Stationary probability vectors of higher-order two-dimensional transition probability tensors

TL;DR

This paper characterizes stationary probability vectors of symmetric transition probability tensors in two dimensions, revealing unique or multiple solutions depending on tensor structure, with implications for understanding higher-order Markov processes.

Contribution

It identifies conditions under which symmetric transition probability tensors have unique or multiple stationary vectors, extending the theory of Markov chains to higher-order tensors.

Findings

Two special tensors have exactly two stationary vectors.

Most symmetric tensors have a unique stationary vector.

Irreducible tensors also have a unique stationary vector.

Abstract

In this paper we investigate stationary probability vectors of higher-order two-dimensional symmetric transition probability tensors. We show that there are two special symmetric transition probability tensors of order $m$ dimension 2, which have and only have two stationary probability vectors; and any other symmetric transition probability tensor of order $m$ dimension 2 has a unique stationary probability vector. As a byproduct, we obtain that any symmetric transition probability tensor of order $m$ dimension 2 has a unique positive stationary probability vector; and that any symmetric irreducible transition probability tensor of order $m$ dimension 2 has a unique stationary probability vector.

Equations

$$0\le p_{i_1{i}_2\cdots i_m}=\text{mbox}\mathrm{Prob}(X_{t+1}=i_1|X_t=i_2,\dots ,X_{t-m+2}=i_m)\le 1$$

$$\sum _{i_1=1}^n{p}_{i_1{i}_2\cdots i_m}=1.$$

$$x^{\ast }\in \left\{x\in {\mathbb{R}}^n:x_i\ge 0\text{mbox}\mathrm{forall}i\in \{1,2,\dots ,n\}\text{mbox}\mathrm{and}\sum _{i=1}^n{x}_i=1\right\}$$

$$\sum _{i_2,\dots ,i_m=1}^n{p}_{i{i}_2\cdots i_m}{x}_{i_2}^{\ast }\cdots x_{i_m}^{\ast }=x_i^{\ast }$$

$$f_1(z_1,z_2):=\sum _{i_2,\dots ,i_m=1}^2{p}_{1i_2\cdots i_m}{z}_{i_2}\cdots z_{i_m}=z_1,$$

$$f_2(z_1,z_2):=\sum _{i_2,\cdots ,i_m=1}^2{p}_{2i_2\cdots i_m}{z}_{i_2}\cdots z_{i_m}=z_2.$$

$$\Delta :=\left\{(x,y{)}^{\top }\in {\mathbb{R}}^2:x\ge 0,y\ge 0,x+y=1\right\}.$$

$$f_1(x,y)=x\text{mbox}\mathrm{and}{f}_2(x,y)=y.$$

$$\displaystyle\begin{array}[]{rcl}f_{1}(x,y)&=&ax^{m-1}+bC_{m-1}^{1}x^{m-2}y+aC_{m-1}^{2}x^{m-3}y^{2}+bC_{m-1}^{3}x^{m-4}y^{3}+\cdots+\\ &&ax^{3}y^{m-4}+bC_{m-1}^{m-3}x^{2}y^{m-3}+aC_{m-1}^{m-2}xy^{m-2}+by^{m-1},\vspace{2mm}\\ f_{2}(x,y)&=&bx^{m-1}+aC_{m-1}^{1}x^{m-2}y+bC_{m-1}^{2}x^{m-3}y^{2}+aC_{m-1}^{3}x^{m-4}y^{3}+\cdots+\\ &&bx^{3}y^{m-4}+aC_{m-1}^{m-3}x^{2}y^{m-3}+bC_{m-1}^{m-2}xy^{m-2}+ay^{m-1}.\end{array}$$

$$\displaystyle\begin{array}[]{rcl}f_{1}(x,y)&=&ax^{m-1}+bC_{m-1}^{1}x^{m-2}y+aC_{m-1}^{2}x^{m-3}y^{2}+bC_{m-1}^{3}x^{m-4}y^{3}+\cdots+\\ &&ax^{4}y^{m-5}+bx^{3}y^{m-4}+aC_{m-1}^{m-3}x^{2}y^{m-3}+bC_{m-1}^{m-2}xy^{m-2}+ay^{m-1},\vspace{2mm}\\ f_{2}(x,y)&=&bx^{m-1}+aC_{m-1}^{1}x^{m-2}y+bC_{m-1}^{2}x^{m-3}y^{2}+aC_{m-1}^{3}x^{m-4}y^{3}+\cdots+\\ &&bx^{4}y^{m-5}+ax^{3}y^{m-4}+bC_{m-1}^{m-3}x^{2}y^{m-3}+aC_{m-1}^{m-2}xy^{m-2}+by^{m-1}.\end{array}$$

