###### Proof.

Equidistribution of continued fraction convergents in
$\mathrm{SL}(2,\mathbb{Z}_{m})$ with an application to local discrepancy

Bence Borda

Graz University of Technology

Steyrergasse 30, 8010 Graz, Austria

Email<EMAIL_ADDRESS>
Keywords: convergent denominators mod $m$, Gauss–Kuzmin problem,
limit law, invariance principle, irrational rotation

Mathematics Subject Classification (2020): 11K50, 37A50, 37E10

###### Abstract

We study the distribution of the sequence of continued fraction convergents
$p_{n}/q_{n}$ to a random irrational number in the group
$\mathrm{SL}(2,\mathbb{Z}_{m})$, and in particular the distribution of
$p_{n}\pmod{m}$ and $q_{n}\pmod{m}$ with a fixed modulus $m$. Improving the
strong law of large numbers due to Szüsz, Moeckel, Jager and Liardet, we
establish the central limit theorem and the law of the iterated logarithm, as
well as the weak and the almost sure invariance principles. As an application,
we find the limit distribution of the maximum and the minimum of the Birkhoff
sum for the irrational rotation with the indicator of an interval as test
function. We also compute the normalizing constant in a classical limit law
for the same Birkhoff sum due to Kesten, and dispel a misconception about its
dependence on the test interval.

##  1 Introduction

The statistical properties of the continued fraction expansion
$\alpha=[0;a_{1},a_{2},\ldots]$ of a random real number $\alpha\in[0,1]$ is a
classical topic in metric number theory. The partial quotients $a_{n}$ form a
weakly dependent sequence of random variables, consequently the sum
$\sum_{n=1}^{N}f(a_{n})$ satisfies various probabilistic limit theorems
depending on the growth rate of the function $f:\mathbb{N}\to\mathbb{R}$. The
asymptotic behavior of the convergents
$p_{n}/q_{n}=[0;a_{1},a_{2},\ldots,a_{n}]$ is also well known. For instance,
we have $\log q_{n}\sim\frac{\pi^{2}}{12\log 2}n$ for a.e. $\alpha$, and $\log
q_{n}$ even satisfies the central limit theorem (CLT) and the law of the
iterated logarithm (LIL). We refer to the monograph [14] for a comprehensive
survey.

The main subject of this paper is the distribution of the sequences
$p_{n}\pmod{m}$ and $q_{n}\pmod{m}$ in $\mathbb{Z}_{m}$ with a fixed integer
$m\geq 2$. One of the first results in the area is due to Szüsz [26, Satz
3.3], who showed that for any $a\in\mathbb{Z}_{m}$,

$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\mathds{1}_{\\{q_{n}\equiv
a\pmod{m}\\}}=\frac{\prod_{p\mid\mathrm{gcd}(a,m)}\left(1-\frac{1}{p}\right)}{m\prod_{p\mid
m}\left(1-\frac{1}{p^{2}}\right)}\quad\textrm{for a.e. }\alpha,$ (1)

where the products are over prime divisors. Contrary to what one might naively
expect, certain residue classes are thus attained more frequently than others.
For instance, only one third of all convergent denominators are even, while
two thirds of them are odd.

The origin of the limit in (1) becomes transparent when we consider the joint
distribution of the quadruple $(p_{n-1},p_{n},q_{n-1},q_{n})\pmod{m}$, and
work in the group $\mathrm{SL}(2,\mathbb{Z}_{m})$. The recursions for $p_{n}$
and $q_{n}$, with the usual convention $p_{0}=0$, $q_{0}=1$, can be written in
matrix form as

$\left(\begin{array}[]{cc}0&1\\\
1&a_{1}\end{array}\right)\left(\begin{array}[]{cc}0&1\\\
1&a_{2}\end{array}\right)\cdots\left(\begin{array}[]{cc}0&1\\\
1&a_{n}\end{array}\right)=\left(\begin{array}[]{cc}p_{n-1}&p_{n}\\\
q_{n-1}&q_{n}\end{array}\right),$

where the right-hand side has determinant $(-1)^{n}$. Taking the previous
formula mod $m$ entrywise leads to the $2\times 2$ matrix with entries in
$\mathbb{Z}_{m}$

$P_{n}=\left(\begin{array}[]{cc}p_{n-1}&p_{n}\\\
q_{n-1}&q_{n}\end{array}\right)\pmod{m}.$

For the rest of the paper, let

$G_{D}=\left\\{\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\,:\,a,b,c,d\in\mathbb{Z}_{m},\,\,ad-bc=D\right\\},\qquad
D\in\mathbb{Z}_{m}^{*}=\\{a\in\mathbb{Z}_{m}\,:\,\mathrm{gcd}(a,m)=1\\},$

and let $G=G_{1}\cup G_{-1}$. Note that $G_{1}=\mathrm{SL}(2,\mathbb{Z}_{m})$
is a normal subgroup of
$\cup_{D\in\mathbb{Z}_{m}^{*}}G_{D}=\mathrm{GL}(2,\mathbb{Z}_{m})$, and
$G_{D}$ is a coset of $G_{1}$. If $m=2$, then
$G_{1}=G_{-1}=\mathrm{SL}(2,\mathbb{Z}_{2})=\mathrm{GL}(2,\mathbb{Z}_{2})$,
otherwise $G_{1}\neq G_{-1}$.

It turns out that the sequence $P_{n}$ equidistributes in $G$, that is, for
any $g\in G$,

$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\mathds{1}_{\\{P_{n}=g\\}}=\frac{1}{|G|}\quad\textrm{for
a.e. }\alpha.$ (2)

The set of possible rows or columns of matrices in $G_{D}$ is

$V=\\{(a,b)\in\mathbb{Z}_{m}^{2}\,:\,\mathrm{gcd}(a,b,m)=1\\}.$

In fact, both rows and both columns of a matrix which is uniformly distributed
on $G_{D}$ with some $D\in\mathbb{Z}_{m}^{*}$ are uniformly distributed on
$V$, see Lemma 4 below. Relation (2) thus immediately implies that for any
$(a,b)\in V$,

$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\mathds{1}_{\\{(p_{n},q_{n})\equiv(a,b)\pmod{m}\\}}=\frac{1}{|V|}\quad\textrm{for
a.e. }\alpha,$ (3)

and the same holds for the pairs $(q_{n-1},q_{n})$ and $(p_{n-1},p_{n})$. For
instance, the “parity type” of the fraction $p_{n}/q_{n}$ can be even/odd,
odd/even or odd/odd; note that the type even/even is impossible as $p_{n}$ and
$q_{n}$ are coprime. The parity type of $p_{n}/q_{n}$ thus equidistributes in
the set $\\{\textrm{even/odd, odd/even, odd/odd}\\}$.

The four entries of a matrix which is uniformly distributed on $G_{D}$,
however, are not uniformly distributed on $\mathbb{Z}_{m}$. Instead, the
distribution of all four entries is the probability measure $\nu$ on
$\mathbb{Z}_{m}$ which assigns the measure

$\nu_{a}=\frac{\prod_{p\mid\mathrm{gcd}(a,m)}\left(1-\frac{1}{p}\right)}{m\prod_{p\mid
m}\left(1-\frac{1}{p^{2}}\right)},\quad a\in\mathbb{Z}_{m}$ (4)

to the singleton $\\{a\\}$, see Lemma 4 below. This explains the value of the
limit in (1), and shows that the same relation holds for $p_{n}$ as well.

Relation (3) for the pair $(q_{n-1},q_{n})$ was first proved by Szüsz [26,
Satz 3.2] by following Lévy’s approach to the Gauss–Kuzmin problem about the
mixing properties of the Gauss map, but his work seems to have gone mostly
unnoticed by the ergodic theory community. Moeckel [19] used the close
relationship between continued fractions and the geodesic flow on the modular
surface, and almost proved (3); we say almost, as Moeckel worked with
$\mathrm{PSL}(2,\mathbb{Z}_{m})$ instead of $\mathrm{SL}(2,\mathbb{Z}_{m})$,
consequently he only showed equidistribution in the factor $V/\sim$ with the
equivalence relation $(a,b)\sim(-a,-b)$. See [12] for a more recent account
and generalizations of Moeckel’s approach. Finally, relation (2) and
consequently also (3) were proved by Jager and Liardet [15] using the
ergodicity of a certain skew product over the Gauss map.

In the terminology of probability theory, relations (1), (2) and (3)
correspond to the strong law of large numbers. The main goal of the present
paper is to extend these results to more precise limit theorems, such as the
CLT and the LIL. In fact, we will even establish the weak and the almost sure
invariance principles. We refer to Billingsley [4] for a general introduction
to invariance principles and their relation to the ordinary and the functional
CLT and LIL.

Throughout, $\alpha$ is a random variable with distribution $\mu$, a Borel
probability measure on $[0,1]$ which is absolutely continuous with respect to
the Lebesgue measure $\lambda$. In some of our results we assume that $\mu$
has a Lipschitz density, that is,

$\left|\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(x)-\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(y)\right|\leq
L|x-y|\quad\textrm{for all }x,y\in[0,1]$ (5)

with some constant $L\geq 0$. The most important example is the Gauss measure
$\mu_{\mathrm{Gauss}}(A)=\frac{1}{\log 2}\int_{A}\frac{1}{1+x}\,\mathrm{d}x$,
$A\subseteq[0,1]$ Borel. The uniform probability measure on a finite set $S$
is denoted by $\mathrm{Unif}(S)$. We write $X\sim\vartheta$ if the random
variable $X$ has distribution $\vartheta$, and $\overset{d}{\to}$ denotes
convergence in distribution.

Given an arbitrary function $f:G\to\mathbb{R}$, we define the constants
$E_{f}$ and $\sigma_{f}\geq 0$ as follows. Let $U\sim\mathrm{Unif}(G)$, and
define $E_{f}=\mathbb{E}f(U)$. Let $\alpha\sim\mu_{\mathrm{Gauss}}$ and
$U_{\pm 1}\sim\mathrm{Unif}(G_{\pm 1})$ be independent random variables, let
$\bar{f}(x)=f(x)-\mathbb{E}f(U_{\pm 1})$ for $x\in G_{\pm 1}$, and define

$\sigma_{f}^{2}=\frac{1}{2}\mathbb{E}\bar{f}(U_{1})^{2}+\sum_{n=1}^{\infty}\mathbb{E}\bar{f}(U_{1})\bar{f}(U_{1}P_{n})+\frac{1}{2}\mathbb{E}\bar{f}(U_{-1})^{2}+\sum_{n=1}^{\infty}\mathbb{E}\bar{f}(U_{-1})\bar{f}(U_{-1}P_{n}).$
(6)

######  Theorem 1.

Fix an integer $m\geq 2$, and let $f:G\to\mathbb{R}$ be arbitrary.

  1. (i)

The right-hand side of (6) is finite and nonnegative.

  2. (ii)

Let $\alpha\sim\mu$ with $\mu\ll\lambda$. Then the process $\sum_{1\leq n\leq
tN}f(P_{n})$, $t\in[0,1]$ satisfies the functional CLT

$\frac{\sum_{1\leq n\leq
tN}f(P_{n})-E_{f}tN}{\sqrt{N}}\overset{d}{\to}\sigma_{f}W(t)$

in the Skorokhod space $\mathcal{D}[0,1]$, where $W(t)$, $t\in[0,1]$ is a
standard Wiener process.

  3. (iii)

Let $\alpha\sim\mu$ with $\mu\ll\lambda$, and assume (5). Without changing its
distribution, the process $\sum_{1\leq n\leq t}f(P_{n})$, $t\geq 0$ can be
redefined on a richer probability space so that

$\sum_{1\leq n\leq
t}f(P_{n})-E_{f}t=\sigma_{f}W(t)+O(t^{1/2-\eta})\quad\textrm{a.s.}$

with a universal constant $\eta>0$, where $W(t)$, $t\geq 0$ is a standard
Wiener process.

Theorem 1 (ii) immediately implies that the sum $\sum_{n=1}^{N}f(P_{n})$
satisfies the CLT

$\frac{\sum_{n=1}^{N}f(P_{n})-E_{f}N}{\sqrt{N}}\overset{d}{\to}\mathcal{N}(0,\sigma_{f}^{2}),$

where $\mathcal{N}(a,\sigma^{2})$ denotes the normal distribution with mean
$a$ and variance $\sigma^{2}$ (or the constant $a$ in case $\sigma=0$),
whereas Theorem 1 (iii) implies the LIL

$\limsup_{N\to\infty}\frac{\sum_{n=1}^{N}f(P_{n})-E_{f}N}{\sqrt{2N\log\log
N}}=\sigma_{f}\quad\textrm{for a.e. }\alpha.$

Theorem 1 applied to functions $f$ supported on $G_{1}$ resp. $G_{-1}$
describes the quantitative equidistribution of $P_{2n}$ resp. $P_{2n-1}$ in
$G_{1}$ resp. $G_{-1}$.

Given an arbitrary function $f:V\to\mathbb{R}$, applying Theorem 1 to the map
$\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\mapsto f(b,d)$ shows
that the sum $\sum_{n=1}^{N}f((p_{n},q_{n})\pmod{m})$ satisfies the CLT

$\frac{\sum_{n=1}^{N}f((p_{n},q_{n})\pmod{m})-EN}{\sqrt{N}}\overset{d}{\to}\mathcal{N}(0,\sigma^{2})$

and the LIL

$\limsup_{N\to\infty}\frac{\sum_{n=1}^{N}f((p_{n},q_{n})\pmod{m})-EN}{\sqrt{2N\log\log
N}}=\sigma\quad\textrm{for a.e. }\alpha,$

where $E=\mathbb{E}f(U)$ with $U\sim\mathrm{Unif}(V)$, and the constant
$\sigma\geq 0$ depends only on $f$. The same holds for the pairs
$(q_{n-1},q_{n})$ and $(p_{n-1},p_{n})$.

Given an arbitrary function $f:\mathbb{Z}_{m}\to\mathbb{R}$, the correct
centering term is defined using an auxiliary random variable $U\sim\nu$, that
is, $E=\mathbb{E}f(U)=\sum_{a\in\mathbb{Z}_{m}}\nu_{a}f(a)$. Then the sum
$\sum_{n=1}^{N}f(q_{n}\pmod{m})$ satisfies the CLT

$\frac{\sum_{n=1}^{N}f(q_{n}\pmod{m})-EN}{\sqrt{N}}\overset{d}{\to}\mathcal{N}(0,\sigma^{2})$

and the LIL

$\limsup_{N\to\infty}\frac{\sum_{n=1}^{N}f(q_{n}\pmod{m})-EN}{\sqrt{2N\log\log
N}}=\sigma\quad\textrm{for a.e. }\alpha$

with a suitable constant $\sigma\geq 0$ depending only on $f$. The same holds
for $p_{n}$.

The key ingredient in the proof of Theorem 1 is a version of the
Gauss–Kuzmin–Lévy theorem, see Theorem 8 below. As a corollary, in Lemma 14 we
show that the sequence $P_{n}$ is $\psi$-mixing with exponential rate. The
weak and the almost sure invariance principles then follow from general
results of Philipp and Stout [22] on the partial sums of weakly dependent
random variables.

