A Nonlinear Transform for the Diagonalization of the Bernoulli-Laplace Diffusion Model and Orthogonal Polynomials
aa r X i v : . [ m a t h - ph ] D ec A NONLINEAR TRANSFORM FOR THE DIAGONALIZATION OFTHE BERNOULLI-LAPLACE DIFFUSION MODEL ANDORTHOGONAL POLYNOMIALS
CHJAN C. LIM
AND
WILLIAM PICKERING ∗ Abstract.
The Bernoulli-Laplace model describes a diffusion process of two types of particlesbetween two urns. To analyze the finite-size dynamics of this process, and for other constructiveresults we diagonalize the corresponding transition matrix and calculate explicitly closed-form ex-pressions for all eigenvalues and eigenvectors of the Markov transition matrix T BL . This is done bya new method based on mapping the eigenproblem for T BL to the associated problem for a linearpartial differential operator L BL acting on the vector space of homogeneous polynomials in threeindeterminates. The method is applicable to other Two Urns models and is relatively easy to usecompared to previous methods based on orthogonal polynomials or group representations.
1. Introduction.
The Bernoulli-Laplace (BL) model arise from diffusion theoryand is related to the shuffling of cards [8]. Symmetries of the permutation group S N appear naturally in this model and other random walks on groups. Previous solutionsof this model have appeared in Diaconis and Shashahani [7] and in the works ofKarlin and MacGregor [10]. Group representations are used explicitly in the first; thederivation of a non-standard inner product or equivalently a measure for orthogonalpolynomials which are related to the eigensolutions of the BL model appears in thesecond.In this paper, we give a third way for deriving exact solutions of all the eigen-vectors of the BL model, through a nonlinear transform that triangularizes and thendiagonalizes the transition matrix T BL . In brief, our method associates a specificlinear partial differential operator (LPDO) L BL that acts on the vector space of ho-mogeneous polynomials, G, to the matrix T BL . The L BL inherits the symmetries ofthe BL model; it encodes the tri-diagonal singly-stochastic (column sums are equal to1 ) and anti-symmetric structure of T BL . The components of the (right) eigenvectors(in view of the equal column sums of T BL ) of T BL is encoded in the coefficients ofthe homogeneous polynomial G. A classical theory for the symmetries of such LP-DOs have been formulated in terms of the Lie algebra of symmetry operators K thatcommutes with L BL (cf. [13]).It turns out and we exploit in our method, that the symmetries of L BL appeariu the form of suitable linear and nonlinear transformations P on the independentvariables x, y, etc. or indeterminates of G. The ease of use of this method residesin the transparent or explicit way to find these transformations P. Our algorithm iscompleted by associating the transformed LPDO, L ′ BL , back to what turns out to bea triangular matrix T ′ BL ; in other words, the transformation P for L BL encodes asimilarity transformation that triangularizes T BL , i.e., P T BL P − = T ′ BL , which isthen solved directly for its eigenvalues and right eigenvectors.Here, we give a summary of the Urn models to which the BL model is related asan extension. The Ehrenfest model and the Polya Urn models are two of the earlysolvable models in the literature [9]. They appear as two of the exactly solved casesin Friedman’s formulation of Urn models where precisely one urn and balls of twocolors are drawn and replaced with additions [9]. A dual formulation of Friedman’s ∗ Department of Mathematical Sciences, Rensselaer Polytechnic Institute, 110 8th Street, Troy,New York 12180, USA 1
This manuscript is for review purposes only. rn models was introduced in a series of recent papers [16, 14]: instead of balls of twocolors and one urn, the dual formulation uses two urns and one-colored balls. Thelatter is more convenient for modelling of certain network science models [3], suchas the Voter model where two balls are drawn and returned to the two urns withprescribed probabilities that depend on the order in which they are drawn. This isbecause many of the network science models are irreversible Markov chain models[4] which have absorbing states. Their transition matrices T , unlike T BL for the BLmodel, are not symmetric, in an essential sense, that is, there are no non-standardinner products for R N in which these matrices T have a symmetric form.Using a new method based on diagonalization of transition matrices [15], we solvedexactly the eigenvectors of several well-known models, including the Ehrenfest model,the Voter model [18, 12, 11, 5, 17, 2], the Moran model for genetic drift, and theNaming game models [19], [20]. Most of these models are