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Test2-Solutions-S09

Test2-Solutions-S09 - 1 22 marks In a particular soccer...

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Unformatted text preview: 1. ( 22 marks ) In a particular soccer league, a study has shown that 40% of the players are below average (with risk parameter & 1 ) and 60% are above average (with risk parameter & 2 ). Let X j represent the number of goals scored in game j by a chosen soccer player ( j = 1 ; 2 ; ::: ). It is assumed that Pr ( X j = x j j & = & 1 ) = e & 1 x j ! , x j = 0 ; 1 ; ::: and Pr ( X j = x j j & = & 2 ) = 1 4 e & 1 x j ! + 3 4 e & 2 2 x j x j ! , x j = 0 ; 1 ; ::: Conditional on the risk parameter & , the r.v.&s X 1 ; X 2 ; ::: are all independent. A given soccer player have scored 1 goal in game 1 and 2 goals in game 2 . (a) ( 6 marks ) Compute the hypothetical means and process variances for both types of soccer player. Solution: For the hypothetical means, ¡ ( & 1 ) = E [ X j j & = & 1 ] = 1 X x j =0 x j & e & 1 x j ! ¡ = 1 , and ¡ ( & 2 ) = E [ X j j & = & 2 ] = 1 X x j =0 x j & 1 4 e & 1 x j ! + 3 4 e & 2 2 x j x j ! ¡ = 1 4 1 X x j =0 x j e & 1 x j ! + 3 4 1 X x j =0 x j e & 2 2 x j x j ! = 1 4 (1) + 3 4 (2) = 1 : 75 . For the process variances, v ( & 1 ) = V ar ( X j j & = & 1 ) = E ¢ X 2 j j & = & 1 £ & ( ¡ ( & 1 )) 2 = 1 X x j =0 x 2 j e & 1 x j ! & (1) 2 = 2 (1) & (1) 2 = 1 , 1 and v ( & 2 ) = V ar ( X j j & = & 2 ) = E & X 2 j j & = & 2 ¡ & ( ¡ ( & 2 )) 2 = 1 X x j =0 x 2 j ¢ 1 4 e & 1 x j !...
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