# On Risk-Sensitive Decision
Making Under Uncertainty

Chung-Han Hsieh,∗ Member, IEEE and Yi-Shan Wong∗∗ This paper is partially
supported by the Ministry of Science and Technology (MOST), Taiwan. under
Grants: MOST111-2221-E-007-124-.
∗Chung-Han Hsieh is with the Department of Quantitative Finance, National
Tsing Hua University, Hsinchu 300044, Taiwan. E-mail<EMAIL_ADDRESS>∗∗Yi-Shan Wong was with the Department of Quantitative Finance, National Tsing
Hua University, Hsinchu 300044, and is currently with the Actuarial Department
I, Cathay Life Insurance, Taipei 106436, Taiwan. (E-mail:
yishan.wong13@gmail.com).

###### Abstract

This paper studies a risk-sensitive decision-making problem under uncertainty.
It considers a decision-making process that unfolds over a fixed number of
stages, in which a decision-maker chooses among multiple alternatives, some of
which are deterministic and others are stochastic. The decision-maker’s
cumulative value is updated at each stage, reflecting the outcomes of the
chosen alternatives. After formulating this as a stochastic control problem,
we delineate the necessary optimality conditions for it. Two illustrative
examples from optimal betting and inventory management are provided to support
our theory.

###### Index Terms:

Stochastic Systems, Decision Making Under Uncertainty, Risk-Sensitive
Optimization, Variance Regularization.

##  I Introduction

Optimal decision-making under uncertainty is critical across various
disciplines, including engineering, management, economics, and finance. The
standard approach to addressing this issue is by applying expected utility
theory, which formulates the problem as stochastic programming; see [1, 2].

Early paradigms for this type of problem include expected utility theory in
[3], the Markowitz mean-variance portfolio optimization [4], Kelly’s criterion
[5, 6], and the prospect theory [7]. Subsequent studies have explored
different aspects of this topic. For instance, [8] analyzed the rebalancing
frequency in log-optimal portfolio selection, [9] examined the Kelly betting
problem in a finite stage framework. Studies by [10, 11] investigated the
foundations of using Expected Logarithmic Growth (ELG) and the variance of
logarithmic growth as preferences in portfolio management problems.

Further studies by [12, 13] delved into ELG portfolios and derived various
optimalities. More recently, [14, 15] explored the frequency-dependent log-
optimal portfolio and its asymptotic log-optimality. Then [16] studied the
frequency-dependent log-optimal portfolio problems with costs. A nonlinear
control theoretic approach is considered in [17]. However, these studies
largely omitted an explicit focus on risk within the optimization goal.

To this end, this paper studies a class of risk-sensitive decision-making
problems under uncertainty. The objective is composed of expected log-growth
and variance of log-growth of the decision maker’s account. We consider a
decision-making process that unfolds over a fixed number of stages, where a
decision-maker chooses among multiple alternatives, some of which are
deterministic and others are stochastic. The decision-maker’s accumulated
value is updated at each stage based on the chosen alternatives. This problem
is modeled as a stochastic control problem, aimed at maximizing a risk-
sensitive objective function. Furthermore, the paper provides the necessary
optimality conditions for this problem.

The rest of the paper is organized as follows. Section II provides the problem
formulation for the risk-sensitive decision-making problem. Section III first
presents an equivalent optimization problem. Then we provide rigorous proof
for the necessary condition for the optimality of the problem. Subsequently,
Section IV illustrates the optimality result with several numerical examples.
Finally, Section V provides concluding remarks and outlines potential future
research directions.

##  II Problem Formulation

Consider a decision maker engaged in a decision-making process that unfolds
over a fixed number of stages, denoted by $N>1$. For $k=0,1,\dots,N-1$, a
decision maker chooses among $m\geq 2$ alternatives, with one being a
deterministic option with a nonnegative payoff $r(k)\geq 0.$ That is, the
outcome is certain and is treated as a degenerate random variable with value
$r(k)$ for all $k$ with probability one. Alternatively, if the $i$th
alternative is stochastic, we take $X_{i}(k)$ as the random payoffs. We assume
that the payoff vectors
$X(k):=\left[X_{1}(k),\,X_{2}(k),\,\dots,\,X_{m}(k)\right]^{\top}$ are i.i.d.
and have a known distribution and have components $X_{i}(\cdot)$ which can be
arbitrarily correlated. These vectors have components satisfying
$X_{\min,i}\leq X_{i}(k)\leq X_{\max,i}$ with predefined limits, where
$-1<X_{\min,i}<0<X_{\max,i}<1$.

###  II-A Notation

Throughout the paper, we take $\mathbb{R}$ to be the set of real numbers and
$\mathbb{R}_{+}$ to be the set of nonnegative real numbers. We denote by 1 the
vector with all components to be equal to $1$ and denote by
$e_{i}\in\mathbb{R}^{m}$ to be the unit vector having 1 at the $i$th
component. All random objects are defined in a probability space
$(\Omega,\mathcal{F},\mathbb{P})$ with $\Omega$ being the sample space,
$\mathcal{F}$ being the information set, and $\mathbb{P}$ being the
probability measure. For $a,b\in\mathbb{R}^{n}$, the term $\langle a,b\rangle$
means the standard inner product $a^{\top}b$.

###  II-B Linear Decision Policy

Take $V(k)$ to be the decision maker’s accumulated value and the $i$th
feedback gain $K_{i}(k)\in[0,1]$ represents the fraction of the account
allocated to the $i$th asset for $i=1,\dots,m$. Said another way, the $i$th
controller is a linear feedback of the form $u_{i}(k)=K_{i}(k)V(k).$ For each
$k$, we take $K_{i}(k):=K_{i}$ and impose a unit simplex constraint

$K\in{\mathcal{K}}:=\left\\{K\in\mathbb{R}^{m}:K_{i}\geq 0\text{ for all
$i$},\;K^{\top}\textbf{1}=1\right\\}$

which is readily verified as a convex set.

###  II-C Value Dynamics

Letting $n\geq 1$ be the number of steps between decisions, at time $k=0$, the
decision-maker begins with initial account $V_{0}:=V(0)$ and initial
investment control $u(0)=\sum_{i=1}^{m}K_{i}V_{0}$ and waits $n$ steps. Then,
when $k=n$, the control is updated to be $u(n)=\sum_{i=1}^{m}K_{i}V(n).$

The decision-maker starts with an initial accumulated value $V(0)>0$. At each
stage, the decision-maker chooses to allocate a fraction of the accumulated
value to each of the $m$ alternatives. This allocation is represented by the
control gain $K$, where each $K_{i}$ is the fraction of the accumulated value
allocated to the $i$th alternative. For decision period $n\geq 1$, the
corresponding account value at stage $n$ is described by the stochastic
recursion $V(n)=\langle K,\mathcal{R}_{n}\rangle V_{0}$ where $\langle
K,\mathcal{R}_{n}\rangle$ represents the total return from all alternatives at
stage $n$ with the random vector $\mathcal{R}_{n}$ having the $i$th component

$\mathcal{R}_{n,i}:=\prod_{k=0}^{n-1}(1+X_{i}(k))$

for $i=1,2,\ldots,m$. It is readily seen that $\mathcal{R}_{n,i}>0$ for all
$n\geq 1$. In the sequel, we may sometimes write $V(n;K)$ to emphasize the
dependence on the feedback gain $K$.

###  II-D Risk-Sensitive Decision Making Problem

For any decision period $n\geq 1$, we consider a risk-sensitive objective
function,111In finance, a utility function $U$ is said to be risk-averse if it
is concave, continuous, and strictly increasing; see [11].  see [10], as
follows: Take

$\displaystyle{U_{n}^{\rho}}(K;X)$
$\displaystyle:=\frac{1}{n}\mathbb{E}\left[\log\frac{V(n;K)}{V_{0}}\right]-\frac{\rho}{2n^{2}}\text{var}\left(\log\frac{V(n;K)}{V_{0}}\right)$
(1)

where $\rho\geq 0$ is the risk-aversion constant, which modulates the degree
of risk sensitivity, with higher values of $\rho$ indicating a greater
aversion to risk. Our goal is to solve the following stochastic control
problem parameterized by feedback gain $K$:

$\displaystyle\begin{aligned} &\sup_{K\in\mathcal{K}}\;U_{n}^{\rho}(K;X)\\\
&{\rm s.t.}\;V(n)=\langle K,\mathcal{R}_{n}\rangle V_{0}\end{aligned}$ (2)

with $\mathcal{R}_{n}=[\mathcal{R}_{n,1},\dots,\mathcal{R}_{n,m}]^{\top}$ with
$\mathcal{R}_{n,i}=\prod_{k=0}^{n-1}(1+X_{i}(k))$ for $i\in\\{1,2,\dots,m\\}$.
That is, the decision-maker must balance the expected logarithmic return
against the risk, represented by the variance of the logarithmic return.

######  Remark.

It should be noted that Problem (2) may be nonconvex since the log-variance
quantity may not be convex. However, if the risk-aversion constant $\rho=0$,
then Problem (2) reduces to the classical expected log-growth maximization
problem, which generally forms a concave program; see [18, 15]. Later in
Section IV, we shall demonstrate some cases where the log-variance is indeed
convex.

##  III Main Results

This section first presents an equivalent problem and then provides the
necessary conditions for the optimality of the problem. Indeed, the next lemma
simplifies Problem (2).

