# Dynamics from Multivariable Longitudinal Data

Authors: Maria Vivien Visaya; David Sherwell
Journal: Journal of Nonlinear Dynamics
 (2014)

---

## Abstract

We introduce a method of analysing longitudinal data inn
≥
1 variables and a population of K
≥
1 observations. Longitudinal data of each observation is exactly coded to an orbit in a two-dimensional state space S
n. At each time, information of each observation is coded to a point (
x
,
y
)
∈
S
n, where x is the physical condition of the observation and y is an ordering of variables. Orbit of each observation in S
n is described by a map that dynamically rearranges order of variables at each time step, eventually placing the most stable, least frequently changing variable to the left and the most frequently changing variable to the right. By this operation, we are able to extract dynamics from data and visualise the orbit of each observation. In addition, clustering of data in the stable variables is revealed. All possible paths that any observation can take in S
n are given by a subshift of finite type (SFT). We discuss mathematical properties of the transition matrix associated to this SFT. Dynamics of the population is a nonautonomous multivalued map equivalent to a nonstationary SFT. We illustrate the method using a longitudinal data of a population of households from Agincourt, South Africa.


---

## Body

## 1. Introduction

Analysis of multivariable longitudinal data involves either statistical or nonstatistical methods. Statistical methods include multivariate Markov chain model [1], regression model [2], and mixed models [3], while some nonstatistical methods involve extraction of dynamical system using state space reconstruction technique [4] or visual methods such as motion charts [5, 6] and parallel coordinate plots [7, 8]. A motion chart shows additional dimensions of the data at different time points, where the size and color of the bubble (among others) are used as variables. PCP represents variables as parallel axes, where a sequence of line segments intersects each axis at a point corresponding to the observation’s value at the associated variable. Both methods aim to identify correlation among variables and identification of clusters and patterns among observations in the data. We present a novel nonstatistical method of analysis that is useful particularly for collecting information of change. State space reconstruction method mainly requires use of delay coordinates from data to come up with models for prediction. Our method does not rely on delay coordinates, and our aim is not prediction. Contrasting to motion charts, there is, in principle, no limitation to the number of variables studied in our method. Contrasting to PCP, for large n and large number of observations, orbits over our two-dimensional state space, or over time, are easily visualized.Here we consider dynamics of multivariable longitudinal data of a population of observations by applying a swap operation on the data of each observation. We suppose that values ofn
≥
1 variables are discretised to m
>
1 bins so that each variable takes value from V
=
{
0,1
,
…
,
m
-
1
}. For a fixed order of variables (e.g., v
0
,
v
1
,
…
,
v
n
-
1) denote by
(1)
M
n
,
m
=
V
n
the space of all m-ary sequences of length n and element x
∈
M by x
=
x
0
x
1
⋯
x
n
-
1 such that x
i is the value of v
i. The m-ary multivariable time series in n
≥
1 variables of an observation k is defined by
(2)
X
k
=
{
X
0
k
,
X
2
k
,
…
,
X
T
k
}
,
where each X
t
k
∈
M
n
,
m. Our method can be used for m-ary valued data, but for illustration, we use m
=
2. We denote ourbinary multivariable longitudinal data D of a population of K observations by the set of binary multivariable time series of observations
(3)
D
=
{
X
1
,
X
2
,
…
,
X
K
}
.
In the general analysis of longitudinal data we see no recognition that data is taken with a purpose. Here we suppose that the value of a binary variable is either favourable or unfavourable to a given purpose. Suppose we would like to investigate the effect of change of households variables, namely, biological mother (BM), household head (HH), and adult death (AD), to a child’s educational progress. Consider n
=
3 binary questions
(4)
Q
0
:
Is the biological mother present?
(
BM
)
Q
1
:
Is the household headed by a minor?
(
HH
)
Q
2
:
Is there an adult death in the household?
(
AD
)
and suppose that answers are coded either favourable = 1 or unfavourable = 0 to our purpose. It is a reasonable hypothesis that the answer ‘‘yes’’ to Q
0 is favourable to child education, while ‘‘yes’’ to Q
1 and Q
0 is unfavourable. Table 1 shows coded data of 4 subjects (households) over 7 time steps. Using parallel coordinate plots (PCP), Figure 1(a) shows the answer of each subject at t
=
0. To illustrate the answer for all time t using PCP, the evolution of line segments, with time, will form surfaces that obscure each other. Figure 1(b) shows the Bratelli diagram [9, 10] of transition of answers from time t to t
+
1, to variable order BM, HH, and AD. Transitions per time interval can be associated with a state transition matrix, where states are analysed on the space
(5)
M
3
=
{
000,001,010,011,100,101,110,111
}
.
This can be used to generate probability matrices for Markov models [11, 12].Table 1
Unfavourable = 0/favourable = 1 coding of four households to questions related to biological mother (BM), household head (HH), and death (AD).
Time
h
1
h
2
h
3
h
4
BM, HH, AD
BM, HH, AD
BM, HH, AD
BM, HH, AD
0
0,1
,
1
0,1
,
1
0,0
,
1
0,1
,
1
1
1,1
,
0
0,0
,
1
0,0
,
1
1,0
,
1
2
0,1
,
1
1,1
,
1
1,0
,
1
1,1
,
1
3
0,1
,
1
1,0
,
0
0,0
,
0
0,0
,
1
4
0,1
,
1
0,1
,
0
0,0
,
1
0,1
,
1
5
0,1
,
1
0,1
,
0
0,0
,
1
1,0
,
1
6
1,1
,
1
0,0
,
0
1,0
,
1
0,0
,
1Visual display of data of four subjects in Table1 (a) at t
=
0 using PCP, where each parallel line is composed of answers 0 and 1 and (b) for all time t
=
0,1
,
…
,
6 using a Bratelli diagram, where each parallel line is composed of states from M
3
=
{
0,1
}
3, the space of binary sequences of length 3.
(a)
(b)Observe that, for question regardingQ
1 (HH), household h
1 has constant favourable answer, while h
3 has constant unfavourable answer. On the other hand, h
4 has constant favourable answer to Q
3 (AD). Our aim is to extract clusters associated with stable variables. Underlying our method is our belief that the set of physical variables in which an observation spends the most time in is important (e.g., HH = 1 for h
1, HH = 0 for h
3, and AD = 1 for h
4) and that among the physical variables themselves the variables that are most often experienced (most probable) by the observations are important. We elegantly expose both most probable variable and value of variable, by a simple process of dynamically reordering variables.There is no a priori indication of any absolute dynamics in data and here it is deterministically imposed. Because longitudinal data is fundamentally defined by change (if nothing changes, cross-sectional data is sufficient), frequency of answer change of variables then becomes a property of interest. A deterministic operation is applied to the multivariable data of each observation at each time step, dynamically reordering position of variables (and their corresponding values) by their stability; that is, the most stable is eventually positioned to the left, and the most frequently changing one is positioned to the right. All possible orders of variables are considered. From this, we introduce the significance state (order of variables) and fitness state (associated values with variables) of an observation. It is in a chosen ordering that the notion of fitness takes an objective and consistent meaning. The idea of fitness and significance is new in literature. Then-dimensional longitudinal data of an observation is represented as a 2-dimensional orbit in fitness-significance axes S
n composed of 2
n
n
! points. Orbits in S
n sufficiently encode the longitudinal data of each observation. Analysis of orbits at the individual and population levels in this space can then follow.This paper is organized as follows. In Section2, we present the method of constructing orbits from multivariable longitudinal data. A detailed theory of the reordering operation applied to data is presented. In Section 3, a deterministic equation of motion that generates all observed orbits, as well as other possible orbits, is presented. In Section 4 we discuss transitions that occur in data. This is captured by a nonstationary SFT. We also discuss dynamics at the population level. An illustration of the method is presented in Section 5. We give concluding remarks in Section 6.


## 2. Background and Preliminaries: Method of Orbits

We suppose that longitudinal data is gathered by first specifying a purpose and then choosing questions that are of interest to purpose. The questionnaire may be designed to a purpose posed in the form ‘‘to investigate the effect of then variables v
0
,
v
2
,
…
,
v
n
-
1 on Z.” Here we will only consider binary-valued questions with responses hypothesized as either favourable or unfavourable to the purpose. A favourable answer is coded 1 and 0 otherwise. Our longitudinal data is the response to a set of n
≥
1 questions (associated with variables) surveyed from a population P composed of K
≥
1 observations over T periods.Denote thequestionnaire by
(6)
Q
=
{
Q
0
,
Q
1
,
…
,
Q
n
-
1
}
,
the set containing n questions. Let I
n
=
{
0,1
,
…
,
n
-
1
} be an index set of n elements, and let time t
=
0,1
,
…
,
T. Let j, i
j
∈
I
n and let x
j
∈
{
0,1
}. For each observation k
=
1,2
,
…
,
K, denote a reordering of questions in Q at time t by
(7)
Q
t
k
=
{
Q
i
0
t
k
,
Q
i
1
t
k
,
…
,
Q
i
(
n
-
1
)
t
k
}
:
=
{
Q
i
0
,
Q
i
1
,
…
,
Q
i
n
-
1
}
t
k
,
the concatenation of answers to Q
t
k by
(8)
x
t
k
=
x
0
t
k
x
1
t
k
⋯
x
(
n
-
1
)
t
k
=
(
x
j
t
k
)
j
=
0
n
-
1
:
=
(
x
0
x
1
⋯
x
(
n
-
1
)
)
t
k
,
and the concatenation of question indices of Q
t
k by
(9)
y
t
k
=
i
0
t
k
i
1
t
k
⋯
i
(
n
-
1
)
t
k
=
(
i
j
t
k
)
j
=
0
n
-
1
:
=
(
i
0
i
1
⋯
i
(
n
-
1
)
)
t
k
.Example 1.
Coded data of observationk for t
=
0 and t
=
1 is shown in column 2 of Table 2. Suppose we (arbitrarily) reorder the questions at t
=
0 to {
Q
1
,
Q
0
,
Q
2
}. Then x
0
k
=
010 and y
0
k
=
102. Similarly, if Q
1
k
=
{
Q
2
,
Q
0
,
Q
1
}, we have x
1
k
=
001 and y
1
k
=
201. As we are merely rewriting entries from the original data, all information is preserved.Table 2
Data of observationk (column 2), with questions reordered at t
=
0 and t
=
1.
Observationk
Coded answers to
Q
=
{
Q
0
,
Q
1
,
Q
2
}
Concatenationof answers to Q
t
k
Concatenationof indices of Q
t
k
t
=
0
{
0
,
1
,
0
}
100
102
1
{
0
,
0
,
1
}
010
021


### 2.1. Fitness and Significance States

Consider questions in (4) and assign index i to Q
i (i
=
0,1
,
2). Suppose we give more weight (significance) to Q
0 and do this by positioning 0 in the left-end of question order, say, Q
0, Q
1, Q
2, denoted by 012 (or Q
0
,
Q
2
,
Q
1
=
021). As in numbers or decimals, our weighting places the most significant number at the left-end position. All possible (concatenated) answers to 012 are given by
(10)
001,000,010,111,110,101,100
,
011
.
Since 000 has all unfavourable answers we say that it is the leastfit answer, while 111 is thefittest. Note that there are states with the same number of favourable values, for example, 001, 100, and 010. By a suitable weighting of questions, we show below that the lexicographic ordering of answers in (10) is an appropriate ordering of fitness.Table3 illustrates concatenated coded answers to question order 012 of three observations from a population. To Q
0, observation k has constant answer 0 while k
′ has constant answer 1.Table 3
Concatenated coded answers of observations to question orderQ
0
,
Q
1
,
Q
2
=
(
012
).
t
k
k
′
k
′
′
0
010
100
111
1
001
110
001
2
000
101
001
3
010
110
111
4
000
100
001Suppose we arrange answers in (10) lexicographically along an x-axis. Then for question order 012, a one-dimensional dynamics on the x-axis composed of the eight states arises. Answers of k and k
′ to Q
0 seldom change (i.e., they are both constant in Q
0) so the two households k and k
′stay in the regions 0
*
* and 1
*
* of the x-axis, respectively. Recall that Q
0 is the question associated with the significant (left) position of the question order so fitness is biased towards the left position. We can then write x
=
0
*
*
<
x
′
=
1
*
* because the significant variable Q
0 is unfavourable in x and favourable in x
′. This holds true even if x has the same, or more, favourable values as x
′. This argument can be extended to any two elements x
,
x
′
∈
M
n with the same first ℓ entries.In general, not all observations may be stable in the same variable; for example,k
′′ in Table 3 is constant in Q
0, not in Q
2. Moreover, stability of an observation may change in time; it may be stable in variable i over one time interval and then stable in variable i
′ over another time interval. We will not study orbits in a fixed question order alone. We construct a y-axis with states corresponding to question orders. The order of questions per observation becomes a new variable.Definition 2.
Givenn
≥
1 and Q
k
t, thefitness stateandsignificance state of observation k at time t are the sequences
(11)
x
t
k
=
(
x
j
t
)
j
=
0
n
-
1
,
y
t
k
=
(
i
j
t
)
j
=
0
n
-
1
,
respectively. The set of fitness states of length n is called thefitness space of
n
variables defined by
(12)
X
n
:
=
{
(
x
j
)
j
=
0
n
-
1
:
x
j
∈
{
0,1
}
}
and the set of significance states of length n is called thesignificance space of
n
variables defined by
(13)
Y
n
:
=
{
(
i
j
)
j
=
0
n
-
1
∶
i
j
∈
I
n
,
i
j
′
s
distinct
}
.
Elements of both X
n and Y
n are arranged according to the lexicographic ordering (<) of sequences of length n.Definition 3.
The space(14)
S
n
=
{
p
=
(
x
,
y
)
:
x
∈
X
n
,
y
∈
Y
n
}
=
X
n
×
Y
n
is called thechange space for
n
variables.Givenn, we have the cardinalities |
X
n
|
=
2
n, |
Y
n
|
=
n
!, and |
S
n
|
=
N
=
2
n
n
!. A way of labeling state s
j
∈
S
n is via the map
(15)
ψ
:
S
n
⟶
{
1,2
,
…
,
N
}
,
s
j
⟼
ψ
(
s
j
)
=
j
,
g
h
f
f
g
h
f
f
g
j
h
l
k
j
=
1,2
,
…
,
N
.
For convenience, we label states in S
n from left to right, and from top to bottom. If s
j
=
(
x
,
y
) and ψ
(
s
j
)
=
j, we will refer to state s
j as state j and write j
=
(
x
,
y
).Remark 4.
In general, for multivariate data inn variables, withall variables m-ary valued, the space S
m
,
n is composed of m
n
×
n
! states.