irreversible Markov modelswith absorbing states, and have essentially non-symmetric transition matrices in thesense just mentioned. The BL model however, is based on two urns and balls of twocolors. Thus, it is not strictly in the class of Two Urns models to which we recentlyapplied our method. In modifying this method so that it applies to the BL model,we will have shown that the new method is not only easy to use but also flexible inextension to new problems.One of the main points here is the technical simplicity of uncovering the symme-tries of the above LPDOs within our method, through the explicit appearance of theexpressions u = f ( x, y ) in the coefficients of L. We give here the LPDO L V for theVoter model [16, 14] to indicate what we mean: first the propagation equation for thetransition matrix T V is given by a ( m +1) j = p j − a ( m ) j − + (1 − p j ) a ( m ) j + p j +1 a ( m ) j +1 (1.1) p j = j ( N − j ) N ( N −
1) ;(1.2)the eigen-problem for the associated LPDO L V = ( x − y ) G xy acting on the homo-geneous polynomial G ( x, y ) = P j c j x j y N − j (which encodes the components c j of theright-eigenvector of T V ) is given(1.3) ( x − y ) G xy = N ( N − λ − G, which clearly suggests the transformation u = x − y, v = y. Indeed this triangularizedand diagonalized the Voter model and led to its complete solution.Contrast this ease of use with the fact that triangularization and diagonalizationof a given transition matrix of size N has computational complexity O ( N ) . In otherwords, exact integration of the Two Urns models via diagonalization of transitionmatrices are nontrivial problems, that are difficult to solve but once known, the so-lutions are easy to verify. [16, 14] provides a simple method to find such explicitdiagonalization and hence all eigenvectors for a class of transition matrices from theTwo Urns models, even when their transition matrices are essentially non-symmetric.Note that the eigenproblem and diagonalization of symmetric matrices have a lowercomputational complexity.We aim here to highlight this method’s ease of use, relative to the group rep-resentation method and the method of orthogonal polynomials. Moreover, the BLmodel differs significantly from the original Two Urns subclass of models for which This manuscript is for review purposes only. ur method was initially formulated. Thus, we also aim to show that, with the spe-cific introduction of a nonlinear change of independent variables, this method can beapplied to more complex models than the original class of models. Since the BL tran-sition matrix T BL is from a reversible Markov chain with a stationary distribution [4],it is non-symmetric singly-stochastic only in a trivial sense. In other words, there ex-ists (a difficult to find) non-standard inner product for R N , in which T BL is symmetricand hence doubly-stochastic. Karlin and McGregor [10] using their powerful integralrepresentation method of finding an explicit way to symmetrize T by introducing anon-standard inner product (or orthogonal measure) into the problem, have relatedthe right eigenvectors of the transition matrix T BL of the BL model to the orthogonalpolynomials called the dual Hahn polynomials. A third aim of this paper is there-fore to re-derive from the diagonalization of T BL , this non-standard inner product inwhich the dual Hahn polynomials are an orthogonal polynomial system. Note thatthis non-standard inner product, once found, yields a symmetric version of T BL whichis an example of a Jacobi operator that arise in the the classical moments problem[1], and is related to orthogonal polynomials via the Riemann-Hilbert method [6].The paper is organized as follows: section 3 concerns the calculation of the righteigenvectors and eigenvalues of T in closed form by method introduced in [16, 15,14]; in view of the fact that these right eigenvectors are not orthogonal in the usualEuclidean inner-product, section 4 concerns the transformations needed to calculatethe orthonormal system of left eigenvectors of T BL , and also the derivation of the non-standard inner product in which the right eigenvectors are now orthogonal; section 5concerns the elementary proofs, based only on the eigenvectors and eigenvalues of T, for the tight upper and lower bounds for times to stationarity in the BL model [7] ;section 6 concerns a numerically exact evaluation of the expression for the TV normin these bounds using the eigenvectors and eigenvalues of T BL directly, hence slightlysharper estimates for the mixing times of the BL model.Beyond the balanced special case of the BL problem treated in detail in this paper,the same generating function method can be used to prove similar tight bounds formixing times in the other cases.
