Download Introduction to Probability - Solutions Manual by Charles M. Grinstead and J. Laurie Snell PDF

By Charles M. Grinstead and J. Laurie Snell

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Use the solution to Exercise 24 with w = f. 29. For the chain with pattern HTH we have already verified that the conjecture is correct starting in HT. Assume that we start in H. Then the first player will win 8 with probability 1/4, so his expected winning is 2. Thus E(T |H) = 10 − 2 = 8, which is correct according to the results given in the solution to Exercise 28. The conjecture can be verified similarly for the chain HHH by comparing the results given by the conjecture with those given by the solution to Exercise 28.

K ¿From this it follows that N is finite if and only if m < 1, in which case N = 1/(1 − m). 3 1 1. (a) g(t) = (e2t − 1) 2t e2t (2t − 1) + 1 (b) g(t) = 2t2 e2t − 2t − 1 (c) g(t) = 2t2 2t e (ty − 1) + 2et − t − 1 (d) g(t) = t2 2t 2 e (4t − 4t + 2) − 2 (e) (3/8) t3 3. 5. 2 2−t 4 − 3t (b) g(t) = 2(1 − t)(2 − t) 4 λ (c) g(t) = (d) g(t) = , (2 − t)2 λ+t (a) g(t) = (c) (d) (e) 9. 1 2iτ (e − 1) 2iτ e2iτ (2iτ − 1) + 1 k(τ ) = −2τ 2 e2iτ − 2iτ − 1 k(τ ) = −2τ 2 2iτ e (iτ − 1) + 2eiτ − iτ − 1 k(τ ) = −τ 2 2iτ e (−4τ 2 − 4iτ + 2 k(τ ) = (3/8) −iτ 3 (a) k(τ ) = (b) 7.

1/3  1/6 w = (1/4, 1/4, 1/4, 1/4) . From this we see that wf = 0. From the result of Exercise 20 we see that your expected winning starting in GO is the first component of the vector Zf where   15  −30  f =  . 4. 23. (n) Assume that the chain is started in state si . Let Xj step and 0 otherwise. Then (n) Sj (0) = Xj and (1) + Xj equal 1 if the chain is in state si on the nth (2) + Xj (n) E(Xj ) = Pijn . Thus n (n) (n) E(Sj ) = pij . h=0 If now follows then from Exercise 16 that (n) E(Sj ) = wj .

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