Convergence in Distribution of Products of I.I.D. Nonnegative Matrices 1

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1 Journal of Theoretical Probability, Vol. 10. No Convergence in Distribution of Products of I.I.D. Nonnegative Matrices 1 S. Dhar 2 and A. Mukherjea 2 Let (X i ) be a sequence of m x m i.i.d. stochastic matrices with distribution u. Then u" is the distribution of X n X n-1,... X 1. Simple sufficient conditions for the weak convergence of (u") are presented here. An extremely simple (and verifiable) necessary and sufficient condition is provided for m = 3. The method for m = 3 works for m > 3 even though calculations are more involved for higher values of m. We also discuss the purity of the limit distribution for m > 2. KEY WORDS: Random matrices; convergence in distribution; weak convergence. 1. INTRODUCTION In this paper, we consider an old problem of convergence in distribution of products of m x m i.i.d. stochastic matrices (or equivalently, the problem of weak convergence of the convolution iterates u n of a probability measure u on m x m stochastic matrices). Many years back, it was looked into by Rosenblatt, (7) and later, by many other authors. (1) It is well-known that when m 2, u n converges weakly iff u is not the unit mass at the stochastic matrix where the diagonal elements are all zeros. As far as we know, such simple (and complete) results are not known for m > 2. A similar complete result is presented here for m = 3. Our method will also show that with extra work, one can get similar results for higher values of m. Theorem 2 describes the result for m = 3. This result motivated us to obtain a very simple sufficient condition in the general m x m case for the weak convergence of (u"), namely, that (u") converges weakly whenever S u contains a stochastic matrix with at least m 1 positive diagonal elements. This theorem remains true, even when S u consists of m x m nonnegative (where 1 Received June 1, 1995; revised August 14, University of South Florida, Tampa, Florida ( math.usf.edu) /97/ $ 12.50/0 C 1997 Plenum Publishing Corporation

2 376 Dhar and Mukherjea each entry is nonnegative, but the row sums need not be one as in the stochastic matrix case), provided the sequence (u n ) is tight, that is, every weak*-cluster point of (u n ) is a probability measure. As far as we know, all results in this paper are new. We present the m x m case in Section 2 and the 3 x 3 case in Section 3. In Section 4, we deal with the limiting distribution, and describe when it is pure and continuous. At the end of this section, we present an example showing how the limit can be computed in certain non-obvious cases. Finally, we must mention (along with Rosenblatt's work in [7], see also [8]) the paper by Kesten and Spitzer, (2) that considered the convergence problem in nonnegative matrices. Aside from tightness, the conditions imposed (2) were on the measure u. itself, not on the support of u (as in this paper). 2. CONVERGENCE IN THE SET OF m x m STOCHASTIC MATRICES If S is the closed multiplicative semigroup generated by the support S u of a probability measure u on m x m stochastic matrices, then S is compact (in the usual euclidean topology of matrices), so that the convolution sequence u n, defined, as usual, inductively by is tight (that is, weak*-relatively compact). This is, of course, no longer true when S is not compact, as is usually the case when S consists of (nonstochastic) nonnegative matrices. However, when S is not compact, there are results (see Refs. 4 and 5) which provide necessary and sufficient conditions for (u n ) to be tight. We can now present the following convergence theorem. Theorem 1. Let n be a probability measure on the multiplicative semigroup of m x m nonnegative matrices such that the convolution sequence (u n ) is tight. Suppose that S f, contains a matrix which has at least m- 1 positive diagonal entries. Then (u n ) converges weakly. Proof. Suppose that S u contains an m x m matrix y which has at least m 1 positive diagonal entries. Let S = U n-1 S n and K be the kernel of S. Note that if A is the weak limit of 1/n E k = 1 u k, then S l = K. Let us now suppose, if possible, that the sequence (u n ) does not converge weakly. Then by Theorem 2.1 in Ref. 3, there exists an idempotent e and a proper normal subgroup H of the compact group eke, where e is an idempotent element in K such that es u e <= gh for some g e eke H.

