A Uniform Polynomial Rate for the Extreme-Value Law of ϑ-Expansion Digits

1. Introduction and Main Results

Fix $\vartheta \in (0,1]$ with

$${\vartheta }^2={\displaystyle \frac{1}{m}},m\in \mathbb{N},$$

and let $T_{\vartheta }:[0,\vartheta ]\to [0,\vartheta ]$ be the generalized Gauss map

$$T_{\vartheta }\left(x\right)={\displaystyle \frac{1}{x}}-\vartheta ⌊{\displaystyle \frac{1}{\vartheta x}}⌋(x\ne 0),T_{\vartheta }\left(0\right)=0.$$

Every irrational $x\in (0,\vartheta )$ has a $\vartheta$-expansion

$$x={\displaystyle \frac{1}{a_1\vartheta +\frac{1}{a_2\vartheta +\frac{1}{a_3\vartheta +\cdots }}}},a_j=a_j\left(x\right)\in {\mathbb{N}}_m:=\{m,m+1,\dots \}.$$

The metrical theory of these expansions was initiated by Chakraborty and Rao [1]. In the case ${\vartheta }^2=1/m$ the transformation preserves the explicit probability measure

$$d{\gamma }_{\vartheta }\left(x\right)={\displaystyle \frac{\vartheta }{log(1+{\vartheta }^2)}}{\displaystyle \frac{dx}{1+\vartheta x}},0\le x\le \vartheta ,$$

(1)

and the digit process ${\left(a_j\right)}_{j\ge 1}$ is strictly stationary. Sebe, Lascu and Selmi [5] further record an exponential $\psi$-mixing estimate for this process.

Put

$$M_n=\underset{1\le j\le n}{max}{a}_j,L_{\vartheta }=log(1+{\vartheta }^2),c_{\vartheta }={\displaystyle \frac{1}{L_{\vartheta }}}.$$

The extreme-value theory of continued-fraction digits goes back to Galambos [2] and Philipp [4]. For regular continued fractions, quantitative rates were substantially sharpened by Ghosh, Kirsebom and Roy [3] using the Chen–Stein method. For $\vartheta$-expansions, Sebe, Lascu and Selmi [5, Theorems 4.5–4.6] proved the corresponding Frechet law and quantitative error estimates. In the normalization used here their limit law is

$${\gamma }_{\vartheta }(M_n\le ny)⟶exp\left(-{\displaystyle \frac{c_{\vartheta }}{y}}\right),y>0.$$

(2)

The statement of [5, Theorem 4.6] gives a subpolynomial remainder, while its proof also contains polynomial estimates on a restricted, n-dependent level range. Accordingly, the point of the present note is not the existence of a polynomial estimate by itself. Rather, we obtain one explicit polynomial Kolmogorov bound that holds uniformly over the entire half-line $y>0$, including the small-level regime in which the effective Poisson parameter grows like $logn$.

The argument uses two particularly simple features of the $\vartheta$-system: the marginal tail can be computed exactly from (1), and distant blocks are multiplicatively close to independent at an exponential rate. We deliberately use a block argument rather than a Chen–Stein argument. The latter would require a two-sided conditional mixing estimate not supplied by the standard one-sided $\psi$-mixing statement.

Our main result is the following.

Theorem 1.1

Let ${\vartheta }^2=1/m$, $m\in \mathbb{N}$. There exists $C_{\vartheta }<\infty$ such that, for all $n\ge 3$,

$$\underset{y>0}{sup}\left|{\gamma }_{\vartheta }(M_n\le ny)-exp\left(-{\displaystyle \frac{c_{\vartheta }}{y}}\right)\right|\le C_{\vartheta }{\displaystyle \frac{{(logn)}^2}{\sqrt{n}}}.$$

(3)

Moreover, for every fixed $\delta >0$ there exists $C_{\vartheta ,\delta }<\infty$ such that

$$\underset{y\ge \delta }{sup}\left|{\gamma }_{\vartheta }(M_n\le ny)-exp\left(-{\displaystyle \frac{c_{\vartheta }}{y}}\right)\right|\le C_{\vartheta ,\delta }{\left({\displaystyle \frac{logn}{n}}\right)}^{1/2}.$$

(4)

The first estimate is the contribution emphasized here: it is a single polynomial Kolmogorov bound valid on the full half-line. The second estimate is included as a sharper bound on fixed level ranges, but it is not used as a separate priority claim. The extra logarithmic factor in (3) comes from pushing the block comparison into levels for which the effective Poisson parameter is of order $logn$.

The proof rests on a quantitative comparison between ${\gamma }_{\vartheta }(M_n\le u)$ and $e^{-n\tau \left(u\right)}$, where

$$\tau \left(u\right):={\gamma }_{\vartheta }(a_1>u).$$

The exact tail formula also gives an immediate i.i.d. benchmark.

