Discrete-time stochastic processes

8 the expected aggregate reward as a function of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: scale factor) and is the only non-negative non-zero vector (within a scale factor) that satisfies ∏π ≤ π [A]. Proof: A left eigenvector of [A] is a right eigenvector (transposed) of [A]T . The graph corresponding to [A]T is the same as that for [A] with all the arc directions reversed, so that all pairs of nodes still communicate and [A]T is irreducible. Since [A] and [A]T have the same eigenvalues, the corollary is just a restatement of the theorem. 154 CHAPTER 4. FINITE-STATE MARKOV CHAINS Corollary 4.2. Let ∏ be the largest real eigenvalue of an irreducible matrix and let the right and left eigenvectors of ∏ be ∫ > 0 and π > 0. Then, within a scale factor, ∫ is the only non-negative right eigenvector of [A] (i.e., no other eigenvalues have non-negative eigenvectors). Similarly, within a scale factor, π is the only non-negative left eigenvector of [A]. Proof: Theorem 4.6 asserts that ∫ is the unique right eigenvector (within a scale factor) of the largest real eigenvalue ∏, so suppose that u is a right eigenvector of some other eigenvalue µ. Letting π be the left eigenvector of ∏, we have π [A]u = ∏π u and also π [A]u = µπ u . Thus π u = 0. Since π > 0 , u cannot be non-negative and non-zero. The same argument shows the uniqueness of π . Corollary 4.3. Let [P ] be a stochastic irreducible matrix (i.e., the matrix of a recurrent Markov chain). Then ∏ = 1 is the largest real eigenvalue of [P ], e = (1, 1, . . . , 1)T is the right eigenvector of ∏ = 1, unique within a scale factor, and there is a unique probability vector π > 0 that is a left eigenvector of ∏ = 1. Proof: Since each row of [P ] adds up to 1, [P ]e = e . Corollary 4.2 asserts the uniqueness of e and the fact that ∏ = 1 is the largest real eigenvalue, and Corollary 4.1 asserts the uniqueness of π . The proof above shows that every stochastic matrix, whether irreducible or not, has an eigenvalue ∏ = 1 with e = (1, . . . , 1)T as a right eigenvector. In general, a stochastic matrix with r recurrent classes has r independent non-negative right eigenvectors and r independent non-negative left eigenvectors; the left eigenvectors can be taken as the steadystate probability vectors within the r recurrent classes (see Exercise 4.14). The following corollary, proved in Exercise 4.13, extends corollary 4.3 to unichains. Corollary 4.4. Let [P ] be the transition matrix of a unichain. Then ∏ = 1 is the largest real eigenvalue of [P ], e = (1, 1, . . . , 1)T is the right eigenvector of ∏ = 1, unique within a scale factor, and there is a unique probability vector π ≥ 0 that is a left eigenvector of ∏ = 1; πi > 0 for each recurrent state i and πi = 0 for each transient state. Corollary 4.5. The largest real eigenvalue ∏ of an irreducible matrix [A] ≥ 0 is a strictly increasing function of each component of [A]. Proof: For a given irreducible [A], let [B ] satisfy [B ] ≥ [A], [B ] 6= [A]. Let ∏ be the largest real eigenvalue of [A] and ∫ > 0 be the corresponding right eigenvector. Then ∏∫ = [A]∫ ≤ [B ]∫ , but ∏∫ 6= [B ]∫ . Let µ be the largest real eigenvalue of [B ], which is also irreducible. If µ ≤ ∏, then µ∫ ≤ ∏∫ ≤ [B ]∫ , and µ∫ 6= [B ]∫ , which is a contradiction of property 1 in Theorem 4.6. Thus, µ > ∏. We are now ready to study the asymptotic behavior of [A]n . The simplest and cleanest result holds for [A] > 0. We establish this in the following corollary and then look at the case of greatest importance, that of a stochastic matrix for an ergodic Markov chain. More general cases are treated in Exercises 4.13 and 4.14. 4.4. PERRON-FROBENIUS THEORY 155 Corollary 4.6. Let ∏ be the largest eigenvalue of [A] > 0 and let π and ∫ be the positive left and right eigenvectors of ∏, normalized so that π ∫ = 1. Then lim n→1 [A]n = ∫π. ∏n (4.26) Proof*: Since ∫ > 0 is a column vector and π > 0 is a row vector, ∫ π is a positive matrix of the same dimension as [A]. Since [A] > 0, we can define a matrix [B ] = [A] − α∫ π which is positive for small enough α > 0. Note that π and ∫ are left and right eigenvectors of [B ] with eigenvalue µ = ∏ − α. We then have µn∫ = [B ]n∫ , which when pre-multiplied by π yields XX n (∏ − α)n = π [B ]n∫ = πi Bij ∫j . i j n where Bij is the i, j element of [B ]n . Since each term in the above summation is positive, n n we have (∏ − α)n ≥ πi Bij ∫j , and therefore Bij ≤ (∏ − α)n /(πi ∫j ). Thus, for each i, n ∏−n = 0, and therefore lim n −n = 0. Next we use a convenient j , limn→1 Bij n→1 [B ] ∏ matrix identity: for any eigenvalue ∏ of a matrix [A], and any corresponding right and left eigenvectors ∫ and π , normalized so that π ∫ = 1, we have {[A] − ∏∫ π }n = [A]n − ∏n∫ π (see Exercise 4.12). Applying the same identity to [B ], we have {[B ] − µ∫ π }n = [B ]n − µn∫ π . Finally, since [B ] = [A] − α∫ π , we have [B ] − µ∫ π = [A] − ∏∫ π , so that [A]n − ∏n∫ π = [B ]n − µn∫ π . (...
View Full Document

This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online