Singular Distribution Functions for Random Variables with Stationary Digits

Abstract

Let F be the cumulative distribution function (CDF) of the base-q expansion $\sum _{n=1}^\infty X_n q^{-n}$, where $q\ge 2$ is an integer and $\{X_n\}_{n\ge 1}$ is a stationary stochastic process with state space $\{0,\ldots ,q-1\}$. In a previous paper we characterized the absolutely continuous and the discrete components of F. In this paper we study special cases of models, including stationary Markov chains of any order and stationary renewal point processes, where we establish a law of pure types: F is then either a uniform or a singular CDF on [0, 1]. Moreover, we study mixtures of such models. In most cases expressions and plots of F are given.

Appendix

In this appendix we verify the theorems, the propositions, and the corollary in Sections 2–5 which remain to be proven, and we establish some related results. It is convenient to introduce the notation

$$\begin{aligned} (0.x_1\dots x_n)_q:= (0.x_1\dots x_n 0 0 \dots )_q \end{aligned}$$

for $x_1,\dots ,x_n\in \{0,\dots ,q-1\}$.

1.1 Proof of Proposition 1

Let $x=(0.x_1x_2\ldots )_q\in [0,1]$, where the non-terminating expansion is chosen if $x\in \mathbb Q_q$. Then $y=(0.y_1y_2\dots )_q$ is strictly less than x if and only if $y_n<x_n$ for the first index n where $y_n\ne x_n$, and thus by the law of total probability,

$$\begin{aligned} F(x)-\mathrm P\left( X=x\right)&=\mathrm P\left( \sum _{n=1}^\infty X_nq^{-n}<\sum _{n=1}^\infty x_nq^{-n}\right) \\&=\sum _{n=1}^\infty \mathrm P(X_1=x_1,\ldots ,X_{n-1}=x_{n-1},X_n<x_n)\\&=\sum _{n\ge 1:\,x_n\ge 1}\sum _{k=0}^{x_n-1}p(x_1,\ldots ,x_{n-1},k). \end{aligned}$$

Hence (4) is verified and (4) implies (5).

1.2 Proof of Proposition 2

Let $x=(0.x_1x_2\ldots )_q\in [0,1]\setminus \mathbb Q_q$ (by stationarity, F is continuous at x if $x\in \mathbb Q_q$, cf. Remark 2). Then, by monotonicity of probabilities,

$$\mathrm P(X=x)=\lim _{n\rightarrow \infty }p(x_1,\ldots ,x_n),$$

and so Proposition 2 follows immediately.

1.3 Proof of Proposition 3

The first part of Proposition 3 follows immediately from the mixture representation of F as given by (I)–(III) in Section 1.1, where c is the probability of obtaining the case (I) (i.e., $F=cF_1+(1-c)\tilde{F}$ where $\tilde{F}$ is a singular CDF on [0, 1]). For the second part of Proposition 3 we only consider the case $m\ge 1$ as the case $m=0$ follows from similar arguments. Then the following lemma will be useful.

Lemma 1

Let $A\subset \mathbb {R}$ and suppose that $f:A\mapsto \mathbb {R}$ is differentiable at $x\in A$ and $\{a_n\}_{n\ge 1},\{b_n\}_{n\ge 1}$ $\subset A\setminus \{x\}$ are sequences converging to x such that there exists $c>0$ with $\vert a_n-b_n\vert \ge c\max \{\vert a_n-x\vert ,\vert b_n-x\vert \}$ for all $n\in \mathbb {N}$. Then

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{f(a_n)-f(b_n)}{a_n-b_n} =f'(x). \end{aligned}$$

Proof

We have

$$\begin{aligned}&\Big \vert \frac{f(a_n)-f(b_n)}{a_n-b_n} -f'(x)\Big \vert \le \\&\frac{\vert f(a_n)-f(x)-(a_n-x)f'(x)\vert + \vert f(x)-f(b_n)-( x-b_n)f'(x)\vert }{\vert a_n-b_n\vert } \le \\&\frac{1}{c} \Big \vert \frac{f(a_n)-f(x)}{a_n-x}-f'(x)\Big \vert +\frac{1}{c}\Big \vert \frac{f(x)-f(b_n)}{x-b_n}-f'(x)\Big \vert , \end{aligned}$$

where the first inequality follows from the triangle inequality and the second from the assumption on $\vert a_n-b_n\vert$. As the right hand side above goes to 0 for $n\rightarrow \infty$, the proof is complete.