Corollary 1.2

Let ${\left(a_j^{*}\right)}_{j\ge 1}$ be i.i.d. with the same marginal law as $a_1$ and put $M_n^{*}={max}_{1\le j\le n}{a}_j^{*}$. For every fixed $\delta >0$,

$$\underset{y\ge \delta }{sup}\left|\mathbb{P}(M_n^{*}\le ny)-exp\left(-{\displaystyle \frac{c_{\vartheta }}{y}}\right)\right|=O_{\vartheta ,\delta }\left(n^{-1}\right).$$

In addition,

$$\mathbb{P}(M_n^{*}\le u)={(1-\tau \left(u\right))}^n=exp\left(-n\tau \left(u\right)-{\displaystyle \frac{1}{2}}n\tau {\left(u\right)}^2+O\left(n\tau {\left(u\right)}^3\right)\right)$$

uniformly whenever $n\tau \left(u\right)=O\left(1\right)$.

Organization.Section 2 recalls the exact marginal tail and the exponential mixing estimate. Section 3 proves a quantitative block comparison valid for a range of Poisson parameters. Section 4 derives the global and locally uniform rates in Theorem 1.1 and the i.i.d. benchmark.

2. Preliminaries

For $k\in {\mathbb{N}}_m$, the rank-one cylinder is

$$C_k=\{x:a_1\left(x\right)=k\}=\left({\displaystyle \frac{1}{\vartheta (k+1)}},{\displaystyle \frac{1}{\vartheta k}}\right],$$

up to immaterial endpoint conventions. Consequently, for every real $u\ge m$,

$$\{a_1>u\}=\left(0,{\displaystyle \frac{1}{\vartheta (\lfloor u\rfloor +1)}}\right].$$

(5)

Integrating (1) gives the exact tail.

Lemma 2.1

(Exact marginal tail). For $u\ge m$,

$$\tau \left(u\right)={\displaystyle \frac{1}{L_{\vartheta }}}log\left(1+{\displaystyle \frac{1}{\lfloor u\rfloor +1}}\right).$$

(6)

In particular,

$$\tau \left(u\right)={\displaystyle \frac{c_{\vartheta }}{u}}+O_{\vartheta }\left(u^{-2}\right),u\to \infty .$$

(7)

Proof. 

By (5) and (1),

$$\tau \left(u\right)={\displaystyle \frac{1}{L_{\vartheta }}}{\int }_0^{1/\left(\vartheta \right(\lfloor u\rfloor +1\left)\right)}{\displaystyle \frac{\vartheta dx}{1+\vartheta x}}={\displaystyle \frac{1}{L_{\vartheta }}}log\left(1+{\displaystyle \frac{1}{\lfloor u\rfloor +1}}\right).$$

Since ${(\lfloor u\rfloor +1)}^{-1}=u^{-1}+O\left(u^{-2}\right)$ and $log(1+z)=z+O\left(z^2\right)$ as $z\downarrow 0$, (7) follows. □

We use the exponential $\psi$-mixing estimate established for the stationary $\vartheta$-digit process in [5, Lemma 4.4]: there exist constants $K_{\vartheta }<\infty$ and $q_{\vartheta }\in (0,1)$ such that, whenever

$$A\in \sigma (a_1,\dots ,a_k),B\in \sigma (a_{k+r},a_{k+r+1},\dots ),$$

one has

$$\left|{\gamma }_{\vartheta }(A\cap B)-{\gamma }_{\vartheta }\left(A\right){\gamma }_{\vartheta }\left(B\right)\right|\le K_{\vartheta }{q}_{\vartheta }^r{\gamma }_{\vartheta }\left(A\right){\gamma }_{\vartheta }\left(B\right).$$

(8)

In particular, by stationarity,

$${\gamma }_{\vartheta }(a_i>u,a_j>u)\le (1+K_{\vartheta })\tau {\left(u\right)}^2,i\ne j.$$

(9)

The following elementary estimate will be used for a single block.

Lemma 2.2

(One-block estimate). There exists $C_{\vartheta }<\infty$ such that, for every $b\in \mathbb{N}$ and $u\ge m$,

$$\left|{\gamma }_{\vartheta }(M_b\le u)-{(1-\tau \left(u\right))}^b\right|\le C_{\vartheta }{b}^2\tau {\left(u\right)}^2.$$

(10)

Proof. 