$$\displaystyle\begin{array}[]{rcl}f_{1}(x,y)&=&p_{1\cdots 11}x^{m-1}+p_{1\cdots 12}C_{m-1}^{1}x^{m-2}y+p_{1\cdots 122}C_{m-1}^{2}x^{m-3}y^{2}\\ &&+p_{1\cdots 1222}C_{m-1}^{3}x^{m-4}y^{3}+\cdots+p_{11112\cdots 2}x^{3}y^{m-4}\\ &&+p_{1112\cdots 2}C_{m-1}^{m-3}x^{2}y^{m-3}+p_{112\cdots 2}C_{m-1}^{m-2}xy^{m-2}+p_{12\cdots 2}y^{m-1},\vspace{2mm}\\ f_{2}(x,y)&=&p_{21\cdots 11}x^{m-1}+p_{21\cdots 12}C_{m-1}^{1}x^{m-2}y+p_{21\cdots 122}C_{m-1}^{2}x^{m-3}y^{2}\\ &&+p_{21\cdots 1222}C_{m-1}^{3}x^{m-4}y^{3}+\cdots+p_{21112\cdots 2}x^{3}y^{m-4}\\ &&+p_{2112\cdots 2}C_{m-1}^{m-3}x^{2}y^{m-3}+p_{212\cdots 2}C_{m-1}^{m-2}xy^{m-2}+p_{22\cdots 2}y^{m-1}=y.\end{array}$$

$$\displaystyle\begin{array}[]{l}p_{1\cdots 11}+p_{1\cdots 12}=1,\;p_{1\cdots 12}+p_{1\cdots 122}=1,\;p_{1\cdots 122}+p_{1\cdots 1222}=1,\;\ldots,\\ \qquad p_{11112\cdots 2}+p_{1112\cdots 2}=1,\;p_{1112\cdots 2}+p_{112\cdots 2}=1,\;p_{112\cdots 2}+p_{12\cdots 2}=1,\vspace{2mm}\\ p_{21\cdots 11}+p_{21\cdots 12}=1,\;p_{21\cdots 12}+p_{21\cdots 122}=1,\;p_{21\cdots 122}+p_{21\cdots 1222}=1,\;\ldots,\\ \qquad p_{21112\cdots 2}+p_{2112\cdots 2}=1,\;p_{2112\cdots 2}+p_{212\cdots 2}=1,\;p_{212\cdots 2}+p_{22\cdots 2}=1,\end{array}$$

$$\displaystyle\begin{array}[]{l}p_{1\cdots 11}=p_{1\cdots 122}=\cdots=p_{11112\cdots 2}=p_{112\cdots 2}=a,\vspace{2mm}\\ p_{1\cdots 12}=p_{1\cdots 1222}=\ldots=p_{1112\cdots 2}=p_{12\cdots 2}=b,\vspace{2mm}\\ p_{21\cdots 11}=p_{21\cdots 122}=\cdots=p_{21112\cdots 2}=p_{212\cdots 2}=b,\vspace{2mm}\\ p_{21\cdots 12}=p_{21\cdots 1222}=\ldots=p_{2112\cdots 2}=p_{22\cdots 2}=a.\end{array}$$

$$g_1(x):=f_1(x,1-x)\text{mbox}\mathrm{and}{g}_2(y):=f_2(1-y,y).$$

$$C_n^0+C_n^2+\cdots +C_n^n=C_n^1+C_n^3+\cdots +C_n^{n-1}=2^{n-1};$$

$$C_n^0+C_n^2+\cdots +C_n^{n-1}=C_n^1+C_n^3+\cdots +C_n^n=2^{n-1}.$$

$$g_1(x)=\frac{a-b}{2}(2x-1{)}^{m-1}+b.$$

$$g_1(x)=\frac{a-b}{2}(2x-1{)}^{m-1}+a.$$

$$g_1(x)=(a-b)(m-1)\sum _{t\in \{1,3,\dots ,m-2\}}\left\{\frac{2^{m-t-1}}{m-t}{C}_{m-2}^{t-1}{x}^{m-t}-\frac{2^{m-t-2}}{m-t-1}{C}_{m-2}^t{x}^{m-t-1}\right\}+b.$$

