interpretation of boxcox with lambda equal 0 I am working on this non linear data set, and running my Box-Cox I find that the best value to use is $\lambda = 0$.
If I understand correctly, $\lambda =2$ implies $Y^2$. Similarly, $\lambda = -0.5$ corresponds to $1 \over \sqrt Y$. However how would I work with a null lambda, all $Y$ values being made 0 makes little sense? NOT 0, 1
Any explanation would be much useful.
 A: Box-Cox transformation is defined as
$$ y^{(\lambda)} =
\begin{cases}
\frac{y^\lambda - 1}{\lambda}   & \text{if } & \lambda \ne 0, \\
\log(y) & \text{if } & \lambda=0.
\end{cases} $$
For further details check the original paper that introduced it:

Box, G. E., & Cox, D. R. (1964). An analysis of transformations.
Journal of the Royal Statistical Society. Series B (Methodological),
211-252.

So there is really nothing to interpret, it's simply a log transformation.

Also notice that $x^0 = 1$, so it would rather be all ones, rather than all zeros.
A: If you consider   non-null $y$ values, the coefficient of variation of the Box-Cox transformation for a real $\lambda$-power near $0$ goes like:
$$ \frac{\exp \left( \lambda\ \log y\right)−y^0}{\lambda−0}≈\frac{1+\lambda\log y−1}{\lambda}$$
as $\lambda$ tends to $0$. So the limit is $\log y$. So in a way, a constant behaves like a limit behaviour of the logarithm, or the other way around. 
The logarithm is somehow dimensionless, or $0$-homogenous, like a constant. A personal and physical interpretation can be found at What is the logarithm of a kilometer? Is it a dimensionless number?. In short:


*

*If you consider $y$ as a non-zero constant, $y^0=1$,

*If you consider $y$ as a realisation of a  continuous variable, you inherit a variational context, and the natural limit is a logarithm.