2. Transition matrix of the BL model.
Let the transition matrix T BL bedefined so that ( T BL ) ij = P r { n w ( m + 1) = i | n w ( m ) = j } , so that the sum of eachcolumn is 1. In the general BL model for balls of two colors and two urns, N , N ,N w , N b are fixed parameters satisfying the constraints(2.1) N w + N b = N = N + N where N equals total number of balls in the model. For i = 0 , ..., N w ≤ N , (where byabuse of notation i stands for both the row label of transposed matrix T tBL and thenumber of white balls in urn 1, n w ) , the transition probabilities are explicitly givenby p i = Pr { n w ( t + 1) = i + 1 | n w ( t ) = i } = ( N − i ) ( N w − i ) N N (2.2) q i = Pr { n w ( t + 1) = i − | n w ( t ) = i } = i ( N b − ( N − i )) N N (2.3) r i = Pr { n w ( t + 1) = i | n w ( t ) = i } = 1 − q i − p i . (2.4). This manuscript is for review purposes only. . Diagonalization - Right eigenvectors of the general BL model.
In[16, 14], we developed an explicit method for exactly integrating or solving a 5-parameters subclass of a class of Two Urns models which is parametrized by six realparameters. Our method is based on a relationship between certain banded stochasticmatrices T (such as tridiagonal and pentadiagonal non-symmetric transition matricesof markov chain models) and the LPDOs acting on the vector space of homogeneouspolynomials, G ( x, y ) of finite order in two indeterminates. The symmetries of theLPDO, L, associated with a given non-symmetric singly stochastic matrix from thissolvable subclass of the Two Urns models, are identified and used explicitly to trans-form from the original indeterminates ( independent variables x, y say) to suitable newvariables (such as u = f ( x, y ) , v = g ( x, y )). In the new variables u, v, the transformedLPDO, L ′ , acts on the (again homogeneous of same order as G ( x, y )) polynomial H ( u, v ) . We have shown in [16, 14] that the transformed eigen-problem(3.1) L ′ [ H ( u, v )] = λ ( N ) H ( u, v )for a well-defined subclass of such Two Urns problems is equivalent (via the inverseof the original relationship between banded matrix and LPDO) to the eigen-problemfor a triangular matrix, which can then be solved explicitly for both right and lefteigenvectors. In other words, at the end of this brief summary, the symmetries ofLPDO L inherited from the original banded stochastic matrix T, generate an explicitsimilarity transformation, P, such that(3.2) ST S − = D where D is diagonal, and S contains the eigenvectors of T. This method can be formalized as an Algorithm as follows: Given the input of asingly stochastic transition matrix T of size N + 1,(I) Choose a suitable homogeneous polynomial of finite degree N , G that has thecomponents c i of a right eigenvector of T as coefficients of the monomials x i y j z k ; partof this choice is the number of indeterminates in G . For example, the Voter model ofsize N (number of balls) with a transition matrix T V ( N ) which is a N + 1 by N + 1real matrix, requires a homogeneous polynomial G V of degree N in the indeterminates x, y because there are two urns.(II) Associate the recursion inplicit in given Markov matrix T to a LPDO, L which acts on the homogeneous polynomial G ; the basic elements of this associationscheme are the standard linear differential operators for increasing, decreasing and notchanging the numbers of balls in each urn (which correspond in the example below tothe probabilities p, q, r prescibed by the transition matrix), and a set of multiplicationtype linear operators that correspond to shifts.(III) A transformation to new independent variables,(for instance, u = f ( x, y, z ) ,v = g ( x, y, z ) , w = h ( x, y, z )) is chosen to satisfy two conditions:( A ) the transformed polynomial H ( u, v, w ) = H ( f ( x, y, z ) , g ( x, y, z ) , h ( x, y, z )) = G ( x, y, z )is a homogeneous polynomial of the same finite degree as G ;( B ) u = f ( x, y, z ) , v = g ( x, y, z ) , w = h ( x, y, z ) is a transformation based on thesymmetries of L (cf. [13]), that is, the combinations f ( x, y, z ) , g ( x, y, z ) , h ( x, y, z ) ofthe original variables x, y, z appear naturally in the coefficients of the LPDO, L. This manuscript is for review purposes only. hese conditions ( A ) and ( B ) are clearly not sufficient to ensure the transformedLPDO eigenproblem(3.3) L ′ [ H ( u, v, w )] = λ ( N ) H ( u, v, w )is associated with a similar triangular matrix T ′ which explicitly yields all its eigen-vectors b i . That they are sufficient has to be proved either in each problem to whichwe apply the Algorithm, or for a class of models as in the case of the Two Urnsmodels.