Counterexamples to the conjecture on stationary probability vectors of the second-order Markov chains

Linear Algebra and its Applications 507 (2016) 153–157

Contents lists available at ScienceDirect

Linear Algebra and its Applications www.elsevier.com/locate/laa

Counterexamples to the conjecture on stationary probability vectors of the second-order Markov chains Mansoor Saburov ∗ , Nur Atikah Yusof Faculty of Science, International Islamic University Malaysia, P.O. Box, 25200, Kuantan, Pahang, Malaysia

a r t i c l e

i n f o

Article history: Received 28 November 2015 Accepted 6 June 2016 Available online xxxx Submitted by P. Semrl MSC: 15A69 15A18

a b s t r a c t It was conjectured in the paper “Stationary probability vectors of higher-order Markov chains” (Li and Zhang, 2015 [7]) that if the set of stationary vectors of the second-order Markov chain contains k-interior points of the (k −1)-dimensional face of the simplex Ωn then every vector in the (k − 1)-dimensional face is a stationary vector. In this paper, we provide counterexamples to this conjecture. © 2016 Elsevier Inc. All rights reserved.

Keywords: Transition hypermatrix The second-order Markov chain Stationary probability vector

1. The m-order Markov chains A discrete-time m-order Markov chain is a stochastic process with a sequence of random variables {Xt , t = 0, 1, 2 . . .} , * Corresponding author. E-mail address: [email protected] (M. Saburov). http://dx.doi.org/10.1016/j.laa.2016.06.012 0024-3795/© 2016 Elsevier Inc. All rights reserved.

154

M. Saburov, N.A. Yusof / Linear Algebra and its Applications 507 (2016) 153–157

which takes on values in a discrete finite state space [n] = {1, . . . , n} for a positive integer n, such that pi,i1 ···im = Pr (Xt+1 = i|Xt = i1 , Xt−1 = i2 , . . . , X1 = it , X0 = it+1 ) = Pr (Xt+1 = i|Xt = i1 , · · · , Xt−m+1 = im ) , where i, i1 , · · · , im , · · · , it+1 ∈ [n] and n 

pi,i1 ···im ≥ 0,

pi,i1 ···im = 1,

1 ≤ i, i1 , · · · , im ≤ n.

i=1

Namely, the current state of the chain depends on m past states (see [1,2]). Here n P = (pi,i1 ···im )i,i1 ,...,im =1 is called the transition hypermatrix of an m-order Markov chain {Xt , t = 0, 1, 2 . . .}. Let x(t) = (x1 (t), · · · , xn (t))T be the tth-distribution of the discrete-time m-order Markov chain {Xt , t = 0, 1, 2 . . .}. Note that P is an (m +1)-order stochastic hypermatrix governing the transition of states in the m-order Markov chain according to the following rule 

xi (t + 1) =

pi,i1 ···im xi1 (t) · · · xim (t),

i = 1, · · · , n.

1≤i1 ,··· ,im ≤n

Denote by  Ωn =

x = (x1 , · · · , xn )T : x1 , · · · , xn ≥ 0,

n 

 xi = 1

(1)

i=1

the standard simplex consisting of probability vectors in Rn . A polynomial operator P : Ωn → Ωn associated with the (m + 1)-order stochastic n hypermatrix P = (pi,i1 ···im )i,i1 ,...,im =1 (P(x))i =



pi,i1 ···im xi1 · · · xim ,

i = 1, · · · , n,

(2)

1≤i1 ,··· ,im ≤n

is called a nonlinear Markov operator (see [3]). It is clear that a set of all stationary distributions of the m-order Markov chain is nothing but a set of all fixed points of the nonlinear Markov operator (2). Due to Brouwer’s fixed-point theorem, the set Fix(P) = {x ∈ Ωn : P(x) = x} is nonempty. If m = 2 then P : Ωn → Ωn is a quadratic stochastic operator which has an incredible application in population genetics (see [4–6]).