3 Convergence in Distribution of Products, of l.l.d. Nonnegative Matrices 377 Now we exploit the structure theory of compact groups of nonnegative matrices as given in Ref. 6. Since e is idempotent, there is a partition of {1,2,..., m} such that where p is the rank of the matrix e; also, Also, it is given in Ref. 6 that there is an isomorphism from the compact group eke to a subgroup of the group of permutations on {1,2,..., p}. According to this isomorphism, if M e eke and n is the permutation, which is the isomorphic image of M, then the (C k1 x C k2 )-block of M is an all-zero block, whenever n(k 1 ) +k 2. Now yes f and y has at least m 1 positive diagonal elements. Let M = eye. Let card T = u, where T is one of the members of the partition corresponding to e as defined earlier. Then since and e ii > 0 whenever i T, it is clear that M ii > 0 for at least m 1 u different values of i. Note that if u = m 1, then p = 1, and whenever p = 1, the rank of all matrices in K is one; consequently, the compact group eke must then be a singleton and therefore, since then it cannot have a proper normal subgroup, the sequence (u") must converge. Thus, we can assume that p> 1 and u<m 1. Now we observe that for any k 1,k 2, (k 1 =k 2 ), 1 < k 1 <p, 1 <k 2 <p, in the (C k1 u C k2 ) block of M, there is at most one diagonal element which is zero. Let n be the permutation on {1,2,..., p} corresponding to M in the previously-mentioned isomorphism. If n is not the identity permutation, then there will be k 1,k 2 such that k 1 =k 2, 1 < k1 <p, 1 <k 2 <p and p(k 1 ) = k 2 =k 1 so that n(k 2 )=k 2. Thus, the (C k1 x C k1 ) block as well as the (C k2 x C k2 ) block of M consists of only zero elements. This contradicts that the (C K1 u C k2 ) x (C k1 u C k2 ) block of M has at most one diagonal element which is zero. Thus, the isomorphic image of M must be the identity permutation, so that M must be an idempotent. But M=eye e es u e, so that es u e contains the identity of the group eke (note that es u ece De = eke), and this means that es u e cannot be contained in a coset of a proper normal subgroup of eke. It follows that the sequence (u") must converge weakly.

4 378 Dhar and Mukherjea Note that we cannot relax the assumption in the theorem from m 1 to m 1. A trivial example is when m = 2 and u is the unit mass on the matrix (0 1). 3. CONVERGENCE IN THE SET OF 3 x 3 STOCHASTIC MATRICES Let S u be contained in 3 x 3 stochastic matrices and let S be as before. First, let us assume that K, the kernel of S consists of only rank 2 matrices. [Recall that (see Ref. 1) K= the set of matrices in S with the minimal rank.] Let P 1, P 2, P 3, P 4, P 5, P 6 be the six 3 x 3 permutation matrices. Then there are exactly 2 (distinct) types of 3 x 3 (rank two) idempotent stochastic matrices, which are where 0 <a <1. By this, we mean that given any other 3x3 rank two idempotent stochastic matric matrix C, there is a permutation matrix P i such that C is either P -1 AP i or P -1 BP i, where A and B are of the forms given earlier. Recall (see Ref. 6) that eke is a compact group of stochastic matrixes of rank 2 (e being an idempotent in K), and as such, it is either the singleton {e} or {e, f}, where {e, f} is isomorphic to S 2 (the group of permutations on two elements). It is not difficult to prove the following assertions: When e = A as in (i) before, then When e = B as in (ii) previously, then Proposition 1. Suppose G is a (multiplicative) group of mxm stochastic matrices with k! elements such each matrix in G has rank k, where 1 < k <m. Let e be the identity of G, Then, if H is also such a group of m x m stochastic matrices each with rank k such that H has k! elements and e (the identity of G) is also the identity of H, then G = H.