(IV) Using the transformation in step (III), derive the corresponding transformedLPDO, L ′ that acts on the transformed polynomial H ( u, v, w ).(V) Without explicitly calculating the transformed matrix T ′ which is associ-ated with the transformed LPDO, L ′ in step (IV), check that the transformed eigen-problem for L ′ is indeed a recursion system for the transformed eigenvectors b i thatcan be solved explicitly, i.e., it is equivalent to a triangular linear system of equations.Solve for the eigenvalues and then the eigenvector components b i , and if requiredtransform back to the original components c i . These are the main outputs of theAlgorithm.(VI) Use the eigenvectors in step (V) to diagonalize the original matrix if neces-sary. This is the end of the ALgorithm.Now we apply the Algorithm to the BL model. Given the transition matrix ofthe BL model (cf. section 2), it will be obvious that three independent variables(instead of the two before) should be used to formulate the BL problem. In step(I) of the Algorithm, we adopt the anzatz that the LP DO, L BL , associated withthe above N by N matrix T BL , now acts on a homogeneous polynomial G ( x, y, z ) inthree indeterminates, x, y, z. We encode the entries c k ( i ), i = 0 , ..., N w of the k − th eigenvector of the transition matrix for the BL model as follows:(3.4) G ( k ) ( x, y, z ) = N w X i =0 c k ( i ) x i y N − i z N w − i . where i = number of white balls in urn 1 (also denoted n w ). The choice of threeindependent variables to encode the components of an eigenvector of T BL in thehomogeneous polynomial G ( x, y, x ) is now made obvious by this explicit expressionfor G. In step (II) of the Algorithm, we derive from the original eigen-problem for tran-sition matrix T BL , an LPDO, L BL , that acts on G ( k ) . Towards that aim, we note, inparticular, the entries for p i and q i in T BL correspond respectively to the followinglinear differential operators with coefficients that are monomials in x, y, and z,L p = yzG ( k ) yz N N (3.5) L q = N b xN N G ( k ) x − xyN N G ( k ) xy , (3.6)where G ( k ) yz = ∂ ∂y∂z G ( k ) for example. In addition, it is part of the association schemethat multiplication in the LPDO (cf. [13]) by the coefficient xyz (resp. yzx ) represents This manuscript is for review purposes only. own (resp. up) shifts in the index i within the discrete recursion equations of theoriginal eigen-problem for matrix T BL.
The L BL associated with the eigen-problemof the tridiagonal Markov matrix T BL is given by: L BL [ G ( k ) ] = N N ( λ k − G ( k ) (3.7) L BL [ G ( k ) ] ≡ ( x − yz ) G ( k ) yz + y ( x − yz ) G ( k ) xy − N b ( x − yz ) G ( k ) x . (3.8)In step(III) of the Algorithm, we note that the symmetries of L BL with respectto transformations of its independent variables, is expressed in the factor ( x − yz ) inits coefficients. This suggests the transformation to the new independent variables(3.9) u = x − yz, y = y, z = z. Since the transformed homogeneous polynomial is now given by(3.10) H ( k ) ( u, y, z ) = G ( k ) ( x ( u, yz ) , y, z ) = X i b ki u i y N − i z N w − i in terms of the (new) components b ki of the k − th right eigenvector, this transformationclearly satisfies both necessary conditions (A) and (B) in step (III) of the Algorithm.To prove that it is sufficient for our purpose of obtaining the eigenvalues andeigenvectors exactly and for diagonalization, we proceed by direct calculations.In step (IV) using the following obvious identities for the transformation of partialderivatives ∂ x = ∂ u (3.11) ∂ y = ∂ y − z∂ u (3.12) ∂ z = ∂ z − y∂ u (3.13) ∂ xy = ∂ yu − z∂ u (3.14) ∂ yz = ∂ yz − y∂ yu − z∂ uz + yz∂ u − ∂ u (3.15)the transformed LPDO, L ′ BL in H ( k ) , k = 0 , ..., N w is N N ( λ k − H ( k ) = L ′ BL [ H ( k ) ] , (3.16) L ′ BL [ H ( k ) ] = − N b u∂ u H ( k ) + yu (cid:0) ∂ yu − z∂ u (cid:1) H ( k ) (3.17) + u (cid:0) ∂ yz − y∂ yu − z∂ uz + yz∂ u − ∂ u (cid:1) H ( k ) (3.18) = u ( ∂ yz − z∂ uz ) H ( k ) − ( N b + 1) u∂ u H ( k ) (3.19)In step (V), by reversing the derivation of the original L BL through the associationscheme [13], this L ′ BL in H is shown to be equivalent to the following triangular systemfor the (right) eigen-problem of the transformed matrix T ′ BL : N N ( λ k − b ki (3.20) = ( N − i + 1) ( N w − i + 1) b ki − − i ( N w − i ) b ki − ( N b + 1) ib ki . (3.21)We have therefore verified the sufficiency of the transformation where u = x − yz for triangularizing (and later diagonalizing) T BL . This triangular system implies therecursion(3.22) b ki = ( N − i + 1) ( N w − i + 1) b ki − N N ( λ k −