######  Lemma 3.1.

For $n\geq 1$, the risk-sensitive objective function $U_{n}^{\rho}(K;X)$
satisfies

$\displaystyle{U_{n}^{\rho}}(K;X)$
$\displaystyle=\frac{1}{n}\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]-\frac{\rho}{2n^{2}}\mathbb{E}\left[\left(\log\langle
K,\mathcal{R}_{n}\rangle\right)^{2}\right]$
$\displaystyle\qquad+\frac{\rho}{2n^{2}}\left(\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]\right)^{2}.$

###### Proof.

We shall proceed with a straightforward calculation. Note that

$\displaystyle{U_{n}^{\rho}}(K;X)$
$\displaystyle=\frac{1}{n}\mathbb{E}\left[\log\frac{V(n)}{V(0)}\right]-\frac{\rho}{2n^{2}}\text{var}\left(\log\frac{V(n)}{V(0)}\right)$
$\displaystyle=\frac{1}{n}\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]-\frac{\rho}{2n^{2}}\text{var}\left(\log\langle
K,\mathcal{R}_{n}\rangle\right)$
$\displaystyle=\frac{1}{n}\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]-\frac{\rho}{2n^{2}}\mathbb{E}\left[\left(\log\langle
K,\mathcal{R}_{n}\rangle\right)^{2}\right]$
$\displaystyle\qquad+\frac{\rho}{2n^{2}}\left(\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]\right)^{2},$

which is desired. ∎

While Problem (2) may be nonconvex, see Appendix, the following lemma
indicates that the necessary condition for optimality is possible to
establish.

######  Lemma 3.2 (Necessary Optimality Conditions).

If the feedback vector $K^{*}\in\mathcal{K}$ is optimal to the stochastic
control problem

$\displaystyle\begin{aligned} &\sup_{K\in\mathcal{K}}\;U_{n}^{\rho}(K;X)\\\
&{\rm s.t.}\;V(n)=\langle K,\mathcal{R}_{n}\rangle V_{0}\end{aligned}$ (3)

then for $i=1,2,\ldots,m$, we have

$\displaystyle\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]-\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\cdot\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]$
$\displaystyle+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]=1,\text{ if $K_{i}^{*}>0$ }$
$\displaystyle\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]-\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]$
$\displaystyle+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]\leq 1,\text{ if $K_{i}^{*}=0$. }$

###### Proof.

With the aid of Lemma 3.1, we rewrite Problem (LABEL:problem:_risk-
sensitive_problem) as an equivalent constrained minimization problem described
as follows:

$\displaystyle\min_{K}-\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]+\frac{\rho}{2n}\mathbb{E}\left[(\log\langle
K,\mathcal{R}_{n}\rangle)^{2}\right]$
$\displaystyle\qquad-\frac{\rho}{2n}\left(\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]\right)^{2}$ $\displaystyle{\rm
s.t.}\;K^{\top}\mathbf{1}-1=0;$ $\displaystyle\;\;\;\;\;-K^{\top}e_{i}\leq
0,\;i=1,2,\ldots,m$

where $e_{i}\in\mathbb{R}^{m}$ is unit vector having 1 at the $i$th component.
Consider the Lagrangian

$\displaystyle\mathcal{L}(K,\lambda,\mu)$
$\displaystyle:=-\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]+\frac{\rho}{2n}\mathbb{E}\left[(\log\langle
K,\mathcal{R}_{n}\rangle)^{2}\right]$
$\displaystyle\qquad-\frac{\rho}{2n}\left(\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]\right)^{2}$
$\displaystyle\qquad+\lambda(K^{\top}\mathbf{1}-1)-\mu^{\top}K.$

Then the Karush-Kuhn-Tucker (KKT) Conditions implies that if $K^{*}$ is a
local maximum then there exists a scalar $\lambda\in\mathbb{R}^{1}$ and a
vector $\mu\in\mathbb{R}^{m}$ with component $\mu_{j}\geq 0$ such that, for
$i=1,2,\ldots,m$,

$\displaystyle-\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]\left(1+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]\right)$
$\displaystyle\qquad+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]+\lambda-\mu_{i}=0$ (4) $\displaystyle
K^{*\top}\mathbf{1}-1=0$ (5) $\displaystyle\mu_{i}K_{i}^{*}=0$ (6)

From Equation (4), we obtain

$\displaystyle\mu_{i}$
$\displaystyle=-\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{K^{*\top}\mathcal{R}_{n}}\right]\left(1+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]\right)$
$\displaystyle\qquad+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\cdot\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]+\lambda.$ (7)

for $i=1,2,\ldots,m$. Using the fact that $\mu_{i}K_{i}^{*}=0$, for
$i=1,2,\dots,n$, we take weighted sum of Equation (III) and have

$\displaystyle\sum_{i=1}^{m}\mu_{i}K_{i}^{*}$
$\displaystyle\quad=\sum_{i=1}^{m}K_{i}^{*}\left(-\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]\left(1+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]\right)\right)$
$\displaystyle\quad+\sum_{i=1}^{m}K_{i}^{*}\left(\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\cdot\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]+\lambda\right)=0.$

Since

$\sum_{i=1}^{m}K_{i}^{*}\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]=\mathbb{E}\left[\frac{\langle
K^{*},\mathcal{R}_{n}\rangle}{\langle K^{*},\mathcal{R}_{n}\rangle}\right]=1,$

and $\sum_{i=1}^{m}K_{i}^{*}=1$, we obtain

$\displaystyle-1-\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]+\lambda=0.$ (8)

Thus, we have $\lambda=1$ and substitute it into Equation (III). This tells us
that for $i=1,2,\ldots,m$,

$\displaystyle\mu_{i}$
$\displaystyle=-\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]\left(1+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]\right)$
$\displaystyle\qquad+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\cdot\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]+1.$

The fact $\mu_{i}K_{i}^{*}=0$ implies that if $K_{i}^{*}>0$, then $\mu_{i}=0$
and

$\displaystyle\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]-\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\cdot\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]$
$\displaystyle\qquad+\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\right]\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]=1.$

If $K_{i}^{*}=0$, then $\mu_{i}\geq 0$, which implies that

$\displaystyle\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]-\frac{\rho}{n}\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{n}\rangle\cdot\frac{\mathcal{R}_{n,i}}{\langle
K^{*},\mathcal{R}_{n}\rangle}\right]$
$\displaystyle\qquad+\frac{\rho}{n}\mathbb{E}\left[\log(K^{*\top}\mathcal{R}_{n})\right]\mathbb{E}\left[\frac{\mathcal{R}_{n,i}}{K^{*\top}\mathcal{R}_{n}}\right]\leq
1.$

Hence, the proof is complete. ∎

######  Remark.

While Lemma 3.2 only provides necessary conditions for optimality, it is
possible to construct an example that extends these to both necessary and
sufficient conditions. For an illustration, see Section IV to follow.

##  IV Illustrations

This section provides two examples to validate our findings on the impact of
the risk-aversion coefficient. The first is on optimal betting and the second
is on retail inventory management.

###  IV-A Optimal Betting

We now provide an example in optimal betting and illustrate how the risk term
$\rho$ affects the optimal feedback vector $K^{*}$. We consider a two-
alternative decision-making process with one risk-less alternative having zero
interest rate; i.e., $X_{1}(k)=0$ with probability one; and the other is a
risky alternative with i.i.d. payoffs
$X_{2}(k)\in\\{-\frac{1}{2},\frac{1}{2}\\}$ with probability

$\mathbb{P}\left(X_{2}(k)=\frac{1}{2}\right):=p\in\left(\frac{1}{2},\frac{3}{4}\right).$

With this setting, the log-variance becomes

$\displaystyle v(K)$ $\displaystyle:={\rm
var}\left(\log\frac{V(n;K)}{V_{0}}\right)$
$\displaystyle=\mathbb{E}\left[\left(\log\langle
K,\mathcal{R}_{n}\rangle\right)^{2}\right]-\left(\mathbb{E}\left[\log\langle
K,\mathcal{R}_{n}\rangle\right]\right)^{2}$
$\displaystyle=p(\log(1+K_{2}/2))^{2}+(1-p)\log(1-K_{2}/2)^{2}$
$\displaystyle\qquad-(p\log(1+K_{2}/2)+(1-p)\log(1-K_{2}/2)^{2}.$

A lengthy but straightforward calculation leads to the second-order derivative
of $v(K)$:

$\frac{\partial^{2}v(K)}{\partial
K_{2}^{2}}=\frac{16p(1-p)(2+K_{2}\log\frac{2+K_{2}}{2-K_{2}})}{\left(K_{2}^{2}-4\right)^{2}}\geq
0,$

which shows that the log-variance $v(K)$ is convex. Therefore, the risk-
sensitive objective (1) becomes a concave function in $K_{2}$, which implies
that Problem (LABEL:problem:_risk-sensitive_problem) is a concave program.
Therefore, Lemma 3.2 becomes a necessary and sufficient condition for
optimality. Specifically, for $n=1$, with the aids of Lemma 3.2, we have

$\displaystyle\mathbb{E}\left[\frac{\mathcal{R}_{1,2}}{\langle
K^{*},\mathcal{R}_{1}\rangle}\right]-\rho\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{1}\rangle\cdot\frac{\mathcal{R}_{1,2}}{\langle
K^{*},\mathcal{R}_{1}\rangle}\right]$
$\displaystyle+\rho\mathbb{E}\left[\log\langle
K^{*},\mathcal{R}_{1}\rangle\right]\mathbb{E}\left[\frac{\mathcal{R}_{1,2}}{\langle
K^{*},\mathcal{R}_{1}\rangle}\right]=1.$

We define the left-hand side as a function of $p,K_{2}$, and $\rho$, call it
$f(p,K_{2},\rho)$. Then, a straightforward calculation leads to

$\displaystyle f(p,K_{2},\rho)$
$\displaystyle=-1+\frac{2p}{1+K_{2}^{*}}-\frac{2p\rho\log(1+K_{2}^{*})}{1+K_{2}^{*}}$
$\displaystyle+\frac{2p\rho\left(p\log(1+K_{2}^{*})+(1-p)\log(1-K_{2}^{*})\right)}{1+K_{2}^{*}}$
$\displaystyle=0.$

Note that, for example, if $\rho=0$, the function $f$ has a zero-crossing
solution $K_{2}^{*}=2(2p-1)$, which reduces to the classical Kelly’s solution;
see [5, 19]. As a second example, if $p=0.6$, then $K_{2}^{*}=0.4$ as
$\rho=0$. If we increase $\rho$ to $0.1$, we obtain the $K_{2}^{*}\approx
0.3646$, slightly smaller than $0.4$, implying that investors will reduce the
weight as the risk aversion constant $\rho$ rises. When $\rho$ rises from $0$
to $1$, $K_{2}^{*}$ falls by around $0.2$ to $0.2035$; see Figures 1 and 2.