Write $A_i=\{a_i>u\}$. Bonferroni’s inequalities and (9) give

$$b\tau \left(u\right)-C_{\vartheta }{b}^2\tau {\left(u\right)}^2\le {\gamma }_{\vartheta }\left({\displaystyle \bigcup _{i=1}^b}{A}_i\right)\le b\tau \left(u\right).$$

On the other hand, $1-{(1-\tau )}^b=b\tau +O\left(b^2{\tau }^2\right)$. Subtracting the two relations yields (10). □

3. A Quantitative Block Comparison

For integers $b,g\ge 1$, split $\{1,\dots ,n\}$ into consecutive retained blocks $I_1,\dots ,I_s$ of length b, separated by gaps of length g; a final remainder has length less than $b+g$. Put

$$B_j=\{\underset{i\in I_j}{max}{a}_i\le u\},\lambda =n\tau \left(u\right).$$

Proposition 3.1

(Block comparison). There exists $C_{\vartheta }<\infty$ such that, whenever $u\ge m$ and $b\tau \left(u\right)\le 1/2$,

$$\begin{array}{cc}\hfill \left|{\gamma }_{\vartheta }(M_n\le u)-e^{-\lambda }\right|\le C_{\vartheta }(& \lambda {\displaystyle \frac{g}{b}}+\lambda {\displaystyle \frac{b+g}{n}}+{\displaystyle \frac{n{q}_{\vartheta }^g}{b}}\hfill \\ \hfill & +{\displaystyle \frac{b{\lambda }^2}{n}}+{\displaystyle \frac{{\lambda }^2}{n}}).\hfill \end{array}$$

(11)

Proof. 

Deleting the gaps and the final remainder changes the no-exceedance event only if an exceedance occurs at one of at most $sg+b+g$ deleted sites. Hence

$$0\le {\gamma }_{\vartheta }\left({\displaystyle \bigcap _{j=1}^s}{B}_j\right)-{\gamma }_{\vartheta }(M_n\le u)\le (sg+b+g)\tau \left(u\right)\le C\left(\lambda {\displaystyle \frac{g}{b}}+\lambda {\displaystyle \frac{b+g}{n}}\right).$$

(12)

Successive use of (8) across the gaps, followed by a telescoping sum, gives

$$\left|{\gamma }_{\vartheta }\left({\displaystyle \bigcap _{j=1}^s}{B}_j\right)-{\displaystyle \prod _{j=1}^s}{\gamma }_{\vartheta }\left(B_j\right)\right|\le C_{\vartheta }s{q}_{\vartheta }^g\le C_{\vartheta }{\displaystyle \frac{n{q}_{\vartheta }^g}{b}}.$$

(13)

By stationarity all retained blocks have the same law. Lemma 2.2 and the inequality $|x^s-y^s|\le s|x-y|$ for $x,y\in [0,1]$ imply

$$\left|{\displaystyle \prod _{j=1}^s}{\gamma }_{\vartheta }\left(B_j\right)-{(1-\tau \left(u\right))}^{sb}\right|\le C_{\vartheta }s{b}^2\tau {\left(u\right)}^2\le C_{\vartheta }{\displaystyle \frac{b{\lambda }^2}{n}}.$$

(14)

Since $0\le n-sb\le sg+b+g$, another elementary telescoping estimate gives

$$\left|{(1-\tau \left(u\right))}^{sb}-{(1-\tau \left(u\right))}^n\right|\le C\left(\lambda {\displaystyle \frac{g}{b}}+\lambda {\displaystyle \frac{b+g}{n}}\right).$$

(15)

Finally, because $b\tau \left(u\right)\le 1/2$ implies $\tau \left(u\right)\le 1/2$,

$$\left|{(1-\tau \left(u\right))}^n-e^{-n\tau \left(u\right)}\right|\le Cn\tau {\left(u\right)}^2=C{\displaystyle \frac{{\lambda }^2}{n}}.$$

(16)

Combining (12)–(16) proves (11). □

Remark 3.2

(Why we do not use Chen–Stein here). For regular continued fractions, Ghosh, Kirsebom and Roy [3] obtained a sharper quantitative Poisson approximation by the Chen–Stein method. A direct transplantation of that argument to the present setting would require control of the conditional law of one exceedance given the two-sided $\sigma$-field generated by observations outside a symmetric neighbourhood. The estimate (8) is a one-sided past–future mixing statement and does not by itself provide that conditional estimate. Proposition 3.1 avoids this issue completely.

