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stat2303_assignment01_ans

# stat2303_assignment01_ans - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT2303/2803 PROBABILITY MODELLING/STOCHASTIC MODELS ASSIGNMENT 1 SOLUTION 1. From (1), 𝑃𝑃 ( 𝐴𝐴𝐴𝐴 | 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴 | 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴 | 𝐶𝐶 ) (1) 𝑃𝑃 ( 𝐴𝐴𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴 | 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) ⇔ 𝑃𝑃 ( 𝐴𝐴 | 𝐴𝐴𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴 | 𝐶𝐶 ) (2) And, 𝑃𝑃 ( 𝐴𝐴𝐴𝐴 | 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴 | 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴 | 𝐶𝐶 ) (1) 𝑃𝑃 ( 𝐴𝐴𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴 | 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) ⇔ 𝑃𝑃 ( 𝐴𝐴 | 𝐴𝐴𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴𝐶𝐶 ) 𝑃𝑃 ( 𝐶𝐶 ) = 𝑃𝑃 ( 𝐴𝐴 | 𝐶𝐶 ) (3) So, (1) (2) (3) 2. 𝑃𝑃 ( 𝑌𝑌 = 1) = � 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑌𝑌 = 1) 𝑥𝑥 = 1 9 + 1 3 + 1 9 = 5 9 𝑃𝑃 ( 𝑌𝑌 = 2) = � 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑌𝑌 = 2) 𝑥𝑥 = 1 9 + 0 + 1 18 = 3 18 𝑃𝑃 ( 𝑌𝑌 = 3) = � 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑌𝑌 = 3) 𝑥𝑥 = 0 + 1 6 + 1 9 = 5 18 𝐸𝐸 ( 𝑋𝑋 | 𝑌𝑌 = 1) = � 𝑥𝑥 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑌𝑌 = 1) 𝑃𝑃 ( 𝑌𝑌 = 1) 𝑥𝑥 = 9 5 1 1 9 + 2 1 3 + 3 1 9 �� = 2 𝐸𝐸 ( 𝑋𝑋 | 𝑌𝑌 = 2) = � 𝑥𝑥 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑌𝑌 = 2) 𝑃𝑃 ( 𝑌𝑌 = 2) 𝑥𝑥 = 18 3 1 1 9 + 2(0) + 3 1 18 �� = 5 3 𝐸𝐸 ( 𝑋𝑋 | 𝑌𝑌 = 3) = � 𝑥𝑥 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑌𝑌 = 3) 𝑃𝑃 ( 𝑌𝑌 = 3) 𝑥𝑥 = 18 5 1(0) + 2 1 6 + 3 1 9 �� = 12 5 3. 𝑃𝑃 ( 𝑌𝑌 = 2) = � � 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑍𝑍 = 𝑧𝑧 , 𝑌𝑌 = 2) 𝑧𝑧 𝑥𝑥 = 1 16 + 1 4 = 5 16

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𝑃𝑃 ( 𝑋𝑋 = 1| 𝑌𝑌 = 2) = 𝑃𝑃 ( 𝑋𝑋 = 1, 𝑌𝑌 = 2) 𝑃𝑃 ( 𝑌𝑌 = 2) = 1 16 + 0 5 16 = 1 5 𝑃𝑃 ( 𝑋𝑋 = 2| 𝑌𝑌 = 2) = 𝑃𝑃 ( 𝑋𝑋 = 2, 𝑌𝑌 = 2) 𝑃𝑃 ( 𝑌𝑌 = 2) = 0 + 1 4 5 16 = 4 5 𝐸𝐸 ( 𝑋𝑋 | 𝑌𝑌 = 2) = 1 1 5 + 2 4 5 = 9 5 𝑃𝑃 ( 𝑌𝑌 = 2, 𝑍𝑍 = 1) = � 𝑃𝑃 ( 𝑋𝑋 = 𝑥𝑥 , 𝑌𝑌 = 2, 𝑍𝑍 = 1) 𝑥𝑥 = 1 16 + 0 = 1 16 𝑃𝑃 ( 𝑋𝑋 = 1| 𝑌𝑌 = 2, 𝑍𝑍 = 1) = 1
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