Suppose 8 out of 20 data points in a weighted leastsquares p

Suppose 8 out of 20 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 12 points. One method is to weight the 12 points by a factor of 1 and the other 8 by a factor of 1/6. A second method is to weight the 12 points by a factor of 6 and the other 8 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)^T WA_x= (WA)TWy. How is the weight matrix W\' for the second case related to the weight matrix for the first case? w\' = 6W Write the normal equation for the second case in terms of the weight matrix W. (W\'A)^TW\'Ax = (W\'A)^TW\'y (6Wa)^T (6WA)x = (6WA)^T (6W)y Now use the properties of scalar multiplication to collect all the coefficients. (36)(WA)^TWAx = (36)(WA)^T Wy (Simplify your answers.) Compare this equation to the normal equation for the first case. (WA)^TWAx = (WA)^TWy. How are their solutions related? The solutions for the second case are the same because the normal equation in the second case is a nonzero multiple of the equation from the first case.

Solution

Given that there are 20 data points. 8 of these twenty data points have a less reliable y value such that it is to be weighted one sixth in these 8 eight point in comparison to the weights of the other 12 points.

Method 1:

w1 = w2 = .... = w8 = 1/6

w9 = w10 = .... = w20 = 1

Method 2:

w1\' = w2\' = .... = w8\' = 1

w9\' = w10\' = .... = w20\' = 6

Diagonal elements of matrix W will be given by w1,w2...w20. Diagonoal elements of matrix W

will be given by w1\',w2\',....,w20\'. Therefore, on comparing the two matrices, we get

W\'=6W

Normal equation for second case is:

(W\'A)^TW\'Ax = (W\'A)^TW\'y

On plugging the expression for W\' in terms of W, we get:

(6WA)^T(6W)Ax = (6WA)^T(6W)y

36(WA)^T(W)Ax = 36(WA)^T(W)y

(WA)^T(W)Ax = (WA)^T(W)y

This is the normal equation for first method. Therefore, we can see that the final result for both the methods is the same.

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