1) + i ( N w − i ) + ( N b + 1) i This manuscript is for review purposes only. hich can be solved directly.For nontrivial eigensolutions for k = 0 , ..., N w , the denominator in b ki must vanish,yielding the following exact expressions for the eigenvalues, λ k = 1 − k (1 − k + N w + N b ) N N (3.23) = 1 − k (1 − k + N ) N N (3.24) = 1 − k ( N − k + 1) N N (3.25)In the case N = N w , λ = 1(3.26) λ = 1 − NN N . (3.27)The eigenvectors (in the transformed variables of H ) are given explicitly by: b ki = i Y j = k +1 ( N − j + 1) ( N w − j + 1) N N ( λ k −
1) + j ( N w + N b + 1 − j )(3.28) = i Y j = k +1 ( j − N −
1) ( j − N w − − k ( N + 1 − k ) + j ( N + 1 − j )(3.29) = i Y j = k +1 − ( j − N −
1) ( j − N w − j − k ) ( j + k − N − − i − k ( k − N ) i − k ( k − N w ) i − k ( i − k )!(2 k − N ) i − k . (3.31)Using these coefficients in the definition for H gives(3.32) H ( k ) = N ′ X i = k ( − i − k ( k − N ) i − k ( k − N w ) i − k ( i − k )!(2 k − N ) i − k u i y N − i z N w − i We summarize the consequences of the above steps of the Algorithm on the BLmodel in the following theorem:
Theorem
In the above Algorithm for the BL model, for any size N of themodel, the LPDO, L ′ BL , after the transformation (3.9) on the independent variables,is equivalent to a triangular linear system (3.21) which has (right) eigenvectors givenby (3.31) and eigenvalues (3.25). The (right) eigenvectors of the original BL matrix T BL are in turn given by (3.38). The next tofinal step left in this part of the paper is step (VI) in the Algorithm, to invert theabove similarity transformation to obtain explicitly the closed-form expressions forthe original components of the right-eigenvectors c k ( i ) of T BL.
For this purpose, let h ( k ) ( u ) = H ( k ) ( u, , g ( k ) ( x ) = G ( k ) ( x, ,
1) = H ( k ) ( x − , ,
1) = h ( k ) ( x − x − k F ( k − N , k − N w ; 2 k − N ; 1 − x ) . This manuscript is for review purposes only. sing the hypergeometric identity,(3.33) F ( a, b ; c ; 1 − z ) ∝ F ( a, b ; a + b − c + 1; z )and the fact that any multiple of an eigenvector remains an eigenvector, we take thepolynomial for the right eigenvector components to be(3.34) g ( k ) ( x ) = ( x − k F ( k − N , k − N w ; N − N w + 1; x )whose coefficients are the original components c k ( i ) of the k − th right eigenvector cor-responding to λ k prior to the transformation above. These expressions are equivalentto the dual Hahn polynomials [10].For the hypergeometric representation of the eigenvectors to be well defined, werequire N ≥ N w . There is no loss in generality with this assumption, because we canrelabel N ↔ N and N w ↔ N b so that the assumption holds. From the solution for g ( k ) , we expand in x l ,g ( k ) ( x ) = X n (cid:18) kn (cid:19) ( − k − n x n X i ( k − N ) i ( k − N w ) i ( N − N w + 1) i i ! x i (3.35) = X i X n (cid:18) kn (cid:19) ( − k − n ( k − N ) i ( k − N w ) i ( N − N w + 1) i i ! x i + n (3.36) = ( − k X i "X n (cid:18) kn (cid:19) ( − n ( k − N ) i − n ( k − N w ) i − n ( N − N w + 1) i − n ( i − n )! x i (3.37)to find the explicit form for the components of the k − th right eigenvectors:.(3.38) c ki = X n (cid:18) kn (cid:19) ( − n ( k − N ) i − n ( k − N w ) i − n ( N − N w + 1) i − n ( i − n )!Notice that the solution is the kth order backwards difference of the components ofthe hypergeometric coefficients of g .The above treatment of the eigen-problem by transforming via symmetries, theindependent variables of the associated LPDO, L BL , is equivalent to a similaritytransformation of the transition matrix T BL . Let w = Pv for some transformationmatrix P . Then, the eigen-problem for w is given by PT BL P − w = λ w . Theabove calculations is equivalent to the matrix P such that the new matrix T ′ BL = PTP − is lower triangular. The last step in this section is to diagonalize T BL. . We do this by diagonalizing the matrix triangular matrix T ′ BL = WΛW − . Here, Λ = diag ( λ , . . . , λ N ) and W are the eigenvectors of T ′ BL . The components of theseeigenvectors are b i corresponding to eigenvalue λ k . Diagonalization of T ′ BL allows usto explicitly diagonalize the original transition matrix as(3.39) T BL = P − WΛW − P . Note that the matrix of eigenvectors is given by P − W .