In fact, we also prove that the sequence $(a_{n},P_{n})$ is $\psi$-mixing with
exponential rate, from which various probabilistic limit theorems follow for
the sum $\sum_{n=1}^{N}f(a_{n},P_{n})$ depending on the growth rate of $f$ in
its first variable. As an application, in Section 2 we find the limit
distribution of the maximum and the minimum of certain Birkhoff sums for the
circle rotation. In Section 3 we gather the necessary facts about the group
$\mathrm{SL}(2,\mathbb{Z}_{m})$. The proofs are given in Sections 4, 5 and 6.

##  2 An application to local discrepancy

The circle rotation $x\mapsto x+\alpha\pmod{\mathbb{Z}}$ on
$\mathbb{R}/\mathbb{Z}$ with a given irrational $\alpha$ is perhaps the
simplest discrete time dynamical system. The system is uniquely ergodic,
consequently Birkhoff sums satisfy
$\sum_{n=1}^{N}f(n\alpha+\beta)=N\int_{0}^{1}f(x)\,\mathrm{d}x+o(N)$ for any
starting point $\beta$ and any $1$-periodic function $f$ which is Riemann
integrable on $[0,1]$. The remainder term $o(N)$ depends sensitively on the
continued fraction expansion of $\alpha$ and the function $f$, and satisfies
various probabilistic limit theorems, see [11] for a survey.

A classical example is $f(x)=\mathds{1}_{[0,r]}(\\{x\\})-r$ with a fixed
$r\in(0,1)$, the centered indicator function of the interval $[0,r]$ extended
with period $1$, where $\\{\cdot\\}$ denotes the fractional part. The first
limit theorem for the corresponding Birkhoff sum is due to Kesten [17, 18],
who proved that if $(\alpha,\beta)$ is a random variable uniformly distributed
on the unit square, then

$\frac{\sum_{n=1}^{N}\mathds{1}_{[0,r]}(\\{n\alpha+\beta\\})-rN}{\sigma\log
N}\overset{d}{\to}\mathrm{Cauchy}.$ (7)

Here “Cauchy” denotes the standard Cauchy distribution, with density function
$1/(\pi(1+x^{2}))$. Because of the random starting point $\beta$, the same
holds for any subinterval of $[0,1]$ of length $r$.

Kesten gave a complicated but explicit formula for the normalizing constant
$\sigma>0$, see Section 4. He showed that the value of $\sigma$ is the same
for all irrational $r$, but his formula involves $r$ if $r$ is rational. This
apparent dependence of $\sigma$ on $r$ in the rational case has been cited by
several authors. Disproving this long-held view, we show that the dependence
is illusory, and (7) holds with the same value of $\sigma$ for both rational
and irrational $r$.

######  Theorem 2.

Kesten’s limit law (7) holds with $\sigma=1/(3\pi)$ for all $r\in(0,1)$.

The function $f(x)=\\{x\\}-1/2$ also leads to the same limit law [17]: if
$(\alpha,\beta)$ is a random variable uniformly distributed on the unit
square, then

$\frac{\sum_{n=1}^{N}(\\{n\alpha+\beta\\}-1/2)}{\sigma^{\prime}\log
N}\overset{d}{\to}\mathrm{Cauchy}.$ (8)

In Section 4 we show that (8) holds with $\sigma^{\prime}=1/(4\pi)$. The
question whether (7) and (8) hold with a fixed starting point $\beta$ was
already raised by Kesten, and remains open.

Remarkable random behavior of the two Birkhoff sums mentioned above with a
fixed quadratic irrational $\alpha$ was first proved by Beck [3], and later
generalized by Bromberg and Ulcigrai [7]. The special case $r=1/2$ is
sometimes called the deterministic random walk [1, 2]. Higher dimensional
analogues of (7) are due to Dolgopyat and Fayad [9, 10].

In this paper, we consider the Birkhoff sum
$S_{N,r}(\alpha)=\sum_{n=1}^{N}\mathds{1}_{[0,r]}(\\{n\alpha\\})-rN$ with a
rational $r$, that is, when the starting point is $\beta=0$, and we find the
limit law of its maximum and minimum. Let $\mathrm{Stab}(1,\pm 1)$ denote the
standard stable law of index $1$ and skewness parameter $\pm 1$, whose
characteristic function is $\exp(-|x|(1\pm(2i/\pi)\mathrm{sgn}(x)\log|x|))$,
and let $\otimes$ denote the product of two measures.

######  Theorem 3.

Let $\alpha\sim\mu$ with $\mu\ll\lambda$, and let $r\in(0,1)$ be a fixed
rational. Then

$\left(\frac{\displaystyle{\max_{0\leq
N<M}S_{N,r}(\alpha)-E_{M}}}{{\frac{1}{2\pi}\log
M}},\frac{\displaystyle{\min_{0\leq
N<M}S_{N,r}(\alpha)+E_{M}}}{{\frac{1}{2\pi}\log
M}}\right)\overset{d}{\to}\mathrm{Stab}(1,1)\otimes\mathrm{Stab}(1,-1)\quad\textrm{as
}M\to\infty,$

where $E_{M}=(1/\pi^{2})\log M\log\log M-c(r)\log M$ with an explicit constant
$c(r)$ depending only on $r$.

In particular,

$\frac{\displaystyle{\max_{0\leq
N<M}S_{N,r}(\alpha)-E_{M}}}{\frac{1}{2\pi}\log
M}\overset{d}{\to}\mathrm{Stab}(1,1)\qquad\textrm{and}\qquad\frac{\displaystyle{\min_{0\leq
N<M}S_{N,r}(\alpha)+E_{M}}}{\frac{1}{2\pi}\log
M}\overset{d}{\to}\mathrm{Stab}(1,-1).$

The main motivation for deducing the joint limit law in Theorem 3 is that it
immediately implies

$\frac{\displaystyle{\max_{0\leq
N<M}|S_{N,r}(\alpha)|-E_{M}}}{\frac{1}{2\pi}\log M}\overset{d}{\to}\vartheta,$

where $\vartheta$ is the distribution of $\max\\{X,Y\\}$, with
$X,Y\sim\mathrm{Stab}(1,1)$ independent. Note that the cumulative distribution
function of $\vartheta$ is the square of that of $\mathrm{Stab}(1,1)$.

The constant $c(r)$ in Theorem 3 depends only on the denominator of $r$ in its
reduced form. If $r=l/m$ with some coprime integers $l$ and $m$, then
$c(r)=(6\theta_{m}+\gamma+\log(2\pi))/\pi^{2}$, where $\gamma$ is the
Euler–Mascheroni constant, and111Throughout the paper, we use the convention
$0\log 0=0$.

$\theta_{m}=\sum_{a\in\mathbb{Z}_{m}}\nu_{a}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right)\log\left(\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right)\right).$
(9)

We rely on an explicit formula for the maximum and the minimum of
$S_{N,r}(\alpha)$ in terms of the sequences $q_{n}\pmod{m}$ and $a_{n}$ due to
Roçadas and Schoissengeier [24], see (20) in Section 6. Theorem 3 then follows
from the fact that the sequence $(a_{n},P_{n})$ is $\psi$-mixing with
exponential rate, as established in Lemma 14. It would be interesting to see
whether Theorem 3 holds with a fixed irrational $r$ as well. The maximum and
the minimum of the Birkhoff sum $\sum_{n=1}^{N}(\\{n\alpha\\}-1/2)$ satisfies
the same limit law as in Theorem 3 with suitable normalizing constants, but
the proof is based on the theory of quantum modular forms instead [5].

Relation (7) and Theorem 3 concern $|S_{N,r}(\alpha)|$, which is sometimes
called the local discrepancy of the Kronecker sequence $\\{n\alpha\\}$ at
$r\in(0,1)$. Taking the supremum over all subintervals $I\subseteq[0,1]$ leads
to the notion of discrepancy:

$D_{N}(\alpha)=\sup_{I\subseteq[0,1]}\left|\sum_{n=1}^{N}\mathds{1}_{I}(\\{n\alpha\\})-\lambda(I)N\right|.$

Kesten [16] proved that if $\alpha\sim\lambda$, then

$\begin{split}\frac{D_{N}(\alpha)}{\log N\log\log
N}&\to\frac{2}{\pi^{2}}\quad\textrm{in measure, as }N\to\infty,\\\
\frac{\displaystyle{\max_{0\leq N<M}D_{N}(\alpha)}}{\log M\log\log
M}&\to\frac{3}{\pi^{2}}\quad\textrm{in measure, as }M\to\infty.\end{split}$

See also [25]. It remains a challenging open problem to show whether

$\frac{D_{N}(\alpha)-(2/\pi^{2})\log N\log\log N}{\log
N}\quad\textrm{and}\quad\frac{\displaystyle{\max_{0\leq
N<M}D_{N}(\alpha)-(3/\pi^{2})\log M\log\log M}}{\log M}$

have nondegenerate limit distributions.

##  3 The group $\mathrm{SL}(2,\mathbb{Z}_{m})$

In this section, we recall some basic facts about the group
$\mathrm{SL}(2,\mathbb{Z}_{m})$. Fix an integer $m\geq 2$, let $G_{D}$ be the
set of $2\times 2$ matrices with entries in $\mathbb{Z}_{m}$ of determinant
$D\in\mathbb{Z}_{m}^{*}$, and let
$V=\\{(a,b)\in\mathbb{Z}_{m}\,:\,\mathrm{gcd}(a,b,m)=1\\}$, as in the
Introduction. Let $\nu$ be the probability measure on $\mathbb{Z}_{m}$ defined
in (4).

######  Lemma 4.

Let $D\in\mathbb{Z}_{m}^{*}$. We have $|G_{D}|=m|V|=m^{3}\prod_{p\mid
m}\left(1-\frac{1}{p^{2}}\right)$. If
$U=\left(\begin{array}[]{cc}u_{1}&u_{2}\\\
u_{3}&u_{4}\end{array}\right)\sim\mathrm{Unif}(G_{D})$, then
$(u_{1},u_{2}),(u_{3},u_{4}),(u_{1},u_{3}),(u_{2},u_{4})\sim\mathrm{Unif}(V)$,
and $u_{1},u_{2},u_{3},u_{4}\sim\nu$.

###### Proof.

One readily checks that the value of $\mathrm{gcd}(a,b,m)$ is invariant under
multiplying the row vector $(a,b)\in\mathbb{Z}_{m}^{2}$ by a matrix in
$\mathrm{GL}(2,\mathbb{Z}_{m})$ from the right. In particular, the group
$\mathrm{GL}(2,\mathbb{Z}_{m})$ acts on the set $V$ by multiplication.

First, let $D=1$. Recall that given $a,b,e\in\mathbb{Z}_{m}$, the linear
congruence $ax+by=e$ has a solution $x,y\in\mathbb{Z}_{m}$ if and only if
$\mathrm{gcd}(a,b,m)\mid e$. In particular, given any $(a,b)\in V$, there
exist $c,d\in\mathbb{Z}_{m}$ such that $ad-bc=1$. Therefore we can obtain any
$(a,b)\in V$ by multiplying the row vector $(1,0)\in V$ by a suitable matrix
in $G_{1}$, showing that $G_{1}$ acts transitively on $V$. It immediately
follows that if $U\sim\mathrm{Unif}(G_{1})$, then its two rows are uniformly
distributed on $V$. Since transposition is a bijection of $G_{1}$, the two
columns of $U$ are also uniformly distributed on $V$.

The stabilizer of the row vector $(1,0)\in V$ under the action of $G_{1}$ is
$\left\\{\left(\begin{array}[]{cc}1&0\\\
a&1\end{array}\right)\,:\,a\in\mathbb{Z}_{m}\right\\}$, which has size $m$.
Hence $|G_{1}|=m|V|$.

Now fix $a\in\mathbb{Z}_{m}$. Since
$\mathrm{gcd}(a,b,m)=\mathrm{gcd}(b,\mathrm{gcd}(a,m))$, we have

$|\\{b\in\mathbb{Z}_{m}\,:\,(a,b)\in
V\\}|=\frac{m}{\mathrm{gcd}(a,m)}\varphi(\mathrm{gcd}(a,m))=m\prod_{p\mid\mathrm{gcd}(a,m)}\left(1-\frac{1}{p}\right),$

where $\varphi$ is the Euler totient function. Given a divisor $d\mid m$, we
have $|\\{a\in\mathbb{Z}_{m}\,:\,\mathrm{gcd}(a,m)=d\\}|=\varphi(m/d)$, hence

$|V|=\sum_{a\in\mathbb{Z}_{m}}\frac{m}{\mathrm{gcd}(a,m)}\varphi(\mathrm{gcd}(a,m))=\sum_{d\mid
m}\frac{m}{d}\varphi(d)\varphi\left(\frac{m}{d}\right)=m^{2}\prod_{p\mid
m}\left(1-\frac{1}{p^{2}}\right).$

The last step can be seen using the prime factorization of $m$. Since the rows
and columns of $U$ are uniformly distributed on $V$, the previous two formulas
show that each entry $u_{i}$ of $U$ attains $a\in\mathbb{Z}_{m}$ with
probability $\nu_{a}$. This finishes the proof for the case $D=1$. The claims
for general $D\in\mathbb{Z}_{m}^{*}$ follow immediately from the fact that
$G_{D}$ is a coset of $G_{1}$ in $\mathrm{GL}(2,\mathbb{Z}_{m})$. ∎

######  Lemma 5.

We have

$\left\\{\prod_{i=1}^{4}\left(\begin{array}[]{cc}0&1\\\
1&x_{i}\end{array}\right)\,:\,x_{1},x_{2},x_{3},x_{4}\in\mathbb{Z}_{m}\right\\}=G_{1}.$

Let $m^{\prime}=\prod_{p\mid m}p$ denote the greatest square-free divisor of
$m$. More precisely, for any $g\in G_{1}$ there exist at least
$\varphi(m^{\prime})$ elements $y\in\mathbb{Z}_{m^{\prime}}$ such that
whenever $x_{4}\in\mathbb{Z}_{m}$ satisfies $x_{4}\equiv y\pmod{m^{\prime}}$,
the equation $\prod_{i=1}^{4}\left(\begin{array}[]{cc}0&1\\\
1&x_{i}\end{array}\right)=g$ has a solution in the variables
$x_{1},x_{2},x_{3}\in\mathbb{Z}_{m}$.

###### Proof.

Fix $a,b,c,d\in\mathbb{Z}_{m}$, $ad-bc=1$, and consider the equation
$\prod_{i=1}^{4}\left(\begin{array}[]{cc}0&1\\\
1&x_{i}\end{array}\right)=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)$. Multiplying the four matrices, we find that this
matrix equation is equivalent to the system of four equations

$\displaystyle 1+x_{2}x_{3}$ $\displaystyle=a,$ $\displaystyle
x_{2}+x_{4}+x_{2}x_{3}x_{4}$ $\displaystyle=b,$ $\displaystyle
x_{1}+x_{3}+x_{1}x_{2}x_{3}$ $\displaystyle=c,$ $\displaystyle
1+x_{1}x_{2}+x_{1}x_{4}+x_{3}x_{4}+x_{1}x_{2}x_{3}x_{4}$ $\displaystyle=d.$

Substituting $x_{2}x_{3}=a-1$ into the second and third equation, we see that
$x_{2}=b-ax_{4}$ and $x_{3}=c-ax_{1}$. We are left with two variables
$x_{1},x_{4}$, and using the assumption $ad-bc=1$, the system turns out to be
equivalent to the single equation

$ax_{1}x_{4}-bx_{1}-cx_{4}+d-1=0.$

It will thus be enough to prove that there exist at least
$\varphi(m^{\prime})$ elements $y\in\mathbb{Z}_{m^{\prime}}$ such that
whenever $x_{4}\in\mathbb{Z}_{m}$ satisfies $x_{4}\equiv y\pmod{m^{\prime}}$,
then $ax_{4}-b\in\mathbb{Z}_{m}^{*}$. Indeed, then the remaining linear
congruence $(ax_{4}-b)x_{1}=cx_{4}-d+1$ has a solution in the variable
$x_{1}\in\mathbb{Z}_{m}$.