4. Proofs of the Main Results

Proof 

(Proof of Theorem 1.1). Choose a constant $A_{\vartheta }>0$ so large that

$$q_{\vartheta }^{\lceil A_{\vartheta }logn\rceil }\le n^{-3},n\ge 3,$$

and put

$$g_n=\lceil A_{\vartheta }logn\rceil .$$

Locally uniform bound. Fix $\delta >0$ and let $y\ge \delta$, $u=ny$. By Lemma 2.1,

$$\lambda =n\tau \left(ny\right)\le C_{\vartheta ,\delta }$$

for all sufficiently large n. Take

$$b_n=⌊\sqrt{n{g}_n}⌋.$$

Then $b_n\tau \left(ny\right)=O_{\vartheta ,\delta }\left(\sqrt{g_n/n}\right)$, so the hypothesis of Proposition 3.1 holds for large n. Substitution in (11) yields

$$\underset{y\ge \delta }{sup}\left|{\gamma }_{\vartheta }(M_n\le ny)-e^{-n\tau \left(ny\right)}\right|\le C_{\vartheta ,\delta }\sqrt{{\displaystyle \frac{logn}{n}}}.$$

(17)

Moreover, (7) gives uniformly for $y\ge \delta$,

$$\left|n\tau \left(ny\right)-{\displaystyle \frac{c_{\vartheta }}{y}}\right|\le {\displaystyle \frac{C_{\vartheta ,\delta }}{n}}.$$

Since $|e^{-x}-e^{-z}|\le |x-z|$ for $x,z\ge 0$, combining this estimate with (17) proves (4).

Global bound. Set

$${\delta }_n={\displaystyle \frac{c_{\vartheta }}{4logn}}$$

for all sufficiently large n, and now take

$$b_n=\lfloor \sqrt{n}\rfloor .$$

For $y\ge {\delta }_n$, Lemma 2.1 gives

$$\lambda =n\tau \left(ny\right)\le C_{\vartheta }logn,b_n\tau \left(ny\right)\le C_{\vartheta }{\displaystyle \frac{logn}{\sqrt{n}}}\le {\displaystyle \frac{1}{2}}.$$

Using $g=g_n$ and $b=b_n$ in Proposition 3.1, we obtain

$$\underset{y\ge {\delta }_n}{sup}\left|{\gamma }_{\vartheta }(M_n\le ny)-e^{-n\tau \left(ny\right)}\right|\le C_{\vartheta }{\displaystyle \frac{{(logn)}^2}{\sqrt{n}}}.$$

(18)

Also, uniformly for $y\ge {\delta }_n$,

$$\left|n\tau \left(ny\right)-{\displaystyle \frac{c_{\vartheta }}{y}}\right|\le {\displaystyle \frac{C_{\vartheta }}{n{y}^2}}\le C_{\vartheta }{\displaystyle \frac{{(logn)}^2}{n}},$$

so (18) gives the required bound on this range.

It remains to consider $0<y<{\delta }_n$. By monotonicity,

$${\gamma }_{\vartheta }(M_n\le ny)\le {\gamma }_{\vartheta }(M_n\le n{\delta }_n).$$

At $y={\delta }_n$ we have $c_{\vartheta }/y=4logn$, while Lemma 2.1 gives $n\tau \left(n{\delta }_n\right)=4logn+O({(logn)}^2/n)$. Hence (18), evaluated at ${\delta }_n$, implies

$${\gamma }_{\vartheta }(M_n\le n{\delta }_n)\le n^{-4}+C_{\vartheta }{\displaystyle \frac{{(logn)}^2}{\sqrt{n}}}.$$

Furthermore,

$$exp\left(-{\displaystyle \frac{c_{\vartheta }}{y}}\right)\le exp\left(-{\displaystyle \frac{c_{\vartheta }}{{\delta }_n}}\right)=n^{-4}.$$

Thus the same $O_{\vartheta }({(logn)}^2/\sqrt{n})$ bound holds for $0<y<{\delta }_n$. Enlarging the constant to cover the finitely many remaining n proves (3). □

Proof 

(Proof of Corollary 1.2). For the i.i.d. model,

$$\mathbb{P}(M_n^{*}\le u)={(1-\tau \left(u\right))}^n.$$

If $y\ge \delta$, then $\tau \left(ny\right)=O_{\vartheta ,\delta }(1/n)$ and hence

$${(1-\tau \left(ny\right))}^n=e^{-n\tau \left(ny\right)}\left(1+O_{\vartheta ,\delta }\left(n^{-1}\right)\right).$$

Lemma 2.1 also gives $n\tau \left(ny\right)=c_{\vartheta }/y+O_{\vartheta ,\delta }\left(n^{-1}\right)$, proving the first assertion. The second follows from the Taylor expansion $log(1-z)=-z-z^2/2+O\left(z^3\right)$, uniformly when z is sufficiently small. □

Remark 4.1

(A possible sharper rate). The block method is not expected to be optimal. If one can establish a uniform two-sided conditional mixing estimate strong enough to control the Chen–Stein $b_3$ term for the exceedance indicators, the regular-continued- fraction strategy of [3] suggests a locally uniform rate of order $\left(\log n\right)/n$. We do not assume such an estimate here.