4. Symmetrizing transform, orthogonal measure and dual Hahn poly-nomials.
For transition matrix T BL , let Z be given by (where we drop the subscriptBL herein, i.e., T = T BL )(4.1) Z ij = √ π j T ij √ π i . This manuscript is for review purposes only. ecall the detailed balance of T and its stationary distribution, given by T ij π j = T ji π i [4] follows from the reversibility and ergodicity of the BL model. Note that Z is thesymmetric version of the transition matrix: Z ij = 1 √ π j π j T ij √ π i (4.2) = 1 √ π j T ji √ π i (4.3) = Z ji . (4.4)Therefore, Z has an orthonormal set of left eigenvectors, w Tk . Let W be a matrixwhose columns are w k . The spectral decomposition of Z by left eigenvectors is givenby(4.5) Z = W Λ W T . By the definition of Z ij , the transformation from T to Z can be expressed as(4.6) D − T D = Z, where D is a diagonal matrix whose diagonal entries are √ π i . So, arbitrary powersof T is given by(4.7) T m = DW Λ m ( D − W ) T . Defining a new transformation by S, (4.8) T m = S Λ m S − . Since W has a specific normalization, we can equate S with DW after applying theappropriate normalization for S . That is, for diagonal matrix ∆, we take(4.9) S ∆ = DW.
We can choose any normalization for the right eigenvectors given in S , and ∆ willproperly renormalize them. Here, we solve for ∆ and W by appealing to the orthog-onality of W :(4.10) W T W = ∆ S T D − S ∆ = I. Therefore(4.11) S T D − S = ∆ − . Computing the matrix multiplication on the left side yields the diagonal entriesof ∆ denoted by ∆ k given by(4.12) ∆ − k = N X i =0 π i c k ( i ) This manuscript is for review purposes only. n terms of the right eigenvectors of T BL . Now that we have ∆, we have the repre-sentations for both the left-eigenvectors w k ( i ) and right-eigenvectors v k ( i ) of Z givenby w k ( i ) = ∆ k √ π i c k ( i )(4.13) v k ( i ) = 1 √ π i w k ( i ) = ∆ k π i c k ( i )(4.14)in terms of the right-eigenvectors c k ( i ) of the original T BL that was obtained by ourmethod in section 2.From Eq. (4.11), we also have an explicit formula for S − given by(4.15) S − = ∆ S T D − . So, by Eq. (4.8), we have(4.16) T m = S Λ m ∆ S T D − . Computing the matrix multiplication gives the following spectral decomposition(4.17) T ( m ) ij = 1 π j N X k =0 ∆ k λ mk c k ( i ) c k ( j ) . as the explicit representation of P r { n ( m ) = i | n (0) = j } in the BL model.Since T = I , take m = 0 in Eq. (4.17) to find the stationary distribution of theBL model(4.18) π j δ ij = N X k =0 ∆ k c k ( i ) c k ( j ) . Note that this is the orthogonality relation for the right-eigenvectors of T BL withorthogonal measure ∆ k given in Eq. (4.17). We have derived the orthogonal measure∆ k in which the dual Hahn are an orthogonal polynomial system [10].We summarize these results on the derivation of a non-standard inner product ororthogonal measure in which the original transition matrix of the BL model becomesa symmetric real matrix and the (right) eigenvectors are the system of orthogonaldual Hahn polynomials: Theorem
The orthogonal measure in (4.12) symmetrizes T BL , is related tothe (left) and (right) eigenvectors of T BL by (4.13) and (4.14), and yields the spectraldecomposition (4.17).
