STA3007_Asg1_sol

STA3007_Asg1_sol - Suggested Solution to STA 3007...

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Unformatted text preview: 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 = ,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 = 1 1- α α 1 α 1- α P 2 = 1 α 2 + (1- α ) 2 2 α (1- α ) 1 2 α (1- α ) α 2 + (1- α ) 2 P 3 = 1 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 } = [ 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 = 1 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 01 1- β β 10 α 1- α 11 1- β β     1 4. P =   1 2 . 7 . 2 . 1 1 . 3 . 5 . 2 2 1   P 2 =   1 2 . 55 . 24 . 21 1 . 36 . 31 . 33 2 1   Pr { X 3 = 0 | X = 0 ,T > 3 } = Pr ( X 3 = 0 ,T > 3 | X = 0) Pr ( T > 3 | X = 0) = Pr ( { X 3 = 0 } , {{ X 3 = 0 } ∪ { X 3 = 1 }}| X = 0) Pr ( {{ X 3 = 0 } ∪ { X 3 = 1 }}| X = 0) = Pr ( X 3 = 0 | X = 0) Pr ( X 3 = 0 | X = 0) + Pr ( X 3 = 1 | X = 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)] = . 457 . 687 = . 6652 5. Transition probability matrix for the end-of-period inventory level5....
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This note was uploaded on 05/21/2011 for the course STA 3007 taught by Professor Kb during the Spring '11 term at CUHK.

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STA3007_Asg1_sol - Suggested Solution to STA 3007...

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