### 2.2. The Method of Orbits

We define the dynamics of observations taken from a survey ofn
≥
1 questions. Let ℕ
0 be the set of nonnegative integers, let 𝒞
n
=
𝒫
(
I
n
) be the power set of I
n, and let x
j
* be such that
(16)
x
j
*
=
{
1
if
x
j
=
0
0
if
x
j
=
1
.Definition 5.
LetΔ
∈
𝒞
n and let j
∈
Δ.(a)
The mapϕ
Δ
:
S
n
→
S
n is defined by
(17)
ϕ
Δ
(
x
0
x
1
⋯
x
j
⋯
x
n
-
1
,
y
)
=
(
x
0
x
1
⋯
x
j
*
⋯
x
n
-
1
,
y
)
.
(b)
The mapϕ
J
:
S
n
→
S
n is defined by
(18)
ϕ
J
(
x
0
x
1
⋯
x
j
⋯
x
n
-
1
,
i
0
i
1
⋯
i
j
⋯
i
n
-
1
)
=
(
x
0
x
1
⋯
x
j
-
1
x
j
+
1
⋯
x
n
-
1
x
j
,
i
0
i
1
⋯
i
j
-
1
i
j
+
1
⋯
i
n
-
1
i
j
)
.Let j
,
j
′
∈
Δ. Then x
j and x
j
′ both change values under ϕ
Δ. If j
<
j
′, then ϕ
J is first applied to j
′; that is,
(19)
ϕ
J
(
x
0
x
1
⋯
x
j
⋯
x
j
′
⋯
x
n
-
1
,
i
0
i
1
⋯
i
j
⋯
i
j
′
⋯
i
n
-
1
)
=
(
x
0
x
1
⋯
x
j
-
1
x
j
+
1
⋯
x
j
′
-
1
x
j
′
+
1
⋯
x
n
-
1
x
j
′
x
j
,
w
w
w
w
w
i
0
i
1
⋯
i
j
-
1
i
j
+
1
⋯
i
j
′
-
1
i
j
′
+
1
⋯
i
n
-
1
i
j
′
i
j
)
.Definition 6.
Leti
j
∈
I
n. Consider questionnaire Q with Q
i
j
∈
Q. For each observation k, let f
i
j
k be the frequency of change in answer value of Q
i
j over the observation period. Suppose
(20)
0
<
f
i
0
k
<
f
i
1
k
<
⋯
<
f
i
j
k
<
⋯
<
f
i
n
-
1
k
.
Inequality (20) is called theobservation frequency relation and question order y
0
k
=
i
0
i
1
⋯
i
n
-
1 is theinitial significance state of observation
k. If f
i
j
k
=
f
i
j
+
1
k and f
i
j
P
<
f
i
j
+
1
P at the population level, then choose question order i
j
i
j
+
1. If f
i
j
P
=
f
i
j
+
1
P, then choose question order as in the questionnaire (6). Theinitial fitness state  x
0
k
=
x
0
x
1
⋯
x
n
-
1 is such that x
j is the value of i
j in y
0
k. Theinitial state of
k is the ordered pair p
0
k
=
(
x
0
k
,
y
0
k
).By choosing the initial significancey
0
k given in Definition 6, we start the orbit in its most-likely significance state. This facilitates convergence to clusters (where they exist) and is useful for short data sets. Other strategies (e.g., using order of (6) or random choice) will nonetheless converge to question order according to (20).Remark 7.
Longitudinal data is only of interest where change occurs; else cross-sectional surveys are adequate. We are interested in longitudinal data that give nontrivial information of change about the population; that is,f
i
j
P
≠
0. Otherwise, question i
j may be deleted as any such property becomes an identifier of subpopulations of possible interest for analysis in its own right.Definition 8.
For each observationk, define thechange set at time  t by
(21)
Δ
t
k
(
∈
𝒞
n
)
=
{
j
:
Q
i
j
∈
Q
t
k
changes answer value at time
t
+
1
}
.
Let p
0
k
=
(
y
0
k
,
x
0
k
) be the initial state of k. Denote by p
t
k
=
(
x
t
k
,
y
t
k
) the state of k at time t
≥
0. Thechange map   φ
:
(
ℕ
0
,
𝒞
n
,
S
n
)
→
S
n is such that
(22)
φ
(
t
,
Δ
t
k
,
p
t
k
)
:
=
φ
[
Δ
t
k
]
(
p
t
k
)
=
(
ϕ
J
∘
ϕ
Δ
t
k
)
(
p
t
k
)
=
p
t
+
1
k
.
The set Δ
t
k
∈
𝒞
n is given by the longitudinal data for each t
≥
0 and is a useful ordered listing of questions that change answer values from time t to t
+
1. For each k, the nonautonomous map φ
[
Δ
t
k
] defines an evolutionary process that displaces the most frequently (resp., slowly) changing answers and corresponding questions to the right (resp., left).Definition 9.
Given initial statep
0
k of k, define the state of   k at time t
≥
1 by
(23)
p
t
=
φ
[
Δ
t
-
1
]
(
⋯
(
φ
[
Δ
0
]
(
p
0
)
)
)
.
Theforward orbit of
k
under
φ is defined by
(24)
𝒪
φ
(
k
)
=
{
p
t
k
}
t
∈
ℕ
0
.


### 2.3. Algorithm for Building the Orbit of an Observation inS
n

We give a simple algorithm to determine the statesp
t
>
0
k that comprise the orbit of an observation k from longitudinal data.Step 1.
Determine initial question ordery
0
k and initial state p
0
k
=
(
x
0
k
,
y
0
k
) of k and plot p
0
k in S
n.Step 2.
Identify from the data ofk the question that changes answer at time t
=
1, say Q
i
j. Swap both i
j (in y
0
k) and corresponding answer x
j (in x
0
k) to the right of y
0
k and x
0
k, respectively, and change x
j to x
j
*, where x
j
*
=
0 if x
j
=
1 and 0 otherwise. This new question order and answer order give the next state p
1
k
=
(
x
1
k
,
y
1
k
) of k. Suppose both Q
i
j and Q
i
j change answers at t
=
1. If j
<
j
′, then sequentially swap to the right i
j and i
j
′ (resp., x
j and x
j
′) of the question order (resp., answer order), starting with i
j
′ (resp., x
j
′). Change x
j to x
j
* and x
j
′ to x
j
′
*. Plot the point p
1
k in S
n and directed edge from p
0
k to p
1
k.Step 3.
Repeat Step2, updating state p
t
-
1
k
→
p
t
k and t
→
t
+
1 for time t
=
1,2
,
…
,
T
-
1.Visualization and analysis of orbits of observations inS
n allow capturing information of change in longitudinal data. We illustrate in Figure 2 an orbit in S
3. The useful distance on S
n is given by the discrete metric; that is, d
(
p
,
p
′
)
=
1 if p
≠
p
′ and zero otherwise. The visualized distance between points in S
n has no interpretation so we may represent S
n by a regularly spaced point.Figure 2
State transitions of observationk (from Table 4) in S
3.Remark 10.
The set of all orderings of variables is captured inS
n. For fixed question order and x
,
x
′
∈
X
n, x
<
x
′ means that fitness state x
′ is fitter than x. Each level in the significance axis is question order under frequency ordering. The significance axis informs us which variables are weighted most strongly at each time, where significance is ordered from left to right. Clearly, reordering of variables is one among many families of operations; for example, swapping can be done by swapping changing variable to the left end. This operation however does not reveal clusters.


### 2.4. Transitions inS
n

We now analyze possible state transitions which an observationk can take in S
n.Definition 11.
Letp
=
(
x
,
y
) and p
′
=
(
x
′
,
y
′
) be in S
n, and let Δ
∈
𝒞
n.(a)
SupposeΔ is such that φ
[
Δ
]
(
p
)
=
p
′. Then there is atransition from p to p
′ under Δ, defined by p
→
[
Δ
]
p
′.
(b)
Supposep
→
[
Δ
]
p
′. If y
=
y
′ and x
≠
x
′, then there is ahorizontal transition from p to p
′. If x
=
x
′ and y
≠
y
′, then there is avertical transition from p to p
′.
(c)
Supposep
→
[
Δ
]
p
′. If, in addition, Δ is such that p
′
→
Δ
p, then the transition from p to p
′ isreversible, and one writes p
↔
[
Δ
]
p
′.We haveself-transitions if Δ
=
∅ (no change), the empty set. In general, horizontal transitions denote change in the right-most variable, while vertical transitions denote change in the last two variables.Letp
,
p
′
∈
S
n. Given an observation k, if p
t
k
∈
p and p
t
+
1
k
∈
p
′, then there is Δ
t
∈
𝒞
n such that p
→
[
Δ
t
]
p
′. We use the symbol “∈” to denote that there may be other observations in p or p
′ at times t and t
+
1, respectively. State transitions of observation k in S
n are visualised as a sequence of directed edges.Example 12.
Consider data of observationk in column 2 of Table 4. The asterisks denote changing answers in the next time step. The frequencies of change are f
0
k
=
2, f
1
k
=
3, and f
2
k
=
1, so the initial question order of k is y
0
k
=
201. And initial fitness state is x
0
k
=
010 (Definition 6). At t
=
0, we apply φ
[
Δ
0
k
] to initial state p
0
k
=
(
010,201
). Questions i
1
=
0 and i
2
=
1 in Q
0
k change answers in the next time step so Δ
0
k
=
{
1,2
} (not {
0,1
}). Applying ϕ
Δ
0
k to (
010,201
) changes x
2
=
0 and x
1
=
1 to x
2
*
=
1 and x
1
*
=
0, respectively. Next, ϕ
J is first applied to both x
2
* and i
2 by moving each to the right (they are already rightmost), followed by moving x
1
* and i
1 to the right end. Hence, we have (
x
,
y
)
1
k
=
(
010,210
). For t
=
1 and t
=
2, verify that Δ
1
k
=
{
1
} and Δ
2
k
=
{
0,1
,
2
}, respectively. At each time t the bold numbers in the significance column are the question indices in Q that change answers at t
+
1. The orbit of k in S
3 is shown in Figure 2. The vertical transition from state 11 to 3 denotes two changing answers, while a transition from state 9 to 32 denotes three changing answers.Table 4
States of observationk.
Unitk
Coded answers to
Fitness
Significance
Δ
t
k
Q
=
{
Q
0
,
Q
1
,
Q
2
}
x
t
k
y
t
k
t
=
0
{
1
*
,
0
*
,
0
}
010
201
{
1,2
}
1
{
0
,
1
*
,
0
}
010
210
{
1
}
2
{
0
*
,
0
*
,
0
*
}
000
201
{
0,1
,
2
}
3
{
1
,
1
,
1
}
111
102Example 13.
Figures3(a) and 3(b) illustrate the associated orbits in S
3 of the four households given in Table 1. Observe that orbit of h
3 stays strictly on the left half of S
3, particularly in subset where Q
1
(
HH
)
=
0, while h
1 and h
4 stay in the right half subset, where Q
1
(
HH
)
=
1 and Q
2
(
AD
)
=
1, respectively.(a) Orbits of the four subjects in Table1. A red edge denotes a transition to the left, a green edge is a transition to the right, and a blue edge is a self-transition.
(a)
(b)Definition 14.
LetΔ
∈
𝒞
n and let j
∈
Δ.The return map  φ
~
[
Δ
]
:
S
n
→
S
n is such that
(25)
φ
~
[
Δ
]
(
x
0
x
1
⋯
x
j
⋯
x
n
-
1
,
i
0
i
1
⋯
i
j
⋯
i
n
-
1
)
=
(
x
0
x
1
⋯
x
j
-
1
x
n
-
1
*
x
j
⋯
x
n
-
2
,
w
w
i
0
i
1
⋯
i
j
-
1
i
n
-
1
i
j
⋯
i
n
-
2
)
.
That is, φ
~
[
Δ
] first inserts x
n
-
1 and i
n
-
1 to the position, where x
j and i
j are located, followed by changing x
n
-
1 to the new value x
n
-
1
*. If j
,
j
′
∈
Δ and j
<
j
′ then φ
~
[
Δ
] is first applied to x
j and i
j.
Trivially, for anyn
≥
1, p
∈
S
n, and Δ
∈
𝒞
n,
(26)
φ
~
[
Δ
]
(
φ
[
Δ
]
(
p
)
)
=
φ
[
Δ
]
(
φ
~
[
Δ
]
(
p
)
)
=
p
.Theorem 15.
Letp
∈
S
n. Define the image set and preimage set of p over 𝒞
n by
(27)
𝒥
p
=
⋃
Δ
∈
𝒞
n
{
p
′
∈
S
n
:
φ
[
Δ
]
(
p
)
=
p
′
}
,
𝒥
~
p
=
⋃
Δ
∈
𝒞
n
{
p
~
∈
S
n
:
φ
[
Δ
]
(
p
~
)
=
p
}
,
respectively. For each Δ
∈
𝒞
n, (a)
there exists a uniquep
′
∈
S
n such that φ
[
Δ
]
(
p
)
=
p
′. Moreover, |
𝒥
p
|
=
2
n;
(b)
there exists a uniquep
~
∈
S
n such that φ
[
Δ
]
(
p
~
)
=
p. Moreover, |
𝒥
~
p
|
=
2
n.Proof.
(a) Letx
=
x
0
x
1
⋯
x
n
-
1 be the fitness state of p
∈
S
n. The image of x under both maps ϕ
Δ and ϕ
J is unique so p
′ is unique. Note that |
𝒞
n
|
=
2
n and the set of images of x under φ
[
Δ
], Δ
∈
𝒞
n, is the set of distinct binary numbers of length n, which can be associated with 2
n distinct states in S
n. This is a bijection between elements of 𝒞
n and the binary numbers of length n, so |
𝒥
p
|
=
2
n.
(b) ForΔ
∈
𝒞
n, choose p
~
=
φ
~
[
Δ
]
(
p
), where φ
~
[
Δ
] is the return map. The proofs of the uniqueness of p
~ and the cardinality of 𝒥
~
p follow a similar argument as in (a).Theorem 16.
Letn
≥
2 and p
∈
S
n. If Δ
=
{
n
-
m
,
n
-
1
}, then φ
[
Δ
]
m
(
p
)
=
p for m
=
2,3
,
…
,
n.Proof.
Fixn and m, where 2
≤
m
≤
n. Let a
=
x
0
x
1
⋯
x
(
n
-
m
-
1
) and b
=
i
0
i
1
⋯
i
(
n
-
m
-
1
) be sequences of length (
n
-
m
) so that
(28)
p
=
(
a
x
(
n
-
m
)
x
(
n
-
m
+
1
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
,
w
w
b
i
(
n
-
m
)
i
(
n
-
m
+
1
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
)
.
Let Δ
=
{
n
-
m
,
n
-
1
}. Observe that, for any integer ℓ
>
0, a and b are fixed under φ
[
Δ
]
ℓ. Now
(29)
φ
[
Δ
]
(
p
)
=
(
a
x
(
n
-
m
+
1
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
*
,
w
w
a
x
(
n
-
m
+
1
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
*
b
i
(
n
-
m
+
1
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
i
(
n
-
m
)
)
φ
[
Δ
]
2
(
p
)
=
(
a
x
(
n
-
m
+
2
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
w
×
x
(
n
-
m
+
1
)
*
,
b
i
(
n
-
m
+
2
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
w
×
a
x
(
n
-
m
+
2
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
i
(
n
-
m
)
i
(
n
-
m
+
1
)
)
⋮
φ
[
Δ
]
ℓ
(
p
)
=
(
a
x
(
n
-
m
+
ℓ
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
⋯
x
(
n
-
m
+
ℓ
-
2
)
w
×
x
(
n
-
m
+
ℓ
-
1
)
*
,
b
i
(
n
-
m
+
ℓ
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
w
×
i
(
n
-
m
)
⋯
i
(
n
-
m
+
ℓ
-
2
)
i
(
n
-
m
+
ℓ
-
1
)
)
.
For ℓ
=
m
-
1, φ
[
Δ
]
m
-
1
(
p
)
=
(
a
x
(
n
-
1
)
*
x
n
-
m
⋯
x
n
-
2
*
,
b
i
(
n
-
1
)
i
n
-
m
⋯
i
n
-
2
). Hence, φ
[
Δ
]
(
φ
[
Δ
]
m
-
1
(
p
)
)
=
φ
[
Δ
]
m
(
p
)
=
p.Corollary 17.
For anyn
≥
1, p
∈
S
n, and Δ
∈
{
{
n
-
1
}
,
{
n
-
2
,
n
-
1
}
,
I
n
}, the transition from p to p
′
=
φ
[
Δ
]
(
p
) is reversible.The next theorem states some nonallowable transitions inS
n.Theorem 18.
Let(30)
y
=
i
0
i
1
⋯
i
n
-
2
i
n
-
1
,
y
-
=
i
n
-
1
i
-
1
i
-
2
⋯
i
-
n
-
2
i
-
n
-
1
,
i
-
j
≠
i
j
f
o
r
j
∈
I
n
∖
{
0
}
,
y
′
=
i
0
′
i
1
′
⋯
i
n
-
2
′
i
n
-
1
,
i
j
′
≠
i
j
f
o
r
j
∈
I
n
∖
{
n
-
1
}
.
Define
(31)
S
y
=
{
p
∈
S
n
:
y
i
s
t
h
e
s
i
g
n
i
f
i
c
a
n
c
e
s
t
a
t
e
o
f
p
}
,
S
y
-
=
{
p
∈
S
n
:
y
-
i
s
t
h
e
s
i
g
n
i
f
i
c
a
n
c
e
s
t
a
t
e
o
f
p
}
,
S
y
′
=
{
p
∈
S
n
:
y
′
i
s
t
h
e
s
i
g
n
i
f
i
c
a
n
c
e
s
t
a
t
e
o
f
p
}
.
(a)
Letn
≥
2. The only transitions in S
y, aside from self-transitions, are horizontal transitions. Let n
≥
3.
(b)
There is no transition fromS
y to S
y
-.
(c)
There is no transition betweenS
y and S
y
′.Proof.
(a) It is easy to show that any transition between distinct points with the same significance state is underΔ
=
{
n
-
1
}. In particular, horizontal transitions are between pairs p
=
(
x
p
,
y
) and p
ˇ
=
(
x
p
ˇ
,
y
), where x
p
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1 and x
p
ˇ
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1
*.
(b) Letp
∈
S
y. The sets Δ
=
I
n and Δ
=
{
0,1
,
…
,
n
-
2
} are the only elements of 𝒞
n such that i
n
-
1 is moved to the leftmost position of the question order of φ
[
Δ
]
(
p
), in which case question order must be i
n
-
1
i
n
-
2
⋯
i
0, different from y
-.
(c) As in (b),Δ
=
∅ and Δ
=
{
n
-
1
} are the only elements of 𝒞
n such that i
n
-
1 remains in its position under φ
[
Δ
]. The rest of the proof follows a similar argument as in (b).