5. Bounds of mixing times - elementary proofs.
We will discuss first thecase N = N = N/ , for our method gave the eigenvalues of the BL model to be λ k = 1 − k ( N − k +1) N , λ = 1 − N . A heuristic estimate of the number of switches q needed to mix the colors is thus,(5.1) (1 − N ) q ≃ e − qN = 1 N , This manuscript is for review purposes only. nd therefore q = N log N gives an idea of how many switches or time steps areneeded until the variation distance between ρ q and π is of order O (1 /N ) . The lowerbound is obtained along the lines of Diaconis et al, that is, by an application ofthe Chebyshev’s inequality. However, all estimates of the relevant mean values andvariance needed to apply the Chebyshev’s inequality are constructed explicitly fromthe properties of the eigenvalues and eigenvectors of the BL model. We will state thetheorems below but since the proofs are similar to those in [7], we have provided thedetails in an appendix.
Theorem
For m = N log N + ( c − log 28 ) N, for c > , there is a universalconstant A > such that E π [ k ρ m ( j ; . ) − π k V ] ≤ Ae − c . Theorem If m = N ln N − cN , then || ρ m − π || V ≥ − e c
6. Exact calculations of mixing times.
Given that we can calculate ρ m ( i, j )exactly by Eq. (4.17), and the stationary distribution is given by(6.1) π i = (cid:0) N w i (cid:1)(cid:0) N − N w N − i (cid:1)(cid:0) NN (cid:1) , the total variational distance can be exactly computed for all time steps by(6.2) k ρ − π k V = 12 N/ X i =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i N/ X k =1 λ mk v k ( i ) v k (0) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) This solution is shown in Figure 1, with the upper bound given in Figure 2. . . . . . Time T o t a l V a r i a t i on Fig. 1 . Plot of the exact solution for the total variational distance for N = N = N w = N b =100 . This manuscript is for review purposes only. .0 0.2 0.4 0.6 0.8 1.0 1.2 . . . c T V Total VariationUpper Bound
Fig. 2 . Plot of the exact solution for the total variational distance, and the upper bound givenin Theorem 5.1
Appendix A. Proofs of Theorems 5.1 and 5.2.
The eigenvalues and eigen-vectors of the BL problem can be used to construct an upper bound on the variationdistance between the probability distribution after m . As we proved in section 3, theright-eigenvectors of the original matrix T BL are not orthogonal in the standard innerproduct of R N but a different inner product weighted by ∆ k can be used to deriveorthogonality of a related system of right-eigenvectors v j of T = T BL . A.1. Upper bound.
Let j = 0 to define ρ m ( i ; 0) = π i + π i N/ X k =0 λ mk v k ( i ) v k (0)where Pr { n w ( m ) = i | n w (0) = j } = T mij = ρ m ( j ; i )(A.1) = N/ X k =0 π i λ mk v k ( i ) v k ( j )(A.2) = π i + π i N/ X k =1 λ mk v k ( i ) v k ( j ) . (A.3)Then, ρ ( i ; j ) = δ ij , and hence(A.4) N/ X k =1 v k ( i ) = 1 π i − < π i , implies that This manuscript is for review purposes only. ρ m ( i ; 0) − π i k V = 12 N/ X i =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i N/ X k =1 λ mk v k ( i ) v k (0) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (A.5) ≤ λ m N/ X i =0 π i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N/ X k =1 v k ( i ) v k (0) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (A.6) ≤
12 ( N ) − / e − c N/ X i =0 π i N/ X k =1 v k ( i ) / N/ X k =1 v k (0) / (A.7) ≤
12 (
N π ) − / e − c N/ X i =0 ( π i ) / (A.8)Using(A.9) N/ X i =0 ( π i ) / = O ( N / )and averaging over initial data n w (0) = j, j = 0 , ..., N/ , we have E π [ k ρ m ( j ; i ) − π i k V ] ≤ N/ X j =0 π j k ρ m ( j ; . ) − π k V (A.10) ≤ N − / e − c N/ X j =0 ( π j ) / N/ X i =0 ( π i ) / (A.11) ≤ Ae − c (A.12)for some A independent of N. This proves theorem 5.1.
