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hw1sol - 36-226 Summer 2010 Homework 1 Due June 30 1 Math...

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36-226 Summer 2010 Homework 1 Due June 30 1. Math review. (a) ln n Y i =1 e a b x i x i ! ! = n X i =1 ln e a b x i x i ! = n X i =1 a + x i ln b - x i X j =1 ln j = na + ln b × n X i =1 x i - n X i =1 x i X j =1 ln j (b) f ( x | a, n ) = a ln x + ( n - a ) ln(1 - x ) ∂f ( x | a, n ) ∂x = a x - n - a 1 - x 0 a x = n - a 1 - x a - ax = nx - ax x = a n (critical point) 2 f ( x | a, n ) ∂x 2 = - a x - n - a (1 - x ) 2 < 0 Therefore, a n maximizes f ( x | a, n ). (c) n X i =1 ( X i - μ ) 2 = n X i =1 ( X i - ¯ X + ¯ X - μ ) 2 = n X i =1 (( X i - ¯ X ) + ( ¯ X - μ )) 2 = n X i =1 ( X i - ¯ X ) 2 + ( ¯ X - μ ) 2 - 2( X i - ¯ X )( ¯ X - μ ) = n X i =1 ( X i - ¯ X ) 2 + n X i =1 ( ¯ X - μ ) 2 - 2( ¯ X - μ ) n X i =1 ( X i - ¯ X ) = n X i =1 ( X i - ¯ X ) 2 + n ( ¯ X - μ ) - 2( ¯ X - μ ) × 0 = n X i =1 ( X i - ¯ X ) 2 + n ( ¯ X - μ ) 1
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2. Probability review. (a) E [ X 1 ] = 1 β Z 0 x exp {- x/β } = 1 β - βx exp {- x/β } 0 + β Z 0 exp( - x/β ) (integration by parts) = 0 + β = β. Now recall that V ar [ X 1 ] = E [ X 2 1 ] - E [ X 1 ] 2 , so we need to find E [ X 2 1 ]. E [ X 2 1 ] = 1 β Z 0 x 2 exp {- x/β } dx = 1 β - βx 2 exp {- x/β } 0 + 2 β Z 0 x exp {- x/β } dx (integration by parts) = 0 + 2 β 2 (using the same procedure as for E [ X 1 ]) Therefore, V ar [ X 1 ] = E [
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