Given any $p\mid m$, the congruence $ax-b\equiv 0\pmod{p}$ has at most one
solution in the variable $x\in\mathbb{Z}_{p}$. Indeed, if $p\nmid a$, then
this follows from the fact that $\mathbb{Z}_{p}$ is a field. If $p\mid a$,
then the assumption $ad-bc=1$ implies that $p\nmid b$, and there are no
solutions. By the Chinese remainder theorem, there are at least $\prod_{p\mid
m}(p-1)=\varphi(m^{\prime})$ elements $y\in\mathbb{Z}_{m^{\prime}}$ such that
$ay-b\not\equiv 0\pmod{p}$ for all $p\mid m$. Whenever
$x_{4}\in\mathbb{Z}_{m}$ satisfies $x_{4}\equiv y\pmod{m^{\prime}}$, we have
$ax_{4}-b\not\equiv 0\pmod{p}$ for all $p\mid m$, and in particular,
$ax_{4}-b\in\mathbb{Z}_{m}^{*}$. ∎

######  Remark.

It follows that

$\left\\{\prod_{i=1}^{5}\left(\begin{array}[]{cc}0&1\\\
1&x_{i}\end{array}\right)\,:\,x_{1},x_{2},x_{3},x_{4},x_{5}\in\mathbb{Z}_{m}\right\\}=G_{-1}.$

In fact, we could even prescribe the value of $x_{5}$. The number of factors 4
resp. 5 needed to generate $G_{1}$ resp. $G_{-1}$ is sharp: it is easy to
check that whenever $m>2$,

$\left\\{\prod_{i=1}^{2}\left(\begin{array}[]{cc}0&1\\\
1&x_{i}\end{array}\right)\,:\,x_{1},x_{2}\in\mathbb{Z}_{m}\right\\}\neq
G_{1}\quad\textrm{and}\quad\left\\{\prod_{i=1}^{3}\left(\begin{array}[]{cc}0&1\\\
1&x_{i}\end{array}\right)\,:\,x_{1},x_{2},x_{3}\in\mathbb{Z}_{m}\right\\}\neq
G_{-1}.$

For the sake of completeness, we mention that in the case $m=2$ the group
$G_{1}=G_{-1}$ can be generated using $3$ factors.

##  4 The normalizing constants in Kesten’s limit laws

Kesten [18, Section 4] gave the following explicit formula for the value of
$\sigma$ in (7). Let

$V(x,u,y)=\frac{2}{\pi^{2}}\sum_{k=1}^{\infty}\frac{\sin(2\pi kx)\sin(\pi
ku)\cos(2\pi ky)}{k^{2}},$

and let $F(u)=\int_{0}^{1}\int_{0}^{1}|V(x,u,y)|\,\mathrm{d}x\,\mathrm{d}y$.
Then

$\sigma=\left\\{\begin{array}[]{ll}(6/\pi)\sum_{a\in\mathbb{Z}_{m}}\nu_{a}F(al/m)&\textrm{if
$r=l/m$ with some coprime integers $l$ and $m$},\\\
(6/\pi)\int_{0}^{1}F(u)\,\mathrm{d}u&\textrm{if $r$ is
irrational}.\end{array}\right.$

Kesten even gave the hint to use the classical Fourier series

$\sum_{k=1}^{\infty}\frac{\cos(2\pi
kx)}{2\pi^{2}k^{2}}=\frac{1}{2}\\{x\\}^{2}-\frac{1}{2}\\{x\\}+\frac{1}{12}$
(10)

in order to compute $F(u)$, although he did not follow through his own advice.
This is exactly what we shall do.

###### Proof of Theorem 2.

Trigonometric identities show that

$\begin{split}&4\sin(2\pi kx)\sin(\pi ku)\cos(2\pi ky)=\\\ &\cos(2\pi
k(x+y-u/2))+\cos(2\pi k(x-y-u/2))-\cos(2\pi k(x+y+u/2))-\cos(2\pi
k(x-y+u/2)).\end{split}$

Letting $B(x)=\\{x\\}^{2}/2-\\{x\\}/2+1/12$ denote the second Bernoulli
polynomial, (10) thus gives

$V(x,u,y)=B(x+y-u/2)+B(x-y-u/2)-B(x+y+u/2)-B(x-y+u/2).$

Using the fact that $(x,y)\mapsto(x+y,x-y)$ is a measure preserving map of
$\mathbb{R}^{2}/\mathbb{Z}^{2}$, we obtain

$\begin{split}F(u)&=\int_{0}^{1}\int_{0}^{1}|B(x-u/2)+B(y-u/2)-B(x+u/2)-B(y+u/2)|\,\mathrm{d}x\,\mathrm{d}y\\\
&=\int_{0}^{1}\int_{0}^{1}|B(x)+B(y)-B(x+u)-B(y+u)|\,\mathrm{d}x\,\mathrm{d}y.\end{split}$

In particular, $F(u)$ is $1$-periodic. Assuming now $x,y,u\in[0,1)$, we have

$B(x)-B(x+u)=\frac{1}{2}(\\{x\\}-\\{x+u\\})(\\{x\\}+\\{x+u\\}-1)=\left\\{\begin{array}[]{ll}-\frac{u}{2}(2x+u-1)&\textrm{if
}0\leq x<1-u,\\\ \frac{1-u}{2}(2x+u-2)&\textrm{if }1-u\leq
x<1,\end{array}\right.$

and a similar formula holds for $B(y)-B(y+u)$. Elementary calculations yield

$\begin{split}\int_{0}^{1-u}\int_{0}^{1-u}|B(x)+B(y)-B(x+u)-B(y+u)|\,\mathrm{d}x\,\mathrm{d}y&=u\int_{0}^{1-u}\int_{0}^{1-u}|x+y+u-1|\,\mathrm{d}x\,\mathrm{d}y\\\
&=\frac{u(1-u)^{3}}{3},\end{split}$

and

$\begin{split}\int_{1-u}^{1}\int_{1-u}^{1}|B(x)+B(y)-B(x+u)-B(y+u)|\,\mathrm{d}x\,\mathrm{d}y&=(1-u)\int_{1-u}^{1}\int_{1-u}^{1}|x+y+u-2|\,\mathrm{d}x\,\mathrm{d}y\\\
&=\frac{u^{3}(1-u)}{3}.\end{split}$

Further,

$\begin{split}\int_{0}^{1-u}\int_{1-u}^{1}|B(x)+B(y)-B(x+u)-B(y+u)|\,\mathrm{d}x\,\mathrm{d}y&=\int_{0}^{1-u}\int_{1-u}^{1}|(1-u)x-uy-(1-u)^{2}|\,\mathrm{d}x\,\mathrm{d}y\\\
&=\frac{u^{2}(1-u)^{2}}{3},\end{split}$

and by symmetry the integral on $[0,1-u]\times[1-u,1]$ is the same. Therefore

$F(u)=\frac{u(1-u)^{3}}{3}+\frac{u^{3}(1-u)}{3}+\frac{2u^{2}(1-u)^{2}}{3}=\frac{u(1-u)}{3},\quad
u\in[0,1),$

and by periodicity, $F(u)=\\{u\\}(1-\\{u\\})/3$, $u\in\mathbb{R}$. For
irrational $r\in(0,1)$ we thus have

$\sigma=\frac{6}{\pi}\int_{0}^{1}\frac{u(1-u)}{3}\,\mathrm{d}u=\frac{1}{3\pi},$

as claimed.

Now let $r=l/m$ with some coprime integers $l$ and $m$. Since $\nu_{a}$ is
invariant under multiplication by $l\in\mathbb{Z}_{m}^{*}$, we have

$\sigma=\frac{2}{\pi}\sum_{a\in\mathbb{Z}_{m}}\nu_{a}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right).$

It remains to show that $\sigma$ does not depend on the denominator $m$
either.

Let $\mathcal{P}$ denote the set of prime divisors of $m$. For any
$I\subseteq\mathcal{P}$, let

$w(I)=\prod_{p\in I}\left(1-\frac{1}{p}\right)=\sum_{J\subseteq
I}\frac{(-1)^{|J|}}{\prod_{p\in J}p},\quad\textrm{and}\quad
s(I)=\sum_{\begin{subarray}{c}a\in\mathbb{Z}_{m}\\\ \prod_{p\in I}p\mid
a\end{subarray}}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right).$

In particular, $\nu_{a}=w(I)/(m\prod_{p\mid m}(1-1/p^{2}))$ with
$I=\\{p\in\mathcal{P}\,:\,p\mid a\\}$. By inclusion-exclusion,

$\sum_{a\in\mathbb{Z}_{m}}\nu_{a}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right)=\frac{1}{m\prod_{p\mid
m}\left(1-\frac{1}{p^{2}}\right)}\sum_{I\subseteq\mathcal{P}}s(I)\sum_{J\subseteq
I}(-1)^{|I\backslash J|}w(J).$ (11)

Observe that given any divisor $d\mid\mathrm{gcd}(a,m)$,

$\sum_{\begin{subarray}{c}a\in\mathbb{Z}_{m}\\\ d\mid
a\end{subarray}}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right)=\sum_{k=1}^{m/d-1}\frac{kd}{m}\left(1-\frac{kd}{m}\right)=\frac{1}{6}\left(\frac{m}{d}-\frac{d}{m}\right).$

In particular,

$s(I)=\frac{1}{6}\left(\frac{m}{\prod_{p\in I}p}-\frac{\prod_{p\in
I}p}{m}\right).$

Another application of inclusion-exclusion shows that

$\sum_{J\subseteq I}(-1)^{|I\backslash J|}w(J)=\frac{(-1)^{|I|}}{\prod_{p\in
I}p}.$

The previous two formulas simplify (11) to

$\begin{split}\sum_{a\in\mathbb{Z}_{m}}\nu_{a}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right)&=\frac{1}{m\prod_{p\mid
m}\left(1-\frac{1}{p^{2}}\right)}\sum_{I\subseteq\mathcal{P}}\frac{1}{6}\left(\frac{m}{\prod_{p\in
I}p}-\frac{\prod_{p\in I}p}{m}\right)\frac{(-1)^{|I|}}{\prod_{p\in I}p}\\\
&=\frac{1}{6\prod_{p\mid
m}\left(1-\frac{1}{p^{2}}\right)}\sum_{I\subseteq\mathcal{P}}\frac{(-1)^{|I|}}{\prod_{p\in
I}p^{2}}=\frac{1}{6}.\end{split}$ (12)

Note that we used $\sum_{I\subseteq\mathcal{P}}(-1)^{|I|}=0$. Therefore
$\sigma=1/(3\pi)$ for rational $r$ as well, as claimed. ∎

The explicit formula for $\sigma^{\prime}$ in (8) is [17]

$\sigma^{\prime}=\frac{6}{\pi^{3}}\int_{0}^{1}\int_{0}^{1}\left|\sum_{k=1}^{\infty}\frac{\sin(2\pi
kx)\sin(2\pi ky)}{k^{2}}\right|\,\mathrm{d}x\,\mathrm{d}y.$

A trigonometric identity and the Fourier series (10) lead to

$\sum_{k=1}^{\infty}\frac{\sin(2\pi kx)\sin(2\pi
ky)}{k^{2}}=\sum_{k=1}^{\infty}\frac{\cos(2\pi k(x-y))-\cos(2\pi
k(x+y))}{2k^{2}}=\pi^{2}(B(x-y)-B(x+y)),$

hence

$\begin{split}\sigma^{\prime}&=\frac{6}{\pi}\int_{0}^{1}\int_{0}^{1}|B(x-y)-B(x+y)|\,\mathrm{d}x\,\mathrm{d}y=\frac{6}{\pi}\int_{0}^{1}\int_{0}^{1}|B(x)-B(y)|\,\mathrm{d}x\,\mathrm{d}y\\\
&=\frac{3}{\pi}\int_{0}^{1}\int_{0}^{1}|x^{2}-x-y^{2}+y|\,\mathrm{d}x\,\mathrm{d}y.\end{split}$

The two diagonals partition the unit square into four right triangles. The
sign of $x^{2}-x-y^{2}+y=(x-y)(x+y-1)$ is constant on these triangles, making
them convenient domains for integrating the function $|x^{2}-x-y^{2}+y|$.
Elementary calculations then yield $\sigma^{\prime}=1/(4\pi)$, as claimed.

##  5 The sequence $P_{n}$

###  5.1 Preliminaries

Let $\mathcal{B}$ denote the family of Borel subsets of $[0,1]$. Writing each
$\alpha\in[0,1]$ in continued fraction form $\alpha=[0;a_{1},a_{2},\ldots]$,
the partial quotients $a_{k}$ are thus measurable functions on
$([0,1],\mathcal{B})$. Let $\mathcal{A}_{k}^{\ell}$ denote the
$\sigma$-algebra generated by $a_{i}$, $k\leq i\leq\ell$, and similarly let
$\mathcal{A}_{k}^{\infty}$ be the $\sigma$-algebra generated by $a_{i}$,
$i\geq k$. By convention, $\mathcal{A}_{1}^{k}$ with $k=0$ is the trivial
$\sigma$-algebra. Let $T:[0,1)\to[0,1)$, $Tx=\\{1/x\\}$ if $x\neq 0$, $T0=0$
denote the Gauss map. Note that
$\mathcal{A}_{k+1}^{\infty}=\\{T^{-k}B\,:\,B\in\mathcal{B}\\}$, where $T^{-k}$
denotes the preimage with respect to the $k$th iterate of $T$.

Given a Borel probability measure $\mu$ on $[0,1]$, the partial quotients
$a_{k}$ become random variables, and we interpret, say,
$\mu(a_{k}=5)=\mu(\\{\alpha\in[0,1]\,:\,a_{k}=5\\})$ as the probability of the
event $a_{k}=5$. We also use the conditional probability notation $\mu(B\mid
A)=\mu(A\cap B)/\mu(A)$. Let

$P_{k}=\prod_{i=1}^{k}\left(\begin{array}[]{cc}0&1\\\
1&a_{i}\end{array}\right)\pmod{m}\qquad\textrm{and}\qquad
P_{k,\ell}=\prod_{i=k}^{\ell}\left(\begin{array}[]{cc}0&1\\\
1&a_{i}\end{array}\right)\pmod{m},$

with the convention that $P_{k+1,k}$ is the identity element of the group
$G_{1}$. Let $p_{k}/q_{k}=[0;a_{1},a_{2},\ldots,a_{k}]$ be the convergents.
Recall the identity $q_{k-1}/q_{k}=[0;a_{k},a_{k-1},\ldots,a_{1}]$.

We start with two preparatory lemmas.

######  Lemma 6.

For any $A,A^{\prime}\in\mathcal{A}_{1}^{k}$ and
$B\in\mathcal{A}_{k+1}^{\infty}$ such that
$\lambda(A),\lambda(A^{\prime}),\lambda(B)>0$, we have

$\frac{1}{2}\leq\frac{\lambda(B\mid A)}{\lambda(B\mid A^{\prime})}\leq 2.$

In particular, $\frac{1}{2}\lambda(A)\lambda(B)\leq\lambda(A\cap B)\leq
2\lambda(A)\lambda(B)$.

