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How to understand that the solution to least squares problem transformed with Box-Cox Transformation, is a generalized mean with $h(x)=x^lambda$?

0

$begingroup$

The least squares problem $min_a sum_i^n (x_i-a)^2$ is sometimes solved using transformed variables, that is, solving $min_a sum_i^n [h(x_i)-h(a)]^2$. The solution to this latter problem is $a=h^{-1}(frac{1}{n}sum_i^n h(x_i))$.

If the least squares problem is transformed with the Box-Cox Transformation $$h_lambda(x)=left{begin{matrix}
frac{x^lambda -1}{lambda},text{ if }lambda neq 0\
log x,text{ if }lambda =0
end{matrix}right.$$
we can substitute $h$ by $h_lambda$ to the solution we obtained in the begining, and get $a=h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))$.

Furthermore, Berger and Casella (1992) say that for any monotone, continuous function $h$, it is easy to verify that if $g(x) = ah(x) + b$, where $a$ and $b$ are constants that do not depend on $x$ and $a neq 0$, then $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = g^{-1}(frac{1}{n}sum_i^n g(x_i))$. So, the solution is a generalized mean with $h(x)=x^lambda$, i.e. $$h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))=h^{-1}(frac{1}{n}sum_i^n h(x_i)), text{ where }h(x)=x^lambda.$$

In my opinion (to understand the process of Berger and Casella (1992)), using the notation in the last paragraph, we can let $a=b=frac{1}{lambda}$, $h(x)=x^lambda$, and then $g(x)=ah(x) + b$ is equal to $h_lambda(x)$. Further, the conclusion $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = g^{-1}(frac{1}{n}sum_i^n g(x_i))$ corresponds to $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))$.

However, what it have discussed is only the case of Box-Cox Transformation with $lambda neq 0$. How about the case with $lambda = 0$? Why does it omit that?

This is also the exercise 7.16(c) of the book Statistical Inference 2nd edition, where it gives the PROPOSITION that the solution is a generalized mean with $h(x)=x^lambda$. Again it omits the case with $lambda = 0$. So, is this PROPOSITION rigorous? Or, we don't need to discuss the case with $lambda = 0$?

statistics transformation least-squares means maximum-likelihood

share|cite|improve this question

edited Mar 22 at 12:25

Tim Xu

asked Mar 22 at 12:20

Tim XuTim Xu

63

$endgroup$

add a comment | 

0

$begingroup$

The least squares problem $min_a sum_i^n (x_i-a)^2$ is sometimes solved using transformed variables, that is, solving $min_a sum_i^n [h(x_i)-h(a)]^2$. The solution to this latter problem is $a=h^{-1}(frac{1}{n}sum_i^n h(x_i))$.

If the least squares problem is transformed with the Box-Cox Transformation $$h_lambda(x)=left{begin{matrix}
frac{x^lambda -1}{lambda},text{ if }lambda neq 0\
log x,text{ if }lambda =0
end{matrix}right.$$
we can substitute $h$ by $h_lambda$ to the solution we obtained in the begining, and get $a=h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))$.

Furthermore, Berger and Casella (1992) say that for any monotone, continuous function $h$, it is easy to verify that if $g(x) = ah(x) + b$, where $a$ and $b$ are constants that do not depend on $x$ and $a neq 0$, then $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = g^{-1}(frac{1}{n}sum_i^n g(x_i))$. So, the solution is a generalized mean with $h(x)=x^lambda$, i.e. $$h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))=h^{-1}(frac{1}{n}sum_i^n h(x_i)), text{ where }h(x)=x^lambda.$$

In my opinion (to understand the process of Berger and Casella (1992)), using the notation in the last paragraph, we can let $a=b=frac{1}{lambda}$, $h(x)=x^lambda$, and then $g(x)=ah(x) + b$ is equal to $h_lambda(x)$. Further, the conclusion $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = g^{-1}(frac{1}{n}sum_i^n g(x_i))$ corresponds to $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))$.

However, what it have discussed is only the case of Box-Cox Transformation with $lambda neq 0$. How about the case with $lambda = 0$? Why does it omit that?

This is also the exercise 7.16(c) of the book Statistical Inference 2nd edition, where it gives the PROPOSITION that the solution is a generalized mean with $h(x)=x^lambda$. Again it omits the case with $lambda = 0$. So, is this PROPOSITION rigorous? Or, we don't need to discuss the case with $lambda = 0$?

statistics transformation least-squares means maximum-likelihood

share|cite|improve this question

edited Mar 22 at 12:25

Tim Xu

asked Mar 22 at 12:20

Tim XuTim Xu

63

$endgroup$

add a comment | 

$begingroup$

The least squares problem $min_a sum_i^n (x_i-a)^2$ is sometimes solved using transformed variables, that is, solving $min_a sum_i^n [h(x_i)-h(a)]^2$. The solution to this latter problem is $a=h^{-1}(frac{1}{n}sum_i^n h(x_i))$.

If the least squares problem is transformed with the Box-Cox Transformation $$h_lambda(x)=left{begin{matrix}
frac{x^lambda -1}{lambda},text{ if }lambda neq 0\
log x,text{ if }lambda =0
end{matrix}right.$$
we can substitute $h$ by $h_lambda$ to the solution we obtained in the begining, and get $a=h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))$.

Furthermore, Berger and Casella (1992) say that for any monotone, continuous function $h$, it is easy to verify that if $g(x) = ah(x) + b$, where $a$ and $b$ are constants that do not depend on $x$ and $a neq 0$, then $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = g^{-1}(frac{1}{n}sum_i^n g(x_i))$. So, the solution is a generalized mean with $h(x)=x^lambda$, i.e. $$h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))=h^{-1}(frac{1}{n}sum_i^n h(x_i)), text{ where }h(x)=x^lambda.$$

In my opinion (to understand the process of Berger and Casella (1992)), using the notation in the last paragraph, we can let $a=b=frac{1}{lambda}$, $h(x)=x^lambda$, and then $g(x)=ah(x) + b$ is equal to $h_lambda(x)$. Further, the conclusion $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = g^{-1}(frac{1}{n}sum_i^n g(x_i))$ corresponds to $h^{-1}(frac{1}{n}sum_i^n h(x_i)) = h_lambda^{-1}(frac{1}{n}sum_i^n h_lambda(x_i))$.

However, what it have discussed is only the case of Box-Cox Transformation with $lambda neq 0$. How about the case with $lambda = 0$? Why does it omit that?

This is also the exercise 7.16(c) of the book Statistical Inference 2nd edition, where it gives the PROPOSITION that the solution is a generalized mean with $h(x)=x^lambda$. Again it omits the case with $lambda = 0$. So, is this PROPOSITION rigorous? Or, we don't need to discuss the case with $lambda = 0$?

statistics transformation least-squares means maximum-likelihood

share|cite|improve this question

edited Mar 22 at 12:25

Tim Xu

asked Mar 22 at 12:20

Tim XuTim Xu

63

$endgroup$

share|cite|improve this question

edited Mar 22 at 12:25

Tim Xu

asked Mar 22 at 12:20

Tim XuTim Xu

63

share|cite|improve this question

edited Mar 22 at 12:25

Tim Xu

asked Mar 22 at 12:20

Tim XuTim Xu

63

asked Mar 22 at 12:20

Tim XuTim Xu

63

asked Mar 22 at 12:20

Tim XuTim Xu

63