STA3007_0506_t07_ssol

STA3007_0506_t07_ssol - STA 3007 Applied Probability...

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STA 3007 Applied Probability 2005 Tutorial 7 Suggested Solution 1. Revision i. Yes. ii. P = 0 1 2 0 0 + + 1 + 0 + 2 + + 0 P 2 = 0 1 2 0 + + + 1 + + + 2 + + + and row sums equal 1. Therefore, P is a regular Markov matrix. iii. Since row sum of row 2 is not 1, it is not a regular Markov matrix. iv. Pr { Z k = i | Z k - 1 ,Z k - 2 ,...,Z 0 } = Pr { Z k i | Z k - 1 ,Z k - 2 ,...,Z 0 } - Pr { Z k i - 1 | Z k - 1 ,Z k - 2 ,...,Z 0 } = Pr { max ( X k ,Y k ) i | max ( X k - 1 ,Y k - 1 ) ,...,max ( X 0 ,Y 0 ) } - Pr { max ( X k ,Y k ) i - 1 | max ( X k - 1 ,Y k - 1 ) ,...,max ( X 0 ,Y 0 ) } = Pr { X k i,Y k i | ( X k - 1 ,Y k - 1 ) ,..., ( X 0 ,Y 0 ) } - Pr { X k i - 1 ,Y k ) i - 1 | ( X k - 1 ,Y k - 1 ) ,..., ( X 0 ,Y 0 ) } = Pr { Z k i | Z k - 1 } - Pr { Z k i - 1 | Z k - 1 } = Pr { Z k = i | Z k - 1 } 1
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v. (a) Denote X k as the type of exam in the kth step. State Space: { 1 , 2 , 3 } Pr { X k +1 = 1 | X k = 1 } = Pr { X k +1 = 1 | X k = 1 ,Bad } Pr { Bad | X k = 1 } + Pr { X k +1
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STA3007_0506_t07_ssol - STA 3007 Applied Probability...

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