5 Convergence in Distribution of Products, of I.I.D. Nonnegative Matrices 379 Proof. We use Theorem 4.1 in Ref. 6. By this theorem, there is an isomorphism from G (and one also from H) onto the group of permutations on {1, 2,..., k} such that for each g e G, if n is the corresponding permutation and if is the usual partition of {1, 2,..., m] for the idempotent stochastic matrix e such that then strictly positive block with each row sum 1 and g CsXCu is a block of all zeros (when u = t). Let us consider g e G such that n is the corresponding permutation. We will then show that g is uniquely determined by e (the identity of G) and n. First, let i e C s, 1<s < t. Let n(s) = t. Then ij = 0 if j EC t. If j e C t, then Let i e T and j e C s. Let n -1 (v) = s. Then and

6 380 Dhar and Mukherjca so that for i e T and j e C s, In other words, the identity e of G and the permutation on {1, 2,..., k} uniquely determines each elements in G. Thus, G = H. Let us now assume that (u n ) does not converge weakly. In what follows, we will try to identify S u to the maximum extent possible using a theorem in Ref. 3. By this theorem, then there exists some idempotent e e K such that es u e must be a singleton and coincide with the nonidentity element in eke. [Note that in this case, eke must be a 2-element group, and with no loss of generality, we can assume that either (3.1) or (3.2) occurs. If eke were a singleton, then it would be {e} and consequently, (u n ) must converge. ] We separate the discussion into two cases. Case 1. Suppose that (3.1) occurs. In this case, and for every x e S^, where 0 < a < 1. Let us first assume that 0 < a < 1. In this case, it is easily verified by direct computations that for x e S u to satisfy (3.3), x must be of the form so that in this case, we must have:

7 Convergence in Distribution of Products, of I.I.D. Nonnegative Matrices 381 Let us now assume a = 1, so that for every x e S u, Again, by direct computations, it is easily verified that in this case, Let us now simplify the set on the right further. Choose two elements y 1, y 2 in S u from the set on the right such that Then it follows that which means that since ey 1 y 2 e e KeKe. This means that if there is an element y 1 es U where w < 1, then every matrix in S U must be of the form Thus, when, we have:

8 382 Dhar and Mukherjea Either or Let us now consider the last possibility in this case, namely, when Note that if then T -1 et is the same idempotent matrix we just considered. In other words, when (3.8) holds, we have either or To summarize, in this case, there are exactly three possibilities; either S u consists of only matrices of the form given in (3.10) (or (3.4) or (3.7)) or S u consists of only matrices of the form given in (3.6) or S u consists of only matrices of the form given in (3.9).

9 Convergence in Distribution of Products, of I.I.D. Nonnegative Matrices 383 Note that when (3.10) or (3.4) or (3.7) occurs, then we have for all n >1, and When (3.6) occurs, then we have for all n > 1, and Finally, when (3.9) ) occurs, then we have for all n > 1, and Case 2. Suppose that (3.2) occurs so that e = B and

10 384 Dhar and Mukherjea When a = 0, Note that if then T -1 et is exactly the idempotent in (3.8). It follows that when (3.18) occurs, either or Let us take a = 1 in (3.17). Then we have: eke = {e, f }, where Note that this was covered in Case 1 (see (3.8)), and in this case, either (3.9) or (3.10) occurs. Finally we consider in (3.17), 0 < a < 1. In this case, it follows after direct computations that which is the same as (3.19) (or (3.9)).

11 Convergence in Distribution of Products, of I.I.D. Nonnegativc Matrices 385 Note that when (3.20) occurs, then we have for all n > 1: and Our final observations before we state our theorem in the case when the rank of the mtrices in K is two are the following. Let us define the permutation matrices P 1, P 2, and P 3 as follows: Let A and B be the idempotent matrices (with rank two) as defined in the beginning of this section. Then every other type of rank two idempotent stochastic matrix can be obtained as follows: Also, note that in the preceding discussion, we came up with eaxtly four distinct possibilities (including both Cases 1 and 2) for S u, namely, (3.4), (3.6), (3.9), and (3.20), when (u n ) does not converge weakly. It was also noted in (3.11)-(3.16), (3.23), and (3.24) that in each of these four possibilities (namely when (3.4), (3.6), (3.9) or (3.20) occurs), the sets U n=1 S 2n and U n=1 S 2n+1 are contained in two disjoint closed sets so that whenever (3.4), (3.6), (3.9), or (3.20) occurs, the sequence (u n ) cannot converge weakly.

