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STAT100B_HW4S

# STAT100B_HW4S - STAT 100B Homework 4 Solutions Denise Tsai...

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STAT 100B: Homework 4 Solutions Denise Tsai February 24, 2011 1. For the simplest regression model y i = βx i + i , i = 1 , ..., n , derive the least squares estimate of β . Solution: We can find ˆ β by minimizing R ( β ) = n X i =1 2 i = n X i =1 ( y i - βx i ) 2 Set the derivative of R with respect to β equal to zero: R 0 ( β ) = 2 n X i =1 ( y i - βx i )( - x i ) = 0 Then, n X i =1 x i y i = β n X i =1 ( x i ) 2 ˆ β = n i =1 x i y i n i =1 x 2 i (1) 2. For the simple regression y i = β 0 + β 1 x i + i , i = 1 , ..., n , derive the least squares estimates of β 0 and β 1 . Show that the least squares problem for the simple regression can be transformed into the least squares problem for the simplest regression. Solution: We can find ˆ β 0 and ˆ β 1 by minimizing R ( β 0 , β 1 ) = n X i =1 2 i = n X i =1 [ y i - ( β 0 + βx i )] 2 1

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Set the partial derivative of R with respect to β 0 equal to zero: ∂R ∂β 0 = 2 n X i =1 ( y i - β 0 - β 1 x i )( - 1) = 0 Then, n X i =1 y i = 0 + β 1 n X i =1 x i n ¯ y = 0 + 1 ¯ x ˆ β 0 = ¯ y - ˆ β 1 ¯ x (2) Set the partial derivative of R with respect to β 1
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STAT100B_HW4S - STAT 100B Homework 4 Solutions Denise Tsai...

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