###### Proof.

Let $b_{1},b_{2},\ldots,b_{k+n}\in\mathbb{N}$ be arbitrary, and let
$p_{\ell}/q_{\ell}=[0;b_{1},\ldots,b_{\ell}]$, $\ell=k-1,k$. The set
$\\{a_{1}=b_{1},a_{2}=b_{2},\ldots,a_{k+n}=b_{k+n}\\}$ is an interval with
endpoints $\frac{p_{k}r+p_{k-1}}{q_{k}r+q_{k-1}}$ and
$\frac{p_{k}r^{*}+p_{k-1}}{q_{k}r^{*}+q_{k-1}}$, where
$r=[b_{k+1};b_{k+2},\ldots,b_{k+n}]$ and
$r^{*}=[b_{k+1};b_{k+2},\ldots,b_{k+n}+1]$. Using the identity
$q_{k}p_{k-1}-q_{k-1}p_{k}=(-1)^{k}$, the Lebesgue measure of this interval
simplifies to

$\lambda(a_{1}=b_{1},a_{2}=b_{2},\ldots,a_{k+n}=b_{k+n})=\frac{|r-r^{*}|}{(q_{k}r+q_{k-1})(q_{k}r^{*}+q_{k-1})}.$

The set $\\{a_{1}=b_{1},a_{2}=b_{2},\ldots,a_{k}=b_{k}\\}$ is also an
interval, with endpoints $\frac{p_{k}+p_{k-1}}{q_{k}+q_{k-1}}$ and
$\frac{p_{k}}{q_{k}}$, and has Lebesgue measure

$\lambda(a_{1}=b_{1},a_{2}=b_{2},\ldots,a_{k}=b_{k})=\frac{1}{q_{k}(q_{k}+q_{k-1})}.$

Hence

$\lambda(a_{k+1}=b_{k+1},\ldots,a_{k+n}=b_{k+n}\mid
a_{1}=b_{1},\ldots,a_{k}=b_{k})=|r-r^{*}|\frac{1+\frac{q_{k-1}}{q_{k}}}{\left(r+\frac{q_{k-1}}{q_{k}}\right)\left(r^{*}+\frac{q_{k-1}}{q_{k}}\right)}.$

Here $r,r^{*}$ do not depend on $b_{1},\ldots,b_{k}$. Therefore for any
$b_{1}^{\prime},\ldots,b_{k}^{\prime}\in\mathbb{N}$, we have

$\frac{\lambda(a_{k+1}=b_{k+1},\ldots,a_{k+n}=b_{k+n}\mid
a_{1}=b_{1},\ldots,a_{k}=b_{k})}{\lambda(a_{k+1}=b_{k+1},\ldots,a_{k+n}=b_{k+n}\mid
a_{1}=b_{1}^{\prime},\ldots,a_{k}=b_{k}^{\prime})}=\frac{\frac{1+\frac{q_{k-1}}{q_{k}}}{\left(r+\frac{q_{k-1}}{q_{k}}\right)\left(r^{*}+\frac{q_{k-1}}{q_{k}}\right)}}{\frac{1+\frac{q_{k-1}^{\prime}}{q_{k}^{\prime}}}{\left(r+\frac{q_{k-1}^{\prime}}{q_{k}^{\prime}}\right)\left(r^{*}+\frac{q_{k-1}^{\prime}}{q_{k}^{\prime}}\right)}},$

where
$p_{\ell}^{\prime}/q_{\ell}^{\prime}=[0;b_{1}^{\prime},\ldots,b_{\ell}^{\prime}]$,
$\ell=k-1,k$.

We claim that the right hand side of the previous formula lies in the interval
$[1/2,2]$. Letting

$g(x)=\frac{1+x}{\left(r+x\right)\left(r^{*}+x\right)},\qquad x\in[0,1],$

it will be enough to show that $g(x)/g(y)\leq 2$ for all $x,y\in[0,1]$. Note
that $r,r^{*}\geq 1$, and that $g^{\prime}(x)\geq
0\Leftrightarrow-1-\sqrt{(r-1)(r^{*}-1)}\leq x\leq-1+\sqrt{(r-1)(r^{*}-1)}$.

Case 1. Assume that $(r-1)(r^{*}-1)\leq 1$. Then $g(x)$ is decreasing on
$[0,1]$, hence

$\frac{g(x)}{g(y)}\leq\frac{g(0)}{g(1)}=\frac{(r+1)(r^{*}+1)}{2rr^{*}}\leq 2.$

Case 2. Assume that $1<(r-1)(r^{*}-1)\leq 2$. Then $g(x)$ attains its minimum
at $x=1$ and its maximum at $x=-1+\sqrt{(r-1)(r^{*}-1)}$. Hence

$\frac{g(x)}{g(y)}\leq\frac{g(-1+\sqrt{(r-1)(r^{*}-1)})}{g(1)}=\frac{(r+1)(r^{*}+1)}{2\left(\sqrt{r-1}+\sqrt{r^{*}-1}\right)^{2}}\leq\frac{2+2(r-1)+2(r^{*}-1)+4}{2(r-1+r^{*}-1+2)}\leq
2.$

Case 3. Assume that $2<(r-1)(r^{*}-1)\leq 4$. Then $g(x)$ attains its minimum
at $x=0$ and its maximum at $x=-1+\sqrt{(r-1)(r^{*}-1)}$. Hence

$\frac{g(x)}{g(y)}\leq\frac{g(-1+\sqrt{(r-1)(r^{*}-1)})}{g(0)}=\frac{rr^{*}}{\left(\sqrt{r-1}+\sqrt{r^{*}-1}\right)^{2}}\leq\frac{4+r-1+r^{*}-1+1}{r-1+r^{*}-1+2\sqrt{2}}\leq
2.$

Case 4. Assume that $(r-1)(r^{*}-1)>4$. Then $g(x)$ is increasing on $[0,1]$,
hence

$\frac{g(x)}{g(y)}\leq\frac{g(1)}{g(0)}=\frac{2rr^{*}}{(r+1)(r^{*}+1)}\leq 2.$

This finishes the proof of $g(x)/g(y)\leq 2$ for all $x,y\in[0,1]$. In
particular,

$\frac{1}{2}\leq\frac{\lambda(a_{k+1}=b_{k+1},\ldots,a_{k+n}=b_{k+n}\mid
a_{1}=b_{1},\ldots,a_{k}=b_{k})}{\lambda(a_{k+1}=b_{k+1},\ldots,a_{k+n}=b_{k+n}\mid
a_{1}=b_{1}^{\prime},\ldots,a_{k}=b_{k}^{\prime})}\leq 2.$

By $\sigma$-additivity, for all $A,A^{\prime}\in\mathcal{A}_{1}^{k}$ and all
$B\in\mathcal{A}_{k+1}^{k+n}$ of positive Lebesgue measure,

$\frac{1}{2}\leq\frac{\lambda(B\mid A)}{\lambda(B\mid A^{\prime})}\leq 2.$

Since $\mathcal{A}_{k+1}^{k+n}$, $n\in\mathbb{N}$ generate
$\mathcal{A}_{k+1}^{\infty}$, the same holds for all
$B\in\mathcal{A}_{k+1}^{\infty}$ of positive Lebesgue measure. ∎

######  Lemma 7.

  1. (i)

For any $0\leq n\leq 3$ and $g\in G_{(-1)^{n}}$, we have either
$\lambda(P_{n}=g)=0$ or $\lambda(P_{n}=g)\geq 1/(m+1)^{6}$.

  2. (ii)

For any $g\in G_{1}$, we have

$\lambda(P_{4}=g)\geq\frac{\pi^{6}}{216}\cdot\frac{\varphi(m^{\prime})}{(m+1)^{6}(m^{\prime}+1)(m^{\prime}-\varphi(m^{\prime})+1)}.$

  3. (iii)

For any $n\geq 5$ and $g\in G_{(-1)^{n}}$, we have

$\lambda(P_{n}=g)\geq\frac{\pi^{6}}{432}\cdot\frac{\varphi(m^{\prime})}{(m+1)^{6}(m^{\prime}+1)(m^{\prime}-\varphi(m^{\prime})+1)}.$

###### Proof.

(i) Let $0\leq n\leq 3$. Either $\lambda(P_{n}=g)=0$, or there exist positive
integers $b_{1},b_{2},b_{3}\leq m$ such that

$\lambda(P_{n}=g)\geq\lambda(a_{1}=b_{1},a_{2}=b_{2},a_{3}=b_{3})\geq\frac{1}{(m+1)^{6}}.$

(ii) Let $g\in G_{1}$, and consider the equation

$\prod_{i=1}^{4}\left(\begin{array}[]{cc}0&1\\\
1&b_{i}\end{array}\right)\pmod{m}=g$ (13)

in the variables $b_{1},b_{2},b_{3},b_{4}\in\mathbb{N}$. We have

$\lambda(P_{4}=g)=\sum_{b_{1},b_{2},b_{3},b_{4}}\lambda(a_{1}=b_{1},a_{2}=b_{2},a_{3}=b_{3},a_{4}=b_{4})=\sum_{b_{1},b_{2},b_{3},b_{4}}\frac{1}{q_{4}(q_{4}+q_{3})},$

where the sums are over the set of solutions of the equation (13), and $q_{4}$
resp. $q_{3}$ is the denominator of $[0;b_{1},b_{2},b_{3},b_{4}]$ resp.
$[0;b_{1},b_{2},b_{3}]$. Let $\ell_{1},\ell_{2},\ell_{3}\in\mathbb{N}$. Lemma
5 implies that there exist at least $\varphi(m^{\prime})$ integers $1\leq
b_{4}\leq m^{\prime}$ for which the equation (13) has an integer solution
$(\ell_{i}-1)m<b_{i}\leq\ell_{i}m$, $i=1,2,3$. One readily checks that for all
such solutions of (13), we have $q_{3}\leq\ell_{1}\ell_{2}\ell_{3}(m+1)^{3}$
and $q_{4}\leq\ell_{1}\ell_{2}\ell_{3}(m+1)^{3}b_{4}$. Hence

$\begin{split}\lambda(P_{4}=g)&\geq\sum_{b_{4}}\sum_{\ell_{1},\ell_{2},\ell_{3}=1}^{\infty}\frac{1}{\ell_{1}^{2}\ell_{2}^{2}\ell_{3}^{2}(m+1)^{6}b_{4}(b_{4}+1)}\geq\left(\frac{\pi^{2}}{6}\right)^{3}\frac{1}{(m+1)^{6}}\sum_{j=m^{\prime}-\varphi(m^{\prime})+1}^{m^{\prime}}\frac{1}{j(j+1)}\\\
&=\frac{\pi^{6}}{216}\cdot\frac{\varphi(m^{\prime})}{(m+1)^{6}(m^{\prime}+1)(m^{\prime}-\varphi(m^{\prime})+1)}.\end{split}$

(iii) Let $n\geq 5$ and $g\in G_{(-1)^{n}}$. Lemma 6 and part (ii) show that

$\begin{split}\lambda(P_{n}=g)&=\sum_{h\in
G_{(-1)^{n}}}\lambda(P_{4}=gh^{-1},P_{5,n}=h)\geq\sum_{h\in
G_{(-1)^{n}}}\frac{1}{2}\lambda(P_{4}=gh^{-1})\lambda(P_{5,n}=h)\\\
&\geq\frac{\pi^{6}}{432}\cdot\frac{\varphi(m^{\prime})}{(m+1)^{6}(m^{\prime}+1)(m^{\prime}-\varphi(m^{\prime})+1)}.\end{split}$

∎

###  5.2 A Gauss–Kuzmin–Lévy theorem

In this section, we prove a version of the Gauss–Kuzmin–Lévy theorem, which
will serve as the main tool of this paper. We refer to [14, Chapter 2] for a
comprehensive account of the Gauss–Kuzmin problem. Theorem 8 below generalizes
results of Kesten [17, Lemma 2.5] and Szüsz [26]. They both considered the
pair $(q_{k-1},q_{k})\pmod{m}$ and $\alpha\sim\lambda$, whereas we work with
the matrix $P_{k}$, and allow the distribution of $\alpha$ to have a Lipschitz
density.

######  Theorem 8.

Let $\mu\ll\lambda$, and assume (5). For any $A\in\mathcal{A}_{1}^{k}$, any
$g\in G_{(-1)^{n}}$ and any $B\in\mathcal{A}_{k+n+1}^{\infty}$,

$\left|\mu\left(A\cap\\{P_{k+1,k+n}=g\\}\cap
B\right)-\frac{\mu(A)\mu_{\mathrm{Gauss}}(B)}{|G_{1}|}\right|\leq
C\lambda(A)\lambda(B)e^{-\tau n},$

where, with $m^{\prime}=\prod_{p\mid m}p$ denoting the greatest square-free
divisor of $m$,

$C=4L+3\quad\textrm{and}\quad\tau=\frac{\varphi(m^{\prime})}{12(m+1)^{6}(m^{\prime}+1)(m^{\prime}-\varphi(m^{\prime})+1)}\geq\frac{1}{12(m+1)^{8}}.$
(14)

###### Proof.

Throughout the proof we fix $k\geq 0$ and a set of the form
$A=\\{a_{1}=b_{1},\ldots,a_{k}=b_{k}\\}$, with the convention $A=[0,1]$ if
$k=0$. It will be enough to prove the theorem for this set $A$, as the claim
for a general $A\in\mathcal{A}_{1}^{k}$ then follows by $\sigma$-additivity.

Given a Lipschitz function $F:[0,1]\to\mathbb{R}$, let

$\|F\|_{\mathrm{Lip}}=\sup_{\begin{subarray}{c}x,y\in[0,1]\\\ x\neq
y\end{subarray}}\frac{|F(x)-F(y)|}{|x-y|}=\operatorname*{ess\,sup}_{x\in[0,1]}|F^{\prime}(x)|$

denote the Lipschitz constant. We implicitly use the fact that Lipschitz
functions are a.e. differentiable, and satisfy the fundamental theorem of
calculus.

For any $n\geq 0$ and $g\in G_{(-1)^{n}}$, the measure
$B\mapsto\mu(A\cap\\{P_{k+1,k+n}=g\\}\cap T^{-(k+n)}B)$, $B\in\mathcal{B}$ is
absolutely continuous. Let $f_{n,g}$ denote its density with respect to
$\lambda$. It will be enough to prove that

$\sup_{\begin{subarray}{c}g\in G_{(-1)^{n}}\\\
x\in[0,1]\end{subarray}}\left|f_{n,g}(x)-\frac{\mu(A)}{|G_{1}|(\log
2)(1+x)}\right|\leq C\lambda(A)e^{-\tau n}.$ (15)

Let $F_{n,g}(x)=(\log 2)(1+x)f_{n,g}(x)$, $x\in[0,1]$. We start with the case
$n=0$.

######  Lemma 9.

For any $g\in G_{1}$, the functions $f_{0,g}$ and $F_{0,g}$ are Lipschitz, and
we have

$\|F_{0,g}\|_{\mathrm{Lip}}\leq(\log 2)(3L+2)\lambda(A)\quad\textrm{and}\quad
0\leq F_{0,g}(x)\leq(\log 2)(L+2)\lambda(A).$

###### Proof.

Assumption (5) implies that for all $x\in[0,1]$,

$\left|\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(x)-1\right|=\left|\int_{0}^{1}\left(\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(x)-\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(y)\right)\,\mathrm{d}y\right|\leq\int_{0}^{1}L|x-y|\,\mathrm{d}y\leq\frac{L}{2}.$

In particular,
$\max_{[0,1]}\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}\leq\frac{L}{2}+1$.

