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Unformatted text preview: O.H. Probability and Markov Chains MATH 2561/2571 E09 1 Solving recurrence relations You know at least two different methods for solving linear homogeneous recurrence relations, eg., relations of the form xn+1 = a xn + b xn1 , with some initial conditions x0 , x1 . Method 1. Consider a vector Xn = (xn+1 , xn )T ; then Xn+1 = ie., Xn+1 = AXn , A straightforward induction gives Xn = xn+1 xn = An X0 = An x1 x0 where A = a b 1 0 . xn+2 xn+1 = a xn+1 + b xn xn+1 = a b 1 0 xn+1 xn , n 1, a, b R , (1.1) so we have to compute powers of the matrix A. Using the same approach as in the Markov Chain part of the course, we solve the characteristic equation, det(A  I) = (a  )()  b = 2  a  b = (  a/2)2  (b + a2 /4) = 0 so that a2 a b+ , 2 4 n n and thus1 for every n 0 we have xn = 1 + 2 ; it remains to recall the initial values x0 and x1 to identify the constants and in the usual way. 1,2 = Method 2. By using the initial conditions x0 = x0 , x1 = ax0 + x1  ax0 , and the recurrence relation (1.1), we deduce that the generating function G(s) of (xn )n0 satisfies the equation G(s) = x0 + (x1  ax0 )s + as G(s) + bs2 G(s) , that is, x0 + (x1  ax0 )s . 1  as  bs2 The gen. function G(s) can now be studied (for small s) in the usual way. G(s) = Remarks. The second method works analogously in the nonhomogeneous case, xn+1 = a xn + b xn1 + c , with a fixed constant c R. The higherdimensional cases can be analyzed similarly, as long as one can find all roots of the char. polynomial det(A  I).
1 Of course, if 1 = 2 = , then xn = ( + n) n . 1 ...
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This note was uploaded on 05/12/2010 for the course APPLIED ST 2010 taught by Professor Various during the Spring '10 term at Universidad Nacional Agraria La Molina.
 Spring '10
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