We also consider extreme cases such as $p=0.75$ which yields $K_{2}^{*}=1$
when $\rho=0$ and $K_{2}^{*}\approx 0.5643$ when $\rho=1$; see Figure 3. From
the cases discussed above, we find that $K_{2}^{*}$ is negatively affected by
$\rho$ and is more sensitive to $\rho$ as the probability of profiting becomes
higher.

Figure 1: Optimal Feedback Gain $K_{2}^{*}$ for
$\rho\in\\{0,0.1,0.2,\dots,1\\}$ when $p=0.6$. Figure 2: Zoom In: Optimal
Feedback Gain $K_{2}^{*}$ for $\rho\in\\{0,0.1,0.2,\dots,1\\}$ When $p=0.6$.
Figure 3: Zoom In: Optimal Feedback Gain $K_{2}^{*}$ for
$\rho\in\\{0,0.1,0.2,\dots,1\\}$ When $p=0.75$.

######  Remark.

This example illustrates optimal betting and investment decisions and
underlines its practical relevance. By illustrating how risk aversion $\rho$
influences the allocation of resources among different alternatives, we shed
light on bridging theoretical models and real-world financial decision-making.

###  IV-B Retail Inventory Management

As a second example, we consider an inventory management problem faced by a
retail operation, illustrating the application of a decision-making framework
with an uncertain variable demand. Specifically, the retail operation is
confronted with the challenge of setting the inventory levels for two distinct
categories, encapsulated by feedback gains $K_{1}$ and $K_{2}:=1-K_{1}$ with
$K_{i}\in[0,1]$. These gains represent the proportion of the maximum possible
inventory capacity $I_{\max}$ units, for an upcoming sales period. These
decisions are adjusted in anticipation of demand fluctuations, modeled as
$X_{2}(k)\sim{\rm Uniform}(-1,X_{\max})$ and $X_{1}(k):=0$ with probability
one.

The retailer’s objective is to find the optimal feedback gain vector $K$ that
maximizes a risk-sensitive objective function (LABEL:problem:_risk-
sensitive_problem), integrating the expected logarithmic profits and the
variance of these profits, with the risk aversion parameter $\rho$ modulating
the influence of risk.

Similar to the prior example, we first verify the convexity of the log-
variance. To this end, define an auxiliary random variable
$\mathcal{X}_{n,i}:=\mathcal{R}_{n,i}-1$ for $i\in\\{1,2\\}$ and $n\geq 1$.
Note that

$\displaystyle v(K)$ $\displaystyle={\rm
var}\left(\log\frac{V(n;K)}{V_{0}}\right)$
$\displaystyle=\mathbb{E}[(\log(1+K_{2}\mathcal{X}_{n,2})^{2}]-\mathbb{E}[\log(1+K_{2}\mathcal{X}_{n,2})]^{2}$
$\displaystyle=\int_{-1}^{(1+X_{\max})^{n}-1}(\log(1+K_{2}x))^{2}f_{\mathcal{X}_{n,2}}(x)dx$
$\displaystyle\qquad-\left(\int_{-1}^{(1+X_{\max})^{n}-1}\log(1+K_{2}x)f_{\mathcal{X}_{n,2}}(x)dx\right)^{2}.$

where $f_{\mathcal{X}_{n,2}}$ is the probability density function of induced
payoff $\mathcal{X}_{n}$ obtained in Proposition 0.1 in Appendix.

To illustrate, we take $I_{\max}=1000$ units, symmetric demand volatility
$X_{\max}=1=-X_{\min}$, risk-aversion constant $\rho=1/2$, and $n\geq 1$,
Figure 4 shows the second-order derivative of the log-variance under various
$n\in\\{1,5,10\\}$, which is readily seen that $v(K)$ is a convex function;
hence, Problem (LABEL:problem:_risk-sensitive_problem) is a concave program
and Lemma 3.2 becomes necessary and sufficient conditions. Moreover, by Lemma
3.2, with $n=5$, we obtain $K_{1}^{*}=1$ and $K_{2}^{*}=0$.

Figure 4: Convexity of the Log-Variance: $\frac{\partial^{2}v(K)}{\partial
K_{2}^{2}}$ Versus $K_{2}$.

######  Remark.

Leveraging a uniform distribution for initial demand fluctuations, which
compound into an Erlang-based distribution, underscores the strategic
advantage of predictive inventory management in uncertain environments. The
optimization of $K$ values based on this compounded demand model supports a
focused policy, potentially prioritizing a single product category to manage
risk and enhance returns effectively.

##  V Concluding Remarks

In this paper, we have introduced and examined a new risk-sensitive decision-
making model under uncertainty, formulating the problem as a stochastic
control problem and establishing necessary optimality conditions. Illustrative
examples are given to demonstrate the idea.

Our findings illuminate the intricate balance between maximizing expected log-
growth and mitigating log-variance, providing a robust framework for decision-
makers who face uncertain environments. This work has broad implications in
financial sectors. Moreover, as seen in Section IV, the model’s applicability
extends beyond finance, offering valuable perspectives for risk management
policies in sectors ranging from optimal betting to inventory management.

Despite the contributions of this study, we recognize its limitations,
including the i.i.d. assumptions on the stochastic alternatives. For future
work, one possibility is to explore more generalized models that relax these
assumptions, for example, dive into the variance-regularized optimization
problem, see [20], and seek possible convex surrogates of log-variance so that
it is possible to enhance computational tractability and near-optimality.
Another interesting direction to pursue would be considering other practical
constraints in the model such as risk limits [21] or possible environmental
and social constraints [22].

This appendix studies the probability density function for induced payoffs
$\mathcal{X}_{n}:=\prod_{k=0}^{n-1}(1+X(k))-1=\mathcal{R}_{n}-1$ where $X(k)$
are i.i.d. uniform distributed random variables on $[-1,X_{\max}]$ with
$X_{\max}>0$.

######  Proposition 0.1.

The probability density function for a random variable
$\mathcal{X}_{n}:=\prod_{k=0}^{n-1}(1+X(k))-1\in\mathbb{R}^{1}$ is as follows:
For $-1<z<\left(1+X_{\max}\right)^{n}-1$,

$f_{{\cal
X}_{n}}\left(z\right)=\frac{1}{{{{(1+{X_{\max}})}^{n}}\left({n-1}\right)!}}{\left({\log\frac{{{{(1+{X_{\max}})}^{n}}}}{{1+z}}}\right)^{n-1}};$

Otherwise, $f_{\mathcal{X}_{n}}(z)=0.$

The proof of the proposition above is established with the aid of the
following lemma.

######  Lemma 0.2.

Let $X$ be a uniformly distributed random variable on $[-1,X_{\max}]$ with
$X_{\max}>0$. Define

$Y:=-\log\left(\frac{1+X}{1+X_{\max}}\right).$

Then $Y$ is exponentially distributed with parameter $\lambda=1$; i.e.,
$Y\sim\exp(1)$.

###### Proof.

We begin with computing the cumulative distribution function (CDF) for $Y$.
Note that $F_{Y}(y)=0$ for all $y<0$; therefore, it suffices to consider the
CDF for $y\geq 0$; namely,

$\displaystyle F_{Y}(y)$
$\displaystyle=\mathbb{P}\left(-\log\left({\frac{{1+X}}{1+X_{\max}}}\right)\leq
y\right)$
$\displaystyle=\mathbb{P}\left({X\geq\left({1+{X_{\max}}}\right){e^{-y}}-1}\right)$
$\displaystyle=1-\mathbb{P}\left({X\leq\left({1+{X_{\max}}}\right){e^{-y}}-1}\right).$

Now note that for $y\geq 0$,

$-1\leq\left({1+{X_{\max}}}\right){e^{-y}}-1\leq{X_{\max}}$

and since $X$ is uniform distributed on $[-1,X_{\max}]$, we have

$\displaystyle F_{Y}(y)$
$\displaystyle=1-\frac{{\left(1+X_{\max}\right){e^{-y}}-1+1}}{X_{\max}+1}$
$\displaystyle=1-e^{-y},$

which is the CDF for an exponentially distributed with mean one; i.e.,
$Y\sim\exp(1).$ ∎

###### Proof of Proposition 0.1.

We begin with computing the cumulative distribution function
$F_{\mathcal{X}_{n}}(z)$ for $\mathcal{X}_{n}$. Note that
$F_{\mathcal{X}_{n}}(z)=0$ for $z\leq-1$; hence, it suffices to consider the
CDF for $z>-1$; i.e.,