Note that in Lemma 1 it does not matter from which side the sequences $\{a_n\}_{n\ge 1}$ and $\{b_n\}_{n\ge 1}$ approach x. Thus, letting $n\in \mathbb {N}$ and x be as in Proposition 3, we define $a_n=(0.x_1\dots x_n \xi _1\dots \xi _m)_q+q^{-n-m}$ and $b_n=(0.x_1\dots x_n \xi _1\dots \xi _m)_q$. Observe that $a_n,b_n\in [0,1]\setminus \{x\}$. Then

$$\begin{aligned} q^{-m}\max \{\vert a_n-x\vert ,\vert b_n-x\vert \}\le q^{-n-m} =\left| a_n-b_n\right| , \end{aligned}$$

and hence by Lemma 1,

$$\begin{aligned} F'(x)=\lim _{n\rightarrow \infty } \frac{F(a_n)-F(b_n)}{a_n-b_n}. \end{aligned}$$

Thus, since F is continuous at base-q fractions and both $a_n\in \mathbb Q_q$ and $b_n\in \mathbb Q_q$ for sufficiently large n, we get

$$F'(x)=\lim _{n\rightarrow \infty }q^{n+m}\left( \mathrm P(X<a_n)-\mathrm P(X<b_n)\right) =\lim _{n\rightarrow \infty } q^{n+m} p(x_1,\dots ,x_m,\xi _1,\dots ,\xi _m),$$

whereby the proof is completed.

1.4 Proof of Corollary 1

We only prove Corollary 1 for $m>1$, since the cases $m=0$ and $m=1$ follow from similar arguments. By Proposition 3,

$$\begin{aligned} \lim _{n\rightarrow \infty } q^{n+m-1}p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _{m-1})=F'(x)>0. \end{aligned}$$

This implies $p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _m)>0$ for n sufficiently large. Since also

$$\lim _{n\rightarrow \infty }q^{n+m}p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _m)=F'(x),$$

we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _m)}{p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _{m-1})}=q^{-1}\lim _{n\rightarrow \infty }\frac{q^{n+m}p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _m)}{q^{n+m-1}p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _{m-1})}=q^{-1}. \end{aligned}$$

Thereby (7) is verified.

1.5 Proof of Proposition 4

The proof is straightforward when considering each of the cases $x=0$, $x=1$, $x\in \mathbb Q_q$, and $x\in (0,1)\setminus \mathbb Q_q$. For instance, suppose $x=(0.x_1 x_2\dots )_q\in (0,1)\setminus \mathbb Q_q$. For $n\in \mathbb {N}$, define $y_n:=(0.x_1\dots x_n)_q$ and $z_n:= y_n+q^{-n}$. Then $y_n<x< z_n$ at least for sufficiently large n, and since $y_n\in \mathbb Q_q$ and $z_n\in \mathbb Q_q$, stationarity implies $\mathrm P(X=y_n)=\mathrm P(X=z_n)=0$, and so $F(z_n)-F(y_n)=p(x_1,\dots ,x_n)$. Thereby Proposition 4 is verified in the case where $x=(0.x_1 x_2\dots )_q$ is a non-base-q fraction in (0, 1).

1.6 Proof of Theorem 1

Assume $\{X_n\}_{n\ge 1}$ is a stationary Markov chain of order $m-1$ for which F is not the uniform CDF on [0, 1]. Then it follows from (8) that there must exist $\xi _1,\dots ,\xi _{m}\in \{0,\dots ,q-1\}$ such that $\pi _{\xi _1,\dots ,\xi _{m}}\ne q^{-1}$. Let $x=(0.x_1x_2\dots )_q\in (0,1)\setminus \mathbb Q_q$ where $F'(x)$ exists. If $F'(x)>0$, then we obtain a contradiction:

$$\begin{aligned} q^{-1}=\lim _{n\rightarrow \infty } \frac{p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _m)}{p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _{m-1})}= \pi _{\xi _1,\dots ,\xi _m}\ne q^{-1}, \end{aligned}$$

where the first equality follows from Corollary 1 and the second from the Markov property. Consequently, $F'(x)=0$ and thus F is singular.

