N-1}\right)\right)K_{N}\cdot\delta_{N}\cdot
c^{n-1}_{N}},\quad n\geq 2.$

In particular, for $N=2$, since $\varepsilon_{2,1}=2\log 2-1=0.38629$ and
$\varepsilon_{2,2}=\zeta(2)\log 2-1=0.14018$ it follows that

$\psi_{\rho^{t}_{2}}(1)\leq 0.612302575.$

Also $\psi_{\rho_{N}}(1)=N\log\left(\frac{N}{N-1}\right)-1$,

$\displaystyle\psi_{\rho_{N}}(2)=\max\left\\{\begin{array}[]{ll}\left\lvert\left(1+(N-1)^{2}\zeta(2,N)\right)\log\left(\frac{N}{N-1}\right)-1\right\rvert,\\\
\\\
\left\lvert\left(1+N^{2}\zeta(2,N+1)-N\zeta(2,N)\right)\log\left(\frac{N}{N-1}\right)-1\right\rvert\end{array}\right.$

and

$\psi_{\rho_{N}}(n)\leq\left(\log\left(\frac{N}{N-1}\right)\right)K_{N}\cdot\delta_{N}\cdot
c^{n-2}_{N},\quad n\geq 3.$

In particular, for $N=2$

$\psi_{\rho_{2}}(1)=2\log 2-1=0.38629,\quad\psi_{\rho_{2}}(2)=\zeta(2)\log
2-1=0.14018.$

The doubly infinite sequence $(\overline{a}_{l})_{l\in\mathbb{Z}}$ of extended
incomplete quotients is $\psi$-mixing under the extended measure
$(\overline{\rho}_{N})$, and its $\psi$-mixing coefficients are equal to the
corresponding $\psi$-mixing coefficients under $\rho_{N}$ of
$(a_{n})_{n\in\mathbb{N}+}$.

The proof follows from Proposition 4.1 and Theorem 4.2. As already noted, the
last assertion is obvious by the very definitions of
$(\overline{a}_{l})_{l\in\mathbb{Z}}$ and $\psi$-mixing coefficients.

## Appendix A. Proofs of some results from Section 3

Proof of (3.34) We have

$\displaystyle G^{t}_{N,1}\left(1-\frac{N}{i+1+\theta}\right)$
$\displaystyle=$ $\displaystyle\left\\{\begin{array}[]{ll}\displaystyle
F^{t}_{N,0}\left(1-\frac{N}{i+1}\right),&{0\leq\theta\leq t}\\\ \displaystyle
F^{t}_{N,0}\left(1-\frac{N}{i+1}\right)+P_{N,i+1}(t),&{\theta>t}\end{array}\right.$
$\displaystyle\leq$
$\displaystyle\frac{i+1-N}{i+t}+\frac{t+N-1}{(i+t)(i+1+t)}\leq\frac{1}{N}+\frac{N-1}{N(N+1)}=\frac{2}{N+1}$

and

$\displaystyle G_{N}\left(1-\frac{N}{i+1+\theta}\right)$ $\displaystyle=$
$\displaystyle\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}\log\left(\frac{N(i+\theta)}{(N-1)(i+1+\theta)}\right)$
$\displaystyle\geq$
$\displaystyle\left(\log\left(\frac{N}{N-1}\right)\right)^{-1}\log\left(\frac{N^{2}}{N^{2}-1}\right).$

## References

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