# Where does the Box-Cox Transformation actually come from?

I'm trying to figure out where the actual box-cox transformation comes from. I've looked at the original paper, and some of it's references, but for the most part, it seems that they just drop the transformation without saying why or how they arrived at this particular transformation. In particular, how do we arrive at the transformed variable $$z_i,$$ where $$\tilde{y}^{\lambda-1}$$ is the geometric mean?

$$z_i = \frac{y_i^{\lambda}-1}{\lambda\tilde{y}^{\lambda-1}} \text{ when }\lambda \neq 0 \\ z_i = \frac{\log(y_i)}{\tilde{y}^{\lambda-1}} \text{ when } \lambda=0$$

It seems that parts of this transformation come from the normal log-likelihood, but I'm not sure how to move from one step to the other.

• It has multiple origins. One is in variance stabilization, as explained at stats.stackexchange.com/a/251661/919. Another is in symmetrizing a distribution, as explained at stats.stackexchange.com/a/24236/919 and hinted at in stats.stackexchange.com/a/96684/919. None of it is related to Normal distributions or likelihoods. Could you explain what you mean by "move from one step to the other"? What are the "steps"?
– whuber
Mar 31, 2020 at 19:39
• If you simply start is the notion of using power transformations, you'll discover a problem with zero power, but with a suitable location and scale shift you get the log as the limiting case; as a result you could look at Box-Cox as nothing more than "power transforms normalized to include the log at power 0". Apr 1, 2020 at 2:08
• Thank you for pointing out that it is not related to Normal distributions or likelihoods. From what I understand so far, we want to consider $\frac{y^\lambda-1}{\lambda}$ (rather than just $y^{\lambda}$, since the former goes to $ln\lambda$ rather than 1, as $\lambda$ goes to zero, as is the case for the latter). By "moving from one step to the other", I'm wondering, how do we justify dividing by the geometric mean for the transformation? Apr 1, 2020 at 16:25