Using a notation as in Section 3, assume that $\pi _{x_1,\ldots ,x_m}<1$ for all $x_1,\dots ,x_m\in \{0,\dots , q-1\}$. Consider any $x=(0.x_1x_2\dots )_q\in [0,1]$ and define

$$\begin{aligned} \lambda :=\max _{b_1,\dots ,b_m\in \{0,1,\dots ,q-1\}}\pi _{b_1,\ldots ,b_m} <1. \end{aligned}$$

Then, for any integer $n\ge m$, (9) gives

$$\begin{aligned} p(x_1,\ldots ,x_n)\le \lambda ^{n-m+1}\rightarrow 0\qquad \text{ as } n\rightarrow \infty \text{. } \end{aligned}$$

Hence, Proposition 2 gives that F is continuous at x. This completes the proof of Theorem 1.

1.7 Proof of Proposition 5

For any $m\in \mathbb N$, denote the finite dimensional probabilities of $\{X_n^{(m)}\}_{n\ge 1}$ by

$$\begin{aligned} p_m(x_1,\dots ,x_n)=\mathrm P(X_1^{(m)}=x_1,\dots , X_n^{(m)}=x_n) \end{aligned}$$

for $n\in \mathbb {N}$ and $x_1,\ldots ,x_n\in \{0,\ldots ,q-1\}$. By construction of $\{X_n^{(m)}\}_{n\ge 1}$, we have $p_m(x_1,\dots ,x_n) =p(x_1,\dots ,x_n)$ whenever $n\le m$. For any $x=(0.x_1\ldots x_m)_q$, stationarity implies that $\mathrm P(X=x)=0$, and so it follows from (4) that

$$\begin{aligned} F^{(m)}( (0.x_1\dots x_m)_q)=F( (0.x_1\dots x_m)_q). \end{aligned}$$

(29)

Let $x=(0.x_1x_2\dots )_q\in [0,1]$ be arbitrary. Combining (29) with the fact that $F^{(m)}$ is non-decreasing gives

$$\begin{aligned} F((0.x_1\dots x_m)_q\le F^{(m)}(x) \le F((0.x_1\dots x_m)_q+q^{-m}). \end{aligned}$$

Here, by the continuity of $F_3$, the left and the right hand side expressions of the inequalities converge to F(x) as $m\rightarrow \infty$, so $F^{(m)}$ converges pointwise to F (weak convergence). Hence, since F is a continuous CDF, we obtain (10), cf. Rao (1962).

1.8 Proof of Theorem 2

Let the situation be as in Theorem 2.

The case (I) follows immediately since then the $X_n$’s are independent and uniformly distributed on $\{0,1\}$.

Suppose that $\mathrm P(Z_1=k)=1$ for some $k\in \mathbb N$. Then $Z_0$ is uniformly distributed on $\{1,\ldots ,k\}$. If $\ell \in \{1,\ldots ,k\}$ and $x=(0.x_1x_2\dots )_2=2^{-\ell }/(1-2^{-k})$, then $x= \sum _{m=0}^\infty 2^{-\ell -km}$ and we get from (17) that $p(x_1,\dots ,x_n)=1/k$. Thereby the case (II) is verified.

To show the case (III) assume first that $Z_1$ is not geometrically distributed with mean 2. Equivalently,

$$\mathrm P(Z_1>m\mid Z_1>m-1)\ne {1}/{2}$$

for some $m\in \mathbb {N}$. We show by contradiction that $F'(x)=0$ for all $x\in (0,1)\setminus \mathbb Q_2$ where $F'(x)$ exists. So suppose that $F'(x)>0$ for some $x=(0.x_1x_2\dots )_2\in (0,1)\setminus \mathbb Q_2$. Let $\xi _1=1$ and $\xi _j=0$ for $j\in \{2,\dots ,m+1\}$. Then, for any $n\in \mathbb {N}$,

$$\begin{aligned} \frac{p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _{m+1})}{p(x_1,\dots ,x_n,\xi _1,\dots ,\xi _m)}=\frac{\mathrm {P}(Z_1>m)}{\mathrm {P}(Z_1>m-1)}=\mathrm P(Z_1>m\mid Z_1>m-1)\ne {1}/{2} \end{aligned}$$

using in the first identity (17) and that $Z_1,Z_2,\dots$ are identically distributed. This contradicts Corollary 1 and thus $F'(x)=0$. In conclusion, F is singular if $Z_1$ is not geometrically distributed with mean 2.