$\displaystyle F_{\mathcal{X}_{n}}(z)$
$\displaystyle=\mathbb{P}\left({\prod\limits_{k=0}^{n-1}{(1+X(}k))-1\leq
z}\right)$
$\displaystyle=\mathbb{P}\left({\sum\limits_{k=0}^{n-1}{\log(1+X(k))}\leq\log\left({1+z}\right)}\right).$

Using the identity

$\sum_{k=0}^{n-1}{\log(1+X(k))}=\sum_{k=0}^{n-1}{\log\left({\frac{{1+X(k)}}{1+X_{\max}}}\right)}+n\log\left({1+{X_{\max}}}\right),$

we have

$\displaystyle\mathbb{P}({\cal X}_{n}\leq z)$
$\displaystyle=\mathbb{P}\left({\sum_{k=0}^{n-1}{\log\left({\frac{{1+X\left(k\right)}}{{1+X_{\max}}}}\right)}+n\log\left({1+X_{\max}}\right)\leq\log\left({1+z}\right)}\right)$
$\displaystyle=\mathbb{P}\left({\sum_{k=0}^{n-1}Y(k)\geq
n\log\left(1+X_{\max}\right)-\log\left({1+z}\right)}\right)$

where $Y(k):=-\log\left(\frac{1+X(k)}{1+X_{\max}}\right)$ for
$k=0,1,\dots,n-1$. Applying Lemma 0.2, it follows that $Y(k)\sim\exp(1)$.
Furthermore, since the $Y(k)$ are i.i.d., using the fact that the sum of $n$
i.i.d. random variables with $\exp(1)$ is $Erlang(n,1)$, see [23], we have
$E_{n}:=\sum_{k=0}^{n-1}Y(k)\sim Erlang(n,1).$ Now noting that $Erlang(n,1)$
has CDF $F_{E_{n}}(x)=0$ for $x\leq 0$ and
$F_{E_{n}}(x)=1-\sum_{k=0}^{n-1}\frac{x^{k}}{k!}e^{-x}$ for $x>0$. Hence, we
have

$\displaystyle F_{\mathcal{X}_{n}}(z)$
$\displaystyle=\mathbb{P}\left({\sum_{k=0}^{n-1}Y(k)\geq
n\log\left({1+{X_{\max}}}\right)-\log\left({1+z}\right)}\right)$
$\displaystyle=\mathbb{P}\left({\sum_{k=0}^{n-1}Y(k)\geq\log\left({\frac{{{{\left(1+X_{\max}\right)}^{n}}}}{{1+z}}}\right)}\right)$
$\displaystyle=1-\mathbb{P}\left({\sum_{k=0}^{n-1}Y(k)\leq\log{\frac{{{{\left(1+X_{\max}\right)}^{n}}}}{{1+z}}}}\right).$

Since for $-1<z<(1+X_{\max})^{n}-1$, we have

$\log\left({\frac{{{{\left({1+{X_{\max}}}\right)}^{n}}}}{{1+z}}}\right)>0.$

Therefore, the corresponding CDF for $\mathcal{X}_{n}$ is given as follows:
For $-1<z<{\left({1+X_{\max}}\right)^{n}}-1$, we have

$F_{\mathcal{X}_{n}}(z)=\frac{{1+z}}{{{{\left({1+X_{\max}}\right)}^{n}}}}\sum_{k=0}^{n-1}{\frac{1}{{k!}}{{\left({\log\left({\frac{{{{\left({1+X_{\max}}\right)}^{n}}}}{{1+z}}}\right)}\right)}^{k}}}$

otherwise, $F_{\mathcal{X}_{n}}(z)=0.$ Thus, the associated probability
density function is obtained by taking the derivative of the CDF above; i.e.,
For ${-1<z<{{\left({1+{X_{\max}}}\right)}^{n}}-1}$, we have

${f_{{\cal
X}_{n}}}\left(z\right)=\frac{1}{{{{(1+{X_{\max}})}^{n}}\left({n-1}\right)!}}{\left({\log\frac{{{{(1+{X_{\max}})}^{n}}}}{{1+z}}}\right)^{n-1}}$

otherwise, $f_{\mathcal{X}_{n}}(z)=0.$ ∎

## References

  * [1] A. Shapiro, D. Dentcheva, and A. Ruszczynski, Lectures on Stochastic Programming: Modeling and Theory.  SIAM, 2021.
  * [2] D. Bertsekas, Dynamic Programming and Optimal Control: Volume I, vol. 4.  Athena scientific, 2012.
  * [3] J. Von Neumann and O. Morgenstern, Theory of Games and Economic Behavior.  Princeton University Press, 2007.
  * [4] H. M. Markowitz, “Portfolio Selection,” The Journal of Finance, vol. 7, pp. 77–91, 1952.
  * [5] J. Kelly jr, “A New Interpretation of Information Rate,” The Bell System Technical Journal, 1956.
  * [6] E. O. Thorp, “The Kelly Criterion in Blackjack Sports Betting, and The Stock Market,” Handbook of Asset and Liability Management: Theory and Methodology, vol. 1, p. 385, 2006.
  * [7] D. Kahneman and A. Tversky, “Prospect Theory: An Analysis of Decision under Risk,” Econometrica, vol. 47, no. 2, pp. 263–292, 1979.
  * [8] D. Kuhn and D. G. Luenberger, “Analysis of the Rebalancing Frequency in Log-Optimal Portfolio Selection,” Quantitative Finance, vol. 10, no. 2, pp. 221–234, 2010.
  * [9] M.-E. Wu, H.-H. Tsai, W.-H. Chung, and C.-M. Chen, “Analysis of Kelly Betting on Finite Repeated Games,” Applied Mathematics and Computation, vol. 373, p. 125028, 2020.
  * [10] D. G. Luenberger, “A Preference Foundation for Log Mean-Variance Criteria in Portfolio Choice Problems,” Journal of Economic Dynamics and Control, vol. 17, no. 5-6, pp. 887–906, 1993.
  * [11] D. G. Luenberger, Investment Science.  Oxford university press, 2013.
  * [12] P. H. Algoet and T. M. Cover, “Asymptotic Optimality and Asymptotic Equipartition Properties of Log-Optimum Investment,” The Annals of Probability, vol. 16, no. 2, pp. 876–898, 1988.
  * [13] T. M. Cover and J. A. Thomas, Elements of Information Theory.  Wiley-Interscience, 2006.
  * [14] C.-H. Hsieh, “Necessary and Sufficient Conditions for Frequency-Based Kelly Optimal Portfolio,” IEEE Control Systems Letters, vol. 5, no. 1, pp. 349–354, 2020.
  * [15] C.-H. Hsieh, “On Asymptotic Log-Optimal Portfolio Optimization,” Automatica, vol. 151, p. 110901, 2023.
  * [16] Y.-S. Wong and C.-H. Hsieh, “On Frequency-Based Optimal Portfolio with Transaction Costs,” IEEE Control Systems Letters, 2023.
  * [17] A. V. Proskurnikov and B. R. Barmish, “On the Benefit of Nonlinear Control for Robust Logarithmic Growth: Coin Flipping Games as a Demonstration Case,” IEEE Control Systems Letters, 2023.
  * [18] L. C. MacLean, E. O. Thorp, and W. T. Ziemba, The Kelly Capital Growth Investment Criterion: Theory and Practice.  World Scientific, 2011.
  * [19] C.-H. Hsieh, Contributions to the Theory of Kelly Betting with Applications to Stock Trading: A Control-Theoretic Approach.  PhD thesis, The University of Wisconsin-Madison, 2019.
  * [20] J. Duchi and H. Namkoong, “Variance-Based Regularization with Convex Objectives,” Journal of Machine Learning Research, vol. 20, no. 68, pp. 1–55, 2019.
  * [21] P. Jorion, Value at Risk: the New Benchmark for Managing Financial Risk.  McGraw-Hill, 2007.
  * [22] R. G. Eccles, I. Ioannou, and G. Serafeim, “The Impact of Corporate Sustainability on Organizational Processes and Performance,” Management Science, vol. 60, no. 11, pp. 2835–2857, 2014.
  * [23] J. A. Gubner, Probability and Random Processes for Electrical and Computer Engineers.  Cambridge University Press, 2006.