By construction,

$\int_{0}^{x}f_{0,g}(t)\,\mathrm{d}t=\left\\{\begin{array}[]{ll}\mu(a_{1}=b_{1},\ldots,a_{k}=b_{k},[0;a_{k+1},a_{k+2},\ldots]\leq
x)&\textrm{if }g=1\in G_{1},\\\ 0&\textrm{if }g\neq 1\in
G_{1}.\end{array}\right.$

The set $\\{a_{1}=b_{1},\ldots,a_{k}=b_{k},[0;a_{k+1},a_{k+2},\ldots]\leq
x\\}$ is an interval with endpoints $p_{k}/q_{k}$ and
$(p_{k}+p_{k-1}x)/(q_{k}+q_{k-1}x)$. Differentiating the previous formula with
respect to $x$ thus gives

$f_{0,g}(x)=\left\\{\begin{array}[]{ll}\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}\left(\frac{p_{k}+p_{k-1}x}{q_{k}+q_{k-1}x}\right)\frac{1}{(q_{k}+q_{k-1}x)^{2}}&\textrm{if
}g=1\in G_{1},\\\ 0&\textrm{if }g\neq 1\in G_{1}.\end{array}\right.$

Strictly speaking, the density function $f_{0,g}$ is only defined up to a.e.
equivalence, and Lebesgue’s differentiation theorem yields the previous
formula for a.e. $x$. However, the right-hand side is a Lipschitz function by
assumption, thus $f_{0,g}$ can be chosen to be Lipschitz. The previous formula
thus holds for all $x\in[0,1]$, $F_{0,g}$ is Lipschitz, and for a.e. $x$ we
have

$\begin{split}|F_{0,g}^{\prime}(x)|&=(\log
2)\left|\left(\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}\right)^{\prime}\left(\frac{p_{k}+p_{k-1}x}{q_{k}+q_{k-1}x}\right)\frac{(-1)^{k}(1+x)}{(q_{k}+q_{k-1}x)^{4}}+\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}\left(\frac{p_{k}+p_{k-1}x}{q_{k}+q_{k-1}x}\right)\frac{q_{k}-(2+x)q_{k-1}}{(q_{k}+q_{k-1}x)^{3}}\right|\\\
&\leq\frac{\log
2}{(q_{k}+q_{k-1}x)^{2}}\left(L\frac{1+x}{q_{k}+q_{k-1}x}+\left(\frac{L}{2}+1\right)\frac{|q_{k}-(2+x)q_{k-1}|}{q_{k}+q_{k-1}x}\right)\\\
&\leq\frac{\log
2}{q_{k}^{2}}\left(L\frac{2}{q_{k}+q_{k-1}}+\frac{L}{2}+1\right)\leq\frac{(\log
2)(3L+2)}{q_{k}(q_{k}+q_{k-1})}=(\log 2)(3L+2)\lambda(A).\end{split}$

In particular, $\|F_{0,g}\|_{\mathrm{Lip}}\leq(\log 2)(3L+2)\lambda(A)$, as
claimed. Further,

$0\leq
F_{0,g}(x)\leq\max_{[0,1]}\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}\cdot\frac{2\log
2}{q_{k}(q_{k}+q_{k-1})}\leq(\log 2)(L+2)\lambda(A).$

This finishes the proof of Lemma 9. ∎

We now prove that $f_{n,g}$ can be chosen to be Lipschitz by induction on $n$,
the base case $n=0$ having been established in Lemma 9. Let $n\geq 1$ and
$b\in\mathbb{N}$, and note that (ignoring endpoints)

$[0;a_{k+n+1},a_{k+n+2},\ldots]\leq x\textrm{ and
}a_{k+n}=b\,\Longleftrightarrow\,[0;a_{k+n},a_{k+n+1},\ldots]\in\left[\frac{1}{b+x},\frac{1}{b}\right].$

Using the partition $\\{a_{k+n}=b\\}$, $b\in\mathbb{N}$ thus leads to

$\mu(A\cap\\{P_{k+1,k+n}=g\\}\cap
T^{-(k+n)}[0,x])=\sum_{b=1}^{\infty}\mu\left(A\cap\\{P_{k+1,k+n-1}=gh(b)^{-1}\\}\cap
T^{-(k+n-1)}\left[\frac{1}{b+x},\frac{1}{b}\right]\right),$

where $h(b)=\left(\begin{array}[]{cc}0&1\\\ 1&b\end{array}\right)\pmod{m}$.
Equivalently,

$\int_{0}^{x}f_{n,g}(t)\,\mathrm{d}t=\sum_{b=1}^{\infty}\int_{\frac{1}{b+x}}^{\frac{1}{b}}f_{n-1,gh(b)^{-1}}(t)\,\mathrm{d}t.$

One readily checks that the series on the right-hand side can be
differentiated term by term using the inductive hypothesis that
$f_{n-1,gh(b)^{-1}}$ is Lipschitz, and we obtain

$f_{n,g}(x)=\sum_{b=1}^{\infty}f_{n-1,gh(b)^{-1}}\left(\frac{1}{b+x}\right)\frac{1}{(b+x)^{2}}.$

The right-hand side is easily seen to be Lipschitz, hence $f_{n,g}$ can be
chosen to be Lipschitz, and the previous formula holds for all $x\in[0,1]$.
This finishes the induction. The recursion above should be compared to the
Perron–Frobenius operator of the Gauss map [14, Chapter 2].

The recursion can be written in terms of $F_{n,g}$ as

$F_{n,g}(x)=\sum_{b=1}^{\infty}F_{n-1,gh(b)^{-1}}\left(\frac{1}{b+x}\right)\frac{1+x}{(b+x)(b+1+x)}\qquad\textrm{for
all }x\in[0,1].$ (16)

Taking the derivative leads to

$\begin{split}F_{n,g}^{\prime}(x)=\sum_{b=1}^{\infty}\bigg{(}&F_{n-1,gh(b)^{-1}}^{\prime}\left(\frac{1}{b+x}\right)\frac{-(1+x)}{(b+x)^{3}(b+1+x)}\\\
&+F_{n-1,gh(b)^{-1}}\left(\frac{1}{b+x}\right)\frac{b(b-1)-(1+x)^{2}}{(b+x)^{2}(b+1+x)^{2}}\bigg{)}\qquad\textrm{for
a.e. }x\in[0,1].\end{split}$ (17)

For comparison, note the identities

$\sum_{b=1}^{\infty}\frac{1+x}{(b+x)(b+1+x)}=1\qquad\textrm{and}\qquad\sum_{b=1}^{\infty}\frac{b(b-1)-(1+x)^{2}}{(b+x)^{2}(b+1+x)^{2}}=0.$
(18)

Define $L_{n}=\max_{g\in G_{(-1)^{n}}}\|F_{n,g}\|_{\mathrm{Lip}}$, and

$W_{n}^{-}=\min_{\begin{subarray}{c}g\in G_{(-1)^{n}}\\\
x\in[0,1]\end{subarray}}F_{n,g}(x),\qquad
W_{n}^{+}=\max_{\begin{subarray}{c}g\in G_{(-1)^{n}}\\\
x\in[0,1]\end{subarray}}F_{n,g}(x),\qquad\delta_{n}=W_{n}^{+}-W_{n}^{-}.$

The recursion (16) and the first identity in (18) immediately show that
$W_{n-1}^{-}\leq W_{n}^{-}$ and $W_{n}^{+}\leq W_{n-1}^{+}$, hence
$\delta_{n}\leq\delta_{n-1}$.

######  Lemma 10.

We have $L_{n}\leq(1-\zeta(2)+\zeta(3))L_{n-1}+(1/4)\delta_{n-1}$, where
$\zeta$ is the Riemann zeta function. The sum of the coefficients
$1-\zeta(2)+\zeta(3)=0.5571\ldots$ and $1/4$ is less than $1$.

###### Proof.

The recursion (17) and the second identity in (18) show that

$|F_{n,g}^{\prime}(x)|\leq
L_{n-1}\sum_{b=1}^{\infty}\frac{1+x}{(b+x)^{3}(b+1+x)}+\sum_{b=1}^{\infty}\left|F_{n-1,gh(b)^{-1}}\left(\frac{1}{b+x}\right)-c\right|\frac{|b(b-1)-(1+x)^{2}|}{(b+x)^{2}(b+1+x)^{2}}$

with any $c=c(g,x,y)$ which does not depend on $b$. Choosing
$c=F_{n-1,gh(1)^{-1}}(1/(1+x))$, the $b=1$ term cancels, and by the definition
of $\delta_{n-1}$ we obtain

$|F_{n,g}^{\prime}(x)|\leq
L_{n-1}\sum_{b=1}^{\infty}\frac{1+x}{(b+x)^{3}(b+1+x)}+\delta_{n-1}\sum_{b=2}^{\infty}\frac{|b(b-1)-(1+x)^{2}|}{(b+x)^{2}(b+1+x)^{2}}.$

The derivative of the first series satisfies

$\begin{split}\left(\sum_{b=1}^{\infty}\frac{1+x}{(b+x)^{3}(b+1+x)}\right)^{\prime}&=-\frac{3x+5}{(1+x)^{3}(2+x)^{2}}+\sum_{b=2}^{\infty}\frac{b^{2}-3b-3-(2b+6)x-3x^{2}}{(b+x)^{4}(b+1+x)^{2}}\\\
&\leq-\frac{1}{9}+\sum_{b=4}^{\infty}\frac{b^{2}-3b-3}{b^{4}(b+1)^{2}}=-0.1098\ldots<0.\end{split}$

Therefore the maximum is attained at $x=0$, and

$\sum_{b=1}^{\infty}\frac{1+x}{(b+x)^{3}(b+1+x)}\leq\sum_{b=1}^{\infty}\frac{1}{b^{3}(b+1)}=\sum_{b=1}^{\infty}\left(\frac{1}{b(b+1)}-\frac{1}{b^{2}}+\frac{1}{b^{3}}\right)=1-\zeta(2)+\zeta(3).$

Considering $b=2$ and $b\geq 3$ separately, we check that each term in the
second series attains its maximum at $x=0$ as well, hence

$\sum_{b=2}^{\infty}\frac{|b(b-1)-(1+x)^{2}|}{(b+x)^{2}(b+1+x)^{2}}\leq\sum_{b=2}^{\infty}\frac{b^{2}-b-1}{b^{2}(b+1)^{2}}=\sum_{b=2}^{\infty}\left(\frac{1}{b(b+1)}-\frac{1}{b^{2}}+\frac{1}{(b+1)^{2}}\right)=\frac{1}{4}.$

This finishes the proof of Lemma 10. ∎

For the sake of readability, let
$\kappa_{m}=\varphi(m^{\prime})/((m+1)^{6}(m^{\prime}+1)(m^{\prime}-\varphi(m^{\prime})+1))$.

######  Lemma 11.

We have

$\delta_{n}\leq\left(1-\frac{\pi^{6}}{432}\kappa_{m}\right)\delta_{n-4}+\frac{\pi^{6}}{432}\kappa_{m}L_{n-4}.$

###### Proof.

If $L_{n-4}\geq\delta_{n-4}$, then the desired upper bound is greater or equal
than $\delta_{n-4}$, and the claim follows from the fact that $\delta_{n}$ is
nondecreasing. We may thus assume that $L_{n-4}<\delta_{n-4}$.

Let $g\in G_{1}$ and $B\in\mathcal{B}$ be arbitrary. The partition
$\\{P_{k+n-3,k+n}=h\\}$, $h\in G_{1}$ leads to

$\mu(A\cap\\{P_{k+1,k+n}=g\\}\cap T^{-(k+n)}B)=\sum_{h\in
G_{1}}\mu(A\cap\\{P_{k+1,k+n-4}=gh^{-1}\\}\cap T^{-(k+n-4)}(\\{P_{4}=h\\}\cap
T^{-4}B)).$

By the definition of $f_{n,g}$ and $F_{n,g}$, this is equivalent to

$\int_{B}F_{n,g}(x)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x)=\sum_{h\in
G_{1}}\int_{\\{P_{4}=h\\}\cap
T^{-4}B}F_{n-4,gh^{-1}}(x)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x).$

Fix a pair $(g_{0},x_{0})\in G_{1}\times[0,1]$ at which the minimum
$F_{n-4,g_{0}}(x_{0})=W_{n-4}^{-}$ is attained. Using the bound
$F_{n-4,gh^{-1}}(x)\leq W_{n-4}^{+}$ for all $h\neq g_{0}^{-1}g$ yields

$\int_{B}F_{n,g}(x)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x)\leq
W_{n-4}^{+}\mu_{\mathrm{Gauss}}(B)+\int_{\\{P_{4}=g_{0}^{-1}g\\}\cap
T^{-4}B}\left(F_{n-4,g_{0}}(x)-W_{n-4}^{+}\right)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x).$

Here $F_{n-4,g_{0}}(x)\leq F_{n-4,g_{0}}(x_{0})+L_{n-4}|x-x_{0}|\leq
W_{n-4}^{-}+L_{n-4}$, thus

$\int_{B}F_{n,g}(x)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x)\leq
W_{n-4}^{+}\mu_{\mathrm{Gauss}}(B)+(L_{n-4}-\delta_{n-4})\mu_{\mathrm{Gauss}}(\\{P_{4}=g_{0}^{-1}g\\}\cap
T^{-4}B).$

Note that $L_{n-4}-\delta_{n-4}<0$ by assumption. Lemmas 6 and 7 show that

$\lambda(\\{P_{4}=g_{0}^{-1}g\\}\cap
T^{-4}B)\geq\frac{1}{2}\lambda(P_{4}=g_{0}^{-1}g)\lambda(T^{-4}B)\geq\frac{\pi^{6}}{432}\kappa_{m}\lambda(T^{-4}B),$

and as the density of $\mu_{\mathrm{Gauss}}$ lies between $1/(2\log 2)$ and
$1/\log 2$,

$\mu_{\mathrm{Gauss}}(\\{P_{4}=g_{0}^{-1}g\\}\cap
T^{-4}B)\geq\frac{\pi^{6}}{864}\kappa_{m}\mu_{\mathrm{Gauss}}(B).$

Therefore

$\int_{B}F_{n,g}(x)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x)\leq
W_{n-4}^{+}\mu_{\mathrm{Gauss}}(B)+(L_{n-4}-\delta_{n-4})\frac{\pi^{6}}{864}\kappa_{m}\mu_{\mathrm{Gauss}}(B).$

A similar proof shows the lower bound

$\int_{B}F_{n,g}(x)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x)\geq
W_{n-4}^{-}\mu_{\mathrm{Gauss}}(B)-(L_{n-4}-\delta_{n-4})\frac{\pi^{6}}{864}\kappa_{m}\mu_{\mathrm{Gauss}}(B).$

As the previous two formulas hold for all $B\in\mathcal{B}$, we have

$W_{n-4}^{-}-(L_{n-4}-\delta_{n-4})\frac{\pi^{6}}{864}\kappa_{m}\leq
F_{n,g}(x)\leq
W_{n-4}^{+}+(L_{n-4}-\delta_{n-4})\frac{\pi^{6}}{864}\kappa_{m},$

hence

$\delta_{n}\leq\delta_{n-4}+(L_{n-4}-\delta_{n-4})\frac{\pi^{6}}{432}\kappa_{m}.$