Suppose next that $\mathrm {P}(Z_1=k)<1$ for all $k\in \mathbb {N}$. If $x\in [0,1]$ has finitely many digits equal to 1, then $\mathrm P(X=x)=0$. If x has infinitely many digits equal to 1, then (17) gives

$$\begin{aligned} \lim _{n\rightarrow \infty }p(x_1,\ldots ,x_n)\le \prod _{n:\ x_n=1} \sup _{m\in \mathbb N}\mathrm P(Z_1=m)=0, \end{aligned}$$

since $\sup _{n\in \mathbb N}\mathrm P(Z_1=n)<1$. Hence, by Proposition 2, F is continuous.

Consequently, F is singular continuous in case (III).

Finally, Theorem 2(IV) follows immediately from Proposition 4 using (17) and the definition of the distribution of $Z_0$.

1.9 Proof of Proposition 7

Let the situation be as in Proposition 7 and let G be given by (21). By Proposition 6(V), there exists $c\in [0,1]$ such that $c=G'(x)=F'(x)-\mathrm {P}(F_\Pi =F_1)$ for Lebesgue almost all $x\in [0,1]$. We will show that $c=0$ by contradiction, whereby Proposition 7 is verified.

Suppose that $c>0$. For any $j\in \mathbb {N}$, define $U_j:= \{\pi \in \Omega \mid F_\pi \ne F_1\}\setminus V_j$. Since $V_j$ increases to $\{\pi \in \Omega \mid F_\pi \ne F_1\}$, there exists $j_0\in \mathbb {N}$ such that $P(\Pi \in U_{j_0})<c$. For $x\in \mathbb {R}$, define

$$\begin{aligned} J(x):=\mathrm {E}(F_\Pi (x)1[\Pi \in U_{j_0}]),\qquad K(x):=\mathrm {E}(F_\Pi (x)1[\Pi \in V_{j_0}]). \end{aligned}$$

Then $G=J+K$. Later on in this proof we will show that K, if not identically zero, must at least be singular, which implies that for Lebesgue almost all $x\in [0,1]$, $J'(x)=G'(x)=c>0$, hence by the definition of J, we have $\mathrm {P}(\Pi \in U_{j_0})>0$. The function $J/\mathrm {P}(\Pi \in U_{j_0})$ is a CDF satisfying (1), so by Proposition 6(V), $c/\mathrm {P}(\Pi \in U_{j_0})=J'/ \mathrm {P}(\Pi \in U_{j_0})\le 1$ almost everywhere on [0, 1]. This is in contradiction with $\mathrm {P}(\Pi \in U_{j_0})<c$, so $c=0$.

Now let us show that $K'=0$ almost everywhere. Clearly, we may assume that $\mathrm {P}(\Pi \in V_{j_0})>0$ so that K is not identically zero. Then $K/\mathrm {P}(\Pi \in V_{j_0})$ is a CDF satisfying (1) and therefore it is differentiable almost everywhere. Let A denote the set of points $x\in [0,1]$ for which both $K'(x)$ exists and the sequence $\{f_{x,n}(\cdot )1[\cdot \in V_{j_0}]\}_{n\ge 1}$ converges pointwise to 0 as $n\rightarrow \infty$ and is dominated by a constant. Then A has Lebesgue measure 1, since, by assumption A is the intersection of two sets of Lebesgue measure 1. Hence we can combine Proposition 6(V) with (20) and Lebesgue’s dominated convergence theorem to obtain

$$\begin{aligned} K'(x)/\mathrm {P}(\Pi \in V_{j_0})=\lim _{n\rightarrow \infty } \mathrm {E}(q^n p_\Pi (x_1,\dots ,x_n)1[\Pi \in V_{j_0}])/\mathrm {P}(\Pi \in V_{j_0})=0, \end{aligned}$$

for all $x=(0.x_1x_2\dots )_q\in A$. Thus K is singular.

1.10 Proof of Theorem 3

Theorem 3(I)–(II) follow directly from Proposition 6(III)–(IV).