ON MOEBIUS MAPS WHICH ARE CHARACTERISED BY THE CONFIGURATION OF THEIR DUAL
MAPS
Fritz Schweiger
FB Mathematik, Universität Salzburg, Hellbrunnerstr. 34, 5020 Salzburg,
Austria
<EMAIL_ADDRESS>

Abstract

Here we consider piecewise fractional linear maps with three branches. The
paper presents a study of invariant measures with densities which can be
written as infinite series. These series either have infinitely many poles or
they sum up to a function with just one pole.
Mathematics Subject Classification (2000) : 11A63, 11K55, 28D05, 37A44
Key words : fractional linear maps, f-expansions, invariant measures

## 0\. Introduction

Since the time when Rényi’s influential paper [3] appeared the search for
densities of invariant measures has seen many papers (a remarkable case of an
unknown density of an invariant maesure is given in [1]). This paper is a
continuation of the investigations in [5] and [8] (for some additions and
amendments see also [6] and [7]). In this paper Moebius maps (= piecewise
fractional linear maps) with three branches are discussed. There is a given
partition $a_{0}=0<a_{1}<a_{2}<a_{3}=1$ of the unit interval $B=[0,1]$ such
that the map $T:[0,1]\to[0,1]$ is bijective from $]a_{k-1},a_{k}[$, $k=1,2,3$
onto $]0,1[$. The inverse map $V_{k}$, $k=1,2,3$ is called an (inverse) branch
and is given as

$V_{k}(x)=\frac{c_{k}+d_{k}x}{a_{k}+b_{k}x},\,k=1,2,3.$

The Jacobians of these maps are denoted as

$\omega_{k}=\omega_{k}(x)=\frac{a_{k}d_{k}-b_{k}c_{k}}{(a_{k}+b_{k}x)^{2}},\,k=1,2,3.$

We will use the same notation for the associated matrices

$V_{k}=\left(\begin{array}[]{cc}a_{k}&b_{k}\\\
c_{k}&d_{k}\end{array}\right),\,k=1,2,3.$

Let $\epsilon=+$ stand for an increasing map and $\epsilon=-$ for a decreasing
map. We call the map $T$ of type $(\epsilon_{1},\epsilon_{2},\epsilon_{3})$,
$\epsilon_{k}\in\\{+,-\\},k=1,2,3$, if $V_{k}$ is increasing or decreasing. If
the parameters satisfy some conditions the map $T$ is ergodic and admits a
$\sigma$-finite invariant measure. The set $B(k)=V_{k}[0,1]=[a_{k-1},a_{k}]$
is called a cylinder. Clearly these conventions also apply to maps with two or
more branches.
The question in these investigations is to give an explicit shape of the
density of the invariant measure. In [5] essentially two possibilities are
discussed. A map $T^{*}:B^{*}\to B^{*}$ is called a dual map if $B^{*}$ is an
interval and $T^{*}$ is a fractional linear map whose branches are given by
the adjoint matrices

$V_{k}^{*}=\left(\begin{array}[]{cc}a_{k}&c_{k}\\\
b_{k}&d_{k}\end{array}\right),\,k=1,2,3.$

Then the function

$h(x)=\int_{B^{*}}\frac{dy}{(1+xy)^{2}}$

is (up to a multiplicative factor) the density of the invariant measure for
$T$. If there is map $\psi(t)=\frac{B+Dt}{A+Bt}$ such that $\psi\circ
T=T^{*}\circ\psi$ (or equivalently $\psi\circ
V_{k}=V_{k}^{*}\circ\psi,\,k=1,2,3$) then the dual is called a natural dual.
In the other case it is called an exceptional dual. If the set ${B^{*}}$
shrinks to a point $\xi$ one can see this case as a natural dual as well as an
exceptional one. In this case the density of the invariant measure is
$h(x)=\frac{1}{(1+\xi x)^{2}}$.
However, there are other examples where we can write down an explicit form of
the density of the invariant measure. The basic tool is the following lemma.

######  Lemma 1.

Let $T:[0,1]\to[0,1]$ be a piecewise fractional linear map with two branches
$V_{\alpha}$ and $V_{\beta}$. Let $h$ be the density of the invariant measure
for the map $T$. Then we define $S:[0,1]\to[0,1]$ as

$Sx=T^{2}x,x\in B(\alpha),Sx=Tx,x\in B(\beta).$

If $g$ is the density of the invariant measure for $S$ then the relation

$h(x)=g(x)+g(V_{\alpha}x)\omega_{\alpha}(x)$

holds.

###### Proof.

This lemma is special case of the use of the so-called jump transformation
(see [4] and [8]). The proof is easy. Note that the inverse branches of $S$
are given as $V_{\beta}x$, $V_{\alpha\alpha}x=V_{\alpha}(V_{\alpha}x)$, and
$V_{\alpha\beta}x=V_{\alpha}(V_{\beta}x)$. Their Jacobians are denoted by
$\omega_{\beta}(x)$,
$\omega_{\alpha\alpha}(x)=\omega_{\alpha}(V_{\alpha}x)\omega_{\alpha}(x)$, and
$\omega_{\alpha\beta}(x)=\omega_{\alpha}(V_{\alpha}x)\omega_{\beta}(x)$. If

$h(x)=g(x)+g(V_{\alpha}x)\omega_{\alpha}(x)$

then

$h(V_{\alpha}x)\omega_{\alpha}(x)=g(V_{\alpha}x)\omega_{\alpha}(x)+g(V_{\alpha\alpha}x)\omega_{\alpha\alpha}(x)$
$h(V_{\beta}x)\omega_{\beta}(x)=g(V_{\beta}x)\omega_{\beta}(x)+g(V_{\alpha\beta}x)\omega_{\alpha\beta}(x).$

Then

$h(V_{\alpha}x)\omega_{\alpha}(x)+h(V_{\beta}x)\omega_{\beta}(x)=$
$g(V_{\alpha}x)\omega_{\alpha}(x)+g(V_{\alpha\alpha}x)\omega_{\alpha\alpha}(x)+g(V_{\beta}x)\omega_{\beta}(x)+g(V_{\alpha\beta}x)\omega_{\alpha\beta}(x)$
$=g(V_{\alpha}x)\omega_{\alpha}(x)+g(x)=h(x).$

∎

######  Remark 1.

Clearly, a similar lemma is true if one exchanges the branches $V_{\alpha}$
and $V_{\beta}$.

######  Remark 2.

A recent application of the jump transformation is its use for the
investigation of the so-called odd-odd continued fraction algorithm see ([2]).

Since the density $h(x)$ of a piecewise fractional linear map with two
branches can be determined explicitly we use the equation

$g(x)=h(x)-g(V_{\alpha}x)\omega_{\alpha}(x)=h(x)-h(V_{\alpha}x)\omega_{\alpha}(x)+g(V_{\alpha\alpha}x)\omega_{\alpha\alpha}(x)-...$

to obtain the density $g(x)$ for some piecewise fractional linear maps with
three branches as a possibly infinite series. We call this map with three
branches constructed as just described the $1$-step extension of the given map
with two branches. To simplify notation the map $T$ has the partition
$0<\frac{1}{2}<1$.
We will discuss three cases of maps $S$ with three branches, which we label by
$\lambda$, $\mu$, and $\nu$, namely the types $(+,+,+)$, $(+,-,-)$, and
$(-,+,+)$. In case $(+,+,+)$ the map $T[0,1]$ has an increasing branch
$V_{\lambda}x=\frac{\lambda x}{1+(2\lambda-1)x}$ and an increasing branch
$V_{\beta}x=\frac{1+(\beta-1)x}{2+(\beta-2)x}$. The $1$-step extension $S$
should be a map with partition $0<\frac{1}{2}<\frac{2}{3}<1$. Then
$V_{\beta}(\frac{1}{2})=\frac{2}{3}$, $\beta=1$, and
$V_{\beta}x=\frac{1}{2-x}$. Furthermore we see that $\mu=\lambda$ and $\nu=1$.
In case $(+,-,-)$ the map $T[0,1]$ has an increasing branch
$V_{\alpha}x=\frac{\alpha x}{1+(2\alpha-1)x}$ and a decreasing branch
$V_{\nu}x=\frac{1+(\nu-1)x}{1+(2\nu-1)x}$. The $1$-step extension $S$ should
be a map with partition $0<\frac{1}{3}<\frac{1}{2}<1$. Then
$V_{\alpha}x=\frac{1}{1+x}$, $\lambda=1$, and $\mu=\nu$.
In the third case $(-,+,+)$ the map $T[0,1]$ has a decreasing branch
$V_{\alpha}x=\frac{1-x}{2+(\alpha-2)x}$ and an increasing branch
$V_{\nu}x=\frac{1+(\nu-1)x}{2+(\nu-2)x}$. Again the $1$-step extension $S$
should be a map with partition $0<\frac{1}{3}<\frac{1}{2}<1$. Then
$V_{\alpha}x=\frac{1-x}{2-x}$, $\mu=1$, and $\nu=\lambda$.
The main result of this paper is that a map with three branches which
satisfies some additional conditions given by the configuration of the dual
map of the $1$-step extension in most cases is a $1$-step extension. The
results are connected with the results of [8] but the focus of this paper is
quite different. One asks how far the configuration of the dual map $S^{*}$
determines the structure of $S$. The short last section is devoted to a
generalization of the notion of $1$-extension.
In the sequel we will label the three branches by the letters $\lambda$,
$\mu$, and $\nu$. The two branches of the map $T$ will be labelled by
$\lambda$ and $\beta$ if the cylinder $B(\lambda)$ is kept as the first
cylinder (the map $T$ jumps over the cylinder $B(\beta)$, which explains the
name jump transformation). In a similar way they will be labelled by $\alpha$
and $\nu$ if the cylinder $B(\nu)$ is kept.

######  Remark 3.

Note that the Greek letters $\alpha$, $\beta$, $\lambda$, $\mu$, and $\nu$
will be used to mark the branch with the same letters as parameters. If a
parameter is given a numerical value then we keep the letter as the label of
the branch, e. g. if $V_{\lambda}x=\frac{\lambda x}{1+(2\lambda-1)x}$ is a
branch belonging to $B(\lambda)=[0,\frac{1}{2}]$ and we choose $\lambda=1$
then we write $V_{\lambda}x=\frac{x}{1+x}$.

For the theoretical background of this study we refer to [4].