This finishes the proof of Lemma 11. ∎

Let $z=1-\zeta(2)+\zeta(3)$. Iterating Lemma 10 five times and using the fact
that $\delta_{n}$ is nondecreasing shows that

$L_{n}\leq z^{5}L_{n-5}+(z^{4}+z^{3}+z^{2}+z+1)\frac{1}{4}\delta_{n-5},$

where the sum of the coefficients is
$z^{5}+(z^{4}+z^{3}+z^{2}+z+1)/4=0.5878\ldots$. Lemmas 11 and 10 yield

$\delta_{n}\leq\left(1-\frac{\pi^{6}}{432}\kappa_{m}\right)\delta_{n-5}+\frac{\pi^{6}}{432}\kappa_{m}\left(zL_{n-5}+\frac{1}{4}\delta_{n-5}\right),$

where the sum of the coefficients is

$1-\frac{\pi^{6}}{432}\kappa_{m}+\frac{\pi^{6}}{432}\kappa_{m}\left(z+\frac{1}{4}\right)=1-0.08584\ldots\cdot(5\kappa_{m})>0.5879.$

We thus have the recursive upper bound
$\max\\{L_{n},\delta_{n}\\}\leq(1-0.08584\cdot(5\kappa_{m}))\max\\{L_{n-5},\delta_{n-5}\\}$.
Lemma 9 implies that $\max\\{L_{i},\delta_{i}\\}\leq(\log 2)(3L+2)\lambda(A)$
for $i=0$, and four applications of Lemma 10 shows that the same holds for
$i=1,2,3,4$. Iterating the recursive upper bound $\lfloor n/5\rfloor$ times
thus leads to

$\max\\{L_{n},\delta_{n}\\}\leq\left(1-0.08584\cdot(5\kappa_{m})\right)^{\lfloor
n/5\rfloor}(\log 2)(3L+2)\lambda(A)\leq C_{0}e^{-0.08584\kappa_{m}n}$

with $C_{0}=e^{0.08584\cdot(5\kappa_{m})}(\log 2)(3L+2)$. By the definition of
$\delta_{n}$, this means that $F_{n,g}(x)$ lies in a given interval of length
$C_{0}e^{-0.08584\kappa_{m}n}$ for all $g\in G_{(-1)^{n}}$ and $x\in[0,1]$.
Then the average value

$\frac{1}{|G_{1}|}\sum_{g\in
G_{(-1)^{n}}}\int_{0}^{1}F_{n,g}(x)\,\mathrm{d}\mu_{\mathrm{Gauss}}(x)=\frac{\mu(A)}{|G_{1}|}$

lies in the same interval, hence $|F_{n,g}(x)-\mu(A)/|G_{1}||\leq
C_{0}e^{-0.08584\kappa_{m}n}$. Formula (15) follows with

$C=\frac{C_{0}}{\log 2}\leq
4L+3\quad\textrm{and}\quad\tau=0.08584\kappa_{m}>\frac{\kappa_{m}}{12}.$

This finishes the proof of Theorem 8. ∎

We now show that the limit relation in Theorem 8 without the exponential rate
remains true for an arbitrary absolutely continuous measure, without assuming
that the density is Lipschitz.

######  Corollary 12.

Let $\mu\ll\lambda$. Then

$\lim_{n\to\infty}\sup_{k\geq
0}\sup_{\begin{subarray}{c}A\in\mathcal{A}_{1}^{k},\,\,B\in\mathcal{A}_{k+n+1}^{\infty}\\\
g\in G_{(-1)^{n}}\end{subarray}}\left|\mu\left(A\cap\\{P_{k+1,k+n}=g\\}\cap
B\right)-\frac{\mu(A)\mu_{\mathrm{Gauss}}(B)}{|G_{1}|}\right|=0.$

###### Proof.

Let $f(x)=\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(x)$ denote the density
function. Using a positive mollifier on the circle group
$\mathbb{R}/\mathbb{Z}$, we deduce that for any $\varepsilon>0$ there exists a
smooth function $f_{\varepsilon}$ on $[0,1]$ such that

$\int_{0}^{1}f_{\varepsilon}(x)\,\mathrm{d}x=1,\qquad\int_{0}^{1}|f(x)-f_{\varepsilon}(x)|\,\mathrm{d}x<\varepsilon,\qquad\min_{x\in[0,1]}f_{\varepsilon}(x)>0.$

Let $\mu_{\varepsilon}$ be the Borel probability measure on $[0,1]$ with
density $f_{\varepsilon}$. In particular,
$|\mu(A)-\mu_{\varepsilon}(A)|<\varepsilon$ for all Borel sets
$A\subseteq[0,1]$. Since $\mu_{\varepsilon}$ has a positive smooth density,
the claim holds for $\mu_{\varepsilon}$ by Theorem 8. As $\varepsilon$ was
arbitrary, the claim holds also for $\mu$. ∎

###  5.3 Weak convergence and mixing properties of $P_{n}$

Theorem 8 and Corollary 12 immediately imply that $P_{2n}$ resp. $P_{2n-1}$
converges to the uniform distribution on $G_{1}$ resp. $G_{-1}$.

######  Corollary 13.

Let $\alpha\sim\mu$ with $\mu\ll\lambda$. Then
$P_{2n}\overset{d}{\to}\mathrm{Unif}(G_{1})$ and
$P_{2n-1}\overset{d}{\to}\mathrm{Unif}(G_{-1})$. Under the assumption (5), we
also have $\max_{g\in G_{(-1)^{n}}}|\mu(P_{n}=g)-1/|G_{1}||\leq Ce^{-\tau n}$,
where $C$ and $\tau$ are as in (14).

By Lemma 4 we thus have
$(p_{n},q_{n})\pmod{m}\overset{d}{\to}\mathrm{Unif}(V)$, and the same holds
for $(q_{n-1},q_{n})$ and $(p_{n-1},p_{n})$. Further,
$q_{n}\pmod{m}\overset{d}{\to}\nu$ and $p_{n}\pmod{m}\overset{d}{\to}\nu$. If
the density function $\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}$ is Lipschitz,
the same hold with exponential rate. See also [8] for the special case
$\mu=\lambda$ of Corollary 13.

Theorem 8 and Corollary 12 also imply certain mixing properties of $P_{n}$.
Let us first recall two classical ways of quantifying mixing, see [6] for more
context. Let $X_{n}$, $n\in\mathbb{N}$ be a sequence of random variables on a
probability space $(\Omega,\mathcal{F},P)$ taking values from a measurable
space. Let $\mathcal{F}_{k}^{\ell}$ denote the $\sigma$-algebra generated by
$X_{i}$, $k\leq i\leq\ell$, and similarly let $\mathcal{F}_{k}^{\infty}$ be
the $\sigma$-algebra generated by $X_{i}$, $i\geq k$. The $\alpha$-mixing (or
strong mixing) coefficients of the sequence $X_{n}$ are defined as222The term
$\alpha$-mixing and the notation $\alpha(\ell)$ are unrelated to the random
real number $\alpha\in[0,1]$.

$\alpha(\ell)=\sup_{k\in\mathbb{N}}\sup_{\begin{subarray}{c}A\in\mathcal{F}_{1}^{k}\\\
B\in\mathcal{F}_{k+\ell}^{\infty}\end{subarray}}\left|P(A\cap
B)-P(A)P(B)\right|,\qquad\ell\in\mathbb{N},$

whereas the $\psi$-mixing coefficients are

$\psi(\ell)=\sup_{k\in\mathbb{N}}\sup_{\begin{subarray}{c}A\in\mathcal{F}_{1}^{k},\,\,P(A)>0\\\
B\in\mathcal{F}_{k+\ell}^{\infty},\,\,P(B)>0\end{subarray}}\left|\frac{P(A\cap
B)}{P(A)P(B)}-1\right|,\qquad\ell\in\mathbb{N}.$

######  Lemma 14.

Let $\alpha\sim\mu$ with $\mu\ll\lambda$, and assume (5). Then the
$\alpha$-mixing coefficients of the sequence $X_{n}=(a_{n},P_{n})$ satisfy
$\alpha(\ell)\ll e^{-\tau\ell}$, $\ell\in\mathbb{N}$, where $\tau$ is as in
(14), and the implied constant depends only on $L$ and $m$. Under the
additional assumption $\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(x)\geq K>0$ for
all $x\in[0,1]$ with some constant $K>0$, we also have $\psi(\ell)\ll
e^{-\tau\ell}$, $\ell\in\mathbb{N}$, with an implied constant depending also
on $K$.

######  Remark.

Exactly as in the proof of Corollary 12, we can prove that under the sole
assumption $\mu\ll\lambda$, we have $\lim_{\ell\to\infty}\alpha(\ell)=0$
without an estimate for the rate.

###### Proof of Lemma 14.

Fix $k,\ell\geq 1$. Observe that $(X_{1},X_{2},\ldots,X_{k})$ is a function of
$(a_{1},a_{2},\ldots,a_{k})$, and that $(X_{k+\ell},X_{k+\ell+1},\ldots)$ is a
function of $(P_{k+\ell-1},a_{k+\ell},a_{k+\ell+1},\ldots)$. In particular,
$\mathcal{F}_{1}^{k}\subseteq\mathcal{A}_{1}^{k}$, and any
$B\in\mathcal{F}_{k+\ell}^{\infty}$ is of the form $B=\cup_{g\in
G_{(-1)^{k+\ell-1}}}\\{P_{k+\ell-1}=g\\}\cap B_{g}$ with
$B_{g}\in\mathcal{A}_{k+\ell}^{\infty}$.

Let $A=\\{a_{1}=b_{1},\ldots,a_{k}=b_{k}\\}$, and let
$h=\prod_{i=1}^{k}\left(\begin{array}[]{cc}0&1\\\
1&b_{i}\end{array}\right)\pmod{m}$. An application of Theorem 8 shows that

$\mu(A\cap\\{P_{k+\ell-1}=g\\}\cap
B_{g})=\mu(A\cap\\{P_{k+1,k+\ell-1}=h^{-1}g\\}\cap
B_{g})=\frac{\mu(A)\mu_{\mathrm{Gauss}}(B_{g})}{|G_{1}|}+O(\lambda(A)\lambda(B_{g})e^{-\tau\ell}).$

By $\sigma$-additivity, the same holds with any $A\in\mathcal{A}_{1}^{k}$. In
particular,

$\mu(\\{P_{k+\ell-1}=g\\}\cap
B_{g})=\frac{\mu_{\mathrm{Gauss}}(B_{g})}{|G_{1}|}+O(\lambda(B_{g})e^{-\tau\ell}).$

The previous two formulas and the fact that $\mu(A)\ll\lambda(A)$ yield

$\left|\mu(A\cap\\{P_{k+\ell-1}=g\\}\cap
B_{g})-\mu(A)\mu(\\{P_{k+\ell-1}=g\\}\cap
B_{g})\right|\ll\lambda(A)\lambda(B_{g})e^{-\tau\ell}.$

It is enough to sum over those $g$ for which $\lambda(P_{k+\ell-1}=g)>0$, as
otherwise $\mu(P_{k+\ell-1}=g)=0$. Thus

$\left|\mu(A\cap
B)-\mu(A)\mu(B)\right|\ll\lambda(A)e^{-\tau\ell}\sum_{\begin{subarray}{c}g\in
G_{(-1)^{k+\ell-1}}\\\
\lambda(P_{k+\ell-1}=g)>0\end{subarray}}\lambda(B_{g}).$

Lemmas 6 and 7 show that each term on the right-hand side satisfies

$\lambda(B_{g})\leq 2\frac{\lambda(\\{P_{k+\ell-1}=g\\}\cap
B_{g})}{\lambda(P_{k+\ell-1}=g)}\ll\lambda(\\{P_{k+\ell-1}=g\\}\cap B_{g}).$

Hence

$\sum_{\begin{subarray}{c}g\in G_{(-1)^{k+\ell-1}}\\\
\lambda(P_{k+\ell-1}=g)>0\end{subarray}}\lambda(B_{g})\ll\lambda(B),$

and we obtain $|\mu(A\cap
B)-\mu(A)\mu(B)|\ll\lambda(A)\lambda(B)e^{-\tau\ell}$. In particular,
$\alpha(\ell)\ll e^{-\tau\ell}$. Under the additional assumption
$\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}(x)\geq K>0$ we have
$\lambda(A)\lambda(B)\ll\mu(A)\mu(B)$, and $\psi(\ell)\ll e^{-\tau\ell}$
follows. ∎

###  5.4 Invariance principles for $P_{n}$

We now find the variance of the sum $\sum_{n=M+1}^{M+N}f(P_{n})$, and then
prove Theorem 1.

######  Lemma 15.

Fix an integer $m\geq 2$, and let $f:G\to\mathbb{R}$ be arbitrary.

  1. (i)

The right-hand side of (6) is finite and nonnegative.

  2. (ii)

Let $\alpha\sim\mu$ with $\mu\ll\lambda$, and assume (5). For any integers
$M\geq 0$ and $N\geq 1$,

$\mathbb{E}\left(\sum_{n=M+1}^{M+N}f(P_{n})-E_{f}N\right)^{2}=\sigma_{f}^{2}N+O(\log(N+1))$

with an implied constant depending only on $L$ and $f$.

###### Proof.

(i) Let $\alpha\sim\mu_{\mathrm{Gauss}}$ and $U_{\pm
1}\sim\mathrm{Unif}(G_{\pm 1})$ be independent random variables. By Corollary
13,

$\left|\mathbb{E}\bar{f}(U_{\pm 1})\bar{f}(U_{\pm
1}P_{n})\right|=\left|\sum_{\begin{subarray}{c}g\in G_{\pm 1}\\\ h\in
G_{(-1)^{n}}\end{subarray}}\bar{f}(g)\bar{f}(gh)\frac{\mu_{\mathrm{Gauss}}(P_{n}=h)}{|G_{1}|}\right|=\left|\sum_{\begin{subarray}{c}g\in
G_{\pm 1}\\\ h\in
G_{(-1)^{n}}\end{subarray}}\bar{f}(g)\bar{f}(gh)\frac{1}{|G_{1}|^{2}}\right|+O(e^{-\tau
n}).$

The remaining sum is zero since $\sum_{g\in G_{\pm 1}}\bar{f}(g)=0$. Hence
$|\mathbb{E}\bar{f}(U_{\pm 1})\bar{f}(U_{\pm 1}P_{n})|\ll e^{-\tau n}$, so
both series in (6) are absolutely convergent. The fact that the right-hand
side of (6) is nonnegative will follow from (ii).

