ma316f03hw10 - MA 316 MATHEMATICAL PROBABILITY...

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MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #10 SOLUTIONS: SECTIONS 4.7, 4.8, and 4.4 FALL 2003 4.76: If X 1 and X 2 are independent random variables with normal distributions N (6 , 1) and N (7 , 1) then the distribution of X 1 - X 2 is N (6 - 7 , 1 + 1) = N ( - 1 , 2) so P ( X 1 > X 2 ) = P ( X 1 - X 2 > 0) = P ( X 1 - X 2 ) - ( - 1) 2 > 1 2 = 1 - Φ(1 / 2) = 0 . 24 . 4.80: If X 1 and X 2 are independent and X 1 is Poisson with mean μ 1 and Y 1 = X 1 + X 2 is Poisson with mean μ > μ 1 then the moment generating functions of X 1 is M X 1 ( t ) = e μ 1 ( e t - 1) and of Y 1 is M Y 1 ( t ) = e μ ( e t - 1) . Since X 2 = Y 1 - X 1 then the moment generating function of X 2 is M X 2 ( t ) = E [ e t ( Y 1 - X 1 ) ] = E [ e tY 1 e t ( - X 1 ) ] = E [ e tY 1 ] E [ e t ( - X 1 ) ] = e μ 1 ( e t - 1) e - μ 1 ( e t - 1) = e ( μ - μ 1 )( e t - 1) which shows that X 2 is Poisson with parameter μ - μ 1 . 4.81: If X 1 and X 2 are independent gamma random variables with parameters α 1 = 3, β 1 = 3 and α 2 = 5, β 2 = 1 and moment generating functions M X 1 ( t ) and M X 2 ( t ), respectively, then by Theorem 2 the moment generating function of Y = 2 X 1 + 6 X 2 is M ( t ) = M X 1 (2 t ) M X 2 (6 t ) = [1 - 3(2 t )] - 3 [1
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