### 2.5. Local Dynamics inS
n

Consider the subset(32)
ℒ
m
(
⊂
𝒮
n
)
=
{
(
c
0
c
1
⋯
c
(
n
-
m
-
1
)
x
(
n
-
m
)
⋯
x
(
n
-
1
)
,
x
j
{
0,1
}
∈
i
j
I
n
w
w
x
j
{
0,1
}
∈
i
j
I
n
ℓ
0
ℓ
1
⋯
ℓ
n
-
m
-
1
i
n
-
m
⋯
i
n
-
1
)
w
w
:
c
j
,
ℓ
j
constant
,
x
j
∈
{
0,1
}
,
i
j
∈
I
n
}
,
where n
-
m answers are constant. Since |
ℒ
m
|
=
m
!
2
m states and |
𝒮
n
|
=
n
!
2
n states, then S
n is composed of 2
n
-
m
n
!
/
m
! subsets of the form ℒ
m. For n
-
1 constant answers (i.e., m
=
1), denote by ℒ
n
-
1
(
c
,
ℓ
) the subset of S
n, where question ℓ has constant answer c. Then any p
∈
ℒ
n
-
1
(
c
,
ℓ
) is given by p
=
(
c
x
1
x
2
⋯
x
n
-
1
,
ℓ
i
1
i
2
⋯
i
n
-
1
), x
j
∈
{
0,1
}, i
j
∈
I
n. Define ℒ
n
-
1
(
c
) by
(33)
ℒ
n
-
1
(
c
)
=
⋃
ℓ
∈
I
n
ℒ
n
-
1
(
c
,
ℓ
)
.
Using (33), we can express S
n as
(34)
𝒮
n
=
ℒ
n
-
1
(
0
)
∪
ℒ
n
-
1
(
1
)
,
n
≥
1
=
ℒ
n
.
We now define the set (as in 𝒞
n) associated with transitions between points in ℒ
m, where m
<
n. Recall from Theorem 15 that transitions in S
n are under elements Δ of the set 𝒫
(
I
n
)
=
𝒞
n. For m
=
1, there are n
-
1 constant answer values so transitions in ℒ
1 are under elements of the set
(35)
𝒟
1
(
n
)
=
𝒫
(
{
n
-
1
}
)
,
n
≥
1
=
{
∅
,
n
-
1
}
.
Similarly, for n
-
2 constant answer values (i.e., m
=
2), transitions in ℒ
2 are under the set
(36)
𝒟
2
(
n
)
=
𝒫
(
{
n
-
2
,
n
-
1
}
)
,
n
≥
2
=
{
∅
,
n
-
1
}
∪
{
n
-
2
,
{
n
-
2
,
n
-
1
}
}
=
𝒟
1
(
n
)
∪
(
𝒫
(
{
n
-
2
,
n
-
1
}
)
∖
𝒟
1
(
n
)
)
.
In general, the dynamics in ℒ
m
⊂
𝒮
n is described by transitions under the set
(37)
𝒟
m
(
n
)
=
𝒫
(
{
n
-
m
,
n
-
m
+
1
,
…
,
n
-
1
}
)
,
n
≥
m
=
𝒟
m
-
1
(
n
)
∪
(
𝒫
(
{
n
-
m
,
n
-
m
+
1
,
…
,
n
-
1
}
)
∖
𝒟
m
-
1
(
n
)
)
.
For m
=
n, we see from (37) that 𝒟
n
(
n
)
=
𝒞
n, as expected. Moreover, the transitions in S
n are given by the transitions in S
n
-
1, together with the additional transitions under the set difference 𝒞
n
∖
𝒟
n
-
1
(
n
).Since|
ℒ
m
|
=
|
𝒮
m
| and |
𝒟
m
(
n
)
|
=
|
𝒞
m
| for m
≤
n and that there is a one-to-one correspondence between transitions in ℒ
m and S
m (from (37)), we can write ℒ
m
=
𝒮
m.Example 19.
Figures4(a) and 4(b) illustrate all possible transitions in S
1 and S
2, respectively. Consider horizontal transitions alternating between states 1 and 2. In Figure 4(a), this denotes alternating 0-1 answer to question 0. On the other hand, transitions alternating between states 1 and 2 in Figure 4(b) denote alternating 0-1 answer to question 0 with constant favourable answer to question 1. All transitions between states in the dashed boxes (containing ℒ
1
(
c
j
,
ℓ
j
)) are under the map φ
[
{
1
}
], where “1" refers to the index on the right end of question order i
0
i
1. In each ℒ
1
(
c
j
,
ℓ
j
), one answer value is constant and is positioned at the left end of the fitness state. The digraph associated with transitions in S
1 is exactly the same as the digraph associated with transitions in each ℒ
1
(
c
j
,
ℓ
j
). Transitions in these subsets ℒ
1
(
c
j
,
ℓ
j
) are under 𝒟
1
(
2
). The set ℒ
1
(
0,1
) is composed of states 1 and 2, ℒ
1
(
1,1
) of states 3 and 4, and so on. Transitions under 𝒞
2
∖
𝒟
1
(
2
)
=
{
{
0
}
,
{
0,1
}
} are shown as well.Transitions in (a)S
1 under 𝒞
1 and (b) S
2 under 𝒞
2. Circles denote self-transition (Δ
=
∅).
(a)
(b)Example 20.
Figure5 illustrates orbits in S
4 that cluster into two. A red edge signifies a movement from right to left, a green edge a movement from left to right, and a blue edge a transition to the same state (self-transition). Instead of analysing in S
4, we can remove Q
3 and consider only the three questions Q
0, Q
1, and Q
2 and analysing orbits in S
3 characterised by Q
3
=
0 and orbits characterised by Q
3
=
1. Note that the frequency change of Q
3 here is f
3
P
=
0.Figure 5
Two clusters inS
4 distinguished by the value of Q
3. The left cluster comprises orbits with constant unfavourable value to Q
3, while the right cluster is characterised by favourable value to Q
3. Analysis of each cluster of orbits can be simplified in S
3.


## 2.1. Fitness and Significance States

Consider questions in (4) and assign index i to Q
i (i
=
0,1
,
2). Suppose we give more weight (significance) to Q
0 and do this by positioning 0 in the left-end of question order, say, Q
0, Q
1, Q
2, denoted by 012 (or Q
0
,
Q
2
,
Q
1
=
021). As in numbers or decimals, our weighting places the most significant number at the left-end position. All possible (concatenated) answers to 012 are given by
(10)
001,000,010,111,110,101,100
,
011
.
Since 000 has all unfavourable answers we say that it is the leastfit answer, while 111 is thefittest. Note that there are states with the same number of favourable values, for example, 001, 100, and 010. By a suitable weighting of questions, we show below that the lexicographic ordering of answers in (10) is an appropriate ordering of fitness.Table3 illustrates concatenated coded answers to question order 012 of three observations from a population. To Q
0, observation k has constant answer 0 while k
′ has constant answer 1.Table 3
Concatenated coded answers of observations to question orderQ
0
,
Q
1
,
Q
2
=
(
012
).
t
k
k
′
k
′
′
0
010
100
111
1
001
110
001
2
000
101
001
3
010
110
111
4
000
100
001Suppose we arrange answers in (10) lexicographically along an x-axis. Then for question order 012, a one-dimensional dynamics on the x-axis composed of the eight states arises. Answers of k and k
′ to Q
0 seldom change (i.e., they are both constant in Q
0) so the two households k and k
′stay in the regions 0
*
* and 1
*
* of the x-axis, respectively. Recall that Q
0 is the question associated with the significant (left) position of the question order so fitness is biased towards the left position. We can then write x
=
0
*
*
<
x
′
=
1
*
* because the significant variable Q
0 is unfavourable in x and favourable in x
′. This holds true even if x has the same, or more, favourable values as x
′. This argument can be extended to any two elements x
,
x
′
∈
M
n with the same first ℓ entries.In general, not all observations may be stable in the same variable; for example,k
′′ in Table 3 is constant in Q
0, not in Q
2. Moreover, stability of an observation may change in time; it may be stable in variable i over one time interval and then stable in variable i
′ over another time interval. We will not study orbits in a fixed question order alone. We construct a y-axis with states corresponding to question orders. The order of questions per observation becomes a new variable.Definition 2.
Givenn
≥
1 and Q
k
t, thefitness stateandsignificance state of observation k at time t are the sequences
(11)
x
t
k
=
(
x
j
t
)
j
=
0
n
-
1
,
y
t
k
=
(
i
j
t
)
j
=
0
n
-
1
,
respectively. The set of fitness states of length n is called thefitness space of
n
variables defined by
(12)
X
n
:
=
{
(
x
j
)
j
=
0
n
-
1
:
x
j
∈
{
0,1
}
}
and the set of significance states of length n is called thesignificance space of
n
variables defined by
(13)
Y
n
:
=
{
(
i
j
)
j
=
0
n
-
1
∶
i
j
∈
I
n
,
i
j
′
s
distinct
}
.
Elements of both X
n and Y
n are arranged according to the lexicographic ordering (<) of sequences of length n.Definition 3.
The space(14)
S
n
=
{
p
=
(
x
,
y
)
:
x
∈
X
n
,
y
∈
Y
n
}
=
X
n
×
Y
n
is called thechange space for
n
variables.Givenn, we have the cardinalities |
X
n
|
=
2
n, |
Y
n
|
=
n
!, and |
S
n
|
=
N
=
2
n
n
!. A way of labeling state s
j
∈
S
n is via the map
(15)
ψ
:
S
n
⟶
{
1,2
,
…
,
N
}
,
s
j
⟼
ψ
(
s
j
)
=
j
,
g
h
f
f
g
h
f
f
g
j
h
l
k
j
=
1,2
,
…
,
N
.
For convenience, we label states in S
n from left to right, and from top to bottom. If s
j
=
(
x
,
y
) and ψ
(
s
j
)
=
j, we will refer to state s
j as state j and write j
=
(
x
,
y
).Remark 4.
In general, for multivariate data inn variables, withall variables m-ary valued, the space S
m
,
n is composed of m
n
×
n
! states.