M. Saburov, N.A. Yusof / Linear Algebra and its Applications 507 (2016) 153–157

155

In the paper [7], it was conjectured that if the set of stationary vectors of 2-order Markov chain contains k-interior points of the (k − 1)-dimensional face of the simplex Ωn then every vector in the (k − 1)-dimensional face is a stationary vector. In this paper, we provide counterexamples to this conjecture. 2. Counterexamples 3

Let P = (pi,jk )i,j,k=1 be a cubic stochastic hypermatrix, i.e., pi,jk ≥ 0 and p1,jk + p2,jk + p3,jk = 1 for any 1 ≤ i, j, k ≤ 3. Without loss of generality, we may assume that pi,jk = pi,kj for any 1 ≤ i, j, k ≤ 3. Otherwise, we may define a cubic stochastic  3 pi,jk + pi,kj such that psi,jk = psi,kj for hypermatrix P s = psi,jk where psi,jk := 2 i,j,k=1 any 1 ≤ i, j, k ≤ 3 and the associated quadratic stochastic operators QP and QP s are the same, i.e., QP (x) = QP s (x) for any x ∈ Ω2 . Let QP : Ω2 → Ω2 be a quadratic stochastic operator associated with P = 3 (pi,jk )i,j,k=1 ⎧ 2 2 2 ⎪ ⎪ ⎨(QP (x))1 = p1,11 x1 + p1,22 x2 + p1,33 x3 + 2p1,12 x1 x2 + 2p1,13 x1 x3 + 2p1,23 x2 x3 QP : (QP (x))2 = p2,11 x21 + p2,22 x22 + p2,33 x23 + 2p2,12 x1 x2 + 2p2,13 x1 x3 + 2p2,23 x2 x3 ⎪ ⎪ ⎩(Q (x)) = p x2 + p x2 + p x2 + 2p x x + 2p x x + 2p x x P 3,11 1 3,22 2 3,33 3 3,12 1 2 3,13 1 3 3,23 2 3 3 We say that P > 0 if pi,jk > 0 for any 1 ≤ i, j, k ≤ 3. In this case, Q : Ω2 → Ω2 is called the positive quadratic stochastic operator. Our aim is to provide an example for the positive quadratic stochastic operator Q : Ω2 → Ω2 which has exactly three fixed points in the interior set intΩ2 = {x ∈ Ω2 : x1 x2 x3 > 0} of the simplex Ω2 , i.e., |Fix(Q) ∩ intΩ2 | = 3. In order to accomplish our aim, we use the following strategy: instead of finding fixed points of the given positive quadratic stochastic operator, we try to find a positive quadratic stochastic operator for which the given points are its fixed points. Let A = (0.1, 0.2, 0.7)T , B = (0.4, 0.3, 0.3)T and C = (0.59, 0.31, 0.1)T be interior points of the simplex Ω2 . We define a positive quadratic stochastic operator Q0 : Ω2 → Ω2 as follows ⎧ ⎪ ⎪(Q0 (x))1 = ⎪ ⎪ ⎨(Q (x)) = 0 2 Q0 : ⎪ (Q (x)) ⎪ 0 3 = ⎪ ⎪ ⎩

232873 2 4717 2 207 7 3 1 2 319300 x1 + 10300 x2 + 63860 x3 + 5 x1 x2 + 5 x1 x3 + 50 x2 x3 27 2 1 2 3 2 470171 378421 158157 100 x1 + 2 x2 + 20 x3 + 814300 x1 x2 + 407150 x1 x3 + 814300 x2 x3 54 433 27037 2 18409 191589 2 2 79825 x1 + 10300 x2 + 31930 x3 + 814300 x1 x2 + 407150 x1 x3 + 1454157 814300 x2 x3

A straightforward calculation shows that A, B, C are fixed points of the positive quadratic stochastic operator Q0 : Ω2 → Ω2 . It is clear that if each point of the set intΩ2 is a fixed point of Q0 : Ω2 → Ω2 then due to continuity of Q0 and intΩ2 = Ω2 , each point of the simplex Ω2 must be a fixed point.