$$\displaystyle\begin{array}[]{rcl}g_{1}(x)&=&ax^{m-1}+bC_{m-1}^{1}x^{m-2}(1-x)+aC_{m-1}^{2}x^{m-3}(1-x)^{2}\\ &&+bC_{m-1}^{3}x^{m-4}(1-x)^{3}+\cdots+ax^{3}(1-x)^{m-4}\\ &&+bC_{m-1}^{m-3}x^{2}(1-x)^{m-3}+aC_{m-1}^{m-2}x(1-x)^{m-2}+b(1-x)^{m-1}.\end{array}$$

$$g_1(x)=\sum _{t=1}^m{\alpha }_{m-t}{x}^{m-t},$$

$${\alpha }_{m-t}=\sum _{s\in \{0,2,\dots ,m-t-1\}}\left\{C_{m-1}^{t-1+s}{C}_{t-1+s}^{s}a-C_{m-1}^{t+s}{C}_{t+s}^{s+1}b\right\};$$

$${\alpha }_{m-t}=\sum _{s\in \{2,4,\dots ,m-t-2\}}\left\{C_{m-1}^{t+s}{C}_{t+s}^{s}b-C_{m-1}^{t+s+1}{C}_{t+s+1}^{s+1}a\right\}+C_{m-1}^{m-1}{C}_{m-1}^{m-t-1}b.$$

$$C_{m-1}^{t-1+s}{C}_{t-1+s}^s$$

$${\alpha }_{m-t}$$

$${\alpha }_{m-t}=2^{m-t-1}(b-a)\frac{m-1}{m-t}{C}_{m-2}^{t-1}.$$

$$\frac{m-1}{m-t}{C}_{m-2}^{t-1}=C_{m-1}^{t-1},$$

$$g_1(x)$$

$$\displaystyle\begin{array}[]{rcl}g_{1}(x)&:=&ax^{m-1}+bC_{m-1}^{1}x^{m-2}(1-x)+aC_{m-1}^{2}x^{m-3}(1-x)^{2}\\ &&+bC_{m-1}^{3}x^{m-4}(1-x)^{3}+\cdots+ax^{4}(1-x)^{m-5}+bx^{3}(1-x)^{m-4}\\ &&+aC_{m-1}^{m-3}x^{2}(1-x)^{m-3}+bC_{m-1}^{m-2}x(1-x)^{m-2}+a(1-x)^{m-1}.\end{array}$$

$${\beta }_{m-t}=\sum _{s\in \{0,2,\dots ,m-t-1\}}\left\{C_{m-1}^{t-1+s}{C}_{t-1+s}^{s}a-C_{m-1}^{t+s}{C}_{t+s}^{s+1}b\right\}+C_{m-1}^{m-1}{C}_{m-1}^{m-t-1}a;$$

Full text

Key words:   Transition probability tensor; higher-order Markov chain; stationary probability vector; eigenvalue of tensor.

Mathematics Subject Classifications(2000):   15A18; 15A69; 65F15; 60J10; 60J22.

1 Introduction

It is well known that higher-order Markov chains have various applications in many areas [1, 2, 3, 4, 5, 6]. An $(m-1)$-order $n$-dimensional Markov chain is basically characterized by its associated nonnegative tensor $\mathscr{P}$ which is an $m$-order $n$-dimensional tensor with entries $p_{i_{1}i_{2}\cdots i_{m}}\geq 0$ for all $i_{j}\in\{1,2,\ldots,n\}$ and $j\in\{1,2,\ldots,m\}$ satisfying

[TABLE]

where $\{X_{t}:t=0,1,\ldots\}$ represents the stachastic process that takes on $n$ states $\{1,2,\ldots,n\}$, and for any $i_{2},\ldots,i_{m}\in\{1,2,\ldots,n\}$,

[TABLE]

We will use $\mathbb{T}_{m,n}$ to denote the set of all $m$-order $n$-dimensional real tensors. For any $\mathscr{P}=(p_{i_{1}\cdots i_{m}})\in\mathbb{T}_{m,n}$, if the entries $p_{i_{1}i_{2}\ldots i_{m}}$ are invariant under any permutation of their indices, then $\mathscr{P}$ is called a symmetric tensor. A tensor $\mathscr{P}\in\mathbb{T}_{m,n}$ is called a transition probability tensor if it satisfies (1.1) and (1.2). A vector

[TABLE]

is called a stationary probability vector of $\mathscr{P}\in\mathbb{T}_{m,n}$ if

[TABLE]

holds for all $i\in\{1,2,\ldots,n\}$, which is just a $Z_{1}$-eigenvector associated with the $Z_{1}$-eigenvalue 1 [7]. It is also closely related to the $Z$-eigenvector of a tensor [8, 9].