## 2.2. The Method of Orbits

We define the dynamics of observations taken from a survey ofn
≥
1 questions. Let ℕ
0 be the set of nonnegative integers, let 𝒞
n
=
𝒫
(
I
n
) be the power set of I
n, and let x
j
* be such that
(16)
x
j
*
=
{
1
if
x
j
=
0
0
if
x
j
=
1
.Definition 5.
LetΔ
∈
𝒞
n and let j
∈
Δ.(a)
The mapϕ
Δ
:
S
n
→
S
n is defined by
(17)
ϕ
Δ
(
x
0
x
1
⋯
x
j
⋯
x
n
-
1
,
y
)
=
(
x
0
x
1
⋯
x
j
*
⋯
x
n
-
1
,
y
)
.
(b)
The mapϕ
J
:
S
n
→
S
n is defined by
(18)
ϕ
J
(
x
0
x
1
⋯
x
j
⋯
x
n
-
1
,
i
0
i
1
⋯
i
j
⋯
i
n
-
1
)
=
(
x
0
x
1
⋯
x
j
-
1
x
j
+
1
⋯
x
n
-
1
x
j
,
i
0
i
1
⋯
i
j
-
1
i
j
+
1
⋯
i
n
-
1
i
j
)
.Let j
,
j
′
∈
Δ. Then x
j and x
j
′ both change values under ϕ
Δ. If j
<
j
′, then ϕ
J is first applied to j
′; that is,
(19)
ϕ
J
(
x
0
x
1
⋯
x
j
⋯
x
j
′
⋯
x
n
-
1
,
i
0
i
1
⋯
i
j
⋯
i
j
′
⋯
i
n
-
1
)
=
(
x
0
x
1
⋯
x
j
-
1
x
j
+
1
⋯
x
j
′
-
1
x
j
′
+
1
⋯
x
n
-
1
x
j
′
x
j
,
w
w
w
w
w
i
0
i
1
⋯
i
j
-
1
i
j
+
1
⋯
i
j
′
-
1
i
j
′
+
1
⋯
i
n
-
1
i
j
′
i
j
)
.Definition 6.
Leti
j
∈
I
n. Consider questionnaire Q with Q
i
j
∈
Q. For each observation k, let f
i
j
k be the frequency of change in answer value of Q
i
j over the observation period. Suppose
(20)
0
<
f
i
0
k
<
f
i
1
k
<
⋯
<
f
i
j
k
<
⋯
<
f
i
n
-
1
k
.
Inequality (20) is called theobservation frequency relation and question order y
0
k
=
i
0
i
1
⋯
i
n
-
1 is theinitial significance state of observation
k. If f
i
j
k
=
f
i
j
+
1
k and f
i
j
P
<
f
i
j
+
1
P at the population level, then choose question order i
j
i
j
+
1. If f
i
j
P
=
f
i
j
+
1
P, then choose question order as in the questionnaire (6). Theinitial fitness state  x
0
k
=
x
0
x
1
⋯
x
n
-
1 is such that x
j is the value of i
j in y
0
k. Theinitial state of
k is the ordered pair p
0
k
=
(
x
0
k
,
y
0
k
).By choosing the initial significancey
0
k given in Definition 6, we start the orbit in its most-likely significance state. This facilitates convergence to clusters (where they exist) and is useful for short data sets. Other strategies (e.g., using order of (6) or random choice) will nonetheless converge to question order according to (20).Remark 7.
Longitudinal data is only of interest where change occurs; else cross-sectional surveys are adequate. We are interested in longitudinal data that give nontrivial information of change about the population; that is,f
i
j
P
≠
0. Otherwise, question i
j may be deleted as any such property becomes an identifier of subpopulations of possible interest for analysis in its own right.Definition 8.
For each observationk, define thechange set at time  t by
(21)
Δ
t
k
(
∈
𝒞
n
)
=
{
j
:
Q
i
j
∈
Q
t
k
changes answer value at time
t
+
1
}
.
Let p
0
k
=
(
y
0
k
,
x
0
k
) be the initial state of k. Denote by p
t
k
=
(
x
t
k
,
y
t
k
) the state of k at time t
≥
0. Thechange map   φ
:
(
ℕ
0
,
𝒞
n
,
S
n
)
→
S
n is such that
(22)
φ
(
t
,
Δ
t
k
,
p
t
k
)
:
=
φ
[
Δ
t
k
]
(
p
t
k
)
=
(
ϕ
J
∘
ϕ
Δ
t
k
)
(
p
t
k
)
=
p
t
+
1
k
.
The set Δ
t
k
∈
𝒞
n is given by the longitudinal data for each t
≥
0 and is a useful ordered listing of questions that change answer values from time t to t
+
1. For each k, the nonautonomous map φ
[
Δ
t
k
] defines an evolutionary process that displaces the most frequently (resp., slowly) changing answers and corresponding questions to the right (resp., left).Definition 9.
Given initial statep
0
k of k, define the state of   k at time t
≥
1 by
(23)
p
t
=
φ
[
Δ
t
-
1
]
(
⋯
(
φ
[
Δ
0
]
(
p
0
)
)
)
.
Theforward orbit of
k
under
φ is defined by
(24)
𝒪
φ
(
k
)
=
{
p
t
k
}
t
∈
ℕ
0
.


## 2.3. Algorithm for Building the Orbit of an Observation inS
n

We give a simple algorithm to determine the statesp
t
>
0
k that comprise the orbit of an observation k from longitudinal data.Step 1.
Determine initial question ordery
0
k and initial state p
0
k
=
(
x
0
k
,
y
0
k
) of k and plot p
0
k in S
n.Step 2.
Identify from the data ofk the question that changes answer at time t
=
1, say Q
i
j. Swap both i
j (in y
0
k) and corresponding answer x
j (in x
0
k) to the right of y
0
k and x
0
k, respectively, and change x
j to x
j
*, where x
j
*
=
0 if x
j
=
1 and 0 otherwise. This new question order and answer order give the next state p
1
k
=
(
x
1
k
,
y
1
k
) of k. Suppose both Q
i
j and Q
i
j change answers at t
=
1. If j
<
j
′, then sequentially swap to the right i
j and i
j
′ (resp., x
j and x
j
′) of the question order (resp., answer order), starting with i
j
′ (resp., x
j
′). Change x
j to x
j
* and x
j
′ to x
j
′
*. Plot the point p
1
k in S
n and directed edge from p
0
k to p
1
k.Step 3.
Repeat Step2, updating state p
t
-
1
k
→
p
t
k and t
→
t
+
1 for time t
=
1,2
,
…
,
T
-
1.Visualization and analysis of orbits of observations inS
n allow capturing information of change in longitudinal data. We illustrate in Figure 2 an orbit in S
3. The useful distance on S
n is given by the discrete metric; that is, d
(
p
,
p
′
)
=
1 if p
≠
p
′ and zero otherwise. The visualized distance between points in S
n has no interpretation so we may represent S
n by a regularly spaced point.Figure 2
State transitions of observationk (from Table 4) in S
3.Remark 10.
The set of all orderings of variables is captured inS
n. For fixed question order and x
,
x
′
∈
X
n, x
<
x
′ means that fitness state x
′ is fitter than x. Each level in the significance axis is question order under frequency ordering. The significance axis informs us which variables are weighted most strongly at each time, where significance is ordered from left to right. Clearly, reordering of variables is one among many families of operations; for example, swapping can be done by swapping changing variable to the left end. This operation however does not reveal clusters.


## 2.4. Transitions inS
n

We now analyze possible state transitions which an observationk can take in S
n.Definition 11.
Letp
=
(
x
,
y
) and p
′
=
(
x
′
,
y
′
) be in S
n, and let Δ
∈
𝒞
n.(a)
SupposeΔ is such that φ
[
Δ
]
(
p
)
=
p
′. Then there is atransition from p to p
′ under Δ, defined by p
→
[
Δ
]
p
′.
(b)
Supposep
→
[
Δ
]
p
′. If y
=
y
′ and x
≠
x
′, then there is ahorizontal transition from p to p
′. If x
=
x
′ and y
≠
y
′, then there is avertical transition from p to p
′.
(c)
Supposep
→
[
Δ
]
p
′. If, in addition, Δ is such that p
′
→
Δ
p, then the transition from p to p
′ isreversible, and one writes p
↔
[
Δ
]
p
′.We haveself-transitions if Δ
=
∅ (no change), the empty set. In general, horizontal transitions denote change in the right-most variable, while vertical transitions denote change in the last two variables.Letp
,
p
′
∈
S
n. Given an observation k, if p
t
k
∈
p and p
t
+
1
k
∈
p
′, then there is Δ
t
∈
𝒞
n such that p
→
[
Δ
t
]
p
′. We use the symbol “∈” to denote that there may be other observations in p or p
′ at times t and t
+
1, respectively. State transitions of observation k in S
n are visualised as a sequence of directed edges.Example 12.
Consider data of observationk in column 2 of Table 4. The asterisks denote changing answers in the next time step. The frequencies of change are f
0
k
=
2, f
1
k
=
3, and f
2
k
=
1, so the initial question order of k is y
0
k
=
201. And initial fitness state is x
0
k
=
010 (Definition 6). At t
=
0, we apply φ
[
Δ
0
k
] to initial state p
0
k
=
(
010,201
). Questions i
1
=
0 and i
2
=
1 in Q
0
k change answers in the next time step so Δ
0
k
=
{
1,2
} (not {
0,1
}). Applying ϕ
Δ
0
k to (
010,201
) changes x
2
=
0 and x
1
=
1 to x
2
*
=
1 and x
1
*
=
0, respectively. Next, ϕ
J is first applied to both x
2
* and i
2 by moving each to the right (they are already rightmost), followed by moving x
1
* and i
1 to the right end. Hence, we have (
x
,
y
)
1
k
=
(
010,210
). For t
=
1 and t
=
2, verify that Δ
1
k
=
{
1
} and Δ
2
k
=
{
0,1
,
2
}, respectively. At each time t the bold numbers in the significance column are the question indices in Q that change answers at t
+
1. The orbit of k in S
3 is shown in Figure 2. The vertical transition from state 11 to 3 denotes two changing answers, while a transition from state 9 to 32 denotes three changing answers.Table 4
States of observationk.
Unitk
Coded answers to
Fitness
Significance
Δ
t
k
Q
=
{
Q
0
,
Q
1
,
Q
2
}
x
t
k
y
t
k
t
=
0
{
1
*
,
0
*
,
0
}
010
201
{
1,2
}
1
{
0
,
1
*
,
0
}
010
210
{
1
}
2
{
0
*
,
0
*
,
0
*
}
000
201
{
0,1
,
2
}
3
{
1
,
1
,
1
}
111
102Example 13.
Figures3(a) and 3(b) illustrate the associated orbits in S
3 of the four households given in Table 1. Observe that orbit of h
3 stays strictly on the left half of S
3, particularly in subset where Q
1
(
HH
)
=
0, while h
1 and h
4 stay in the right half subset, where Q
1
(
HH
)
=
1 and Q
2
(
AD
)
=
1, respectively.(a) Orbits of the four subjects in Table1. A red edge denotes a transition to the left, a green edge is a transition to the right, and a blue edge is a self-transition.
(a)
(b)Definition 14.
LetΔ
∈
𝒞
n and let j
∈
Δ.The return map  φ
~
[
Δ
]
:
S
n
→
S
n is such that
(25)
φ
~
[
Δ
]
(
x
0
x
1
⋯
x
j
⋯
x
n
-
1
,
i
0
i
1
⋯
i
j
⋯
i
n
-
1
)
=
(
x
0
x
1
⋯
x
j
-
1
x
n
-
1
*
x
j
⋯
x
n
-
2
,
w
w
i
0
i
1
⋯
i
j
-
1
i
n
-
1
i
j
⋯
i
n
-
2
)
.
That is, φ
~
[
Δ
] first inserts x
n
-
1 and i
n
-
1 to the position, where x
j and i
j are located, followed by changing x
n
-
1 to the new value x
n
-
1
*. If j
,
j
′
∈
Δ and j
<
j
′ then φ
~
[
Δ
] is first applied to x
j and i
j.
Trivially, for anyn
≥
1, p
∈
S
n, and Δ
∈
𝒞
n,
(26)
φ
~
[
Δ
]
(
φ
[
Δ
]
(
p
)
)
=
φ
[
Δ
]
(
φ
~
[
Δ
]
(
p
)
)
=
p
.Theorem 15.
Letp
∈
S
n. Define the image set and preimage set of p over 𝒞
n by
(27)
𝒥
p
=
⋃
Δ
∈
𝒞
n
{
p
′
∈
S
n
:
φ
[
Δ
]
(
p
)
=
p
′
}
,
𝒥
~
p
=
⋃
Δ
∈
𝒞
n
{
p
~
∈
S
n
:
φ
[
Δ
]
(
p
~
)
=
p
}
,
respectively. For each Δ
∈
𝒞
n, (a)
there exists a uniquep
′
∈
S
n such that φ
[
Δ
]
(
p
)
=
p
′. Moreover, |
𝒥
p
|
=
2
n;
(b)
there exists a uniquep
~
∈
S
n such that φ
[
Δ
]
(
p
~
)
=
p. Moreover, |
𝒥
~
p
|
=
2
n.Proof.
(a) Letx
=
x
0
x
1
⋯
x
n
-
1 be the fitness state of p
∈
S
n. The image of x under both maps ϕ
Δ and ϕ
J is unique so p
′ is unique. Note that |
𝒞
n
|
=
2
n and the set of images of x under φ
[
Δ
], Δ
∈
𝒞
n, is the set of distinct binary numbers of length n, which can be associated with 2
n distinct states in S
n. This is a bijection between elements of 𝒞
n and the binary numbers of length n, so |
𝒥
p
|
=
2
n.
(b) ForΔ
∈
𝒞
n, choose p
~
=
φ
~
[
Δ
]
(
p
), where φ
~
[
Δ
] is the return map. The proofs of the uniqueness of p
~ and the cardinality of 𝒥
~
p follow a similar argument as in (a).Theorem 16.
Letn
≥
2 and p
∈
S
n. If Δ
=
{
n
-
m
,
n
-
1
}, then φ
[
Δ
]
m
(
p
)
=
p for m
=
2,3
,
…
,
n.Proof.
Fixn and m, where 2
≤
m
≤
n. Let a
=
x
0
x
1
⋯
x
(
n
-
m
-
1
) and b
=
i
0
i
1
⋯
i
(
n
-
m
-
1
) be sequences of length (
n
-
m
) so that
(28)
p
=
(
a
x
(
n
-
m
)
x
(
n
-
m
+
1
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
,
w
w
b
i
(
n
-
m
)
i
(
n
-
m
+
1
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
)
.
Let Δ
=
{
n
-
m
,
n
-
1
}. Observe that, for any integer ℓ
>
0, a and b are fixed under φ
[
Δ
]
ℓ. Now
(29)
φ
[
Δ
]
(
p
)
=
(
a
x
(
n
-
m
+
1
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
*
,
w
w
a
x
(
n
-
m
+
1
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
*
b
i
(
n
-
m
+
1
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
i
(
n
-
m
)
)
φ
[
Δ
]
2
(
p
)
=
(
a
x
(
n
-
m
+
2
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
w
×
x
(
n
-
m
+
1
)
*
,
b
i
(
n
-
m
+
2
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
w
×
a
x
(
n
-
m
+
2
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
i
(
n
-
m
)
i
(
n
-
m
+
1
)
)
⋮
φ
[
Δ
]
ℓ
(
p
)
=
(
a
x
(
n
-
m
+
ℓ
)
⋯
x
(
n
-
2
)
x
(
n
-
1
)
*
x
(
n
-
m
)
⋯
x
(
n
-
m
+
ℓ
-
2
)
w
×
x
(
n
-
m
+
ℓ
-
1
)
*
,
b
i
(
n
-
m
+
ℓ
)
⋯
i
(
n
-
2
)
i
(
n
-
1
)
w
×
i
(
n
-
m
)
⋯
i
(
n
-
m
+
ℓ
-
2
)
i
(
n
-
m
+
ℓ
-
1
)
)
.
For ℓ
=
m
-
1, φ
[
Δ
]
m
-
1
(
p
)
=
(
a
x
(
n
-
1
)
*
x
n
-
m
⋯
x
n
-
2
*
,
b
i
(
n
-
1
)
i
n
-
m
⋯
i
n
-
2
). Hence, φ
[
Δ
]
(
φ
[
Δ
]
m
-
1
(
p
)
)
=
φ
[
Δ
]
m
(
p
)
=
p.Corollary 17.
For anyn
≥
1, p
∈
S
n, and Δ
∈
{
{
n
-
1
}
,
{
n
-
2
,
n
-
1
}
,
I
n
}, the transition from p to p
′
=
φ
[
Δ
]
(
p
) is reversible.The next theorem states some nonallowable transitions inS
n.Theorem 18.
Let(30)
y
=
i
0
i
1
⋯
i
n
-
2
i
n
-
1
,
y
-
=
i
n
-
1
i
-
1
i
-
2
⋯
i
-
n
-
2
i
-
n
-
1
,
i
-
j
≠
i
j
f
o
r
j
∈
I
n
∖
{
0
}
,
y
′
=
i
0
′
i
1
′
⋯
i
n
-
2
′
i
n
-
1
,
i
j
′
≠
i
j
f
o
r
j
∈
I
n
∖
{
n
-
1
}
.
Define
(31)
S
y
=
{
p
∈
S
n
:
y
i
s
t
h
e
s
i
g
n
i
f
i
c
a
n
c
e
s
t
a
t
e
o
f
p
}
,
S
y
-
=
{
p
∈
S
n
:
y
-
i
s
t
h
e
s
i
g
n
i
f
i
c
a
n
c
e
s
t
a
t
e
o
f
p
}
,
S
y
′
=
{
p
∈
S
n
:
y
′
i
s
t
h
e
s
i
g
n
i
f
i
c
a
n
c
e
s
t
a
t
e
o
f
p
}
.
(a)
Letn
≥
2. The only transitions in S
y, aside from self-transitions, are horizontal transitions. Let n
≥
3.
(b)
There is no transition fromS
y to S
y
-.
(c)
There is no transition betweenS
y and S
y
′.Proof.
(a) It is easy to show that any transition between distinct points with the same significance state is underΔ
=
{
n
-
1
}. In particular, horizontal transitions are between pairs p
=
(
x
p
,
y
) and p
ˇ
=
(
x
p
ˇ
,
y
), where x
p
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1 and x
p
ˇ
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1
*.
(b) Letp
∈
S
y. The sets Δ
=
I
n and Δ
=
{
0,1
,
…
,
n
-
2
} are the only elements of 𝒞
n such that i
n
-
1 is moved to the leftmost position of the question order of φ
[
Δ
]
(
p
), in which case question order must be i
n
-
1
i
n
-
2
⋯
i
0, different from y
-.
(c) As in (b),Δ
=
∅ and Δ
=
{
n
-
1
} are the only elements of 𝒞
n such that i
n
-
1 remains in its position under φ
[
Δ
]. The rest of the proof follows a similar argument as in (b).


