# Uncertainty from Box–Cox estimation

Consider the model $$Y_i^{(\lambda)} = \alpha+\beta x_i + \varepsilon_i,\qquad \varepsilon_i,\ (i=1,\ldots,n) \sim \mathrm{i.i.d.}\ N(0,\sigma^2)$$ where \begin{align} y_i^{(\lambda)} & = \begin{cases} \dfrac{y_i^\lambda-1}{\lambda(\operatorname{GM}(y))^{\lambda -1}} , &\text{if } \lambda \neq 0 \\[12pt] \operatorname{GM}(y)\log{y_i} , &\text{if } \lambda = 0 \end{cases} \\[12pt] \text{and } & \operatorname{GM}(y) = (y_1\cdots y_n)^{1/n}\text{ is the geometric mean.} \end{align}

As my estimate of $\lambda$, I use the value that minimizes the sum of square of residuals when $\alpha$ and $\beta$ have likewise been estimated by least squares.

In finding a confidence region for $\alpha$ and $\beta$, how should one take into account the uncertainty in the estimate of the Box–Cox parameter $\lambda$?

• It would simplify things to throw out the rescaling by the GM. In practice this will be handled through the estimates of $\alpha$, $\beta$, and $\sigma$ anyway, so the fits will be identical, but for theoretical work the presence of that GM looks terribly complicating. It's rather bizarre that you multiply the $\log(y_i)$ by the GM, by the way. Does that have an interpretation? – whuber May 24 '13 at 19:46
• The rescaling makes $y_i^{(\lambda)}$ dimensionless, so that it has the same value regardless of what units $y$ is measured in. Without the rescaling, the value of $\lambda$ that minimizes the sum of squares actually depends on the units of measurement---e.g. whether it's feet or inches. The multiplication by the GM is there for a reason I'd have thought was obvious: $\displaystyle\lim_{\lambda\to0} y_i^{(\lambda)}$ $=\operatorname{GM}(y)\log y_i$. ${}\qquad{}$ – Michael Hardy May 24 '13 at 20:02
• It is obvious--but it does not appear statistically meaningful in this context. Using the GM for such normalization appears mathematically ad hoc rather than motivated by a statistical principle. Ordinarily, one uses a Box-Cox transformation for either or both of two reasons: linearization and achieving approximate homoscedasticity. If that's what you're looking for, then rather than rescaling to make it unitless, you might consider optimizing an objective related to those purposes. Two approaches seem appropriate: maximizing likelihood (best) or $R^2$ (both of which are unitless, too). – whuber May 24 '13 at 20:18
• As an aside, if you are using this test in a time series context please be aware that if the errors are not i.i.d. due to outliers or level shifts or time trends or ARIMA structure or parameter changes over time or points in time where the variance of the errors change deterministically or incorrect specification of the transfer between y and x this test yields false positives. – IrishStat May 24 '13 at 20:25
• So you're saying that estimating $\lambda$ by minimizing mean squared error is the wrong thing to do? It seems as if making it unitless in that case would be necessary since otherwise the sum of squares of residuals for one value of lambda is not measure in the same units as for another value, and which value gives you the smaller sum of squares actually depends on the units of measurement. – Michael Hardy May 24 '13 at 20:25

• +1. Is there anything published on how bootstrap performs on problems like this? (That final "8" in "0.4733268" is of course colossally silly: this is where the noise totally overwhelms the signal.) I wonder if anyone's used Bayesian methods in which the prior on $\lambda$ puts all the weight on rational numbers, and gives more weight to those with small denominators? That way things like $0.5$ are automatically favored over things like $0.4733268$, and in physical data there seem to be reasons for doing things that way. – Michael Hardy May 24 '13 at 22:27