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 $\lambda = 0$?


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