## 2.5. Local Dynamics inS
n

Consider the subset(32)
ℒ
m
(
⊂
𝒮
n
)
=
{
(
c
0
c
1
⋯
c
(
n
-
m
-
1
)
x
(
n
-
m
)
⋯
x
(
n
-
1
)
,
x
j
{
0,1
}
∈
i
j
I
n
w
w
x
j
{
0,1
}
∈
i
j
I
n
ℓ
0
ℓ
1
⋯
ℓ
n
-
m
-
1
i
n
-
m
⋯
i
n
-
1
)
w
w
:
c
j
,
ℓ
j
constant
,
x
j
∈
{
0,1
}
,
i
j
∈
I
n
}
,
where n
-
m answers are constant. Since |
ℒ
m
|
=
m
!
2
m states and |
𝒮
n
|
=
n
!
2
n states, then S
n is composed of 2
n
-
m
n
!
/
m
! subsets of the form ℒ
m. For n
-
1 constant answers (i.e., m
=
1), denote by ℒ
n
-
1
(
c
,
ℓ
) the subset of S
n, where question ℓ has constant answer c. Then any p
∈
ℒ
n
-
1
(
c
,
ℓ
) is given by p
=
(
c
x
1
x
2
⋯
x
n
-
1
,
ℓ
i
1
i
2
⋯
i
n
-
1
), x
j
∈
{
0,1
}, i
j
∈
I
n. Define ℒ
n
-
1
(
c
) by
(33)
ℒ
n
-
1
(
c
)
=
⋃
ℓ
∈
I
n
ℒ
n
-
1
(
c
,
ℓ
)
.
Using (33), we can express S
n as
(34)
𝒮
n
=
ℒ
n
-
1
(
0
)
∪
ℒ
n
-
1
(
1
)
,
n
≥
1
=
ℒ
n
.
We now define the set (as in 𝒞
n) associated with transitions between points in ℒ
m, where m
<
n. Recall from Theorem 15 that transitions in S
n are under elements Δ of the set 𝒫
(
I
n
)
=
𝒞
n. For m
=
1, there are n
-
1 constant answer values so transitions in ℒ
1 are under elements of the set
(35)
𝒟
1
(
n
)
=
𝒫
(
{
n
-
1
}
)
,
n
≥
1
=
{
∅
,
n
-
1
}
.
Similarly, for n
-
2 constant answer values (i.e., m
=
2), transitions in ℒ
2 are under the set
(36)
𝒟
2
(
n
)
=
𝒫
(
{
n
-
2
,
n
-
1
}
)
,
n
≥
2
=
{
∅
,
n
-
1
}
∪
{
n
-
2
,
{
n
-
2
,
n
-
1
}
}
=
𝒟
1
(
n
)
∪
(
𝒫
(
{
n
-
2
,
n
-
1
}
)
∖
𝒟
1
(
n
)
)
.
In general, the dynamics in ℒ
m
⊂
𝒮
n is described by transitions under the set
(37)
𝒟
m
(
n
)
=
𝒫
(
{
n
-
m
,
n
-
m
+
1
,
…
,
n
-
1
}
)
,
n
≥
m
=
𝒟
m
-
1
(
n
)
∪
(
𝒫
(
{
n
-
m
,
n
-
m
+
1
,
…
,
n
-
1
}
)
∖
𝒟
m
-
1
(
n
)
)
.
For m
=
n, we see from (37) that 𝒟
n
(
n
)
=
𝒞
n, as expected. Moreover, the transitions in S
n are given by the transitions in S
n
-
1, together with the additional transitions under the set difference 𝒞
n
∖
𝒟
n
-
1
(
n
).Since|
ℒ
m
|
=
|
𝒮
m
| and |
𝒟
m
(
n
)
|
=
|
𝒞
m
| for m
≤
n and that there is a one-to-one correspondence between transitions in ℒ
m and S
m (from (37)), we can write ℒ
m
=
𝒮
m.Example 19.
Figures4(a) and 4(b) illustrate all possible transitions in S
1 and S
2, respectively. Consider horizontal transitions alternating between states 1 and 2. In Figure 4(a), this denotes alternating 0-1 answer to question 0. On the other hand, transitions alternating between states 1 and 2 in Figure 4(b) denote alternating 0-1 answer to question 0 with constant favourable answer to question 1. All transitions between states in the dashed boxes (containing ℒ
1
(
c
j
,
ℓ
j
)) are under the map φ
[
{
1
}
], where “1" refers to the index on the right end of question order i
0
i
1. In each ℒ
1
(
c
j
,
ℓ
j
), one answer value is constant and is positioned at the left end of the fitness state. The digraph associated with transitions in S
1 is exactly the same as the digraph associated with transitions in each ℒ
1
(
c
j
,
ℓ
j
). Transitions in these subsets ℒ
1
(
c
j
,
ℓ
j
) are under 𝒟
1
(
2
). The set ℒ
1
(
0,1
) is composed of states 1 and 2, ℒ
1
(
1,1
) of states 3 and 4, and so on. Transitions under 𝒞
2
∖
𝒟
1
(
2
)
=
{
{
0
}
,
{
0,1
}
} are shown as well.Transitions in (a)S
1 under 𝒞
1 and (b) S
2 under 𝒞
2. Circles denote self-transition (Δ
=
∅).
(a)
(b)Example 20.
Figure5 illustrates orbits in S
4 that cluster into two. A red edge signifies a movement from right to left, a green edge a movement from left to right, and a blue edge a transition to the same state (self-transition). Instead of analysing in S
4, we can remove Q
3 and consider only the three questions Q
0, Q
1, and Q
2 and analysing orbits in S
3 characterised by Q
3
=
0 and orbits characterised by Q
3
=
1. Note that the frequency change of Q
3 here is f
3
P
=
0.Figure 5
Two clusters inS
4 distinguished by the value of Q
3. The left cluster comprises orbits with constant unfavourable value to Q
3, while the right cluster is characterised by favourable value to Q
3. Analysis of each cluster of orbits can be simplified in S
3.


## 3. General Equation of Motion of Dynamics inS
n

This section concerns properties of all possible paths inS
n.Definition 21 (see [13]).
LetX and Y be arbitrary sets. A multivalued map F from X to Y, denoted by F
:
X
⇉
Y, is such that F
(
x
) is assigned a set Y
x
⊂
Y for all x
∈
X.Consider our state spaceS
n. Denote by φ
[
Δ
]
:
𝒮
n
→
𝒮
n the map in (22), where Δ
t is constant. Define
(38)
G
=
{
φ
[
Δ
]
:
Δ
∈
𝒞
n
}
and the multivalued map F
:
𝒮
n
⇉
𝒮
n by
(39)
F
(
p
)
=
⋃
{
g
(
p
)
∣
g
∈
G
}
.
|
F
(
p
)
|
=
2
n so F is a 1 to 2
n map. Under F, the orbit {
p
t
k
}
t
∈
ℕ
0 is such that p
t
+
1
k
∈
F
(
p
t
k
). The multivalued map F can be interpreted as a digraph whose N
=
n
!
2
n vertices are the points in S
n and edge p
→
p
′ if p
′
∈
F
(
p
).Equivalently,F can be defined as a square matrix of size N that encodes all possible paths in S
n.Definition 22.
LetV
=
{
1,2
,
…
,
N
} and let i
,
j
,
∈
V. For p
,
p
′
∈
S
n, let ψ
(
p
)
=
i and ψ
(
p
′
)
=
j. Let F be the multivalued map in (39). Thetheoretical transition matrix for
F
under
n
questions is denoted by T
n
(
th
)
=
(
T
n
,
i
j
(
th
)
), where
(40)
T
n
,
i
j
(
t
h
)
=
{
1
if
p
′
∈
F
(
p
)
0
otherwise
.
T
n
,
i
j
(
th
) captures all theoretically admissible transitions between states in S
n. Each entry T
n
,
i
j
(
th
)
=
1 indicates that there is a transition from state i to j in one step. Hence, T
n
,
i
j
(
th
) gives all possible paths in S
n that any observation can take.Example 23.
From Figure4, we have T
n
(
th
) for n
=
1 and n
=
2, as given below
(41)
T
1
(
th
)
=
[
1
1
1
1
]
T
2
(
th
)
=
[
1
1
0
0
0
1
0
1
1
1
0
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
1
1
1
0
1
0
0
1
0
1
1
1
0
0
0
1
0
1
1
1
0
0
1
0
1
0
0
0
1
1
1
0
1
0
0
0
1
1
]
.
Because the digraph in S
1 is the same as the digraphs in ℒ
1
(
c
j
,
ℓ
j
)
⊂
𝒮
2 (Example 19), the matrix encoding transitions in S
1 is the same as the matrix encoding transitions in ℒ
1
(
c
j
,
ℓ
j
). The 2
×
2 submatrices of entries all equal to 1 (in bold numbers) in T
2
(
th
) correspond to transitions in ℒ
1
(
c
j
,
ℓ
j
). The 2
×
2 zero submatrices in T
2
(
th
) (in bold) denote the nonallowable transitions between states with the same significance (Theorem 18(a)).