When verifying Theorem 3(III) we assume $m\ge 2$ and define without loss of generality $\Omega$ as follows (the case $m=1$ is simpler and follow similar lines as below). Denote the standard simplex in $\mathbb {R}^n$ by

$$\begin{aligned} \Delta _n:= \{(\alpha _1,\dots ,\alpha _{n})\in [0,1]^n\mid \alpha _1+\dots +\alpha _{n}=1\}. \end{aligned}$$

Consider the initial distribution $\{\pi (x_1,\dots ,x_{m-1})\mid x_1,\ldots ,x_{m-1}\in \{0,\ldots ,q-1\}\}$ as a vector in $\Delta _{q^{m-1}}$ and the collection of transition probabilities $\{\pi _{x_1,\dots ,x_m}\mid x_1,\ldots ,x_m\in \{0,\ldots ,q-1\}\}$ as a $q^{m-1}$-tuple of vectors in $\Delta _{q}$. Then $\Delta _{q^{m-1}}\times \Delta _{q}^{q^{m-1}}$ can be identified with the collection of distributions for Markov chains with state space $\{0,\ldots ,q-1\}$ and of order $m-1$. Let $\Omega$ be the subset of those elements of $\Delta _{q^{m-1}}\times \Delta _{q}^{q^{m-1}}$ which correspond to stationary Markov chains of order $m-1$, and let the $\sigma$-algebra on $\Omega$ be induced by the Borel $\sigma$-algebra on $\mathbb {R}^{q^{m-1}+q^{m}}$.

We now verify the requirements of Proposition 7 whereby Theorem 3(III) follows. Define

$$\begin{aligned} \phi (\pi ):= q\prod _{t_1,\dots ,t_{m}\in \{0,\ldots ,q-1\}}\pi _{t_1,\dots ,t_{m}}^{q^{-m}},\qquad \pi \in \Omega , \end{aligned}$$

and let

$$\begin{aligned} V_j=\{\pi \in \Omega \mid \phi (\pi )\le q^{-1/j}\},\qquad j\in \mathbb N. \end{aligned}$$

(30)

Then $V_1\subseteq V_2\subseteq \dots$. Further, let $\pi ^*\in \Omega$ correspond to having all initial probabilities equal to $q^{-m}$ and all transition probabilities equal to $q^{-1}$ (so $F_{\pi ^*}$ is the uniform CDF on [0, 1]). Then $\pi ^*$ is the unique maximizer of $\phi$ and $\phi (\pi ^*)=1$. This follows e.g. with the use of Lagrange multipliers. Furthermore, it follows from (8) that $F_\pi =F_1$ if and only if $\pi =\pi ^*$. Hence

$$\begin{aligned} \bigcup _{j=1}^\infty V_j=\Omega \setminus \{\pi ^*\}=\{\pi \in \Omega \mid F_\pi \ne F_1\}, \end{aligned}$$

which is one requirement of Proposition 7.

To verify the other requirement of Proposition 7 we need some notation. For any $x=(0.x_1x_2\ldots )_q\in [0,1]\setminus \mathbb Q_q$ and any $n\in \mathbb N$, define

$$\begin{aligned} \phi _{x,n}(\pi ):= q\prod _{t_1,\dots ,t_{m}\in \{0,\ldots ,q-1\}}\pi _{t_1,\dots ,t_{m}}^{n_{t_1,\dots ,t_{m}}(x_1,\dots ,x_n)/n},\qquad \pi \in \Omega , \end{aligned}$$

where $n_{t_1,\dots ,t_j}(x_1,\dots ,x_k)$ is the number of times the string $t_1,\dots ,t_j$ appears in the string $x_1,\dots ,x_k$. Then the other requirement of Proposition 7 states that for all $j\in \mathbb N$, Lebesgue almost all $x=(0.x_1x_2\ldots )_q\in [0,1]\setminus \mathbb Q_q$, and all $\pi \in \Omega$ we have that

$$\begin{aligned}&q^n \pi (x_1,\ldots ,x_{m-1})\prod _{j=m}^n \pi _{x_{j-m+1},\ldots ,x_j} 1[\phi (\pi )\le q^{-1/j}]\\ =&\, \pi (x_1,\ldots ,x_{m-1}) \phi _{x,n}(\pi )^n1[\phi (\pi )\le q^{-1/j}] \end{aligned}$$

converges to 0 as $n\rightarrow \infty$ and is less than some number c(x, j). To verify this we recall that $(0.x_1x_2\dots )_q\in [0,1]$ is a normal number (in base q) if for all $j\in \mathbb N$ and all $t_1,\dots ,t_j\in \{0,\ldots ,q-1\}$,

$$\begin{aligned} \lim _{k\rightarrow \infty } {n_{t_1,\dots ,t_j}(x_1,\dots ,x_k)}/{k}=q^{-j}. \end{aligned}$$

