Suppose 5 out of 25 data points in a weighted least-squares problem have a y-measurement that is less reliable than the others, and they are to be weighted one sixth as much as the other 20 points. One method is to weight the 20 points by a factor of 1 and the other 5 by a factor of A second method is to weight the 20 points by a factor of 6 and the other 5 by a factor of 1. Do the two methods produce different results? Explain. Let W represent the weight matrix for the first case. The coefficients for the least-squares line are the solution to the normal equation (WA) WAX= (WA) Wy. How is the weight matrix W' for the second case related to the weight matrix for the first case? w'=0 Write the normal equation for the second case in terms of the weight matrix W. (W'A) TW'AX= (W'A) TW'y 0¹0x=0¹0 y Now use the properties of scalar multiplication to collect all the coefficients. (WA) WAX = ((WA) Wy (Simplify your answers.) Compare this equation to the normal equation for the first case, (WA) WAX = (WA) Wy. How are their solutions related? The solutions for the second case are because the normal equation in the second case
Suppose 5 out of 25 data points in a weighted least-squares problem have a y-measurement that is less reliable than the others, and they are to be weighted one sixth as much as the other 20 points. One method is to weight the 20 points by a factor of 1 and the other 5 by a factor of A second method is to weight the 20 points by a factor of 6 and the other 5 by a factor of 1. Do the two methods produce different results? Explain. Let W represent the weight matrix for the first case. The coefficients for the least-squares line are the solution to the normal equation (WA) WAX= (WA) Wy. How is the weight matrix W' for the second case related to the weight matrix for the first case? w'=0 Write the normal equation for the second case in terms of the weight matrix W. (W'A) TW'AX= (W'A) TW'y 0¹0x=0¹0 y Now use the properties of scalar multiplication to collect all the coefficients. (WA) WAX = ((WA) Wy (Simplify your answers.) Compare this equation to the normal equation for the first case, (WA) WAX = (WA) Wy. How are their solutions related? The solutions for the second case are because the normal equation in the second case
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![Suppose 5 out of 25 data points in a weighted least-squares problem have a y-measurement that s less reliable than the others, and they are to be weighted one sixth as much as the other 20 points. One method is to weight the 20 points by a factor of 1 and the
1
other 5 by a factor of A second method is to weight the 20 points by a factor of 6 and the other 5 by a factor of 1. Do the two methods produce different results? Explain.
Let W represent the weight matrix for the first case. The coefficients for the least-squares line are the solution to the normal equation (WA) WAX = (WA) Wy. How is the weight matrix W' for the second case related to the weight matrix for the first case?
W' =
Write the normal equation for the second case in terms of the weight matrix W.
(W'A) TW'AX = (W'A) TW'y
0'0x=0¹0 y
Now use the properties of scalar multiplication to collect all the coefficients.
(WA) WAX = ((WA) Wy
(Simplify your answers.)
C
Compare this equation to the normal equation for the first case, (WA) WAx=(WA) Wy. How are their solutions related?
The solutions for the second case are
because the normal equation in the second case](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85e06f30-5d78-4642-a613-33cb835cdc30%2F72ec2083-9ed7-4c0f-9d27-cb60b5c8e303%2F3cl51pc_processed.png&w=3840&q=75)
Transcribed Image Text:Suppose 5 out of 25 data points in a weighted least-squares problem have a y-measurement that s less reliable than the others, and they are to be weighted one sixth as much as the other 20 points. One method is to weight the 20 points by a factor of 1 and the
1
other 5 by a factor of A second method is to weight the 20 points by a factor of 6 and the other 5 by a factor of 1. Do the two methods produce different results? Explain.
Let W represent the weight matrix for the first case. The coefficients for the least-squares line are the solution to the normal equation (WA) WAX = (WA) Wy. How is the weight matrix W' for the second case related to the weight matrix for the first case?
W' =
Write the normal equation for the second case in terms of the weight matrix W.
(W'A) TW'AX = (W'A) TW'y
0'0x=0¹0 y
Now use the properties of scalar multiplication to collect all the coefficients.
(WA) WAX = ((WA) Wy
(Simplify your answers.)
C
Compare this equation to the normal equation for the first case, (WA) WAx=(WA) Wy. How are their solutions related?
The solutions for the second case are
because the normal equation in the second case
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