### 3.1. Properties ofT
n
(
th
)

We give some elementary results forT
n
(
th
).Theorem 24.
(a) The digraph𝒢
T
n
(
t
h
) is 2
n-regular; that is, the in- and out-degree of all its vertices is 2
n.
(b) The largest eigenvalue ofT
n
(
t
h
) is λ
max
⁡
(
T
n
(
t
h
)
)
=
2
n, with associated eigenvector v
λ
max
⁡
=
1.
(c) Consider trace(
T
n
(
t
h
)
)
=
N.
(d) Considerdet
⁡
(
T
n
(
t
h
)
)
=
0.Proof.
(a) This is a consequence of Theorem15.
(b) This is a consequence of (a) and the Perron-Frobenius theorem [14].
(c) Letψ
(
p
)
=
i. From the definition of self-transition, φ
[
∅
]
(
p
)
=
p for any p
∈
S
n. Hence, T
n
,
i
i
(
th
)
=
1 for i
=
1,2
,
…
,
N.
(d) Letp
=
(
x
,
y
) and p
′
=
(
x
′
,
y
) be points in S
n such that
(42)
x
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1
,
y
=
i
0
i
1
⋯
i
n
-
2
i
n
-
1
,
x
′
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1
*
,
where
x
n
-
1
*
is
given
in
(
16
)
.
Claim. 𝒥
p
=
𝒥
p
′, where 𝒥
p and 𝒥
p
′ are given in (27).
The case wheren
=
1 is given in Example 23. We prove our claim for n
=
2. By definition of self-transition, φ
[
∅
]
(
p
)
=
p and φ
[
∅
]
(
p
′
)
=
p
′. It is clear that φ
[
{
1
}
]
(
p
)
=
p
′ and φ
[
{
1
}
]
(
p
′
)
=
p. Let
(43)
φ
[
{
0
}
]
(
p
)
=
p
1
,
φ
[
{
0,1
}
]
(
p
)
=
p
2
.
Observe that
(44)
φ
[
{
0
}
]
(
p
′
)
=
φ
[
{
0
}
]
(
φ
[
{
1
}
]
p
)
=
φ
[
0,1
]
(
p
)
=
p
2
,
φ
[
{
0,1
}
]
(
p
′
)
=
φ
[
{
0,1
}
]
(
φ
[
{
1
}
]
p
)
=
φ
[
{
0
}
]
(
φ
[
{
1
}
]
2
(
p
)
)
=
φ
[
{
0
}
]
(
p
)
=
p
1
.
Since p and p
′ have the same image for all Δ
∈
𝒞
2, 𝒥
p
=
𝒥
p
′, as claimed.
Our claim implies that rows ofT
2
(
th
) associated with p and p
′ are the same. Because of these repeating rows, det(
T
2
(
th
)
)
=
0. The proof can be readily extended to the general case and is treated in exactly the same manner.


### 3.2. Subshift of Finite Type

Letw
i, j
∈
V
=
{
1,2
,
…
,
N
}. If observation k is in state s
j
∈
S
n after t time steps, we will write p
t
k
∈
s
j to indicate that more than one observation may be in s
j. We associate with the orbit 𝒪
φ
(
k
) a sequence of symbols w
=
(
w
0
w
1
⋯
w
i
⋯
), where w
i
=
j if p
i
k
∈
s
j. Denote the correspondingsymbol space of one-sided sequences of
T
n
(
th
) by
(45)
Σ
T
n
(
t
h
)
+
=
{
w
=
(
w
0
w
1
⋯
)
∣
T
n
,
w
i
w
i
+
1
(
th
)
=
1
h
j
g
T
n
,
w
i
w
i
+
1
(
t
h
)
∀
i
∈
ℕ
0
,
w
i
∈
V
}
.
Theequation of motion in
S
n is given by the shift map Σ
T
n
(
th
)
:
Σ
T
n
(
th
)
+
→
Σ
T
n
(
th
)
+ defined by
(46)
σ
T
n
(
t
h
)
(
w
)
=
w
′
where
w
i
′
=
w
i
+
1
(
∀
i
≥
0
)
.Definition 25 (see [15]).
The pair(
σ
T
n
(
th
)
,
Σ
T
n
(
th
)
+
) is called thesubshift of finite type (SFT) determined by
T
n
(
th
).The SFT determined byT
n
(
th
) captures the exact detail of all possible itineraries of observations in S
n. Many dynamical properties of an SFT depend on the structure of its associated transition matrix or digraph.Definition 26 (see [16]).
(i) A transition matrixT isirreducible if for each pair (
i
,
j
), there exists ℓ
>
0 so that T
ℓ
>
0. Otherwise, T isreducible. (ii) The digraph G
T
=
(
V
,
E
) associated with T isstrongly connected if it is possible to get from any vertex to any other one by traversing a sequence of edges as directed by T. (iii) If there exists ℓ
′
>
0 such that T
ℓ
>
0 for all ℓ
>
ℓ
′, then T isprimitive. (iv) Let X be any set. A continuous map f
:
X
→
X is(topologically) transitive if, for all nonempty open sets U
,
U
′
⊂
X, there exists ℓ
≥
0 such that f
ℓ
(
U
)
∩
U
′
≠
∅. (v) If there is ℓ
′
> 0 such that f
ℓ
(
U
)
∩
U
′
≠
∅ for all ℓ
≥
ℓ
′, then f is(topologically) mixing.Consider the SFT given by(
σ
T
,
Σ
T
+
). The following results [17, 18] concern dynamical properties of a transitive SFT and algebraic properties of its transition matrix. (i) The shift map Σ
T is transitive (resp., mixing) if and only if T is irreducible (resp., primitive). (ii) T is irreducible if and only if the digraph G
T
=
(
V
,
E
) is strongly connected. In that case we say that V is irreducible. (iii) A nonnegative, irreducible matrix with a positive element on the main diagonal is primitive.Theorem 27.
T
n
(
t
h
) is irreducible.Proof.
We prove by induction onn. From Example 23, T
1
,
i
j
(
th
)
=
1 for all (
i
,
j
) so T
n
(
th
) is irreducible for n
=
1. Assume that T
n
(
th
) is irreducible for n
=
m. Then the digraph G
T
m
(
th
) is strongly connected, and S
m is irreducible. We prove that G
T
n
(
th
) is strongly connected to n
=
m
+
1.
Recall that transitions inS
m
+
1 are under the set 𝒞
m
+
1. From (37), this set can be expressed as 𝒞
m
+
1
=
𝒟
m
(
m
+
1
)
∪
(
𝒞
m
+
1
∖
𝒟
m
(
m
+
1
)
), where 𝒟
m
(
m
+
1
) is the set associated with transitions in the irreducible set ℒ
m
(
c
,
i
j
)
(
=
𝒮
m
), c
∈
{
0,1
} and i
j
∈
I
m
+
1. With the additional transitions under 𝒞
m
+
1
∖
𝒟
m
(
m
+
1
), we show that (i)
the setℒ
m
(
c
)
⊂
𝒮
m
+
1 (given in (33)) is irreducible;
(ii)
there is a path to and from any pairp
,
p
′, where p
∈
ℒ
m
(
0
) and p
′
∈
ℒ
m
(
1
).We prove (i) by showing that there is p
∈
ℒ
m
(
c
,
i
j
) such that, for all i
j
′
≠
i
j, there is a transition from p to p
′
∈
ℒ
m
(
c
,
i
j
′
). By Theorem 16, φ
[
{
0
,
m
}
]
m
+
1
(
p
)
=
p for any p
∈
𝒮
m
+
1. Let
(47)
p
0
=
(
c
c
⋯
c
c
*
,
i
0
i
1
⋯
i
m
-
1
i
m
)
p
1
=
φ
[
{
0
,
m
}
]
(
p
0
)
=
(
c
c
⋯
c
c
*
,
i
1
i
2
⋯
i
m
i
0
)
p
2
=
φ
[
{
0
,
m
}
]
2
(
p
1
)
=
(
c
c
⋯
c
c
*
,
i
2
i
3
⋯
i
0
i
1
)
⋮
p
m
=
φ
[
{
0
,
m
}
]
m
(
p
m
-
1
)
=
(
c
c
⋯
c
c
*
,
i
m
i
0
⋯
i
m
-
2
i
m
-
1
)
.
Observe that all p
j’s have the same fitness states but distinct significance states. In particular, each p
j is contained in a distinct subset ℒ
m
(
c
,
i
j
). This vertical transition allows all p
j
∈
ℒ
m
(
c
,
i
j
) to visit a distinct ℒ
m
(
c
,
i
j
′
) in at most m steps. Hence, ℒ
m
(
c
) is irreducible. To show (ii), take
(48)
p
=
(
00
⋯
0
,
m
(
m
-
1
)
⋯
10
)
∈
ℒ
m
(
0
)
,
p
′
=
(
11
⋯
1,01
⋯
(
m
-
1
)
m
)
∈
ℒ
m
(
1
)
.
Under Δ
=
I
m
+
1, there is a reversible transition p
↔
[
Δ
]
p
′.
From (i), (ii), and the irreducibility ofT
m
(
th
), there is a path between any pair p
,
p
′
∈
𝒮
m
+
1. Hence, G
T
m
+
1
(
th
) is strongly connected and T
m
+
1
(
th
) is irreducible, as desired.Corollary 28.
T
n
(
t
h
) is primitive.Definition 29 (see [15]).
The topological entropy of the shift mapΣ
T
:
Σ
T
+
→
Σ
T
+ is defined by
(49)
h
top
(
σ
T
)
=
lim
⁡
m
→
∞
ln
⁡
(
|
𝒲
(
m
)
|
)
m
,
where 𝒲
(
m
) is the set of allowable sequences of length m
≥
1.A continuous mapf on a compact metric space X is chaotic if h
top
(
f
)
>
0 [17]. The topological entropy for SFTs is given by the following theorem [15].Theorem 30.
LetT be a transition matrix and let Σ
T
:
Σ
T
+
→
Σ
T
+ be the associated SFT. Then
(50)
h
t
o
p
(
σ
T
)
=
ln
⁡
(
λ
max
⁡
)
,
where λ
max
⁡ is the maximum eigenvalue of T.Since(
σ
T
n
(
th
)
,
Σ
T
n
(
th
)
+
) is the associated subshift of the multivalued map F in (39), one has
(51)
h
top
(
F
)
=
h
top
(
σ
T
n
(
th
)
)
=
ln
⁡
(
2
n
)
=
n
ln
⁡
2
>
0
.


## 3.1. Properties ofT
n
(
th
)

We give some elementary results forT
n
(
th
).Theorem 24.
(a) The digraph𝒢
T
n
(
t
h
) is 2
n-regular; that is, the in- and out-degree of all its vertices is 2
n.
(b) The largest eigenvalue ofT
n
(
t
h
) is λ
max
⁡
(
T
n
(
t
h
)
)
=
2
n, with associated eigenvector v
λ
max
⁡
=
1.
(c) Consider trace(
T
n
(
t
h
)
)
=
N.
(d) Considerdet
⁡
(
T
n
(
t
h
)
)
=
0.Proof.
(a) This is a consequence of Theorem15.
(b) This is a consequence of (a) and the Perron-Frobenius theorem [14].
(c) Letψ
(
p
)
=
i. From the definition of self-transition, φ
[
∅
]
(
p
)
=
p for any p
∈
S
n. Hence, T
n
,
i
i
(
th
)
=
1 for i
=
1,2
,
…
,
N.
(d) Letp
=
(
x
,
y
) and p
′
=
(
x
′
,
y
) be points in S
n such that
(42)
x
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1
,
y
=
i
0
i
1
⋯
i
n
-
2
i
n
-
1
,
x
′
=
x
0
x
1
⋯
x
n
-
2
x
n
-
1
*
,
where
x
n
-
1
*
is
given
in
(
16
)
.
Claim. 𝒥
p
=
𝒥
p
′, where 𝒥
p and 𝒥
p
′ are given in (27).
The case wheren
=
1 is given in Example 23. We prove our claim for n
=
2. By definition of self-transition, φ
[
∅
]
(
p
)
=
p and φ
[
∅
]
(
p
′
)
=
p
′. It is clear that φ
[
{
1
}
]
(
p
)
=
p
′ and φ
[
{
1
}
]
(
p
′
)
=
p. Let
(43)
φ
[
{
0
}
]
(
p
)
=
p
1
,
φ
[
{
0,1
}
]
(
p
)
=
p
2
.
Observe that
(44)
φ
[
{
0
}
]
(
p
′
)
=
φ
[
{
0
}
]
(
φ
[
{
1
}
]
p
)
=
φ
[
0,1
]
(
p
)
=
p
2
,
φ
[
{
0,1
}
]
(
p
′
)
=
φ
[
{
0,1
}
]
(
φ
[
{
1
}
]
p
)
=
φ
[
{
0
}
]
(
φ
[
{
1
}
]
2
(
p
)
)
=
φ
[
{
0
}
]
(
p
)
=
p
1
.
Since p and p
′ have the same image for all Δ
∈
𝒞
2, 𝒥
p
=
𝒥
p
′, as claimed.
Our claim implies that rows ofT
2
(
th
) associated with p and p
′ are the same. Because of these repeating rows, det(
T
2
(
th
)
)
=
0. The proof can be readily extended to the general case and is treated in exactly the same manner.