(31)

Since the set of normal numbers in [0, 1] has Lebesgue measure 1 Borel (1909), we can assume that x is normal. By (31) there exists some $n_j\in \mathbb N$ such that for all $t_1,\dots ,t_m\in \{0,\dots ,q-1\}$,

$$\begin{aligned} n_{t_1,\dots ,t_m}(x_1,\dots ,x_n)/n\ge \frac{2j}{(2j+1)q^m} \end{aligned}$$

whenever $n\ge n_j$. Consequently, for all $n\ge n_j$ and $\pi \in V_j$,

$$\begin{aligned} \phi _{x,n}(\pi )\le q\prod _{t_1,\dots ,t_{m}\in \{0,\ldots ,q-1\}}\pi _{t_1,\dots ,t_{m}}^{\frac{2j}{(2j+1)q^m}}= (q\phi (\pi )^{2j})^{1/(2j+1)}\le q^{-1/(2j+1)}<1, \end{aligned}$$

where the second last inequality uses (30). Thereby the other requirement of Proposition 7 follows.

1.11 Proof of Theorem 4

Theorem 4(I)–(II) follow directly from Proposition 6(III)–(IV). Below we verify the requirements of Proposition 7 whereby Theorem 4(III) follows.

Define without loss of generality

$$\begin{aligned} \Omega =\left\{ (\pi _1,\pi _2,\dots )\in \ell ^1(\mathbb {N})\mid \sum _{k=1}^\infty \pi _k=1, \mathrm{and} \sum _{k=1}^\infty k\pi _k<\infty \right\} \end{aligned}$$

and let the $\sigma$-algebra on $\Omega$ be induced by the Borel $\sigma$-algebra on $\ell ^1(\mathbb {N})$ (the space of absolutely summable sequences) equipped with the usual $\ell ^1$-norm. For any $m\in \mathbb {N}$ and $\pi \in \Omega$, define

$$\begin{aligned} \phi _m(\pi ):= 2\prod _{k=1}^m\pi _k^{2^{-k-1}}, \qquad \pi ^{(m)}:= (1-2^{-m})^{-1} \left( 2^{-1},\dots ,2^{-m},0,0,\dots \right) , \end{aligned}$$

and

$$\phi (\pi ):= 2\prod _{k=1}^\infty \pi _k^{2^{-k-1}}, \qquad \pi ^*:=(1/2,1/4,1/8,\ldots ) ,$$

so $\pi ^*\in \Omega$ is the geometric distribution with mean 2. We now verify that

$$\begin{aligned} \pi ^* \text{ is } \text{ the } \text{ unique } \text{ maximizer } \text{ of } \phi \text{. } \end{aligned}$$

(32)

It is easily seen that $\pi ^{(m)}\in \Omega$ is the unique maximizer of $\phi _m$, and since for all $\pi \in \Omega$, $\phi _m(\pi )$ non-increases towards $\phi (\pi )$ as $m\rightarrow \infty$, it follows that

$$\begin{aligned} \phi (\pi )\le \lim _{m\rightarrow \infty }\phi _m\left( \pi ^{(m)}\right) =\lim _{m\rightarrow \infty }2^{(m+2)/2^{m+1}}(1-2^{-m})^{(2^{-m}-1)/2}=1. \end{aligned}$$

Furthermore, $\Omega$ is a convex set and we claim that $\phi$ is a strictly log-concave function: Consider any $\alpha \in (0,1)$ and distinct $\pi ,\tilde{\pi }\in \Omega$, so there exist some $\pi _\ell \not =\tilde{\pi }_\ell$. Since $\ln$ is strictly concave, we have $\ln (\alpha \pi _\ell +(1-\alpha )\tilde{\pi }_\ell )>\alpha \ln \pi _\ell +(1-\alpha )\ln \tilde{\pi }_\ell$ and $\ln (\alpha \pi _k+(1-\alpha )\tilde{\pi }_k)\ge \alpha \ln \pi _k+(1-\alpha )\ln \tilde{\pi }_k$ for $k\not =l$, so

$$\begin{aligned} \ln \phi (\alpha \pi +(1-\alpha )\tilde{\pi })>\alpha \ln \phi (\pi )+ (1-\alpha )\phi (\tilde{\pi }). \end{aligned}$$

Consequently, $\phi \le 1$ is strictly log-concave with

$$\begin{aligned} \phi \left( \pi ^*\right) =2^{1-\sum _{k=1}^\infty k2^{-k-1}}=1, \end{aligned}$$

and so (32) follows.