Transition probability tensors and the associated stationary probability vectors are important issues in studies of higher-order Markov chains [10, 7, 11, 12, 13, 14, 15]. In particular, the uniqueness of the stationary probability vector of transition probability tensors has attracted a lot of interest. Li and Ng [12] proposed some conditions which ensure the uniqueness of the stationary probability vector of transition probability tensors and the linear convergence of the proposed iterative method; Hu and Qi [11] studied the uniqueness of the stationary probability vector of the third order $n$-dimensional positive transition probability tensor, and they proved that an irreducible transition probability tensor of order 3 dimension 2 has a unique stationary probability vector; and Chang and Zhang [7] investigated sufficient conditions for transition probability tensors to ensure the uniqueness of the stationary probability vector by using three different methods: contraction mappings, monotone operators, and the Brouwer index of fixed points.

More recently, Culp, Pearson and Zhang [15] investigated symmetric irreducible transition probability tensors of order 4 dimension 2 and of order 3 dimension 3, and showed that a symmetric irreducible transition probability tensor in these orders and dimensions has a unique stationary probability vector.

In this paper, we give a full characterization on the stationary probability vectors of $m$-order $2$-dimensional symmetric transition probability tensors. We show that there are two special symmetric transition probability tensors of order $m$ dimension 2, which have and only have two stationary probability vectors, where one is $(\frac{1}{2},\frac{1}{2})^{\top}$, and the other one is $(1,0)^{\top}$ or $(0,1)^{\top}$. In particular, we show that if the concerned symmetric transition probability tensor of order $m$ dimension 2 is not one of the above two tensors, then it has a unique stationary probability vector, which is $(\frac{1}{2},\frac{1}{2})^{\top}$. As a byproduct, we obtain that any symmetric irreducible transition probability tensor of order $m$ dimension 2 has a unique stationary probability vector. When $m=4$, such a result was obtained by Culp, Pearson and Zhang in [15].

Throughout this paper, we assume that $m\geq 3$ is an integer number.

2 Main results

Let $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ be a transition probability tensor. Then $z=(z_{1},z_{2})^{\top}\in\mathbb{R}^{2}$ is a stationary probability vector of $\mathscr{P}$ if and only if $z_{1},z_{2}\geq 0$, $z_{1}+z_{2}=1$, and

[TABLE]

[TABLE]

In the following, we denote $x:=z_{1}$ and $y:=z_{2}$, and

[TABLE]

Then, $(x,y)^{\top}\in\mathbb{R}^{2}$ is a stationary probability vector of $\mathscr{P}$ if and only if $(x,y)^{\top}\in\Delta$ and

[TABLE]

Throughout this paper, we denote $a:=p_{1\cdots 11}$ and $b:=p_{21\cdots 11}$.

Lemma 2.1

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric transition probability tensor. Let $f_{1}(\cdot,\cdot)$, $f_{2}(\cdot,\cdot)$, and $\Delta$ be defined by (2.1), (2.2), and (2.3), respectively. Then, for any $(x,y)^{\top}\in\Delta$, we have the following results.

• (i)

If $m$ is an even number, then

[TABLE]

• (ii)

If $m$ is an odd number, then

[TABLE]

Proof. Since $\mathscr{P}$ is a symmetric tensor, the above equalities can be rewritten as

[TABLE]

Suppose that $m$ is an even number. Since $\mathscr{P}$ is a transition probability tensor, it follows that

[TABLE]

and hence,

[TABLE]

Thus, two equalities given in (i) hold from (2.13).

Suppose that $m$ is an odd number. Then two equalities given in (ii) can be showed similarly. We omit them here. $\Box$

Denote

[TABLE]

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric transition probability tensor. Then, it is obvious that (2.4) has a (unique) solution $(x^{*},y^{*})^{\top}\in\Delta$ if and only if $g_{1}(x)-x=0$ has a (unique) solution $x^{*}\in[0,1]$, and $1-x^{*}$ solves $g_{2}(y)-y=0$. Thus, (2.4) can be investigated by considering $g_{1}(x)-x=0$ and $g_{2}(y)-y=0$. In order to give an appropriate reformulation of the function $g_{1}(\cdot)$, we need to use the following combinatorial identity.