12 386 Dhar and Mukherjea Let us also note that the four possibilities (3.4), (3.6), (3.9), and (3.20) will lead to exactly six possibilities when we take into account the transformations p 1, p 2 and p 3. where p i (M) = P -1 MP i, i= 1, 2, 3. Let us now consider the case when the rank of the matrices in K is one, where K is still the kernel of S with S and S u as before. Note that in this case, (u n ) must converge weakly. The reason is the following: If A, and A 2 are two weak cluster points of the sequence (u n ), then A 1 * A 2 = A 2 * A 1 and S A1 u S A2 c K, since lim n- >I u n (G) = 1 for any open set G containing K (by Theorem 1 in Ref. 7). Since K has only rank one stochastic matrices, for any two matrices x and y in K, xy = y so that A 1 * A 2 = A 2 and A 2 * A 1 = A 1. It follows that A 1 = A 2. Now suppose that the rank of the matrices in K is 3. In this case, K= S is a compact group of 3 x 3 stochastic matrices, and as such, S is a finite group which is isomorphic to a subgroup of permutations on {1, 2, 3}. In this case, if S u has more than one element, then (u n ) does not converge weakly iff Note that when S u has only one element x of rank 3 and x generates a finite group, then u n cannot converge unless x is the identity matrix. The preceding discussion leads to the following theorem. Theorem 2. Let u be a probability measure on 3 x 3 stochastic matrices such that S fl has more than one element. Then the sequence u n does not converge weakly iff S u is contained in one of the following seven sets of matrices:

13 Convergence in Distribution of Products, of I.I.D. Nonncgativc Matrices PURITY OF THE LIMIT DISTRIBUTION Suppose that Su is contained in d x d stochastic matrices each with rank d such that the kernel K of S, the closed multiplicative semigroup generated by St, consists of rank one stochastic matrices. Then, as we saw earlier, un converges weakly to a limit probability measure v and Sx = K. Then we claim that either v is discrete or continuous. [Note that v is discrete iff v(e) = 1 for some countable set E, and v is continuous iff v{x} = 0 for each x e S. ] To see this, let us define the set E by Define: v1(a) = v(a re), that we can write v2(a) = v(a r\e'). Then, v = v1 + v2. This means where 0 < a < 1, 0 < B <1, a + B = 1, v11 is a discrete probability measure and v12 is a continuous probability measure. Notice that v12*u is continuous since v12 is continuous and every matrix in Su is invertible. Thus, we have 860 f1 2-8

14 388 Dhar and Mukherjea Now we can decompose v11 * u as follows; where 0 < r1 1, 0 <r2 1< 1, r1+ r2=1. V13, is a discrete probability measure, and v14 is a continuous probability measure. Suppose that a >0 in (4.1). Then we have from (4.2) and (4.3) that so that r1 = 1, r2 = 0, and consequently, Thus, if a > 0 in (4.1), then v=v11. Consequently, either v is discrete or v is continuous. Now notice that every matrix in K has identical rows, where each row is of the form so that the points in K can be regarded as elements in Rd-1. Let md-1 be the (d 1 )-dimensional Lebesgue measure on Rd-1. We now show that when v is continuous, then v is either singular or absolutely continuous with respect to md-1. To prove this, we can write (using the classical Lebesgue decomposition theorem) that where 0 < a < 1, 0 < b < 1, a+ b = 1, v3 J md-1, and v4 < md-1. Note that Also, if y is a d x d inverticle stochastic matrix such that its ith row is and if y* is a (d 1) x (d 1) matrix such that its ith row is

15 Convergence in Distribution of Products, of I.I.D. Nonnegative Matrices 389 then it follows easily that det y* = 0. The reason is that if there exist constants c i (note all zero) such that then, since y is a stochastic matrix, we must also have: so that det y = det y** = 0, where y** is the matrix whose dth row is the same as the dth row of y, and for 1 < i <d 1, (y* *) ij = y ij y dj. Now let us denote, by B*, the set {(b 1, b 2,..., b d-1 ): (b 1, b 2,..., b d-1 ): 1 b 1 b 2... b d-1 )}, is one of the rows of some element in B}, where B is a set of stochastic matrices each with identical rows. Notice that Wherever we write m d-1 (B), we actually mean m d-1 (B*), where B is a set as earlier. If Ey -1 = B, then E = B.y, so that Thus, it follows that v 4 * u < m d-1. We can also write: where 0< c < 1, 0 < d <1, c + d= 1, v 5 1 m d-1 and v 6 < m d-1. From (4.4) and (4.5), we have: so that