(ii) We may assume that $\mathbb{E}f(U_{\pm 1})=0$. Expanding the square leads
to

$\mathbb{E}\left(\sum_{n=M+1}^{M+N}f(P_{n})\right)^{2}=\sum_{n=M+1}^{M+N}\mathbb{E}f(P_{n})^{2}+2\sum_{\ell=1}^{N-1}\sum_{n=M+1}^{M+N-\ell}\mathbb{E}f(P_{n})f(P_{n+\ell}).$
(19)

Corollary 13 shows that
$\mathbb{E}f(P_{n})^{2}=\mathbb{E}f(U_{(-1)^{n}})^{2}+O(e^{-\tau n})$, hence
the first sum in (19) is

$\sum_{n=M+1}^{M+N}\mathbb{E}f(P_{n})^{2}=\left(\frac{1}{2}\mathbb{E}f(U_{1})^{2}+\frac{1}{2}\mathbb{E}f(U_{-1})^{2}\right)N+O(1).$

Let $1\leq R\leq N-1$ be a parameter to be chosen, and consider the second sum
in (19). Lemma 14 implies that the $\alpha$-mixing coefficients of the
sequence $P_{n}$ satisfy $\alpha(\ell)\ll e^{-\tau\ell}$, thus

$|\mathbb{E}f(P_{n})f(P_{n+\ell})-\mathbb{E}f(P_{n})\mathbb{E}f(P_{n+\ell})|\ll
e^{-\tau\ell}.$

Here $|\mathbb{E}f(P_{n+\ell})|\ll e^{-\tau(n+\ell)}$ and
$|\mathbb{E}f(P_{n})|\ll 1$. Hence $|\mathbb{E}f(P_{n})f(P_{n+\ell})|\ll
e^{-\tau\ell}$, and

$\left|\sum_{\ell=R}^{N-1}\sum_{n=M+1}^{M+N-\ell}\mathbb{E}f(P_{n})f(P_{n+\ell})\right|\ll\sum_{\ell=R}^{N-1}Ne^{-\tau\ell}\ll
Ne^{-\tau R}.$

Now let $1\leq\ell\leq R$, and consider

$\mathbb{E}f(P_{n})f(P_{n+\ell})=\sum_{\begin{subarray}{c}g\in G_{(-1)^{n}}\\\
h\in
G_{(-1)^{\ell}}\end{subarray}}f(g)f(gh)\mu\left(\\{P_{n}=g\\}\cap\\{P_{n+1,n+\ell}=h\\}\right).$

Theorem 8 shows that here
$\mu(\\{P_{n}=g\\}\cap\\{P_{n+1,n+\ell}=h\\})=\mu_{\mathrm{Gauss}}(P_{\ell}=h)/|G_{1}|+O(e^{-\tau
n})$. Therefore

$\mathbb{E}f(P_{n})f(P_{n+\ell})=\sum_{\begin{subarray}{c}g\in G_{(-1)^{n}}\\\
h\in
G_{(-1)^{\ell}}\end{subarray}}f(g)f(gh)\frac{\mu_{\mathrm{Gauss}}(P_{\ell}=h)}{|G_{1}|}+O(e^{-\tau
n})=\mathbb{E}f(U_{(-1)^{n}})f(U_{(-1)^{n}}P_{\ell})+O(e^{-\tau n}),$

and

$\begin{split}\sum_{\ell=1}^{R}\sum_{n=M+1}^{M+N-\ell}\mathbb{E}f(P_{n})f(P_{n+\ell})&=\sum_{\ell=1}^{R}\frac{N-\ell}{2}\left(\mathbb{E}f(U_{1})f(U_{1}P_{\ell})+\mathbb{E}f(U_{-1})f(U_{-1}P_{\ell})\right)+O(R)\\\
&=\frac{N}{2}\sum_{\ell=1}^{\infty}\left(\mathbb{E}f(U_{1})f(U_{1}P_{\ell})+\mathbb{E}f(U_{-1})f(U_{-1}P_{\ell})\right)+O(R+Ne^{-\tau
R}).\end{split}$

The previous estimates for the right-hand side of (19) and the definition (6)
of $\sigma_{f}$ lead to

$\mathbb{E}\left(\sum_{n=M+1}^{M+N}f(P_{n})\right)^{2}=\sigma_{f}^{2}N+O(R+Ne^{-\tau
R}),$

and the claim follows by choosing $R\approx\log(N+1)$. ∎

###### Proof of Theorem 1.

Claim (i) was proved in Lemma 15. Now let $\alpha\sim\mu$ with
$\mu\ll\lambda$, and assume (5). Lemmas 14 and 15 show that the sequence of
random variables $f(P_{n})$ is $\alpha$-mixing with exponential rate, and
satisfy
$\mathbb{E}(\sum_{n=M+1}^{M+N}f(P_{n})-E_{f}N)^{2}=\sigma_{f}^{2}N+O(\log(N+1))$
uniformly in $M\geq 0$. Assuming $\sigma_{f}>0$, a general result of Philipp
and Stout [22, Theorem 7.1] on partial sums of nonstationary $\alpha$-mixing
random variables shows that without changing its distribution, the process
$\sum_{1\leq n\leq t}f(P_{n})$ can be redefined on a richer probability space
so that $\sum_{1\leq n\leq t}f(P_{n})-E_{f}t=\sigma_{f}W(t)+O(t^{1/2-\eta})$
a.s. with a universal constant $\eta>0$. This proves (iii) in the case
$\sigma_{f}>0$.

As Philipp and Stout assume $\sigma_{f}>0$ throughout their treatise, for the
sake of completeness we include a proof of (iii) in the case $\sigma_{f}=0$,
i.e. when $\mathbb{E}(\sum_{n=M+1}^{M+N}f(P_{n})-E_{f}N)^{2}\ll\log(N+1)$
uniformly in $M\geq 0$. Fix a small $\varepsilon>0$. An application of the
Chebyshev inequality gives

$\mu\left(\left|\sum_{n=1}^{N^{2}}f(P_{n})-E_{f}N^{2}\right|\geq
N^{1/2+\varepsilon}\right)\ll\frac{\log(N+1)}{N^{1+2\varepsilon}}.$

Since $f(P_{n})$ is bounded and $\alpha$-mixing with exponential rate, for all
$p>2$ we have $\mathbb{E}|\sum_{n=M+1}^{M+N}f(P_{n})-E_{f}N|^{p}\ll N^{p/2}$
uniformly in $M\geq 0$ and $N\geq 1$ [23]. The Erdős–Stechkin inequality [20]
strengthens this to $\mathbb{E}\max_{1\leq k\leq
N}|\sum_{n=M+1}^{M+k}f(P_{n})-E_{f}k|^{p}\ll N^{p/2}$. Choosing a suitably
large $p>2$ thus leads to

$\mu\left(\max_{1\leq k\leq
2N}\left|\sum_{n=N^{2}+1}^{N^{2}+k}f(P_{n})-E_{f}k\right|\geq
N^{1/2+\varepsilon}\right)\ll\frac{N^{p/2}}{N^{p(1/2+\varepsilon)}}\ll\frac{1}{N^{2}}.$

An application of the Borel–Cantelli lemma then shows that

$\left|\sum_{n=1}^{N^{2}}f(P_{n})-E_{f}N^{2}\right|\ll
N^{1/2+\varepsilon}\quad\textrm{and}\quad\max_{1\leq k\leq
2N}\left|\sum_{n=N^{2}+1}^{N^{2}+k}f(P_{n})-E_{f}k\right|\ll
N^{1/2+\varepsilon}\quad\textrm{for a.e. }\alpha.$

In particular, $|\sum_{n=1}^{N}f(P_{n})-E_{f}N|\ll N^{1/4+\varepsilon}$ for
a.e. $\alpha$. This proves (iii) in the case $\sigma_{f}=0$.

The almost sure approximation by a Wiener process in part (iii) immediately
implies the functional CLT under the assumption (5). See Peligrad [21] for a
direct proof of the functional CLT under even weaker mixing assumptions.
Exactly as in the proof of Corollary 12, we can easily remove assumption (5)
on the density from the functional CLT. This proves (ii). ∎

##  6 Limit laws for the local discrepancy

We rely on an explicit formula of Roçadas and Schoissengeier [24], who showed
that for any irrational $\alpha\in[0,1]$, any $r\in(0,1)$ and any $k\geq 0$,

$\begin{split}\max_{0\leq
N<q_{k+1}}S_{N,r}(\alpha)&=\sum_{\begin{subarray}{c}j=0\\\ j\textrm{
even}\end{subarray}}^{k}\left\\{q_{j}r\right\\}\left(\left(1-\\{q_{j}r\\}\right)a_{j+1}+\\{q_{j+1}r\\}-\\{q_{j-1}r\\}\right)+O(1),\\\
\min_{0\leq N<q_{k+1}}S_{N,r}(\alpha)&=-\sum_{\begin{subarray}{c}j=0\\\
j\textrm{
odd}\end{subarray}}^{k}\left\\{q_{j}r\right\\}\left(\left(1-\\{q_{j}r\\}\right)a_{j+1}+\\{q_{j+1}r\\}-\\{q_{j-1}r\\}\right)+O(1)\end{split}$
(20)

with universal implied constants. We give the proof of Theorem 3 after a
preparatory lemma. Let $\theta_{m}$ be as in (9), and recall that $\gamma$
denotes the Euler–Mascheroni constant.

######  Lemma 16.

Let $\alpha\sim\mu$ with $\mu\ll\lambda$, and assume that
$\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}$ is positive and Lipschitz. Let
$l/m\in(0,1)$ be a reduced fraction, and let

$X_{j}=\left\\{q_{j-1}\frac{l}{m}\right\\}\left(1-\left\\{q_{j-1}\frac{l}{m}\right\\}\right)a_{j}.$

Then

$\left(\frac{\sum_{j=1}^{k}X_{2j-1}-A_{k}}{s_{k}},\frac{\sum_{j=1}^{k}X_{2j}-A_{k}}{s_{k}}\right)\overset{d}{\to}\mathrm{Stab}(1,1)\otimes\mathrm{Stab}(1,1),$

where $A_{k}=\frac{1}{6\log 2}k\log k-\frac{1}{6\log
2}\left(6\theta_{m}+\gamma+\log\frac{12\log 2}{\pi}\right)k$ and
$s_{k}=\frac{\pi}{12\log 2}k$.

###### Proof.

Fix real numbers $x_{1},x_{2}$ such that $(x_{1},x_{2})\neq(0,0)$, and let
$Y_{j}=(x_{1}/k)X_{2j-1}+(x_{2}/k)X_{2j}$, $1\leq j\leq k$. Throughout,
implied constants are allowed to depend on $x_{1},x_{2}$. The random variable
$Y_{j}$ is a function of $(a_{2j-1},a_{2j},P_{2j-1},P_{2j})$, therefore by
Lemma 14 the sequence $Y_{j}$ is $\psi$-mixing with exponential rate. One
readily checks that if $x_{1}\neq 0$, then with a suitable constant $B>0$,

$\begin{split}\mathbb{E}(1-\cos
Y_{j})&\geq\mu\left(\frac{\pi}{2}\leq|Y_{j}|\leq\frac{3\pi}{2}\right)\\\
&\gg\mu\left(\\{q_{2j-2}\equiv 1\pmod{m}\\}\cap\left\\{\frac{B}{2}k\leq
a_{2j-1}\leq Bk\right\\}\cap\\{a_{2j}=1\\}\right)\\\
&\gg\mu_{\mathrm{Gauss}}\left(\frac{B}{2}k\leq a_{2j-1}\leq
Bk\right)\gg\frac{1}{k}.\end{split}$

A similar argument shows that $\mathbb{E}(1-\cos Y_{j})\gg 1/k$ holds in the
case $x_{1}=0$, $x_{2}\neq 0$ as well. Further, we have

$\mathbb{E}|e^{iY_{j}}-1|\leq\mathbb{E}\min\\{|Y_{j}|,2\\}\ll\mathbb{E}\min\left\\{\frac{a_{2j-1}+a_{2j}}{k},1\right\\}\ll\frac{\log
k}{k}.$

An application of [13, Lemma 1] thus leads to

$\mathbb{E}\exp\left(i\sum_{j=1}^{k}Y_{j}\right)=\exp\left(\sum_{j=1}^{k}\mathbb{E}(e^{iY_{j}}-1)\right)+O\left(\frac{(\log
k)^{2}}{k}\right).$ (21)

Here $\sum_{j\ll\log k}\mathbb{E}(e^{iY_{j}}-1)=O((\log k)^{2}/k)$, hence it
will be enough to consider the terms $j\gg\log k$.

We can express $Y_{j}$ as $Y_{j}=F(P_{2j-2},a_{2j-1},a_{2j})$ with the
function

$F\left(\left(\begin{array}[]{ll}a&b\\\
c&d\end{array}\right),b_{1},b_{2}\right)=\frac{x_{1}}{k}\left\\{d\frac{l}{m}\right\\}\left(1-\left\\{d\frac{l}{m}\right\\}\right)b_{1}+\frac{x_{2}}{k}\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\left(1-\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\right)b_{2}.$

Theorem 8 yields

$\begin{split}\mathbb{E}(e^{iY_{j}}-1)&=\sum_{\begin{subarray}{c}g\in G_{1}\\\
b_{1},b_{2}\in\mathbb{N}\end{subarray}}(e^{iF(g,b_{1},b_{2})}-1)\mu\left(P_{2j-2}=g,a_{2j-1}=b_{1},a_{2j}=b_{2}\right)\\\
&=\sum_{\begin{subarray}{c}g\in G_{1}\\\
b_{1},b_{2}\in\mathbb{N}\end{subarray}}(e^{iF(g,b_{1},b_{2})}-1)\frac{\mu_{\mathrm{Gauss}}(a_{1}=b_{1},a_{2}=b_{2})}{|G_{1}|}+O(e^{-\tau
j}).\end{split}$

For a fixed $g=\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\in
G_{1}$, we have $F(g,b_{1},b_{2})=t_{1}b_{1}+t_{2}(b_{1})b_{2}$ with

$t_{1}=\frac{x_{1}}{k}\left\\{d\frac{l}{m}\right\\}\left(1-\left\\{d\frac{l}{m}\right\\}\right),\qquad
t_{2}(b_{1})=\frac{x_{2}}{k}\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\left(1-\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\right).$

In [5, Lemma 20] it was shown that for any $t_{1},t_{2}\in(-1/2,1/2)$, we have

$\begin{split}\sum_{b_{1},b_{2}\in\mathbb{N}}\left(e^{i(t_{1}b_{1}+t_{2}b_{2})}-1\right)&\mu_{\mathrm{Gauss}}(a_{1}=b_{1},a_{2}=b_{2})\\\
=&-\frac{1}{\log 2}\left(i\gamma
t_{1}+\frac{\pi}{2}|t_{1}|+it_{1}\log|t_{1}|+i\gamma
t_{2}+\frac{\pi}{2}|t_{2}|+it_{2}\log|t_{2}|\right)\\\
&+O\left(t_{1}^{2}\log\frac{1}{|t_{1}|}+t_{2}^{2}\log\frac{1}{|t_{2}|}+|t_{1}t_{2}|\log\frac{1}{|t_{1}|}\log\frac{1}{|t_{2}|}\right).\end{split}$

The proof actually gives that more generally, for any constant $t_{1}$ and any
sequence $t_{2}(n)$, $n\in\mathbb{N}$ with $t_{1},t_{2}(n)\in(-1/2,1/2)$, we
have

