Show explicitly that the following identity holds under a Simple Linear Regression: sum_(i=1)^n r_i mu_i=0 with residuals r_i=y_i-mu_i and mu_i=beta_0+beta_1 x_i

Davirnoilc

Davirnoilc

Answered question

2022-11-17

Show explicitly that the following identity holds under a Simple Linear Regression:
  i = 1 n r i μ i ^ = 0
with residuals r i = y i μ i ^ and μ i ^ = β 0 ^ + β 1 ^ x i .
my steps:
i = 1 n r i μ i ^ = i = 1 n ( y i μ i ^ ) ( β 0 ^ + β 1 ^ x i ) =   i = 1 n ( y i β 0 ^ β 1 ^ x i ) ( β 0 ^ + β 1 ^ x i ) =   i = 1 n ( β 0 ^ y i + β 1 ^ x i y i β 0 ^ 2 β 0 ^ β 1 ^ x i β 0 ^ β 1 ^ x i β 1 ^ 2 x i 2 ) =   i = 1 n ( β 0 ^ y i + β 1 ^ x i y i ( β 0 ^ + β 1 ^ x i ) 2 ) =   i = 1 n ( y i ( β 0 ^ + β 1 ^ x i ) ( β 0 ^ + β 1 ^ x i ) 2 ) =   i = 1 n ( β 0 ^ + β 1 ^ x i ) ( y i β 0 ^ + β 1 ^ x i ) =   i = 1 n μ i ^ r i
how to proceed?

Answer & Explanation

cismadmec

cismadmec

Beginner2022-11-18Added 22 answers

After the third equality, you should use the linearity of the sum and notice that β ^ 0 and β ^ 1 don't depend on the index i. That way, you should get:
i = 1 n r i μ ^ i = β ^ 0 n y ¯ + β ^ 1 i = 1 n x i y i n β ^ 0 2 2 β ^ 0 β ^ 1 n x ¯ β ^ 1 2 i = 1 n x i 2 .
Now, you should use the fact that β ^ 0 = y ¯ β ^ 1 x ¯ and remember the normal equations, namely the second one:
β ^ 0 n x ¯ + β ^ 1 i = 1 n x i 2 i = 1 n x i y i = 0.
The result now directly follows.

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