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STA3007_Asg1_sol

# STA3007_Asg1_sol - Suggested Solution to STA 3007...

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Suggested Solution to STA 3007 Assignment 1 1. Whether the next step shows a good item or a defective one only depends on the current step. So this process is a Markov Chain.We can define X n = 0 , if the n th item is a good item 1 , if the n th item is a defective item P 00 = α P 01 = 1 - α P 11 = β P 10 = 1 - β Then the probability that the first defective item to appear is the fifth item: Pr { X 5 = 1 , X 4 = 0 , X 3 = 0 , X 2 = 0 , X 1 = 0 } = Pr { X 5 = 1 | X 4 = 0 , X 3 = 0 , X 2 = 0 , X 1 = 0 } Pr { X 4 = 0 , X 3 = 0 , X 2 = 0 , X 1 = 0 } = Pr { X 5 = 1 | X 4 = 0 } Pr { X 4 = 0 , X 3 = 0 , X 2 = 0 , X 1 = 0 } = Pr { X 5 = 1 | X 4 = 0 } Pr { X 4 = 0 | X 3 = 0 , X 2 = 0 , X 1 = 0 } Pr { X 3 = 0 , X 2 = 0 , X 1 = 0 } = Pr { X 5 = 1 | X 4 = 0 } Pr { X 4 = 0 | X 3 = 0 } Pr { X 3 = 0 | X 2 = 0 } Pr { X 2 = 0 | X 1 = 0 } Pr { X 1 = 0 } = P 01 P 00 P 00 P 00 * 1 = (1 - α ) α 3 2. P = 0 1 0 1 - α α 1 α 1 - α P 2 = 0 1 0 α 2 + (1 - α ) 2 2 α (1 - α ) 1 2 α (1 - α ) α 2 + (1 - α ) 2 P 3 = 0 1 0 3 α 2 (1 - α ) + (1 - α ) 3 α 3 + 3 α (1 - α ) 2 1 α 3 + 3 α (1 - α ) 2 3 α 2 (1 - α ) + (1 - α ) 3 Pr { X 5 = 0 | X 0 = 0 } = [ α 2 + (1 - α ) 2 ][3 α 2 (1 - α ) + (1 - α ) 3 ] + [2 α (1 - α )][ α 3 + 3 α (1 - α ) 2 ] = (1 - α )[ α 2 + (1 - α ) 2 ][3 α 2 (1 - α ) + (1 - α ) 3 ] + (1 - α )[2 α 4 + 6 α 2 (1 - α ) 2 ] = (1 - α )[3 α 4 + 4 α 2 (1 - α ) 2 + (1 - α ) 4 ] + (1 - α )[2 α 4 + 6 α 2 (1 - α ) 2 ] = (1 - α )[5 α 4 + 10 α 2 (1 - α ) 2 + (1 - α ) 4 ] 3. When Z n = ( X n - 1 , X n ), Z n +1 = ( X n , X n +1 ). We can see that the second item of Z n equals to the first item of Z n +1 .So the probabilities of (*,0) to (1,*) and (*,1) to (0,*) in one step are impossible. From the probability matrix of X n P = 0 1 0 α 1 - α 1 1 - β β Z n = ( X n - 1 , X n ) is a Markov chain having the four states (0,0),(0,1),(1,0), and (1,1). P = 00 01 10 11 00 α 1 - α 0 0 01 0 0 1 - β β 10 α 1 - α 0 0 11 0 0 1 - β β 1

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4. P = 0 1 2 0 0 . 7 0 . 2 0 . 1 1 0 . 3 0 . 5 0 . 2 2 0 0 1 P 2 = 0 1 2 0 0 . 55 0 . 24 0 . 21 1 0 . 36 0 . 31 0 . 33 2 0 0 1 Pr { X 3 = 0 | X 0 = 0 , T > 3 } = Pr ( X 3 = 0 , T > 3 | X 0 = 0) Pr ( T > 3 | X 0 = 0) = Pr ( { X 3 = 0 } , {{ X 3 = 0 } ∪ { X 3 = 1 }}| X 0 = 0) Pr ( {{ X 3 = 0 } ∪ { X 3 = 1 }}| X 0 = 0) = Pr ( X 3 = 0 | X 0 = 0) Pr ( X 3 = 0 | X 0 = 0) + Pr ( X 3 = 1 | X 0 = 0) = P 3 00 P 3 00 + P 3 01 = (0 . 7)(0 . 55) + (0 . 2)(0 . 36) + (0 . 1)(0) [(0 . 7)(0 . 55) + (0 . 2)(0 . 36) + (0 . 1)(0)] + [(0 . 7)(0 . 24) + (0 . 2)(0 . 31) + (0 . 1)(0)] = 0 . 457 0 . 687 = 0 . 6652 5. Transition probability matrix for the end-of-period inventory level X n : P = - 2 - 1 0 1 2 3 - 2 0 0 0 . 2 0 . 3 0 . 4 0 . 1 - 1 0 0 0 . 2 0 . 3 0 . 4 0 . 1 0 0 0 0 . 2 0 . 3 0 . 4 0 . 1 1 0 . 2 0 . 3 0 . 4 0 . 1 0 0 2 0 0 . 2 0 . 3 0 . 4 0 . 1 0 3 0 0 0 . 2 0 . 3 0 . 4 0 . 1 6. (As the lifetime of the component is not clearly defined in the question, we just check the solution to this questions, but it will not make any contributions to the total mark of each student) If X n = 0, then X n +1 is decided by the remaining life of a new component; If X n > 0, then the component will fail in a certain step that only depends on X n .
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STA3007_Asg1_sol - Suggested Solution to STA 3007...

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