IEOR 165 Lecture 8 June 10 2009

IEOR 165 Lecture 8 June 10 2009 - Lecture Notes IEOR 165 |...

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Lecture Notes IEOR 165 | George Shanthikumar Tu W Th 2-4:30 pm | 3113 Etcheverry = - = Ahat Ybar BhatxbarBhat SxYSxx ~ , * Bhat N b 1Sxx σ2 - % - * , + * A1 α100 CI for b is Bhat zα2 σSxx Bhat zα2 σSxx The above assumes that sigma is known. What if σ is unknown?Then we'dneed to estimate σ2 We know that = + + = + Yk a bxk εk and μk a bxk = + + = + Yk a bxk εk μk εk - = Thus Yk μk εk - = = = EYk uk2 Eεk2 Varεk σ2 = + * : Yhatk Ahat Bhat xk forecast of the mean response So we look at - = - - * Yk Yhatk Yk Ahat Bhat xk : We essentially square this and put it in summation = - = = - - * k 1nYk Yhatk2 k 1nYk Ahat Bhat xk2 = SSR This is the Sum of Squared rediduals (SSR) So essentially, we’ve fit it into a plot: Ahat + Bhat*x So we need to find the expected value of the SSR: = ESSR ? = - * Ahat Ybar Bhat xbar + * = + - Ahat Bhat xk Ybar Bhatxk xbar
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Lecture Notes IEOR 165 | George Shanthikumar Tu W Th 2-4:30 pm | 3113 Etcheverry = = + - = - 1nl 1nYl xk xbarSxxl 1nxl xbarYl = = + - - l 1n1n xk xbarxl xbarSxxYl = = - = + - - SSR k 1nYk l 1n1n xk xbarxl xbarSxxYl2 , , : He did not go through the algebra but if we solve it out then = - ESSR n 2σ2 Under MOM, we can conclude that = - σ2 ESSRn 2 , - After doing a linear regression we divide and are able to find thatSSRn 2 is an UNBIASED ESTIMATOR of σ2 = - → = - , σhat2 SSRn 2 σhat SSRn 2 and when sigma is unknown we can replace sigma with sigma hat In conclusion : - * , + * A quick approximate CI for b is Bhat zα2 σhatSxx Bhat zα2 σhatSxx To improve this approximation, we need to note that the
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This note was uploaded on 07/22/2009 for the course IEOR 165 taught by Professor Shanthikumar during the Summer '08 term at University of California, Berkeley.

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IEOR 165 Lecture 8 June 10 2009 - Lecture Notes IEOR 165 |...

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