## 1\. The case $(+,+,+)$

We study the map $V_{\lambda}x=\frac{\lambda x}{1+(2\lambda-1)x}$ and
$V_{\beta}x=\frac{1}{2-x}$. Then the density of the invariant measure is

$h(x)=\frac{1}{(1-\lambda+(2\lambda-1)x)(1-x)}.$

The density of the invariant measure for the $1$-step extension $S$ is

$g(x)=\frac{1}{1-x}\big{(}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n-1)\lambda+1-((n-2)\lambda+1)x}\big{)}.$

Now we consider maps $S$ on the partition $0<\frac{1}{2}<\frac{2}{3}<1$ with
the three branches

$V_{\lambda}x=\frac{\lambda
x}{1+(2\lambda-1)x},V_{\mu}x=\frac{1+(2\mu-1)x}{2+(3\mu-2)x},V_{\nu}x=\frac{2+(\nu-2)x}{3+(\nu-3)x}.$

The parameters satisfy $0<\lambda\leq 1$, $0<\mu$, and $1\leq\nu$. This map is
piecewise linear if $\lambda=\frac{1}{2}$, $\mu=\frac{2}{3}$, and $\nu=3$. The
dual map is given as

$V^{*}_{\lambda}y=2\lambda-1+\lambda
y,V^{*}_{\mu}y=\frac{3\mu-2+(2\mu-1)y}{2+y},V^{*}_{\nu}y=\frac{\nu-3+(\nu-2)y}{3+2y}.$

A picture of the $1$-step extension suggests the following conditions. If
$\eta$ is a fixed point for $V^{*}_{\nu}$ and $\xi$ is the fixed point for
$V^{*}_{\lambda}$ then we require $V^{*}_{\mu}\eta=V^{*}_{\lambda}\eta$ and
$V^{*}_{\mu}\xi=V^{*}_{\lambda\lambda}\eta$.
We start with the following lemma.

######  Lemma 2.

The fixed point $\xi$ of $V_{\lambda}^{*}$ is given as
$\xi=\frac{2\lambda-1}{1-\lambda}$ (if $\lambda=1$ this means $\xi=\infty$).
The fixed points $\eta$ of $V_{\nu}^{*}$ are $\eta=-1$ and
$\eta=\frac{\nu-3}{2}$.

###### Proof.

The value for $\xi$ is immediate. The values for $\eta$ follow from the
quadratic equation $2\eta^{2}+(5-\nu)\eta+3-\nu=0$. ∎

######  Lemma 3.

If $V^{*}_{\mu}\eta=V^{*}_{\lambda}\eta$ and $\eta=-1$ then $\lambda=\mu$.
If $V^{*}_{\mu}\eta=V^{*}_{\lambda}\eta$ and $\eta=\frac{\nu-3}{2}$ then
$4\mu\nu=\lambda\nu^{2}+2\lambda\nu+\lambda$.

###### Proof.

The condition $V^{*}_{\mu}\eta=V^{*}_{\lambda}\eta$ implies

$\frac{3\mu-2+(2\mu-1)\eta}{2+\eta}=2\lambda-1+\lambda\eta.$

If $\eta=-1$ then $\lambda-1=\mu-1$. Hence we get $\lambda=\mu$.
If $\eta=\frac{\nu-3}{2}$ then we obtain

$\frac{2\mu\nu-\nu-1}{\nu+1}=\frac{\lambda\nu+\lambda-2}{2}$

and the equation

$4\mu\nu=\lambda\nu^{2}+2\lambda\nu+\lambda=\lambda(\nu+1)^{2}.$

∎

######  Lemma 4.

Let $\theta$ be the greater fixed point of $V^{*}_{\mu}$. Then $\theta=\xi$ if
and only of $\lambda^{2}\mu+\lambda=\mu.$

###### Proof.

The equation $\theta^{2}+(3-2\mu)\theta+2-3\mu=0$ has the solutions
$\theta=\frac{2\mu-3+\sqrt{4\mu^{2}+1}}{2}$ and
$\theta=\frac{2\mu-3-\sqrt{4\mu^{2}+1}}{2}$. Since $V^{*}_{\mu}\eta\geq
V^{*}_{\mu}(-1)=\mu-1$ implies $\theta\geq\mu-1$ only the greater root is of
interest.
The condition $\theta=\xi$ means

$\frac{2\mu-3+\sqrt{4\mu^{2}+1}}{2}=\frac{2\lambda-1}{1-\lambda}.$

This is equivalent to

$2\lambda\mu+\lambda-2\mu+1=(1-\lambda)\sqrt{4\mu^{2}+1}.$

Since both sides are non-negative we can take squares and find equivalently
the condition

$\lambda^{2}\mu+\lambda=\mu.$

∎

######  Lemma 5.

Assume $V^{*}_{\mu}\eta=V^{*}_{\lambda}\eta$ and suppose additionally
$V^{*}_{\mu}\xi=V^{*}_{\lambda\lambda}\eta$. If $\lambda=\mu$ then $\nu=1$. If
$\mu>\lambda$ then the given map $T$ has a natural dual and the density of the
invariant measure is $g(x)=\frac{1}{(1+\xi x)^{2}}$.

###### Proof.

We calculate $V^{*}_{\mu}\xi=\lambda\mu+\mu-1$ and
$V^{*}_{\lambda\lambda}\eta=2\lambda^{2}+\lambda-1+\lambda^{2}\eta$. If
$\eta=-1$ then $\lambda\mu+\mu-1=\lambda^{2}+\lambda-1$. Hence we see again
$\lambda=\mu$. If $\eta=\frac{\nu-3}{2}$ then we find

$\nu=\frac{2\lambda\mu+2\mu-\lambda^{2}-2\lambda}{\lambda^{2}}.$

If $\lambda=\mu$ then $\nu=1$ (and $\eta=-1$).
The equation $4\mu\nu=\lambda\nu^{2}+2\lambda\nu+\lambda$ gives

$\nu=\frac{2\mu-\lambda+2\sqrt{\mu^{2}-\lambda\mu}}{\lambda}.$

Since $\nu\geq 1$ the negative sign of the square root is not allowed.
Now we equate

$\frac{2\lambda\mu+2\mu-\lambda^{2}-2\lambda}{\lambda^{2}}=\frac{2\mu-\lambda+2\sqrt{\mu^{2}-\lambda\mu}}{\lambda}$

and we find

$\mu-\lambda=\lambda\sqrt{\mu^{2}-\lambda}.$

Therefore $\mu\geq\lambda$. We eventually deduce

$\mu-\lambda=\lambda^{2}\mu.$

If we want to find a natural dual and a suitable map
$\psi(t)=\frac{B+Dt}{A+Bt}$ the standard method leads to

$\psi(t)=\frac{2\lambda-1+(\lambda\mu+\mu-4\lambda+1)t}{1-\lambda+(2\lambda-1)t}.$

The condition $\psi(0)=\psi(1)$ leads to
$\frac{2\lambda-1}{1-\lambda}=\frac{\lambda\mu+\mu-2\lambda}{\lambda}$ and to
$\mu-\lambda=\lambda^{2}\mu$. Hence $g(x)=\frac{1}{(1+\xi x)^{2}}$. ∎

######  Theorem 1.

Let $S$ be a piecewise fractional linear map of type $(+,+,+)$ which satifies
the conditions

$V^{*}_{\mu}\eta=V^{*}_{\lambda}\eta,\,V^{*}_{\mu}\xi=V^{*}_{\lambda\lambda}\eta.$

If $\xi=\eta$ then $S$ has a natural dual and the density of the invariant
measure is given as

$g(x)=\frac{1}{(1+\xi x)^{2}}.$

If $\xi\neq\eta$ then $\lambda=\mu$ and $\nu=1$. The map is a $1$-step
extension and the density of the invariant measure is given as

$g(x)=\frac{1}{1-x}\big{(}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n-1)\lambda+1-((n-2)\lambda+1)x}\big{)}.$

###### Proof.

The theorem follows from the preceding lemmas. To obtain the density in the
second case one uses that the $n$-th iterate of $V_{\beta}x=\frac{1}{2-x}$ is
given as $V_{\beta}^{n}x=\frac{n-(n-1)x}{n+1-nx}$. ∎

## 2\. The case $(+,-,-)$

We start with the map $V_{\alpha}x=\frac{x}{1+x}$ and
$V_{\nu}x=\frac{1+(\nu-1)x}{1+(2\nu-1)x}$. The density of the associated
invariant measure is

$h(x)=\frac{1}{x(1+(\nu-1)x}.$

Then the density of the invariant measure of the $1$-step extension is given
as

$g(x)=\frac{1}{x}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{1+(n+\nu-1)x}.$

Now we consider the map $S$ with the partition $0<\frac{1}{3}<\frac{1}{2}<1$
and the three branches

$V_{\lambda}x=\frac{\lambda
x}{1+(3\lambda-1)x},\,V_{\mu}x=\frac{1+(\mu-1)x}{2+(3\mu-2)x},V_{\nu}x=\frac{1+(\nu-1)x}{1+(2\nu-1)x}.$

The parameters satisfy $0<\lambda\leq 1$, $0<\mu$, and $0<\nu$. The dual map
is given as

$V^{*}_{\lambda}y=3\lambda-1+\lambda
y,V^{*}_{\mu}y=\frac{3\mu-2+(\mu-1)y}{2+y},V^{*}_{\nu}y=\frac{2\nu-1+(\nu-1)y}{1+y}.$

######  Lemma 6.

The fixed point of $V^{*}_{\lambda}$ is $\xi=\frac{3\lambda-1}{1-\lambda}$
(for $\lambda=1$ this means $\xi=\infty$).

###### Proof.

The proof is immediate. ∎

We define $\eta=V^{*}_{\mu}\xi$ and we require the following conditions:
$V^{*}_{\mu}\xi=V^{*}_{\nu}\xi$ and $\ V^{*}_{\mu}\eta=V^{*}_{\nu\nu}\eta$.

