2021 Feb 7 2020 Nov 22 946pm 2020 Nov 1 PSTAT 126 HW2 Solns PDF 2 - PSTAT 126 Homework#2 Solutions Problem 1 Explain why the coefficient of

2021 Feb 7 2020 Nov 22 946pm 2020 Nov 1 PSTAT 126 HW2 Solns PDF 2

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PSTAT 126: Homework #2 Solutions Problem 1: Explain why the coefficient of determination ࠵? ! satisfies 0 ≤ ࠵? ! ≤ 1. Is the same true of the adjusted coefficient of determination ࠵? " ! ? Give a reason. Solution : ࠵? ! = 1 − $ ! " # !$% (& ! ’& () " # !$% = 1 − (& ! ’& * ! ) " # !$% (& ! ’& () " # !$% , where ࠵? ’ = + , ࠵? - , -.+ . So, by what we are calling the Decomposition of Variation formula, i.e., (࠵? - − ࠵? ’) ! , -.+ = (࠵? - − ࠵? + - ) ! , -.+ + (࠵? ’ − ࠵? + - ) ! , -.+ , we find that ࠵? ! = 1 − (࠵? - − ࠵? ’) ! , -.+ − ∑ (࠵? ’ − ࠵? + - ) ! , -.+ (࠵? - − ࠵? ’) ! , -.+ = (࠵? ’ − ࠵? + - ) ! , -.+ (࠵? - − ࠵? ’) ! , -.+ ≥ 0. Then, another appeal to the Decomposition of Variation formula ensures that ࠵? ! ≤ 1 as well. Moreover, no, the analogous assertion is not true in general for ࠵? " ! , which satisfies ࠵? " ! = 1 − (1 − ࠵? ! ) ,’+ ,’/’+ , where N is the number of samples and M is the number of x-variables. In fact, we can see from this definition that for small ࠵? ! and large N with M close to N, ࠵? " ! should be negative. Problem 2: Show, in the context of Simple Linear Regression, that the residuals ࠵? n , ࠵? = 1, .... , ࠵? , are normally-distributed if the noise/error terms ࠵? n , ࠵? =1, .... , ࠵? , are. Give your reasoning. Solution: We have ࠵? n = ࠵? - − ࠵? + - . We know the ࠵? - are normally-distributed because we are assuming the noise/error terms are. We also know the general fact that linear combinations of independent, normally-distributed random variables are themselves normally-distributed. ࠵? : + = (0 ! ’0̅)(& ! ’& () # !$% (0 ! ’0̅) " # !$% = 2 &’ 2 && , ࠵? : 3 = + , ;∑ ࠵? - − ࠵? : + , -.+ ࠵? - , -.+ = , where ࠵? = + , ࠵? 4 , 4.+ and similarly for ࠵?̅, and the ࠵? - can be regarded as non-random, fixed values. We insert these two equations into ࠵? n = ࠵? - − ࠵? + - using the definition of ࠵? + - seen in the lecture slides and then collect coefficients of like ࠵? - terms. Then it is not hard to see that the fact we cited about linear combinations of independent normally distributed random variables will complete the answer.
Problem 3: For any given, fixed number ࠵? of samples, explain why and in what sense each ࠵? : 5 , ࠵? = 0, ..., M , is an optimal estimator for ࠵? m , ࠵? = 0,..., M , respectively. Solution: You can use the statement of the Gauss-Markov Theorem for this. First that theorem tells us that the ࠵? : 5 , ࠵? = 0, ... , ࠵? , are unbiased estimators for the ࠵? m , ࠵? = 0, ... , ࠵? , respectively. So ࠵?D࠵? : 5 E = ࠵? 5 . The theorem also states that the ࠵? : 5 are of minimum variance among all unbiased, linear estimators for ࠵? 5 . But, this then implies as well that, among all unbiased, linear estimators ࠵? + 5 , the error ࠵?[(࠵? + 5 − ࠵? 5 ) ! ] = ࠵?[(࠵?

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