Now, for any $j\in \mathbb {N}$, define

$$\begin{aligned} V_j:=\{\pi \in \Omega \mid \phi _m(\pi )\le 2^{-1/m} \, \mathrm{whenever} \, m\ge j\}. \end{aligned}$$

(33)

Then $V_1\subseteq V_2\subseteq \dots$. For all $m\in \mathbb {N}$, a direct calculation gives $\phi _m(\pi ^*)=2^{(m+2)/2^{m+1}}>1$, so $\pi ^*\notin \bigcup _{j=1}^\infty V_j$. On the other hand, suppose $\pi \in \Omega \setminus \pi \notin \bigcup _{j=1}^\infty V_j$. Then by (33) we can find an increasing sequence $\{m_k\}_{k\ge 1}$ such that for all $k\in \mathbb {N}$, we have $\phi _{m_k}(\pi )>2^{-1/m_k}$, and so taking the limit as $k\rightarrow$ we obtain $\phi (\pi )\ge 1$. Therefore, by (32), $\pi =\pi ^*$, so

$$\begin{aligned} \bigcup _{j=1}^\infty V_j=\Omega \setminus \{\pi ^*\}. \end{aligned}$$

From Theorem 2(I) and the definition of $\pi ^*$ we obtain that $F_\pi =F_1$ if and only if $\pi =\pi ^*$, and hence

$$\begin{aligned} \bigcup _{j=1}^\infty V_j=\{\pi \in \Omega \mid F_\pi \ne F_1\}, \end{aligned}$$

which is one requirement of Proposition 7.

For any $j\in \mathbb {N}$ and any normal number $x=(0.x_1x_2\dots )_q\in [0,1]$ (cf. (31)), define

$$\begin{aligned} \phi _{x,n}(\pi ):= 2\prod _{k=1}^\infty \pi _k^{n_k(x_1,\ldots ,x_n)/n},\qquad \pi \in \Omega . \end{aligned}$$

By (31) there exists some $n_j\in \mathbb N$ such that for all $k\in \{1,\dots ,j\}$,

$$\begin{aligned} n_{k}(x_1,\dots ,x_n)/n\ge \frac{2j}{(2j+1)2^{k+1}} \end{aligned}$$

(34)

whenever $n\ge n_j$. Hence, for all $n\ge n_j$ and all $\pi \in V_j$,

$$\begin{aligned} \phi _{x,n}(\pi )\le 2\prod _{k=1}^j\pi _{k}^{n_k(x_1,\dots ,x_n)/n}\le 2\prod _{k=1}^j\pi _{k}^{\frac{2j}{(2j+1)2^{k+1}}}= (2\phi _j(\pi )^{2j})^{1/(2j+1)}\le 2^{-1/(2j+1)}, \end{aligned}$$

where the first inequality uses that $\prod _{k=1}^m \pi _{k}^{n_k(x_1,\dots ,x_n)/n}\downarrow \prod _{k=1}^\infty \pi _{k}^{n_k(x_1,\dots ,x_n)/n}$ as $m\rightarrow \infty$, the second inequality uses (34), and the last inequality uses (33). Therefore, for all $n\ge n_j$ and all $\pi \in V_j$, we have $\phi _{x,n}(\pi )<1$. Furthermore, by definition, for all $\pi \in V_j$ and $n\in \mathbb {N}$, we have $\phi _{x,n}(\pi )^n\le 2^n$. So, for any $\pi \in \Omega$, we get that $\{\phi _{x,n}^n1[\pi \in V_j]\}_{n\ge 1}$ is dominated by the constant $2^{n_j}$ and converges pointwise to 0 as $n\rightarrow \infty$. Thereby the other requirement of Proposition 7 holds, and so the proof of Theorem 4 is completed.