######  Lemma 7.

Under these conditions the equality $\mu=\nu$ is equivalent to $\lambda=1$.

###### Proof.

From $\eta=V^{*}_{\mu}\xi=V^{*}_{\nu}\xi$ we deduce

$\xi=\frac{1-2\nu+\eta}{\nu-1-\eta}=\frac{2-3\mu+2\eta}{\mu-1-\eta}.$

If $\eta=-2$ then we obtain $\frac{2\nu+1}{\nu+1}=\frac{3\mu+2}{\mu+1}$ or
$\mu\nu+2\mu+1=0$, a contradiction.
Since $V^{*}_{\nu\nu}y=\frac{2\nu-1+\nu y}{2+y}$ the condition $\
V^{*}_{\mu}\eta=V^{*}_{\nu\nu}\eta$ implies

$\frac{3\mu-2+(\mu-1)\eta}{2+\eta}=\frac{2\nu-1+\nu\eta}{2+\eta}.$

From $\eta+2\neq 0$ we obtain

$3\mu-2+(\mu-1)\eta=2\nu-1+\nu\eta$

which gives

$\eta=\frac{3\mu-2\nu-1}{-\mu+\nu+1}.$

We insert this value into

$\xi=\frac{1-2\nu+\eta}{\nu-1-\eta}=\frac{2-3\mu+2\eta}{\mu-1-\eta}$

and obtain

$\xi=\frac{2\mu\nu-2\nu^{2}+2\mu-3\nu}{\nu^{2}-\mu\nu-2\mu+2\nu}=\frac{3\mu^{2}-3\mu\nu+\mu-2\nu}{-\mu^{2}+\mu\nu-\mu+\nu}=-2+\frac{-2\mu+\nu}{\nu^{2}-\mu\nu-2\mu+2\nu}.$

If $\mu=\nu$ then we get $\xi=\infty$ which means $\lambda=1$. If $\lambda=1$
then $\xi=\infty$ together with $V^{*}_{\mu}\xi=V^{*}_{\nu}\xi$ implies
$\mu=\nu$. ∎

######  Lemma 8.

If $\xi<\infty$ then $\eta=\xi$.

###### Proof.

From $V^{*}_{\mu}\xi=V^{*}_{\nu}\xi$ follows

$(\mu-\nu)\xi^{2}+4(\mu-\nu)\xi+3\mu-4\nu=0.$

Hence

$\xi=-2+\frac{\sqrt{\mu^{2}-\mu\nu}}{\mu-\nu}=-2+\frac{-2\mu+\nu}{\nu^{2}-\mu\nu-2\mu+2\nu}.$

Since $\mu>\nu$ and $\nu^{2}-\mu\nu-\mu+2\nu=(2+\nu)(\nu-\mu)$ we obtain

$\sqrt{\mu^{2}-\mu\nu}=\frac{2\mu-\nu}{2+\nu}.$

This relation is equivalent to

$4\mu^{2}-4\mu\nu+\mu^{2}\nu-\mu\nu^{2}=\nu.$

Since the conditions $\eta=V^{*}_{\mu}\xi=V^{*}_{\nu}\xi$ and $\eta=\xi$ imply
that $\xi$ is a fixed point for $V^{*}_{\nu}$ and for $V^{*}_{\mu}$ we see
that

$\xi=\frac{\nu-2+\sqrt{\nu^{2}+4\nu}}{2}=\frac{\mu-3+\sqrt{\mu^{2}+6\mu+1}}{2}.$

Hence we deduce

$\mu-\nu-1=\sqrt{\nu^{2}+4\nu}-\sqrt{\mu^{2}+6\mu+1},$
$\mu\nu+4\mu+\nu=\sqrt{\nu^{2}+4\nu}\sqrt{\mu^{2}+6\mu+1},$

and eventually

$4\mu^{2}-4\mu\nu+\mu^{2}\nu-\mu\nu^{2}=\nu.$

This shows that this relation is equivalent to $\eta=\xi$. ∎

######  Theorem 2.

Let $S$ be a piecewise fractional linear map of type $(+,-,-)$ which satifies
the conditions $\eta=V^{*}_{\mu}\xi=V^{*}_{\nu}\xi$ and
$V^{*}_{\mu}\eta=V^{*}_{\nu\nu}\eta.$ If $\xi=\eta$ then $S$ has a natural
dual and the density of the invariant measure is given as

$g(x)=\frac{1}{(1+\xi x)^{2}}.$

If $\xi\neq\eta$ then $\nu=\mu$ and $\lambda=1$. The map is a $1$-step
extension and the density of the invariant measure is given as

$g(x)=\frac{1}{x}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{1+(\nu+n-1)x}.$

###### Proof.

The assertione follow from the foregoing discussion. ∎

## 3\. The case $(-,+,+)$

It tuns out that this case is different from the cases $(+,+,+)$ and
$(+,-,-)$. Let $V_{\alpha}x=\frac{1-x}{2-x}$ and
$V_{\nu}x=\frac{1+(\nu-1)x}{2+(\nu-2)x}$. We calculate the density of the
invariant measure as

$h(x)=\frac{1}{(1+(\nu-2)x)(\nu+(1-\nu)x)}.$

One further calculates the iterate
$V_{\alpha}^{n}x=\frac{F_{n-1}-F_{n-3}x}{F_{n+1}-F_{n-1}x}$ where $F_{n}$
denotes the $n$-th Fibonacci number with start values $F_{-2}=1,F_{-1}=0$ ,
and $F_{1}=1$. Surprisingly, the terms but the last term in the partial sum of
the infinite series cancel. Furthermore,
$\lim_{n\to\infty}\frac{F_{n}}{F_{n-2}}=\sigma+1$ where $\sigma$ satisfies
$\sigma^{2}=\sigma+1$, $\sigma=\frac{1+\sqrt{5}}{2}$. Then we obtain

$g(x)=\frac{2-\nu}{1+(\nu-2)x}-\frac{1}{\sigma+1-x}.$

Since the density is up to a multiplcative constant we can also write

$g(x)=\frac{1}{(1+(\nu-2)x)(1+\xi x)},$

where $\xi=\frac{-3+\sqrt{5}}{2}$. Inspection of the dual map

$V^{*}_{\lambda}y=\frac{\nu-3-y}{3+y},V^{*}_{\mu}y=\frac{-1}{3+y},V^{*}_{\nu}y=\frac{\nu-2+(\nu-1)y}{2+y}$

shows that it is an exceptional dual on the interval with endpoints $\nu-2$
and $\xi$.
We now consider the map with three branches

$V_{\lambda}x=\frac{1-x}{3+(\lambda-3)x},\,V_{\mu}x=\frac{1+(\mu-1)x}{3+(2\mu-3)x},\,V_{\nu}x=\frac{1+(\nu-1)x}{2+(\nu-2)x}.$

The dual map is given as

$V^{*}_{\lambda}y=\frac{\lambda-3-y}{3+y},V^{*}_{\mu}y=\frac{2\mu-3+(\mu-1)y}{3+y},V^{*}_{\nu}y=\frac{\nu-2+(\nu-1)y}{2+y}.$

Let $\eta$ be a fixed point of $V^{*}_{\nu}$ and $\xi$ a fixed point of
$V^{*}_{\mu}$. Then $\eta=-1$ or $\eta=\nu-2$ and
$\xi=\frac{\mu-4+\sqrt{\mu^{2}+4}}{2}$.

######  Lemma 9.

Suppose $V^{*}_{\lambda}\eta=V^{*}_{\mu}\eta.$ If $\eta=-1$ then
$\lambda=\mu.$ If $\eta=\nu-2$ our map has an exceptional dual.

###### Proof.

If $\eta=-1$ then we find $\lambda=\mu$. If $\eta=\nu-2$ then we find
$\lambda=\mu\nu.$
If we suspect there is an exceptional dual we have to check
$V^{*}_{\lambda}\xi=V^{*}_{\nu}\xi.$ This equation leads to

$\nu\xi^{2}+(4\nu-\lambda)\xi+3\nu-2\lambda=0.$

If we insert $\lambda=\mu\nu$ we find (since $\nu\neq 0$) the equation

$\xi^{2}+(4-\mu)\xi+3-2\mu=0$

which is the fixed point condition for $\xi$. ∎

The next lemma is similar.

######  Lemma 10.

Suppose $V^{*}_{\lambda}\xi=V^{*}_{\nu}\xi.$ Then we find $\lambda=\mu\nu$.

###### Proof.

The condition $V^{*}_{\lambda}\xi=V^{*}_{\nu}\xi$ leads to

$\nu\xi^{2}+(4\nu-\lambda)\xi+3\nu-2\lambda=0.$

Then we see that

$\xi=\frac{\lambda-4\nu+\sqrt{4\nu^{2}+\lambda^{2}}}{2\nu}=\frac{\mu-4+\sqrt{\mu^{2}+4}}{2}.$

If $\xi$ is not a rational function of the parameters $\lambda,\mu,\nu$, then
we find $4\nu-\lambda=\nu(4-\mu)$ and hence $\lambda=\mu\nu$. If $\xi$ is a
rational function of the parameters $\lambda,\mu,\nu$ we calculate

$-\xi^{2}=\frac{4\nu-\lambda}{\nu}\xi+\frac{3\nu-2\lambda}{\nu}=(4-\mu)\xi+3-2\mu$

Then we see that

$\xi=\frac{2\lambda-2\mu\nu}{-\lambda+\mu\nu}=-2$

if $\lambda\neq\mu\nu$. But $\xi=-2$ is not an allowed value.
If $\lambda=\mu$, then we find

$\xi=\frac{\mu-4\nu+\sqrt{4\nu^{2}+\mu^{2}}}{2\nu}=\frac{\mu-4+\sqrt{\mu^{2}+4}}{2}.$

This equation leads to

$\mu\nu-\mu=\sqrt{4\nu^{2}+\mu^{2}}-\sqrt{\mu^{2}\nu^{2}+4\nu^{2}}$

and eventually to $1+\nu^{2}=2\nu$, hence $\nu=1$. ∎

######  Lemma 11.

If $V^{*}_{\lambda}\xi=V^{*}_{\nu}\xi$ and $\xi=\eta$, then $\lambda=\mu\nu$
and $\lambda=\nu^{2}-1$.