Lemma 2.2

If $n$ is an even number, then

[TABLE]

and if $n$ is an odd number, then

[TABLE]

Now, we derive a simple expression of the function $g_{1}(\cdot)$, which is a key to our discussions later.

Lemma 2.3

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric transition probability tensor, and the function $g_{1}(\cdot)$ is defined by (2.16). Then, the following results hold.

• (i)

If $m$ is an even number, then

[TABLE]

• (ii)

If $m$ is an odd number, then

[TABLE]

Proof. (i) Suppose that $m$ is an even number. In this case, we first show that

[TABLE]

Since $m$ is an even number, it follows from Lemma 2.1 that

[TABLE]

Let

[TABLE]

then it follows from (2.21) that $\alpha_{0}=b$, and

• •

if $t\in\{1,3,\ldots,m-1\}$, then

[TABLE]

• •

if $t\in\{2,4,\ldots,m-2\}$, then

[TABLE]

Since

[TABLE]

it follows that for any $t\in\{1,3,\ldots,m-1\}$,

[TABLE]

where the first equality holds by Lemma 2.2; and for any $t\in\{2,4,\ldots,m-2\}$,

[TABLE]

Thus, (2) holds.

Furthermore, it is easy to see that for any $t\in\{1,2,\ldots,m-1\}$,

[TABLE]

and hence,

[TABLE]

where the last equality follows from the binomial theorem. Thus, we complete the proof of the result in (i).

(ii) Suppose that $m$ is an odd number. In this case, it follows from Lemma 2.1 that

[TABLE]

Let $g_{1}(x)=\sum_{t=1}^{m}\beta_{m-t}x^{m-t}$. Then, it follows from (2.25) that $\beta_{0}=a$, and

• •

if $t\in\{1,3,\ldots,m-2\}$, then

[TABLE]

• •

if $t\in\{2,4,\ldots,m-1\}$, then

[TABLE]

Thus, similar to (i), we can obtain that

[TABLE]

and furthermore,

[TABLE]

which implies that the result in (ii) holds. $\Box$

Lemma 2.4

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric transition probability tensor, and the function $g_{1}(\cdot)$ is defined by (2.16). Then,

[TABLE]

Proof. The desired result follows from Lemma 2.3 directly. $\Box$

The following result is a special case of the one in [15, Theorem 3.1].

Lemma 2.5

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric transition probability tensor. Then, $(\frac{1}{2},\frac{1}{2})^{\top}$ is a stationary probability vector of $\mathscr{P}$.

We now give our main results in this paper.

Theorem 2.1

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric transition probability tensor. If $p_{1\cdots 11}=1$ and $p_{21\cdots 11}=0$, then the corresponding tensor $\mathscr{P}$ is denoted by $\mathscr{P}_{1}$; and if $p_{1\cdots 11}=0$ and $p_{21\cdots 11}=1$, then the corresponding tensor $\mathscr{P}$ is denoted by $\mathscr{P}_{2}$. Then, we have the following results.

• (i)

The transition probability tensor $\mathscr{P}_{1}$ has and only has two stationary probability vectors: $(\frac{1}{2},\frac{1}{2})^{\top}$ and $(0,1)^{\top}$.

• (ii)

The transition probability tensor $\mathscr{P}_{2}$ has and only has two stationary probability vectors: $(\frac{1}{2},\frac{1}{2})^{\top}$ and $(1,0)^{\top}$.

• (iii)

If $\mathscr{P}\neq\mathscr{P}_{1}$ and $\mathscr{P}\neq\mathscr{P}_{2}$, then $\mathscr{P}$ has a unique stationary probability vector: $(\frac{1}{2},\frac{1}{2})^{\top}$.

Proof. Let

[TABLE]

Then, it follows from Lemma 2.4 that

[TABLE]

We divide the proof into the following three parts.

Part 1. Suppose that $a=b$. By (2.27) we have that for any $x\in[0,1]$,

[TABLE]

which implies that the function $h(\cdot)$ defined by (2.26) is strictly decreasing on $[0,1]$. This, together with $h(\frac{1}{2})=0$ by Lemma 2.1, implies that $\frac{1}{2}$ is the unique solution of $h(x)=0$ on $[0,1]$. Furthermore, by Lemma 2.1 it follows that $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique solution of (2.4) on $[0,1]$, i.e., $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique stationary probability vector of the transition probability tensor $\mathscr{P}$.