$\begin{split}&\sum_{b_{1},b_{2}\in\mathbb{N}}\left(e^{i(t_{1}b_{1}+t_{2}(b_{1})b_{2})}-1\right)\mu_{\mathrm{Gauss}}(a_{1}=b_{1},a_{2}=b_{2})\\\
=&-\frac{1}{\log 2}\left(i\gamma
t_{1}+\frac{\pi}{2}|t_{1}|+it_{1}\log|t_{1}|-i\sum_{b_{1},b_{2}\in\mathbb{N}}t_{2}(b_{1})b_{2}R(b_{1},b_{2})+\sum_{b_{1}\in\mathbb{N}}\frac{\frac{\pi}{2}|t_{2}(b_{1})|+it_{2}(b_{1})\log|t_{2}(b_{1})|}{b_{1}(b_{1}+1)}\right)\\\
&+O\left(t_{1}^{2}\log\frac{1}{|t_{1}|}+T_{2}^{2}\log\frac{1}{T_{2}}+|t_{1}|T_{2}\log\frac{1}{|t_{1}|}\log\frac{1}{T_{2}}\right),\end{split}$

where $T_{2}=\sup_{b_{1}\in\mathbb{N}}|t_{2}(b_{1})|$, and

$R(b_{1},b_{2})=\mu_{\mathrm{Gauss}}(a_{1}=b_{1},a_{2}=b_{2})-\frac{1}{b_{1}(b_{1}+1)b_{2}(b_{2}+2)}.$

As we observed in [5, Lemma 20] using telescoping sums,
$\sum_{b_{1},b_{2}\in\mathbb{N}}b_{2}R(b_{1},b_{2})=-\gamma$. In particular,
we obtain a formula for $\mathbb{E}(e^{iY_{j}}-1)$ in the form of an average
over $G_{1}$. Lemma 4 and formula (12) show that

$\begin{split}\frac{1}{|G_{1}|}\sum_{g\in
G_{1}}\left\\{d\frac{l}{m}\right\\}\left(1-\left\\{d\frac{l}{m}\right\\}\right)=\sum_{a\in\mathbb{Z}_{m}}\nu_{a}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right)&=\frac{1}{6},\\\
\frac{1}{|G_{1}|}\sum_{g\in
G_{1}}\left\\{d\frac{l}{m}\right\\}\left(1-\left\\{d\frac{l}{m}\right\\}\right)\log\left(\left\\{d\frac{l}{m}\right\\}\left(1-\left\\{d\frac{l}{m}\right\\}\right)\right)&=\theta_{m}.\end{split}$

Note that the previous two formulas do not depend on $l$ since $\nu_{a}$ is
invariant under multiplication by elements of $\mathbb{Z}_{m}^{*}$. We saw in
the proof of Lemma 4 that $G_{1}$ acts transitively on the set of row vectors
$V$, and that transposition is a bijection of $G_{1}$. Consequently for any
fixed $b_{1}\in\mathbb{N}$, the row vector
$(1,b_{1}\pmod{m})\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)^{\top}$ is uniformly distributed on $V$, and its second
coordinate, $b_{1}d+c\pmod{m}$ has distribution $\nu$. In particular, for any
fixed $b_{1}\in\mathbb{N}$ we have the same averages

$\begin{split}\frac{1}{|G_{1}|}\sum_{g\in
G_{1}}\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\left(1-\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\right)=\sum_{a\in\mathbb{Z}_{m}}\nu_{a}\left\\{\frac{a}{m}\right\\}\left(1-\left\\{\frac{a}{m}\right\\}\right)&=\frac{1}{6},\\\
\frac{1}{|G_{1}|}\sum_{g\in
G_{1}}\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\left(1-\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\right)\log\left(\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\left(1-\left\\{(b_{1}d+c)\frac{l}{m}\right\\}\right)\right)&=\theta_{m}.\end{split}$

After some simplification, we arrive at

$\begin{split}\mathbb{E}(e^{iY_{j}}-1)=&i\frac{\log
k-6\theta_{m}-\gamma}{6\log 2}\cdot\frac{x_{1}}{k}-\frac{\pi}{(12\log
2)k}\left(|x_{1}|+\frac{2i}{\pi}x_{1}\log|x_{1}|\right)\\\ &+i\frac{\log
k-6\theta_{m}-\gamma}{6\log 2}\cdot\frac{x_{2}}{k}-\frac{\pi}{(12\log
2)k}\left(|x_{2}|+\frac{2i}{\pi}x_{2}\log|x_{2}|\right)+O\left(\frac{(\log
k)^{2}}{k^{2}}+e^{-\tau j}\right).\end{split}$

We can compute the right-hand side of (21) by summing over $\log k\ll j\leq
k$, thus

$\begin{split}\mathbb{E}\exp\left(i\sum_{j=1}^{k}Y_{j}\right)=&\exp\bigg{(}i\frac{\log
k-6\theta_{m}-\gamma}{6\log 2}x_{1}-\frac{\pi}{12\log
2}\left(|x_{1}|+\frac{2i}{\pi}x_{1}\log|x_{1}|\right)\\\ &\hskip
28.45274pt+i\frac{\log k-6\theta_{m}-\gamma}{6\log 2}x_{2}-\frac{\pi}{12\log
2}\left(|x_{2}|+\frac{2i}{\pi}x_{2}\log|x_{2}|\right)+O\left(\frac{(\log
k)^{2}}{k}\right)\bigg{)}\\\ &+O\left(\frac{(\log
k)^{2}}{k}\right).\end{split}$

Letting $A_{k}$ and $s_{k}$ be as in the claim and replacing $x_{n}$ by
$\frac{12\log 2}{\pi}x_{n}$, $n=1,2$, we obtain

$\begin{split}\mathbb{E}\exp\bigg{(}i&\frac{\sum_{j=1}^{k}X_{2j-1}-A_{k}}{s_{k}}x_{1}+i\frac{\sum_{j=1}^{k}X_{2j}-A_{k}}{s_{k}}x_{2}\bigg{)}\\\
&=\exp\left(-\left(|x_{1}|+\frac{2i}{\pi}x_{1}\log|x_{1}|\right)\right)\exp\left(-\left(|x_{2}|+\frac{2i}{\pi}x_{2}\log|x_{2}|\right)\right)+O\left(\frac{(\log
k)^{2}}{k}\right).\end{split}$

This implies the pointwise convergence of the characteristic function of the
random vector in the claim to that of
$\mathrm{Stab}(1,1)\otimes\mathrm{Stab}(1,1)$, which proves the desired
convergence in distribution. ∎

###### Proof of Theorem 3.

Fix a reduced fraction $r=l/m\in(0,1)$. We may assume that the density
$\frac{\mathrm{d}\mu}{\mathrm{d}\lambda}$ is positive and Lipschitz. This
assumption can be removed exactly as in the proof of Corollary 12.

For any $M\geq 1$, let $k_{M}^{*}=k_{M}^{*}(\alpha)$ denote the random index
for which $q_{k_{M}^{*}}\leq M<q_{k_{M}^{*}+1}$, and let $k_{M}$ be the odd
integer closest to $\frac{12\log 2}{\pi^{2}}\log M$. Using the fact that $\log
q_{k}$ satisfies the CLT with centering term $\frac{\pi^{2}}{12\log 2}k$ and
scaling term $k^{1/2}$ [14, p. 194], we immediately obtain $\mu(|\log
q_{k_{M}}-\log M|\geq(\log M)^{1/2+\varepsilon})\to 0$ with any
$\varepsilon>0$. Consequently, $\mu(|k_{M}^{*}-k_{M}|\geq(\log
M)^{1/2+\varepsilon})\to 0$. The explicit formula (20) thus yields

$\begin{split}\max_{0\leq
N<M}S_{N,r}(\alpha)&=\sum_{j=0}^{\frac{k_{M}-1}{2}}\\{q_{2j}r\\}\left((1-\\{q_{2j}r\\})a_{2j+1}+\\{q_{2j+1}r\\}-\\{q_{2j-1}r\\}\right)+\xi_{M}(\alpha),\\\
\min_{0\leq
N<M}S_{N,r}(\alpha)&=-\sum_{j=0}^{\frac{k_{M}-1}{2}}\\{q_{2j+1}r\\}\left((1-\\{q_{2j+1}r\\})a_{2j+2}+\\{q_{2j+2}r\\}-\\{q_{2j}r\\}\right)+\xi_{M}^{\prime}(\alpha)\end{split}$
(22)

with error terms $\xi_{M}(\alpha)$, $\xi_{M}^{\prime}(\alpha)$ which, outside
a set of $\mu$-measure $o(1)$, satisfy

$|\xi_{M}(\alpha)|,|\xi_{M}^{\prime}(\alpha)|\leq\sum_{j=k_{M}-(\log
M)^{1/2+\varepsilon}}^{k_{M}+(\log M)^{1/2+\varepsilon}}a_{j}+O(1).$

A classical result of Khintchine [14, p. 204] states that
$\sum_{j=1}^{k}a_{j}/(k\log k)\to 1/\log 2$ in measure. Hence

$\mu\left(\sum_{j=k_{M}-(\log M)^{1/2+\varepsilon}}^{k_{M}+(\log
M)^{1/2+\varepsilon}}a_{j}\geq(\log
M)^{1/2+2\varepsilon}\right)\ll\mu_{\mathrm{Gauss}}\left(\sum_{1\leq j\ll(\log
M)^{1/2+\varepsilon}}a_{j}\geq(\log M)^{1/2+2\varepsilon}\right)\to 0.$

In particular, $\xi_{M}(\alpha),\xi_{M}^{\prime}(\alpha)=o(\log M)$ in
$\mu$-measure, and are thus negligible.

Relation (2) shows that

$\sum_{j=0}^{\frac{k_{M}-1}{2}}\\{q_{2j}r\\}\cdot\\{q_{2j-1}r\\}=\rho
k_{M}+o(k_{M})\quad\textrm{and}\quad\sum_{j=0}^{\frac{k_{M}-1}{2}}\\{q_{2j}r\\}\cdot\\{q_{2j+1}r\\}=\rho
k_{M}+o(k_{M})$

hold for a.e. $\alpha$ with the same constant $\rho$. In particular, the error
terms in the previous formula are $o(\log M)$ in $\mu$-measure, and (22)
simplifies to

$\begin{split}\max_{0\leq
N<M}S_{N,r}(\alpha)&=\sum_{j=0}^{\frac{k_{M}-1}{2}}\\{q_{2j}r\\}(1-\\{q_{2j}r\\})a_{2j+1}+o(\log
M)\quad\textrm{in $\mu$-measure},\\\ \min_{0\leq
N<M}S_{N,r}(\alpha)&=-\sum_{j=0}^{\frac{k_{M}-1}{2}}\\{q_{2j+1}r\\}(1-\\{q_{2j+1}r\\})a_{2j+2}+o(\log
M)\quad\textrm{in $\mu$-measure}.\end{split}$

Lemma 16 yields the desired limit law with centering term

$\frac{1}{6\log
2}\cdot\frac{k_{M}}{2}\log\frac{k_{M}}{2}-\frac{6\theta_{m}+\gamma+\log\frac{12\log
2}{\pi}}{6\log 2}\cdot\frac{k_{M}}{2}=\frac{1}{\pi^{2}}\log M\log\log
M-\frac{6\theta_{m}+\gamma+\log(2\pi)}{\pi^{2}}\log M+O(1)$

and scaling term $\frac{\pi}{12\log 2}\cdot\frac{k_{M}}{2}=\frac{1}{2\pi}\log
M+O(1)$. ∎

## Acknowledgments

The author is supported by the Austrian Science Fund (FWF) project M 3260-N.

## References

  * [1] J. Aaronson and M. Keane: The visits to zero of some deterministic random walks. Proc. London Math. Soc. 44 (1982), 535–553.
  * [2] A. Avila, D. Dolgopyat, E. Duryev and O. Sarig: The visits to zero of a random walk driven by an irrational rotation. Israel J. Math. 207 (2015), 653–717.
  * [3] J. Beck: Probabilistic Diophantine Approximation. Randomness in Lattice Point Counting. Springer Monographs in Mathematics. Springer, Cham, 2014.
  * [4] P. Billingsley: Convergence of Probability Measures. Second Edition. John Wiley & Sons, Inc., New York, 1999.
  * [5] B. Borda: Limit laws of maximal Birkhoff sums for circle rotations via quantum modular forms. arXiv:2303.07796.
  * [6] R. Bradley: Basic properties of strong mixing conditions. A survey and some open questions. Update of, and a supplement to, the 1986 original. Probab. Surv. 2 (2005), 107–144.
  * [7] M. Bromberg and C. Ulcigrai: A temporal central limit theorem for real-valued cocycles over rotations. Ann. Inst. Henri Poincaré Probab. Stat. 54 (2018), 2304–2334.
  * [8] K. Dajani and C. Kraaikamp: A note on the approximation by continued fractions under an extra condition. New York J. Math. 3A (1997/98), 69–80.
  * [9] D. Dolgopyat and B. Fayad: Deviations of ergodic sums for toral translations I. Convex bodies. Geom. Funct. Anal. 24 (2014), 85–115.
  * [10] D. Dolgopyat and B. Fayad: Deviations of ergodic sums for toral translations II. Boxes. Publ. Math. Inst. Hautes Études Sci. 132 (2020), 293–352.
  * [11] D. Dolgopyat and B. Fayad: Limit theorems for toral translations. Hyperbolic dynamics, fluctuations and large deviations, 227–277, Proc. Sympos. Pure Math., 89, Amer. Math. Soc., Providence, RI, 2015.
  * [12] A. Fisher and T. Schmidt: Distribution of approximants and geodesic flows. Ergodic Theory Dynam. Systems 34 (2014), 1832–1848.
  * [13] L. Heinrich: Rates of convergence in stable limit theorems for sums of exponentially $\psi$-mixing random variables with an application to metric theory of continued fractions. Math. Nachr. 131 (1987), 149–165.
  * [14] M. Iosifescu and C. Kraaikamp: Metrical Theory of Continued Fractions. Mathematics and its Applications, 547. Kluwer Academic Publishers, Dordrecht, 2002.
  * [15] H. Jager and P. Liardet: Distributions arithmétiques des dénominateurs de convergents de fractions continues. Nederl. Akad. Wetensch. Indag. Math. 50 (1988), 181–197.
  * [16] H. Kesten: The discrepancy of random sequences $\\{kx\\}$. Acta Arith. 10 (1964/65), 183–213.
  * [17] H. Kesten: Uniform distribution mod 1. Ann. of Math. (2) 71 (1960), 445–471.
  * [18] H. Kesten: Uniform distribution mod 1. II. Acta Arith. 7 (1961/1962), 355–380.
  * [19] R. Moeckel: Geodesics on modular surfaces and continued fractions. Ergodic Theory Dynam. Systems 2 (1982), 69–83.
  * [20] F. Móricz: Moment inequalities and the strong law of large numbers. Z. Wahrscheinlichkeitstheorie und Verw. Gebiete 35 (1976), 299–314.
  * [21] M. Peligrad: Invariance principles for mixing sequences of random variables. Ann. Probab. 10 (1982), 968–981.
  * [22] W. Philipp and W. Stout: Almost sure invariance principles for partial sums of weakly dependent random variables. Mem. Amer. Math. Soc. 2 (1975), no. 161.
  * [23] E. Rio: Asymptotic Theory of Weakly Dependent Random Processes. Translated from the 2000 French edition. Probability Theory and Stochastic Modeling, 80. Springer, Berlin, 2017.
  * [24] L. Roçadas and J. Schoißengeier: On the local discrepancy of $(n\alpha)$-sequences. J. Number Theory 131 (2011), 1492–1497.
  * [25] J. Schoissengeier: A metrical result on the discrepancy of $(n\alpha)$. Glasgow Math. J. 40 (1998), 393–425.
  * [26] P. Szüsz: Verallgemeinerung und Anwendungen eines Kusminschen Satzes. Acta Arith. 7 (1961/62), 149–160.