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Math 554 Introduction to Stochastic Processes Instructor: Alex Roitershtein Iowa State University Department of Mathematics Fall 2015 Solutions to Homework #1 0.4 The corresponding characteristic equation is λ= 1 1 + λ2 + λ3 . 3 Notice that λ = 1 is a solution. To find other two solutions, write 1 1 1 − 3λ + λ2 + λ3 = (λ − 1)(λ2 + aλ − 1). 3 3 Comparing the coefficients in front of λ2 on both the sides, we obtain a = 2. The roots of λ2 + 2λ − 1 = 0 satisfy the equation (λ + 1)2 − 2 = 0. These roots are λ1,2 = −1 ± Thus a general solution of the recursion is √ n √ n f (n) = c1 + c2 −1 + 2 + c3 −1 − 2 . √ 2. Since limn→∞ f (n) = 1, we have f (n) = 1 + c2 −1 + √ n 2 . Finally, since f (0) = 0 we obtain f (n) = 1 − √ n 2−1 . 0.5 It is not hard to verify that f (n) = n2 is a solution. Also, if g(n) is a difference of two solutions as defined in the textbook, then g satisfies the following homogeneous equation: g(n) = 1 g(n + 1) + g(n − 1) . 2 The corresponding characteristic equation is (λ − 1)2 = 0. Thus (guess using the analogy with ODE and verify your guess), g(n) = c1 + c2 n. 1 Thus a general solution for f is given by f (n) = g(n) + c3 n2 = c1 + c2 n + n2 . 0.6 (a) The corresponding characteristic equation is λ2 + λ + 1 = 0. Its roots are √ 1 −1 ± −3 . 2 Thus the general solution of the ODE is given by h √3 √3 i t + c2 sin t . f (x) = e−t/2 c1 cos 2 2 λ1,2 = 0.6 (b) A general solution in the complex numbers is √ n √ n 1 f (n) = c1 λn1 + c2 λn2 = n c1 −1 − i 3 + c2 −1 + i 3 . (1) 2 Indeed, it is easy to see that both f (n) = λn1 and f (n) = λn2 are solutions. Hence, since the equation is linear, f (n) in (1) is also a solution for arbitrary c1 , c2 ∈ C. To conclude that (1) gives the general form of the solution, observe that the sequence f (n) satisfying the recursive equation is entirely determined by it initial values f (0) and f (1). Thus it is enough to check that the system f (0) = c1 + c2 , f (1) = c1 λ1 + c2 λ2 has a unique solution c1 , c2 given the initial (arbitrary complex) values f (0) and√f (1). This is in fact the case since the system has a non-zero determinant 1 · λ2 − 1 · λ1 = i 3. The complex solution given by (1) yields a real-valued sequence if and only if f (0) and f (1) are both reals. The condition on f (0) implies that c1 + c2 is a real number, that is c1,2 = c ± id (2) for some reals c, d. But then √ n √ n c1 −1 − i 3 is the complex dual to c2 −1 + i 3 for all n ∈ N. Hence (1) and (2) give a general real-valued solution for the sequence f (n). Using the Euler’s identity eiθ = cos θ + i sin θ, the solution can be alternatively written as f (n) = (c + id)e−2πni/3 + (c − id)e2πni/3 = 2 c cos(2πn/3) + d sin(2πn/3) if n ≡ 0 mod 3, 2c √ −c + d√3 if n ≡ 1 mod 3, = −c − d 3 if n ≡ 2 mod 3 2 Here c and d are two arbitrary real numbers. 1.1 Let Xn be the number of papers in the pile in the evening of day n. Then Xn is a Markov chain with state space Ω = {0, 1, 2, 3, 4} and transition matrix 1/3 2/3 0 0 0 1/3 0 2/3 0 0 1/3 0 0 2/3 0 1/3 0 0 0 2/3 1 0 0 0 0 It is a matter of interpretation and solutions with Ω = {0, 1, 2, 3, 4, 5} will be fully credited. 1.8 (b) For a vertex V let XV be the expected number of steps until the walker starting at V reaches A next time. We thus want to compute XA . Conditioning on the position of the walker after the first step, we obtain: 1 XA = 1 + XB + X C + X D 3 1 XB = 1 + 0 + XC + XE 3 1 X C = 1 + 0 + XB 2 1 X D = 1 + 0 + XE 2 1 XE = 1 + XB + X D 2 We thus have: XE XD = 1 + 2 XB XC = 1 + , 2 and hence, plugging these expression into the initial system, 5 X B XE + + 3 2 6 4 X B XE = + + 3 6 3 3 X B XE = + + 2 2 4 XA = XB XE The last two equations yield 5XB 6 3XE 4 4 XE + 3 3 3 XB = + . 2 2 = 3 Therefore, XB = 36 , 11 XE = 46 , 11 XC = 29 , 11 XD = 34 , 11 XA = 4. (d) For a vertex V let YV be the probability that the walker starting at V reaches A before they reach C. We thus want to compute YB . Conditioning on the position of the walker after the first step, we obtain: 1 1 + YE 3 1 YB + YD = 2 1 = 1 + YE 2 YB = YE YD We thus have: 1 YE 1 1 + YE = + 3 3 3 YB 1 YB 1 YE 1 + 1 + YE = + + . = YB + YD = 2 2 4 2 4 4 YB = YE and hence, 1 YE + 3 3 = 2YB + 1. YB = 3YE The last two equations yield YB = 1 YE 1 1 + = + 2YB + 1 . 3 3 3 9 Therefore, 4 YB = . 7 (e) In the course of solving (b) we have found XC = 4 29 . 11