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

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 $$\lambda = 0$$?