156

M. Saburov, N.A. Yusof / Linear Algebra and its Applications 507 (2016) 153–157

This is impossible because of Q0 (∂Ω2 ) ⊂ intΩ2 where ∂Ω2 = {x ∈ Ω2 : x1 x2 x3 = 0} is the boundary of the simplex Ω2 . We can define another positive quadratic stochastic operator Q1 : Ω2 → Ω2 as follows

Q1 :

⎧ ⎪ (Q1 (x))1 = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨(Q1 (x))2 = ⎪ ⎪ ⎪ ⎪ (Q1 (x))3 = ⎪ ⎪ ⎪ ⎩

17322871 2 990257 2 1559 13 16 2 22351000 x1 + 2163000 x2 + 13410600 x3 + 10 x1 x2 + 25 x1 x3 224 2 488 2 125 2 703327 19461451 1000 x1 + 1000 x2 + 1000 x3 + 1017875 x1 x2 + 24429000 x1 x3 8271787 + 24429000 x2 x3 4301 117199 2 2 4470200 x1 + 2163000 x2 + 1601951 977160 x2 x3

+

2933179 2 3352650 x3

+

18371 2035750 x1 x2

+

+

11 500 x2 x3

13761989 24429000 x1 x3

A straightforward calculation shows that A, B, C are also fixed points of the positive quadratic stochastic operator Q1 : Ω2 → Ω2 . Now, we can define a family of positive quadratic stochastic operators Qε : Ω2 → Ω2 as Qε (x) = (1 − ε)Q0 (x) + εQ1 (x) for any x ∈ Ω2 and 0 ≤ ε ≤ 1. It is clear that A, B, C are also fixed points of the family of positive quadratic stochastic operators Qε : Ω2 → Ω2 . It is worth mentioning that an example for positive quadratic stochastic operator having three fixed points inside the simplex was also given by Yu. Lyubich in his book [4]. However, Lyubich’s example is wrong. Actually the positive quadratic stochastic operator in Lyubich’s example has a unique fixed point. This issue was addressed in the paper [8]. Here, we provide a simple proof of the following result which was proven in the general setting in [4]. Proposition 1. Any positive quadratic stochastic operator Q : Ω2 → Ω2 has either unique or three fixed points in the simplex, i.e., |Fix(Q) ∩ Ω2 | = 1 or 3. Proof. We shall first prove that |Fix(Q) ∩ Ω2 | ≤ 4. Since Q : Ω2 → Ω2 is a positive quadratic stochastic operator, we have that Q(∂Ω2 ) ⊂ intΩ2 . Therefore, we get that Fix(Q)  intΩ2 . Since x3 = 1 − x1 − x2 , it is clear that Fix(Q) is defined by an intersection of two second order curves f1 (x1 , x2 ) = 0 and f2 (x1 , x2 ) = 0 inside the triangle defined by 0 ≤ x1 + x2 ≤ 1 and 0 ≤ x1 , x2 ≤ 1. Due to Bezout’s theorem, the set Fix(Q) is either finite (with |Fix(Q)| ≤ 4) or infinite. Suppose the set Fix(Q) is infinite. Then an intersection of two second order curves must be a compact infinite set, i.e., an ellipse Ef1 ,f2 . Let e1 = (1, 0, 0)T and c be the center of the ellipse Ef1 ,f2 . Let us consider a straight line L connecting two points e1 and c. It is clear that |L ∩ Ef1 ,f2 | = 2 and L ∩ Ef1 ,f2 ⊂ Fix(Q). Let L ∩ ∂Ω2 = {e1 , q}, where q = (0, q2 , q3 )T with q2 + q3 = 1 and q2 , q3 > 0. Since the straight line L contains two fixed points, the following quadratic equation with respect to t (Q((1 − t)e1 + tq))1 = ((1 − t)e1 + tq)1 must have two roots in the segment [0, 1]. It is clear that