Part 2. Suppose that $1\geq a>b\geq 0$. Let $h^{\prime}(x^{*})=0$, then by (2.27), we have

[TABLE]

We consider the following three cases.

• (a)

If $(a-b)(m-1)=1$, then by (2.28), we have $x^{*}=1$. Since $m\geq 3$, it follows that $1>a>b>0$. Since $g_{1}(1)=a$ by (2.21) and (2.25), it follows by (2.26) that $h(1)=a-1<0$, i.e., 1 is not a solution of $h(x)=0$. Moreover, by (2.27) it follows that the function $h(\cdot)$ is strictly decreasing on $[0,1)$. This, together with $h(\frac{1}{2})=0$, implies that $\frac{1}{2}$ is the unique solution of $h(x)=0$ on $[0,1)$. Thus, $\frac{1}{2}$ is the unique solution of $h(x)=0$ on $[0,1]$. This, together with Lemma 2.5, implies that $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique solution of (2.4) on $[0,1]$, i.e., $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique stationary probability vector of the transition probability tensor $\mathscr{P}$.

• (b)

If $(a-b)(m-1)<1$, then by (2.28), we have $x^{*}>1$. Thus, by (2.27) it follows that the function $h(\cdot)$ is strictly decreasing on $[0,1]$. This, together with $h(\frac{1}{2})=0$, implies that $\frac{1}{2}$ is the unique solution of $h(x)=0$ on $[0,1]$. Furthermore, by Lemma 2.5 it follows that $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique solution of (2.4) on $[0,1]$, i.e., $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique stationary probability vector of the transition probability tensor $\mathscr{P}$.

• (c)

If $(a-b)(m-1)>1$, then by (2.28), we have $\frac{1}{2}<x^{*}\leq 1$. Furthermore, by (2.27) it follows that $h^{\prime}(x)<0$ when $x\in[0,x^{*})$ and $h^{\prime}(x)>0$ when $x\in(x^{*},1]$. On one hand, since the function $h(\cdot)$ is strictly decreasing on $[0,x^{*})$ and $h(\frac{1}{2})=0$ with $\frac{1}{2}\in[0,x^{*})$, it follows that $\frac{1}{2}$ is the unique solution of $h(x)=0$ on $[0,x^{*})$; and in the meantime, we have $h(x^{*})<h(\frac{1}{2})=0$. On the other hand, by (2.21) and (2.25), we have $g_{1}(1)=a$, and hence, by (2.26), $h(1)=a-1\leq 0$. If $a=1$, then since the function $h(\cdot)$ is strictly increasing on $(x^{*},1]$ with $h(1)=0$, it follows that $1$ is the unique solution of $h(x)=0$ on $(x^{*},1]$; and if $a<1$, then since the function $h(\cdot)$ is strictly increasing on $(x^{*},1]$ with $h(1)<0$ and $h(x^{*})<0$, it follows that $h(x)=0$ has no solution on $(x^{*},1]$. Thus, if $\mathscr{P}=\mathscr{P}_{2}$, then it has and only has two stationary probability vectors: $(\frac{1}{2},\frac{1}{2})^{\top}$ and $(1,0)^{\top}$; otherwise, it has a unique stationary probability vector: $(\frac{1}{2},\frac{1}{2})^{\top}$.

Part 3. Suppose that $1\geq b>a\geq 0$. If $m$ is an even number, then for any $x\in[0,1]$,

[TABLE]

and hence, the function $h(\cdot)$ is strictly decreasing on $[0,1]$. In this case, $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique solution of (2.4) on $[0,1]$, i.e., $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique stationary probability vector of the transition probability tensor $\mathscr{P}$.

In the following, we assume that $m$ is an odd number. Let $h^{\prime}(x^{*})=0$, then by (2.27), we have

[TABLE]

We consider the following three cases.