A.2. Lower bound.
In terms of the right-eigenvectors v k , k = 0 , ..., N/ , (withcol sum =1 )Pr { n w ( m ) = i | n w (0) = j } = T mij = ρ m ( i ) if we take j = 0(A.13) = N/ X k =0 π i λ mk v k ( i ) v k ( j )(A.14) = π i + π i N/ X k =0 λ mk v k ( i ) v k ( j )(A.15) This manuscript is for review purposes only. nd E ρ m [ v ( i )] = N/ X i =0 v ( i ) ρ m ( i ) = N/ X i =0 π i v ( i ) N/ X k =0 λ mk v k ( i ) v k (0)(A.16) = N/ X i =0 N/ X k =0 λ mk π i v ( i ) v k ( i ) v k (0)(A.17) = N/ X k =0 v k (0) λ mk N/ X i =0 π i v ( i ) v k ( i )(A.18) = v (0) λ m N/ X i =0 π i v ( i ) v ( i )(A.19) = v (0) λ m = v (0) (cid:18) − N (cid:19) m (A.20) E π [ v ( i )] = 0; var π { v ( i ) } = 1 . (A.21)Next for m = N log N − c N , we get E [ v ] = v (0) √ N e c . A similar calculation gives E ρ m [ v ( i )] = N/ X i =0 v ( i ) ρ m ( i ) = N/ X i =0 π i v ( i ) N/ X k =0 λ mk v k ( i ) v k (0)(A.22) = N/ X i =0 π i v ( i ) N/ X k =0 λ mk v k ( i ) v k (0) = N/ X i =0 N/ X k =0 λ mk π i v ( i ) v k ( i ) v k (0)(A.23) = N/ X k =0 v k (0) λ mk N/ X i =0 π i v ( i ) v k ( i ) = v (0) λ m N/ X i =0 π i v ( i ) v ( i )(A.24) = v (0) λ m ∼ v (0) (cid:18) − N (cid:19) m , (A.25) E π [ v ( i )] =0; var π { v ( i ) } = 1(A.26)Next we deduce v (0) and v (0) from v = Av + B , v ( i ) = C ( N/ − i ), and theorthogonality of v i :(A.27) 1 = N/ X i =0 v ( i ) π i = N/ X i =0 C ( i − N/ π i ∼ C N C = √ N , we fine v (0) ∼ √ N . Furthermore, Av (0) = v (0) − b .Now, we have V ar ρ m { v } = E ρ m [ Av + B ] − N λ m (A.28) =( N − B ) λ m + B − N λ m ∼ B (1 − λ m )(A.29)So with the same normalization as above for v , we deduce V ar { v ′ } is uniformlybounded by constant 2 b, since B = b + O (log N/N ) . Now, by Chebyschev’s inequality,(A.30)
P r π {| v | ≤ k } ≥ − k This manuscript is for review purposes only. nd P r ρ m {| v | ≤ k } ≤ P r ρ m { v ≤ k } (A.31) = P r ρ m { E ρ m [ v ] − v ≥ E ρ m [ v ] − k } (A.32) ≤ P r ρ m {| E ρ m [ v ] − v | ≥ | E ρ m [ v ] − k |} (A.33) = P r ρ m { ( E ρ m [ v ] − v ) ≥ ( E ρ m [ v ] − k ) } (A.34) ≤ V ar ρ m ( v )( E ρ m [ v ] − k ) (A.35) ≤ B ( √ N λ m − k ) (A.36)Thus, if K ⊂ { , ...., N/ } such that | v | ≤ k for k ∈ K, we deduce2 k ρ m − π k V = N/ X i =0 | ρ m (0 , i ) − π i | (A.37) ≥ X K | ρ m (0 , i ) − π i | (A.38) ≥ X K π i − X K ρ m (0 , i )(A.39) ≥ − k − B ( √ N λ m − k ) (A.40)Choose k = d √ Nλ m to obtain2 k ρ m − π k V ≥ − (cid:20) d + B (1 − d ) (cid:21) N λ − m (A.41) ≥ − be c , (A.42)which proves theorem 5.2. REFERENCES[1]
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