## 3.2. Subshift of Finite Type

Letw
i, j
∈
V
=
{
1,2
,
…
,
N
}. If observation k is in state s
j
∈
S
n after t time steps, we will write p
t
k
∈
s
j to indicate that more than one observation may be in s
j. We associate with the orbit 𝒪
φ
(
k
) a sequence of symbols w
=
(
w
0
w
1
⋯
w
i
⋯
), where w
i
=
j if p
i
k
∈
s
j. Denote the correspondingsymbol space of one-sided sequences of
T
n
(
th
) by
(45)
Σ
T
n
(
t
h
)
+
=
{
w
=
(
w
0
w
1
⋯
)
∣
T
n
,
w
i
w
i
+
1
(
th
)
=
1
h
j
g
T
n
,
w
i
w
i
+
1
(
t
h
)
∀
i
∈
ℕ
0
,
w
i
∈
V
}
.
Theequation of motion in
S
n is given by the shift map Σ
T
n
(
th
)
:
Σ
T
n
(
th
)
+
→
Σ
T
n
(
th
)
+ defined by
(46)
σ
T
n
(
t
h
)
(
w
)
=
w
′
where
w
i
′
=
w
i
+
1
(
∀
i
≥
0
)
.Definition 25 (see [15]).
The pair(
σ
T
n
(
th
)
,
Σ
T
n
(
th
)
+
) is called thesubshift of finite type (SFT) determined by
T
n
(
th
).The SFT determined byT
n
(
th
) captures the exact detail of all possible itineraries of observations in S
n. Many dynamical properties of an SFT depend on the structure of its associated transition matrix or digraph.Definition 26 (see [16]).
(i) A transition matrixT isirreducible if for each pair (
i
,
j
), there exists ℓ
>
0 so that T
ℓ
>
0. Otherwise, T isreducible. (ii) The digraph G
T
=
(
V
,
E
) associated with T isstrongly connected if it is possible to get from any vertex to any other one by traversing a sequence of edges as directed by T. (iii) If there exists ℓ
′
>
0 such that T
ℓ
>
0 for all ℓ
>
ℓ
′, then T isprimitive. (iv) Let X be any set. A continuous map f
:
X
→
X is(topologically) transitive if, for all nonempty open sets U
,
U
′
⊂
X, there exists ℓ
≥
0 such that f
ℓ
(
U
)
∩
U
′
≠
∅. (v) If there is ℓ
′
> 0 such that f
ℓ
(
U
)
∩
U
′
≠
∅ for all ℓ
≥
ℓ
′, then f is(topologically) mixing.Consider the SFT given by(
σ
T
,
Σ
T
+
). The following results [17, 18] concern dynamical properties of a transitive SFT and algebraic properties of its transition matrix. (i) The shift map Σ
T is transitive (resp., mixing) if and only if T is irreducible (resp., primitive). (ii) T is irreducible if and only if the digraph G
T
=
(
V
,
E
) is strongly connected. In that case we say that V is irreducible. (iii) A nonnegative, irreducible matrix with a positive element on the main diagonal is primitive.Theorem 27.
T
n
(
t
h
) is irreducible.Proof.
We prove by induction onn. From Example 23, T
1
,
i
j
(
th
)
=
1 for all (
i
,
j
) so T
n
(
th
) is irreducible for n
=
1. Assume that T
n
(
th
) is irreducible for n
=
m. Then the digraph G
T
m
(
th
) is strongly connected, and S
m is irreducible. We prove that G
T
n
(
th
) is strongly connected to n
=
m
+
1.
Recall that transitions inS
m
+
1 are under the set 𝒞
m
+
1. From (37), this set can be expressed as 𝒞
m
+
1
=
𝒟
m
(
m
+
1
)
∪
(
𝒞
m
+
1
∖
𝒟
m
(
m
+
1
)
), where 𝒟
m
(
m
+
1
) is the set associated with transitions in the irreducible set ℒ
m
(
c
,
i
j
)
(
=
𝒮
m
), c
∈
{
0,1
} and i
j
∈
I
m
+
1. With the additional transitions under 𝒞
m
+
1
∖
𝒟
m
(
m
+
1
), we show that (i)
the setℒ
m
(
c
)
⊂
𝒮
m
+
1 (given in (33)) is irreducible;
(ii)
there is a path to and from any pairp
,
p
′, where p
∈
ℒ
m
(
0
) and p
′
∈
ℒ
m
(
1
).We prove (i) by showing that there is p
∈
ℒ
m
(
c
,
i
j
) such that, for all i
j
′
≠
i
j, there is a transition from p to p
′
∈
ℒ
m
(
c
,
i
j
′
). By Theorem 16, φ
[
{
0
,
m
}
]
m
+
1
(
p
)
=
p for any p
∈
𝒮
m
+
1. Let
(47)
p
0
=
(
c
c
⋯
c
c
*
,
i
0
i
1
⋯
i
m
-
1
i
m
)
p
1
=
φ
[
{
0
,
m
}
]
(
p
0
)
=
(
c
c
⋯
c
c
*
,
i
1
i
2
⋯
i
m
i
0
)
p
2
=
φ
[
{
0
,
m
}
]
2
(
p
1
)
=
(
c
c
⋯
c
c
*
,
i
2
i
3
⋯
i
0
i
1
)
⋮
p
m
=
φ
[
{
0
,
m
}
]
m
(
p
m
-
1
)
=
(
c
c
⋯
c
c
*
,
i
m
i
0
⋯
i
m
-
2
i
m
-
1
)
.
Observe that all p
j’s have the same fitness states but distinct significance states. In particular, each p
j is contained in a distinct subset ℒ
m
(
c
,
i
j
). This vertical transition allows all p
j
∈
ℒ
m
(
c
,
i
j
) to visit a distinct ℒ
m
(
c
,
i
j
′
) in at most m steps. Hence, ℒ
m
(
c
) is irreducible. To show (ii), take
(48)
p
=
(
00
⋯
0
,
m
(
m
-
1
)
⋯
10
)
∈
ℒ
m
(
0
)
,
p
′
=
(
11
⋯
1,01
⋯
(
m
-
1
)
m
)
∈
ℒ
m
(
1
)
.
Under Δ
=
I
m
+
1, there is a reversible transition p
↔
[
Δ
]
p
′.
From (i), (ii), and the irreducibility ofT
m
(
th
), there is a path between any pair p
,
p
′
∈
𝒮
m
+
1. Hence, G
T
m
+
1
(
th
) is strongly connected and T
m
+
1
(
th
) is irreducible, as desired.Corollary 28.
T
n
(
t
h
) is primitive.Definition 29 (see [15]).
The topological entropy of the shift mapΣ
T
:
Σ
T
+
→
Σ
T
+ is defined by
(49)
h
top
(
σ
T
)
=
lim
⁡
m
→
∞
ln
⁡
(
|
𝒲
(
m
)
|
)
m
,
where 𝒲
(
m
) is the set of allowable sequences of length m
≥
1.A continuous mapf on a compact metric space X is chaotic if h
top
(
f
)
>
0 [17]. The topological entropy for SFTs is given by the following theorem [15].Theorem 30.
LetT be a transition matrix and let Σ
T
:
Σ
T
+
→
Σ
T
+ be the associated SFT. Then
(50)
h
t
o
p
(
σ
T
)
=
ln
⁡
(
λ
max
⁡
)
,
where λ
max
⁡ is the maximum eigenvalue of T.Since(
σ
T
n
(
th
)
,
Σ
T
n
(
th
)
+
) is the associated subshift of the multivalued map F in (39), one has
(51)
h
top
(
F
)
=
h
top
(
σ
T
n
(
th
)
)
=
ln
⁡
(
2
n
)
=
n
ln
⁡
2
>
0
.


## 4. Dynamics from Data

Given multivariable longitudinal binary data of dimensionn, every observed orbit is an orbit of an SFT determined by T
n
(
th
). Dynamics of real-world longitudinal data however is not often defined by an SFT. Data usually selects certain paths given by T
n
(
th
) and may sometimes stay in a particular subspace of S
n.


### 4.1. Nonstationary SFT

Letp
,
p
′
∈
S
n, ψ
(
p
)
=
i, and ψ
(
p
′
)
=
j. For each t
≥
0, denote by T
n
,
t
(
data
) the matrix that records observed transitions that occur in the longitudinal data from time t to t
+
1, where
(52)
(
T
n
,
t
(
data
)
)
i
j
=
{
1
if
∃
k
∈
K
,
such
that
p
→
[
Δ
t
k
]
p
′
0
otherwise
.
We note that T
n
,
t
(
data
) is defined over an interval of time and that T
n
,
t
(
data
) may vary with time. Some allowable transitions between states given by T
n
(
th
) might not occur in the observed data so T
n
,
t
(
data
)
≠
T
n
(
th
). For the case where T
n
,
t
(
data
) is constant we write T
n
,
t
(
data
)
≡
T
n
(
data
) and we can define the SFT given by the pair (
σ
T
n
(
data
)
,
Σ
T
n
(
data
)
+
).IfT
n
,
t
(
data
) is not constant, then we have a sequence of matrices
(53)
A
=
{
T
n
,
t
(
data
)
}
t
≥
0
.
Given A, define
(54)
Σ
A
(
data
)
=
{
(
T
n
,
t
(
data
)
)
w
t
w
t
+
1
w
=
(
w
0
w
1
⋯
w
𝒯
⋯
)
∈
Σ
T
n
(
t
h
)
+
b
b
j
:
(
T
n
,
t
(
data
)
)
w
t
w
t
+
1
=
1
}
.
We call Σ
A
(
data
) thenonstationary symbolic space restricted by the sequence of matrices
A. The shift takes place as usual in Σ
A
(
data
) and is denoted by σ
A
:
Σ
A
(
data
)
→
Σ
A
(
data
). We refer the reader to [19–21] for a discussion on nonstationary SFT.Definition 31.
The pair(
σ
A
,
Σ
A
(
data
)
) is the nonstationary SFT (NSFT) determined by the sequence of matrices A.Visualization of dynamics of longitudinal data defined by an NSFT is illustrated by a sequence of directed graphs called a Bratteli diagram [9, 10]. Equivalently we may plot orbits of observations in S
n over time. Figure 6(a) illustrates the orbits of a population in a subset of S
3 over time.(a) Orbits of Agincourt households inS
3, over time. (b) Number of orbit visits to each state in S
3, where majority are in ℒ
~, having the favourable property of being headed by an adult (i.e., Q
1
=
1).
(a)
(b)


### 4.2. Population Dynamics

We discuss the longitudinal data of a population ofK
>
1 observations. Aside from all possible paths that an observation can take in S
n, we are also interested in the number of observations on paths. Because it is possible for more than one observation to occupy a state in S
n at a given time, we can consider the number of observations that follow the same transition in S
n.In general, given a transition matrixT (e.g., those in (40) or (52)), standard analysis is to accumulate number of transitions between states, and from this construct the associated stochastic transition matrix. Construction of associated transition and stochastic matrices on states of S
n can then follow as usual. In what follows, let V denote a finite index set with |
V
|
≥
2 and i
,
j
∈
V.Definition 32.
LetH
t
(
i
) be the number of observations in state i at time t. The density matrix at time t is defined by
(55)
D
t
=
(
d
t
,
i
j
)
,
where d
t
,
i
j is the number of observations in state i at time t that go to state j at time t
+
1. The (net)flux of state i at time t is defined by
(56)
F
t
(
i
)
=
H
t
+
1
(
i
)
-
H
t
(
i
)
.
Let ℓ
t
,
i
j
=
d
t
,
j
i
-
d
t
,
i
j, i
,
j
∈
V. The flux in (56) can also be expressed as
(57)
F
t
(
i
)
=
∑
j
ℓ
t
,
i
j
.
From (56) and (57), we have
(58)
H
t
(
i
)
=
∑
j
d
t
,
i
j
,
H
t
+
1
(
i
)
=
∑
j
d
t
,
j
i
.
Let H
t be a row vector whose ith entry is H
t
(
i
). We will refer to H
t as theobserved capacity vector at time t. Given the initial observed capacity vector H
0, there are two methods that we can use to determine H
t
+
1.(i) Nonhomogenous Case. From data, we are encouraged to construct a probability matrix based directly on the density matrices D
t. For each D
t, we construct a time-dependent probability matrix P
t
=
(
P
t
,
i
j
), where
(59)
P
t
,
i
j
=
{
d
t
,
i
j
∑
j
d
t
,
i
j
,
if
∑
j
‍
d
t
,
i
j
≠
0
0
otherwise
.
The capacity vector H
t
+
1 is given by the product
(60)
H
t
+
1
=
H
t
·
P
t
=
H
0
·
∏
m
=
0
t
‍
P
m
.
We show that the capacity H
t
+
1
(
i
) in (56) agrees with the ith entry of (60). It is trivial if ∑
j
‍
d
t
,
i
j
=
0 for all t. Otherwise we have
(61)
H
t
+
1
(
i
)
=
H
t
(
i
)
+
F
t
(
i
)
=
H
t
(
i
)
+
∑
j
‍
(
d
t
,
j
i
-
d
t
,
i
j
)
=
H
t
(
i
)
+
∑
j
‍
(
P
t
,
j
i
∑
i
‍
d
t
,
j
i
-
P
t
,
i
j
∑
j
‍
d
t
,
i
j
)
=
H
t
(
i
)
+
∑
j
‍
P
t
,
j
i
H
t
(
j
)
-
∑
j
‍
P
t
,
i
j
H
t
(
i
)
=
H
t
(
i
)
(
1
-
∑
j
‍
P
t
,
i
j
)
+
∑
j
‍
P
t
,
j
i
H
t
(
j
)
=
∑
j
‍
P
t
,
j
i
H
t
(
j
)
.(ii) Homogenous Case. Let D
t be the density matrix at time t and let
(62)
D
(
data
)
=
∑
t
=
0
T
-
1
D
t
be the accumulated density matrix of the data over the observation period. Define the mean density matrix by
(63)
D
=
1
T
D
(
data
)
.
Suppose D is irreducible. Define the mean probability matrix from D by P
D
=
(
P
D
,
i
j
), where
(64)
P
D
,
i
j
=
D
i
j
∑
j
D
i
j
.
The capacity vector at t
+
1 is given by the product
(65)
H
t
+
1
=
H
t
·
(
P
D
)
=
H
0
·
(
P
D
)
t
.
The probabilities in (64) are based on observed frequencies at which one state follows another. The transition probabilities of a Markov model using states in S
n are then estimated. Markov modeling on the bigger space S
n may give better results in generating data as S
n is a more detailed space than a space where order of variable is fixed (e.g., M
n).Definition 33.
A set of observationsK
′ cluster in S
n if for all k
∈
K
′, there exists R
⊂
S
n and time interval I such that the orbit O
(
k
)
⊂
R for all t
∈
I.An advantage in using the bigger spaceS
n to analyse multivariable longitudinal data is that all possible clusterings in stable variables are revealed. We can also visualize in S
n the shift in clusters from one region R to another region R
′. There is much interesting information that we can gather from a cluster in a subset R of S
n. Clustering in ℒ
m (m
≠
n) defined in (32) involves observations with the same stable n
-
m variables. Subsets ℒ
m are important as they automatically determine which variables are trivial in the analysis. We either reduce analysis from S
n to ℒ
m or replace the n
-
m constant variables with other nonconstant variables that are of interest to purpose. In addition, orbits that spend most (if not all) of the time in L
m have strong correlation among the n
-
m variables.