• (a)

If $(b-a)(m-1)=1$, then by (2.29), we have $x^{*}=0$. Since $m\geq 3$, it follows that $1>b>a>0$. Since $m$ is an odd number, we have $g_{1}(0)=a$ by (2.25), and hence, it follows by (2.26) that $h(0)=a-0>0$, i.e., 0 is not a solution of $h(x)=0$. Moreover, by (2.27) it follows that $h^{\prime}(x)<0$ on $(0,1]$, which implies that the function $h(\cdot)$ is strictly decreasing on $(0,1]$, and hence, $\frac{1}{2}$ is a unique solution of $h(x)=0$ on $(0,1]$. So, $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique stationary probability vector of the transition probability tensor $\mathscr{P}$.

• (b)

If $(b-a)(m-1)<1$, then by (2.29), we have $x^{*}<0$; and by (2.27), we have $h^{\prime}(x)<0$ for any $x\in[0,1]$. Thus, the function $h(\cdot)$ is strictly decreasing on $[0,1]$. This, together with $h(\frac{1}{2})=0$, implies that $\frac{1}{2}$ is the unique solution of $h(x)=0$ on $[0,1]$. Thus, $(\frac{1}{2},\frac{1}{2})^{\top}$ is the unique stationary probability vector of the transition probability tensor $\mathscr{P}$.

• (c)

If $(b-a)(m-1)>1$, then it is easy to see from (2.29) that $0<x^{*}<\frac{1}{2}$. Furthermore, by (2.27), we have $h^{\prime}(x)>0$ when $x\in[0,x^{*})$ and $h^{\prime}(x)<0$ when $x\in(x^{*},1]$. On one hand, since the function $h(\cdot)$ is strictly decreasing on $(x^{*},1]$ and $h(\frac{1}{2})=0$ with $\frac{1}{2}\in(x^{*},1]$, it follows that $\frac{1}{2}$ is the unique solution of $h(x)=0$ on $(x^{*},1]$. Meantime, we have $h(x^{*})>h(\frac{1}{2})=0$. On the other hand, since $m$ is an odd number, it follows by (2.25) that $g_{1}(0)=a$, and hence, by (2.26), $h(0)=a\geq 0$. If $a=0$, then $b=1$, and hence, $\mathscr{P}=\mathscr{P}_{2}$. In this case, [math] is a solution of $h(x)=0$. Otherwise, since the function $h(\cdot)$ is strictly decreasing on $(0,x^{*})$ with $h(0)>0$ and $h(x^{*})>0$, it follows that $h(x)=0$ has no solution on $(0,x^{*})$. Thus, if $\mathscr{P}=\mathscr{P}_{2}$, then it has and only has two stationary probability vectors: $(\frac{1}{2},\frac{1}{2})^{\top}$ and $(1,0)^{\top}$; otherwise, it has a unique stationary probability vector: $(\frac{1}{2},\frac{1}{2})^{\top}$.

Therefore, by combining Part 1 with Part 2 and Part 3, we can obtain the desired results. $\Box$

By Theorem 2.1, we have the following result immediately.

Corollary 2.1

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric transition probability tensor. Then, $\mathscr{P}$ has a unique positive stationary probability vector, which is $(\frac{1}{2},\frac{1}{2})^{\top}$.

Recall that a tensor $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,n}$ is called reducible if there exists a nonempty proper index subset $I\subset\{1,2,\ldots,n\}$ such that

[TABLE]

If $\mathscr{P}$ is not reducible, then it is called irreducible [16]. It is easy to see that both tensors $\mathscr{P}_{1}$ and $\mathscr{P}_{2}$ given in Theorem 2.1 are reducible. Thus, by Theorem 2.1, we have the following result immediately.

Corollary 2.2

Suppose that $\mathscr{P}=(p_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{T}_{m,2}$ is a symmetric irreducible transition probability tensor. Then, $\mathscr{P}$ has a unique stationary probability vector, which is $(\frac{1}{2},\frac{1}{2})^{\top}$.

When $m=4$, Corollary 2.2 is just Theorem 3.1 given in [15].

3 Concluding remarks

In this paper, we gave a full characterization on the stationary probability vectors of $m$-order $2$-dimensional symmetric transition probability tensors. In particular, for any integer number $m\geq 3$, any symmetric irreducible transition probability tensor of order $m$ dimension 2 has a unique stationary probability vector. In our analysis, the “symmetry” of transition probability tensor plays an important role. It is worthy of studying the stationary probability vectors of $m$-order $2$-dimensional transition probability tensors in the absence of symmetry. Moreover, it is also worthy of investigating the uniqueness of the stationary probability vectors of $m$-order $n$-dimensional symmetric transition probability tensors when $m,n\geq 3$.