s330t108soln

s330t108soln - STAT 330 TEST 1 SOLUTIONS 1(a) (i) M (t) = X...

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STAT 330 TEST 1 SOLUTIONS 1( a )( i ) M ( t )= X x =0 e tx θ x e θ x ! = e θ X x =0 ( θ e t ) x x ! = e θ e θ e t by the Exponential Series = e θ ( e t 1 ) for t ( −∞ , ) ( ii ) M 0 ( t d dt e θ e θ e t = e θ d dt e θ e t = e θ e θ e t ¡ θ e t ¢ = θ e θ e θ e t e t E ( X M 0 (0) = θ e θ e θ = θ M 00 ( t θ e θ d dt e θ e t e t = θ e θ h e θ e t e t + e θ e t ¡ θ e t ¢ e t i E ¡ X 2 ¢ = M 00 (0) = θ e θ ¡ e θ + θ e θ ¢ = θ (1 + θ ) Var ( X E ¡ X 2 ¢ [ E ( X )] 2 = θ (1 + θ ) θ 2 = θ ( b ) Since f ( x )=0 for x< 1 and since E ( X ) does not exist this implies the improper integral Z 1 xf ( x ) dx (1) diverges. Since (1) diverges and x k f ( x ) xf ( x ) for all x 1 and k =1 , 2 ,... then by the Comparison Test for Improper Integrals the integral Z 1 x k f ( x ) dx also diverges and thus E ¡ X k ¢ = Z 1 x k f ( x ) dx does not exist for k =2 , 3 . 1
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2 . ( a ) F ( x )= Z x θ 3 4 θ μ 1 u 2 θ 2 du = 3 4 θ μ u u 3 3 θ 2 | x θ = 3 4 θ x x 3 3 θ 2 μ θ + θ 3 3 θ 2 ¶¸ = 3
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This note was uploaded on 04/27/2010 for the course STAT 330 taught by Professor Paulasmith during the Fall '08 term at Waterloo.

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s330t108soln - STAT 330 TEST 1 SOLUTIONS 1(a) (i) M (t) = X...

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