# hwk1 - E3106 Solutions to Homework 1 Columbia University...

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E3106, Solutions to Homework 1 Columbia University September 13, 2005 Exercise 3 . According to the conditions given by the question and the notation used in the above solution, we have p 0 =0 . 8 ,p 1 . 6 2 . 6 3 . 6 4 = 0 . 4 5 . 4 6 . 4 7 . 2. Thus, the transition probability matrix P of the Markov chain is P = 0 . 80 0 00 . 20 0 0 0 . 60 0 00 . 40 0 0 00 . 6 000 0 . 40 0 . 6 0 . 0 . 4 0 . 60 0 . 4 0 . 0 . 4 0 . 6 0 . 2 0 . 8 . Exercise 7 . As in Exercise 2, we use (RR),(DR),(RD),(DD) to represent the 4states0 , 1 , 2 , 3 in Example 4.4 respectively. And we suppose that today is in the n -th state of the Markov chain. The required probability is equal to P ( X n +1 =( DR ) | X n 1 DD )) + P ( X n +1 RR ) | X n 1 DD )) . Solution 1. By direct calculation. P ( X n +1 DR ) | X n 1 DD )) + P ( X n +1 RR ) | X n 1 DD )) = P ( X n +1 DR ) ,X n DD ) | X n 1 DD )) + P ( X n +1 RR ) n DR ) | X n 1 DD )) = P ( X n +1 DR ) | X n DD ) n 1 DD )) P ( X n DD ) | X n 1 DD )) + P ( X n +1 RR ) | X n DR ) n 1 DD )) P ( X n DR ) | X n 1 DD )) = P ( X n +1 DR ) | X n DD )) P ( X n DD ) | X n 1 DD )) + P ( X n +1 RR ) | X n DR )) P ( X n DR ) | X n 1 DD )) =( . 2)( . 8) + ( . 5)( .

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hwk1 - E3106 Solutions to Homework 1 Columbia University...

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