(3)

M. Saburov, N.A. Yusof / Linear Algebra and its Applications 507 (2016) 153–157

157

Q((1 − t)e1 + tq) = (1 − t)2 Q(e1 ) + 2t(1 − t)B(e1 , q) + t2 Q(q) where B : Ω2 × Ω2 → Ω2 is a symmetric bilinear operator associated with Q. Let a = (Q(e1 ))1 , b = (B(e1 , q))1 , and c = (Q(q))1 such that 0 < a, b, c < 1. It follows from (3) that the following quadratic function g(t) = (1 − t)2 a + 2t(1 − t)b + t2 c + t − 1 must have two roots in the segment [0, 1]. Therefore, g(0)g(1) > 0. This contradicts to the fact g(0)g(1) = (a − 1)c < 0. Consequently, Fix(Q) is a finite set such that |Fix(Q)| ≤ 4. Let us consider the mapping F : Ω2 → R3 , F(x) = Q(x) − x (it is called a vector field, see [9]). Let Θ = (0, 0, 0) and F−1 (Θ) = {x : Ω2 : F(x) = Θ} be zeros (critical or singular points) of the vector field F. Since Q(∂Ω2 ) ⊂ intΩ2 , the vector field F(x) = Q(x) − x on the boundary ∂Ω2 of the simplex Ω2 is directed its inside. Hence, the index of the vector field F over the boundary ∂Ω2 of the simplex Ω2 is 1. On the other hand, the index of the critical point of the vector field F is either 1 or −1. Due to Hopf’s index theorem (see [9]), the index of the vector field F on the boundary ∂Ω2 is equal to the sum of indices of the critical points of the vector field F inside the simplex Ω2 . Consequently, the total number of the critical points of the vector field F must be odd. This means that |Fix(Q)| is odd, i.e., |Fix(Q)| = 1 or 3. This completes the proof. 2 Acknowledgements This work has been partially supported by the MOHE grant FRGS14-141-0382. The first author (M.S.) also thanks to the Junior Associate Scheme, Abdus Salam International Centre for Theoretical Physics (ICTP), Trieste, Italy, where this paper was written, for the invitation and hospitality. The authors are also greatly indebted to the anonymous reviewer for several useful comments which improved the presentation of the paper. References [1] A. Raftery, A model of high-order Markov chains, J. R. Statist. Soc. 47 (1985) 528–539. [2] A. Berchtold, A. Raftery, The mixture transition distribution model for high-order Markov chains and non-Gaussian time series, Statist. Sci. 7 (2002) 328–356. [3] V.N. Kolokoltsov, Nonlinear Markov Processes and Kinetic Equations, Cambridge University Press, New York, 2010. [4] Y. Lyubich, Mathematical Structures in Population Genetics, Springer-Verlag Berlin Heidelberg, Berlin, 1992. [5] R. Ganikhodzhaev, F. Mukhamedov, U. Rozikov, Quadratic stochastic operators and processes: results and open problems, Infin. Dimens. Anal. Quantum Probab. Relat. Top. 14 (02) (2011) 279–335. [6] F. Mukhamedov, N. Ganikhodjaev, Quantum Quadratic Operators and Processes, Springer International Publishing, Switzerland, 2015. [7] C.-K. Li, S. Zhang, Stationary probability vectors of higher-order Markov chains, Linear Algebra Appl. 473 (2015) 114–125. [8] M. Saburov, N.A. Yusof, On quadratic stochastic operators having three fixed points, J. Phys. Conf. Ser. 697 (2016) 012012. [9] E. Outerelo, J.M. Ruiz, Mapping Degree Theory, American Mathematical Society, Providence, RI, 2009.