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pset7sol

# pset7sol - 18.06 Problem Set 7 Solutions Problem 1 Do...

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18.06 Problem Set 7 Solutions Problem 1: Do problem 11 from section 8.3. Solution A Markov matrix must conserve probability. Hence the columns must sum to 1: A = . 7 . 1 . 2 . 1 . 6 . 3 . 2 . 3 . 5 (1) The steady state vector x satisfies Ax = x . In otherwords, x is an eigenvector of A with eigenvalue 1. Therefore x solves: ( A - I ) x = - . 3 . 1 . 2 . 1 - . 4 . 3 . 2 . 3 - . 5 x 1 x 2 x 3 = 0 (2) 0 - . 11 . 11 . 1 - . 4 . 3 0 . 11 - . 11 x 1 x 2 x 3 = 0 (3) (4) Take x 3 = 1, then x 2 = 1 and . 1 x 1 = . 4 - . 3 = . 1 so x 1 = 1 and x = (111) T . In general, a symmetric Markov matrix A T = A has a steady solution x = (111) T . This follows from: A is Markov columns of A sum to 1. A T = A rows of A sum to 1. The vector x = (111) T sums the row elements. Hence, Ax = x . Problem 2: Do problem 12 from section 8.3. Solution B = ( A - I ) x = - . 2 . 3 . 2 - . 3 (5) 1

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The eigenvalues satisfy the characteristic equation: ( - . 2 - λ )( - . 3 - λ ) - ( . 2)( . 3) = 0 (6) λ ( λ + . 5) = 0 (7) The eigenvalues are λ = 0 and λ = - . 5. In general, when A is Markov, A - I will have a λ = 0 eigenvalue. Specifically, this follows from the (more general) fact that if μ is an eigenvalue of any matrix A , then μ - c is an eigenvalue of A - cI . Markov matrices have an eigenvalue of 1, hence A - I must have an eigenvalue 1 - 1 = 0. The eigenvectors of B are, x 1 = ( . 3 , . 2) T corresponding to λ = 0, and x 2 = (1 , - 1) T corresponding to λ = - 0 . 5. A general solution to the ODE du dt = ( A - I ) u has the form: u = c 1 e 0 · t x 1 + c 2 e - . 5 t x 2 (8) Here c 1 and c 2 are integration constants (determined by initial values). As t → ∞ , u becomes: u c 1 x 1 (9) Problem 3: Do problem 16 section 8.3. Solution A = . 4 . 2 . 3 . 2 . 4 . 3 . 4 . 4 . 4 (10) Since A is a Markov matrix, we know λ = 1 is an eigenvalue. In addition, detA = 0, so λ = 0 must also be an eigenvalue (ie det ( A - 0 · I ) = 0 ). The third eigenvalue λ = 0 . 2 can be found by i ) inspection, ii ) using MATLAB , iii ) by direct computation. Using MATLAB, we can diagonalize A = S Λ S - 1 , where the columns of S are eigenvectors of A: S = . 5145 . 7071 - . 4082 . 5145 - . 7071 - . 4082 . 686 0 . 8165 (11) Λ = 1 0 0 0 . 2 0 0 0 0 (12) 2
Now A k = S Λ k S - 1 , so: A k u 0 = ( S Λ S - 1 ) k u 0 (13) = S Λ k S - 1 u 0 (14) (15) Since S - 1 = . 5831 . 5831 . 5831 . 7071 - . 7071 0 - . 4899 - . 4899 . 7348 (16) then S - 1 1 0 0 = . 5831 . 7071 - . 4899 (17) When we expand out A k u 0 , we have: A k u 0 = . 5831(1) k . 5145 . 5145 . 686 + . 7071(0 . 2) k . 7071 - . 7071 0 + ( - . 4899)(0) k - . 4082 - . 4082 . 8165 (18) Hence, as k → ∞ , A k u 0 . 5831( . 5145 , . 5145 , . 686) T . Similarly, for u 0 = (100 , 0 , 0) T = 100(1 , 0 , 0) T

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