# ch02Em - Chapter 2 Inference in Regression Analysis Math...

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Unformatted text preview: Chapter 2. Inference in Regression Analysis Math Stat Result: If Y i ∼ N ( μ i ,σ 2 i ) and Y i ’s are independent, and a 1 ,a 2 ,...,a n are known constants then n summationdisplay i =1 a i Y i ∼ N parenleftBigg n summationdisplay i =1 a i μ i , n summationdisplay i =1 a 2 i σ 2 i parenrightBigg . Thus, a linear combination of independent normal random variables is itself a normal random variable. Theorem: b and b 1 are linear combinations of the Y i ’s. That is, we can write b 1 = n summationdisplay i =1 k i Y i and b = n summationdisplay i =1 l i Y i where k 1 ,...,k n and l 1 ,...,l n are known constants. Proof: Recall S XX = ∑ n i =1 ( X i − ¯ X ) 2 . So b 1 = 1 S XX n summationdisplay i =1 ( X i − ¯ X )( Y i − ¯ Y ) = 1 S XX bracketleftBigg n summationdisplay i =1 ( X i − ¯ X ) Y i − ¯ Y n summationdisplay i =1 ( X i − ¯ X ) bracketrightBigg = 1 S XX n summationdisplay i =1 ( X i − ¯ X ) Y i = n summationdisplay i =1 parenleftbigg X i − ¯ X S XX parenrightbigg Y i = n summationdisplay i =1 k i Y i with k i = X i − ¯ X S XX b = ¯ Y − b 1 ¯ X = 1 n n summationdisplay i =1 Y i − ¯ X n summationdisplay i =1 k i Y i = n summationdisplay i =1 parenleftbigg 1 n − k i ¯ X parenrightbigg Y i = n summationdisplay i =1 l i Y i with l i = 1 n − k i ¯ X. Thus, b and b 1 are linear combinations of the Y i ’s and, hence, they are normal variates. What about their means and variances? Theorem: Under SLR model with normal errors: b 1 ∼ N parenleftbigg β 1 , σ 2 S XX parenrightbigg and b ∼ N parenleftbigg β , σ 2 n ∑ i X 2 i S XX parenrightbigg . We are first interested in ∑ i k i , ∑ i k i X i and ∑ i k 2 i . n summationdisplay i =1 k i = n summationdisplay i =1 X i − ¯ X S XX = 1 S XX n summationdisplay i =1 ( X i − ¯ X ) = 0 n summationdisplay i =1 k i X i = n summationdisplay i =1 X i − ¯ X S XX X i = 1 S XX S XX = 1 n summationdisplay i =1 k 2 i = 1 S 2 XX n summationdisplay i =1 ( X i − ¯ X ) 2 = 1 S XX . Proof: Since b 1 = ∑ n i =1 k i Y i , we get E ( b 1 ) = n summationdisplay i =1 k i E ( Y i ) = n summationdisplay i =1 k i ( β + β 1 X i ) . Because ∑ i k i = 0 and ∑ i k i X i = 1, this is E ( b 1 ) = β n summationdisplay i =1 k i + β 1 n summationdisplay i =1 k i X i = β 1 . Therefore, β 1 is an unbiased estimator of b 1 . With ∑ i k 2 i = 1 /S XX , we get V ar ( b 1 ) = V ar parenleftBigg n summationdisplay i =1 k i Y i parenrightBigg = n summationdisplay i =1 k 2 i V ar ( Y i ) = σ 2 n summationdisplay i =1 k 2 i = σ 2 S XX . Showing b ∼ N parenleftBig β , σ 2 n ∑ i X 2 i S XX parenrightBig is basically the same. Example: 93 house prices in G’ville sold Dec. 1995 (Good example for the project. Look up http://www.fsboingainesville.com/ for more and current data)....
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ch02Em - Chapter 2 Inference in Regression Analysis Math...

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