16 390 Dhar and Mukherjea Thus, if a > 0, then c = 1, d =0, and v 3 = v 5 = v 3 * u. This means that v 3 * u n = v 3 (for n > 1) so that v = v 3 * v = v 3. Thus, v is either continuous singular or absolutely continuous. Finally, we show that when S has at least two rank one stochastic matrices, and S u consists of only weakly ergodic stochastic matrices (with full rank), then v is continuous. To prove this, notice that S,, is the kernel K of S and has more than one point. Suppose that for every x e S u, the limit points of (x") have rank one, and if possible, let Then there exists x 0 e K such that v{x 0 } = d. Since v = v * u, we have: Thus, for (u)-almost all points y in S u. This means that since {z v({x 0 } z -1 ) «x} is open Va real. Now if Vy e S u, {x 0 } = {x 0 } y -1, then x 0 S u =x 0 so that x 0 S = x 0 ; this means that x 0 K=x 0. But x 0 z = z whenever z is a rank one stochastic matrix so that x 0 K=x 0 means that K= {x 0 }, singleton. This is a contradiction. Thus, 3y e S u B {x 0 } = (x 0 ) y -1. But then The reason for this is the following. Suppose x 0 y m = x 0 for some m >1. Then x 0 y mn = x 0 Vn > 1. Let y mnk -> u. Then u has rank 1, by hypothesis. Thus, u = x 0. Also, x 0 y mn+1 =x 0 y or (x 0 y) y mn = x 0 y implying that u = (x 0 y)u = x 0 y. This contradicts that {x 0 } ={ x 0 } y -1 Thus, we have an infinite number of singleton sets ({x 0 } y ") n=1 such that each has v-measure S. This contradicts that v(s) = 1. Hence, v must be continuous. Thus, we have proven the following theorem. Theorem 3. Let u be a probability measure on d x d stochastic matrices (d > 2) such that S u consists of only invertible matrices and the

17 Convergence in Distribution of Products, of I.I.D. Nonnegative Matrices 391 kernel K of S, the closed multiplicative semigroup generated by S u, consists of rank one matrices. Then, u n converges weakly to a probability measure v, where v has eactly one of the following three properties: (1) v is discrete, that is, v(e) = 1 for some countable set E; (2) v is continuous and singular with respect to the (d- 1)-dimensional Lebesgue measure m d-1 on K (considered as a subset of R d-1 ); (3) v < m d-1. Also, if, furthermore, the matrices in S u are weakly ergodic and v is not the unit mass at a singleton, then v is continuous. We end this paper with an interesting example where the kernel is neither a group nor does it consist of rank one matrices. Let u be a probability measure on 3 x 3 stochastic matrices such that where 0 < b< 1, and Then it is easily verified that the kernel K of S( = U n=1 = i S n ) is a two point group (when a = 5), and is a four point (completely simple) semigroup (when a =1) given by K= {e, g, g', e'}, where By our results in Section 3, u n converges weakly to a probability measure L We will now explicitly determine L When a =1, K is a two point group and A is, of course, the uniform distribution on K. The situation is less trivial when a = 1 and 0 < a < 1. We will utilize the fact that and the structure of idempotent probability measures. Now let u{g} = b and u{x} =1 b. Notice that

18 392 Dhar and Mukherjea so that by Theorem 2.8, p. 79 in Ref. 1, and L = L 1 x L 2 x L 3, where L 1 (e) = 1, L 2 (e) = L 2 (g) =1, L 3 is a probability measure on {e, e'}. Thus, we have: and since e' = (e, e, e'), so that Consequently, Since L * u = L, NOTE ADDED IN PROOF Recently the authors have been informed that Theorem 2 of this paper has also been obtained independently (using a different method) by Santanu Chakravarty and B. V. Rao of the Indian Statistical Institute. It may be noted, however, that the method here works also for n x n stochastic matrices with n > 3.

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