## 4.1. Nonstationary SFT

Letp
,
p
′
∈
S
n, ψ
(
p
)
=
i, and ψ
(
p
′
)
=
j. For each t
≥
0, denote by T
n
,
t
(
data
) the matrix that records observed transitions that occur in the longitudinal data from time t to t
+
1, where
(52)
(
T
n
,
t
(
data
)
)
i
j
=
{
1
if
∃
k
∈
K
,
such
that
p
→
[
Δ
t
k
]
p
′
0
otherwise
.
We note that T
n
,
t
(
data
) is defined over an interval of time and that T
n
,
t
(
data
) may vary with time. Some allowable transitions between states given by T
n
(
th
) might not occur in the observed data so T
n
,
t
(
data
)
≠
T
n
(
th
). For the case where T
n
,
t
(
data
) is constant we write T
n
,
t
(
data
)
≡
T
n
(
data
) and we can define the SFT given by the pair (
σ
T
n
(
data
)
,
Σ
T
n
(
data
)
+
).IfT
n
,
t
(
data
) is not constant, then we have a sequence of matrices
(53)
A
=
{
T
n
,
t
(
data
)
}
t
≥
0
.
Given A, define
(54)
Σ
A
(
data
)
=
{
(
T
n
,
t
(
data
)
)
w
t
w
t
+
1
w
=
(
w
0
w
1
⋯
w
𝒯
⋯
)
∈
Σ
T
n
(
t
h
)
+
b
b
j
:
(
T
n
,
t
(
data
)
)
w
t
w
t
+
1
=
1
}
.
We call Σ
A
(
data
) thenonstationary symbolic space restricted by the sequence of matrices
A. The shift takes place as usual in Σ
A
(
data
) and is denoted by σ
A
:
Σ
A
(
data
)
→
Σ
A
(
data
). We refer the reader to [19–21] for a discussion on nonstationary SFT.Definition 31.
The pair(
σ
A
,
Σ
A
(
data
)
) is the nonstationary SFT (NSFT) determined by the sequence of matrices A.Visualization of dynamics of longitudinal data defined by an NSFT is illustrated by a sequence of directed graphs called a Bratteli diagram [9, 10]. Equivalently we may plot orbits of observations in S
n over time. Figure 6(a) illustrates the orbits of a population in a subset of S
3 over time.(a) Orbits of Agincourt households inS
3, over time. (b) Number of orbit visits to each state in S
3, where majority are in ℒ
~, having the favourable property of being headed by an adult (i.e., Q
1
=
1).
(a)
(b)


## 4.2. Population Dynamics

We discuss the longitudinal data of a population ofK
>
1 observations. Aside from all possible paths that an observation can take in S
n, we are also interested in the number of observations on paths. Because it is possible for more than one observation to occupy a state in S
n at a given time, we can consider the number of observations that follow the same transition in S
n.In general, given a transition matrixT (e.g., those in (40) or (52)), standard analysis is to accumulate number of transitions between states, and from this construct the associated stochastic transition matrix. Construction of associated transition and stochastic matrices on states of S
n can then follow as usual. In what follows, let V denote a finite index set with |
V
|
≥
2 and i
,
j
∈
V.Definition 32.
LetH
t
(
i
) be the number of observations in state i at time t. The density matrix at time t is defined by
(55)
D
t
=
(
d
t
,
i
j
)
,
where d
t
,
i
j is the number of observations in state i at time t that go to state j at time t
+
1. The (net)flux of state i at time t is defined by
(56)
F
t
(
i
)
=
H
t
+
1
(
i
)
-
H
t
(
i
)
.
Let ℓ
t
,
i
j
=
d
t
,
j
i
-
d
t
,
i
j, i
,
j
∈
V. The flux in (56) can also be expressed as
(57)
F
t
(
i
)
=
∑
j
ℓ
t
,
i
j
.
From (56) and (57), we have
(58)
H
t
(
i
)
=
∑
j
d
t
,
i
j
,
H
t
+
1
(
i
)
=
∑
j
d
t
,
j
i
.
Let H
t be a row vector whose ith entry is H
t
(
i
). We will refer to H
t as theobserved capacity vector at time t. Given the initial observed capacity vector H
0, there are two methods that we can use to determine H
t
+
1.(i) Nonhomogenous Case. From data, we are encouraged to construct a probability matrix based directly on the density matrices D
t. For each D
t, we construct a time-dependent probability matrix P
t
=
(
P
t
,
i
j
), where
(59)
P
t
,
i
j
=
{
d
t
,
i
j
∑
j
d
t
,
i
j
,
if
∑
j
‍
d
t
,
i
j
≠
0
0
otherwise
.
The capacity vector H
t
+
1 is given by the product
(60)
H
t
+
1
=
H
t
·
P
t
=
H
0
·
∏
m
=
0
t
‍
P
m
.
We show that the capacity H
t
+
1
(
i
) in (56) agrees with the ith entry of (60). It is trivial if ∑
j
‍
d
t
,
i
j
=
0 for all t. Otherwise we have
(61)
H
t
+
1
(
i
)
=
H
t
(
i
)
+
F
t
(
i
)
=
H
t
(
i
)
+
∑
j
‍
(
d
t
,
j
i
-
d
t
,
i
j
)
=
H
t
(
i
)
+
∑
j
‍
(
P
t
,
j
i
∑
i
‍
d
t
,
j
i
-
P
t
,
i
j
∑
j
‍
d
t
,
i
j
)
=
H
t
(
i
)
+
∑
j
‍
P
t
,
j
i
H
t
(
j
)
-
∑
j
‍
P
t
,
i
j
H
t
(
i
)
=
H
t
(
i
)
(
1
-
∑
j
‍
P
t
,
i
j
)
+
∑
j
‍
P
t
,
j
i
H
t
(
j
)
=
∑
j
‍
P
t
,
j
i
H
t
(
j
)
.(ii) Homogenous Case. Let D
t be the density matrix at time t and let
(62)
D
(
data
)
=
∑
t
=
0
T
-
1
D
t
be the accumulated density matrix of the data over the observation period. Define the mean density matrix by
(63)
D
=
1
T
D
(
data
)
.
Suppose D is irreducible. Define the mean probability matrix from D by P
D
=
(
P
D
,
i
j
), where
(64)
P
D
,
i
j
=
D
i
j
∑
j
D
i
j
.
The capacity vector at t
+
1 is given by the product
(65)
H
t
+
1
=
H
t
·
(
P
D
)
=
H
0
·
(
P
D
)
t
.
The probabilities in (64) are based on observed frequencies at which one state follows another. The transition probabilities of a Markov model using states in S
n are then estimated. Markov modeling on the bigger space S
n may give better results in generating data as S
n is a more detailed space than a space where order of variable is fixed (e.g., M
n).Definition 33.
A set of observationsK
′ cluster in S
n if for all k
∈
K
′, there exists R
⊂
S
n and time interval I such that the orbit O
(
k
)
⊂
R for all t
∈
I.An advantage in using the bigger spaceS
n to analyse multivariable longitudinal data is that all possible clusterings in stable variables are revealed. We can also visualize in S
n the shift in clusters from one region R to another region R
′. There is much interesting information that we can gather from a cluster in a subset R of S
n. Clustering in ℒ
m (m
≠
n) defined in (32) involves observations with the same stable n
-
m variables. Subsets ℒ
m are important as they automatically determine which variables are trivial in the analysis. We either reduce analysis from S
n to ℒ
m or replace the n
-
m constant variables with other nonconstant variables that are of interest to purpose. In addition, orbits that spend most (if not all) of the time in L
m have strong correlation among the n
-
m variables.


## 5. Application

We illustrate our method using then
=
3 demographic binary questions taken from Agincourt [22]. With purpose𝒫: to investigate the effect of household change on a child’s educational progress, the three questions in (4) are associated with household change and are all of interest to 𝒫. Table 5 gives the question (indexed by 0, 1, and 2, resp.) with associated variables (BM, HH, and AD), favourable/unfavourable coding, and frequency of change.Table 5
Agincourt questionnaire with corresponding coded answer and frequency of answer change in the population.
Question
Answer
Code
f
i
P
Q
0: Is the biological mother present (>6 months)? (BM)
Yes
1
760
Q
1: Is the household headed by a minor? (HH)
Yes
0
4
Q
2: Is there an adult death in the household? (AD)
Yes
0
119The longitudinal data consists ofK
=
188 observations with no missing data. Each household is surveyed each year, for nine years [22], and the population is constant throughout the observation period; that is, no households enter nor leave the population. Using (20), the frequency relation in the population P is f
1
P
≪
f
2
P
≪
f
0
P. The number of unfavourable = 0 answers (over 1692 answers) is Q
0
:
844
(
49.9
%
), Q
1
:
2
(
0.1
%
), and Q
2
:
71
(
4.2
%
). We expect that this, together with the very small value of f
1
P, will give clustering in the question order 1*
*, where Q
1
=
1.Figure6(a) illustrates orbits of households in S
3 over time, where a red edge denotes a transition to the left, a green edge is a transition to the right, and a blue edge is a self-transition. Most of the orbits are in
(66)
ℒ
~
=
{
21,22,23,24,29,30,31,32
}
=
{
(
x
,
y
)
:
x
=
1
*
*
,
y
=
1
*
*
}
and are characterised by the favourable property of being headed by an adult (i.e., Q
1
=
1). Of the 179 households that stay in ℒ
~, 159 distinct orbits are found. However, we see from Figure 6(b) that the number of state visits in S
3 is dominant in states 23
=
(
110
,
120
) and 24
=
(
111
,
120
) so dynamics principally takes place in these two states.State 23 is characterised by the most stable variable HH = 1, followed by AD = 1 and then by BM = 0. Similarly, state 24 is characterised by the most stable variable HH = 1, followed by AD = 1 and then by BM = 1. Orbits that stay strictly in states 23 and 24 are associated with households headed by an adult and with no adult death during the survey period. Those idle in 23 (resp., 24) have mothers away (resp., present) during the survey period. Households that transition from 23 to 24 are characterized by mother out- to in-migration, while transitions from 24 to 23 denote in- to out-migration.With regard to purpose, we define educational default by the number of years that a child has failed in his school life. A household with a child who has failed 4 years or more in his school life is classified as defaulting. Orbits in Figure6(a) are then split into defaulting and nondefaulting populations, as shown in Figures 7(a) and 7(b), respectively.Orbits inS
3 of (a) defaulting and (b) nondefaulting households.
(a)
(b)Note that if we include a fourth question(67)
Q
3
:
is
the
household
defaulting
(=0)
or
nondefaulting
(=1)?
then orbits will be plotted in S
4, as seen in Figure 5. Orbits that cluster on the left are the defaulting households (Q
4
=
0), while those on the right are nondefaulting (Q
4
=
1).Figure8 gives H
t
(
i
), the number of observed defaulting and nondefaulting households at each state i
=
23,24 from t
=
1999 to t
=
2007. It is clear that, for both subpopulations, there is an exchange of numbers between states 23 (BM-out) and 24 (BM-in). Figures 9(a) and 9(b) illustrate the orbits of defaulting and nondefaulting households over time, and the accumulated number of visits in S
4, respectively.Figure 8
Number of defaulting and nondefaulting households in states 23 and 24 fort
=
1999 to t
=
2007.(a) Orbits of defaulting (left cluster) and nondefaulting (right cluster) households inS
4, over time. (b) Accumulated number of visits to each state of S
4.
(a)
(b)Table6 gives the (accumulated) number of state transitions i
→
j, i
,
j
∈
{
23,24
} for both subpopulations. Using odds ratio, we have the value 1.59 (>1) for transition 23
→
23, which means that once mothers from defaulting households out-migrate, they take long to return to their households. Out-migrating mothers are then more likely to be away from home in the defaulting households. For 24
→
24 we have odds ratio of 0.70 (<1) indicating that in-migrating mothers are more likely to stay at home in the nondefaulting households. Both odds ratios have significant P values and these support our hypothesis that mother out-migration places child educational progression at risk. We have no strong evidence that mother out-migration (24
→
23) is more likely in the defaulting households nor that in-migration (23
→
24) is more likely in nondefaulting households.Table 6
Total transitions from statei
→
j
(
i
,
j
=
23,24
) in defaulting and nondefaulting household, together with the odds ratios (95% confidence interval and P value).
State
i
→
j
Defaulting
Nondefaulting
Odds ratio (95% CI;P value)
2
3
→
23
256
63
1.59 (1.17  to  2.17;P = 0.0031)
23
→
24
238
84
1.00 (0.75  to   1.33;P = 0.9925)
24
→
23
260
96
0.94 (0.72  to  1.24;P = 0.6653)
24
→
24
214
99
0.70 (0.53  to   0.92;P = 0.0110)


## 6. Concluding Remarks

By defining a specific swap operation on multivariable longitudinal data, we have imposed dynamics on data. No approximations are made and orbit of an observation in the 2-dimensional spaceS
n can be decoded back to the observation’s original data. By including the order of variables in our analysis, we have introduced new states, along with an objective criterion of relative fitness and identification of subshift. We have only considered binary-valued variables but the method may be generalized to m-ary valued multivariable longitudinal data. Although the method of orbits is nonstatistical, it can however be used to aid in statistical analysis.The advantage in plotting orbits inS
n is that information of change can be extracted directly from the visualized orbits. In addition, orbits in S
n facilitate the identification of cause and effect, of preceding events that might uniquely and usefully associate with change. Frequency of change is a property taken directly from the data so that, under our reordering operation, any practitioner with the same favourable and unfavourable coding will find the same ordering of questions and arrive at the same analysis of longitudinal data.



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*Source: 901838-2014-03-19.xml*