Suggested Solutions of Assignment 2

Suggested Solutions of Assignment 2 - THE UNIVERSITY OF...

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Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1306 INTRODUCTORY STATISTICS (SEMESTER 1 2008/2009) Suggestion Solution of Assignment 2 Question 1 (a) M Y ( t ) = E ( e tY ) = Z 1 e ty dy = & e ty t ¡ 1 = e t & 1 t : (b) Since e t = 1 + t + t 2 2 + t 3 6 + ¡¡¡ ; we have M Y ( t ) = e t & 1 t = 1 + t 2 + t 2 6 + ¡¡¡ : Hence, E ( Y ) = dM Y ( t ) dt ¢ ¢ ¢ ¢ t =0 = 1 2 ; E ( Y 2 ) = d 2 M Y ( t ) dt 2 ¢ ¢ ¢ ¢ t =0 = 1 3 ; and V ar ( Y ) = E ( Y 2 ) & [ E ( Y )] 2 = 1 3 & £ 1 2 ¤ 2 = 1 12 : Question 2 (a) M Z ( t ) = e & 3 4 t M Y £ t 4 ¤ = e & 3 4 t e 3 t 4 e 8 t 2 16 = e t 2 2 : (b) The &rst moment of Z is E ( Z ) = dM Z ( t ) dt ¢ ¢ ¢ ¢ t =0 1 = te t 2 2 & & & t =0 = 0 . The second moment of Z is E ( Z 2 ) = d 2 M Z ( t ) dt 2 & & & & t =0 = e t 2 2 + t 2 e t 2 2 & & & t =0 = 1 : Hence, the variance of Z is V ar ( Z ) = E ( Z 2 ) & [ E ( Z )] 2 = 1 : Question 3 (a) When y ¡ a , F Y ( y j a < Y ¡ b ) = P ( Y ¡ a and a < Y ¡ b ) P (...
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Suggested Solutions of Assignment 2 - THE UNIVERSITY OF...

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