If $\eta=-1$ then $-1=\frac{\mu-4+\sqrt{\mu^{2}+4}}{2}$ leads to $\mu=0$, a
contradiction. If $\eta=\nu-2$ then from the equation

$\nu-2=\frac{\mu-4+\sqrt{\mu^{2}+4}}{2}$

follows $\nu^{2}=1+\mu\nu=\lambda+1$.

######  Example 1.

(a) We take $\lambda=3$, $\mu=\frac{3}{2}$, and $\nu=2$. In this (linear) case
we find $\xi=\eta=0$ and $h(x)=1$.
(b) If we take $\lambda=8$, $\mu=\frac{8}{3}$, and $\nu=3$, then we have
$\xi=\eta=1$ and $h(x)=\frac{1}{(1+x)^{2}}$.

######  Lemma 12.

If $V^{*}_{\lambda}\eta=V^{*}_{\mu}\eta$ and
$V^{*}_{\lambda}\xi=V^{*}_{\nu}\xi$ Then $\eta=-1$ implies $\xi\neq\eta$,
$\lambda=\mu$, and $\nu=1$.

This follows from Lemma 10 and Lemma 11.

######  Theorem 3.

Suppose $V^{*}_{\lambda}\eta=V^{*}_{\mu}\eta$ and
$V^{*}_{\lambda}\xi=V^{*}_{\nu}\xi$ . If $\eta=-1$ then $\lambda=\mu$ and
$\nu=1$. If $\eta=\nu-2$, then we find $\lambda=\mu\nu$. It is a $1$-step
extension if $\mu=1$ (and therefore $\lambda=\nu)$.
In both cases the map with three branches has an exceptional dual. The density
of its invariant measure is given as

$g(x)=\frac{1}{(1+(\nu-2)x)(1+\xi x)}$

where $\xi$ is the rightmost fixed point of $V^{*}_{\mu}$.

###### Proof.

The proof follows directly from the foregoing discussion. ∎

Note that in the case $\eta=\nu-2$ the conditions
$V^{*}_{\lambda}\eta=V^{*}_{\mu}\eta$ and $V^{*}_{\lambda}\xi=V^{*}_{\nu}\xi$
are equivalent. The case of a $1$-step extension is not visible from the
conditions of the theorem. Since the case $(-,+,+)$ was not considered in the
section on exceptional duals in [5] a proof has been included.

######  Example 2.

If $V^{*}_{\lambda}\eta=V^{*}_{\mu}\eta$ and $\eta=-1$ then $\nu\neq 1$ is
possible. An example is given by $\lambda=\mu=\nu=2$. Then we have

$V^{*}_{\lambda}y=\frac{-1-y}{3+y},V^{*}_{\mu}y=\frac{1+y}{3+y},V^{*}_{\nu}y=\frac{y}{2+y}.$

Then $V^{*}_{\lambda}(-1)=V^{*}_{\mu}(-1)=0$. We find $\xi=\sqrt{2}-1$ but
$V^{*}_{\lambda}(\sqrt{2}-1)\neq V^{*}_{\nu}(\sqrt{2}-1)$.

## 4\. $n$-step extensions

The construction of the $1$-step extension of a map with 2 branches which uses
$V_{\beta}$ leads to a map with 3 branches This construction can be iterated
using $V_{\beta}^{2}=V_{\beta\beta}$ and leads to a map with 5 branches which
we call the $2$-step extension. By iterating this process we construct the
$n$-step extension using $V_{\beta}^{n}$ which is map with $2^{n}+1$ branches.
We denote the Jacobian of $V_{\beta}^{n}$ by $\omega_{\beta}^{n}$.
From the equality $g(x)=h(x)-g(V_{\beta}x)\omega_{\beta}(x)$ we found

$g(x)=h(x)-h(V_{\beta}x)\omega_{\beta}(x)$
$+h(V_{\beta}^{2}x)\omega_{\beta}^{2}(x)-h(V_{\beta}^{3}x)\omega_{\beta}^{3}(x)+h(V_{\beta}^{4}x)\omega_{\beta}^{4}(x)-h(V_{\beta}^{5}x)\omega_{\beta}^{5}(x)+....$

Let $g_{n}(x)$ be the density of the invariant measure of the $n$-step
extension we have the relation
$g(x)=g_{n-1}(x)-g_{n}(V_{\beta}^{n}x)\omega_{\beta}^{n}(x)$ where
$g_{0}(x)=h(x)$ and $g_{1}(x)=g(x)$. Then we can express $g_{n}(x)$ as a
series which uses the function $h(x)$ only. We find

$g_{2}(x)=h(x)-h(V_{\beta}x)\omega_{\beta}(x)$
$+h(V_{\beta}^{4}x)\omega_{\beta}^{4}(x)-h(V_{\beta}^{5}x)\omega_{\beta}^{5}(x)+h(V_{\beta}^{8}x)\omega_{\beta}^{8}(x)-h(V_{\beta}^{9}x)\omega_{\beta}^{9}(x)+....$

and

$g_{3}(x)=h(x)-h(V_{\beta}x)\omega_{\beta}(x)$
$+h(V_{\beta}^{8}x)\omega_{\beta}^{8}(x)-h(V_{\beta}^{9}x)\omega_{\beta}^{9}(x)+h(V_{\beta}^{16}x)\omega_{\beta}^{16}(x)-h(V_{\beta}^{17}x)\omega_{\beta}^{17}(x)+....$

The following theorem is immediate.

######  Theorem 4.

The sequence $(g_{n}(x))$ is convergent and

$g_{\infty}(x)=\lim_{n\to\infty}g_{n}(x)=h(x)-h(V_{\beta}x)\omega_{\beta}(x)=h(V_{\alpha}x)\omega_{\alpha}(x).$

######  Remark 4.

The function $g_{\infty}(x)=h(V_{\alpha}x)\omega_{\alpha}(x)$ is the density
of the jump transformation $U$ defined by

$Ux=T^{n}x,\,T^{j}x\in B(\beta),0\leq j\leq n-2,\,T^{n-1}x\in B(\alpha).$

This result can be found in [4]. However, the direct proof is easy.

$\sum_{n=0}^{\infty}g_{\infty}(V_{\beta}^{n}V_{\alpha}x)\omega_{\beta}^{n}(V_{\alpha}x)\omega_{\alpha}(x)$
$=\sum_{n=0}^{\infty}h(V_{\beta}^{n}V_{\alpha}x)\omega_{\beta}^{n}(V_{\alpha}x)\omega_{\alpha}(x)-\sum_{n=1}^{\infty}h(V_{\beta}^{n}V_{\alpha}x)\omega_{\beta}^{n}(V_{\alpha}x)\omega_{\alpha}(x)$
$=h(V_{\alpha}x)\omega_{\alpha}(x)=g_{\infty}(x).$

######  Remark 5.

In the case (-,+,+) the series for $g(x)$ reduces to
$g(x)=g_{1}(x)=\frac{1}{(1+(\nu-2)x)(1+\xi x)}$. However, $g_{2}(x)$ is an
infinite series. If $\nu=1$ then $h(x)=\frac{1}{1-x}$ and
$g(x)=g_{1}(x)=\frac{1}{(1-x)(1+\xi x)}$: We find

$g_{2}(x)=\frac{1}{1-x}-\frac{1}{2-x}+\frac{2}{5-2x}-\frac{5}{13-5x}+\frac{13}{34-13x}-\frac{34}{89-34x}+....$

## References

  * [1] Dajani, K., Kraaikamp, C., Langeveld, N. D. S. 2015 Continued fraction expansions with variable numerators. Ramanujan J. 37 (2015), no. 3, 617–639.
  * [2] Dong Han Kim, Seul Bee Lee, and Lingmin Liao 2022: Odd-odd continued fraction algorithm. Monatshefte Mathematik 198 (2022), 323-344
  * [3] Rényi, A. 1957: Representations for real numbers and their ergodic properties, Acta Math. Acad. Sci. Hungar. 8 (1957), 477–493
  * [4] Schweiger, F. 1995: Ergodic Theory of Fibred Systems and Metric Number Theory, Oxford Science Publications, New York and The Clarendon Press, Oxford, 1995
  * [5] Schweiger, F. 2006: Differentiable equivalence of fractional linear maps. Dynamics & stochastics, 237–247, IMS Lecture Notes Monogr. Ser., 48, Inst. Math. Statist., Beachwood, OH, 2006.
  * [6] Schweiger, F. 2017: Invariant measures for continued fractions with variable numerators. Integers 17 (2017), Paper No. A56, 6 pp.
  * [7] Schweiger, F. 2018a: Invariant measures of piecewise fractional linear maps and piecewise quadratic maps. Int. J. Number Theory 14 (2018), 1559-1572
  * [8] Schweiger, F. 2018b: Invariant measures for Moebius maps with three branches J. Number Theory 184 (2018), 206–215.