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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 $\lambda$?

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    $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Commented May 24, 2013 at 19:46
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    $\begingroup$ 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: $\displaystyle\lim_{\lambda\to0} y_i^{(\lambda)}$ $=\operatorname{GM}(y)\log y_i$. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented May 24, 2013 at 20:02
  • $\begingroup$ 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). $\endgroup$
    – whuber ♦
    Commented May 24, 2013 at 20:18
  • $\begingroup$ 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. $\endgroup$
    – IrishStat
    Commented May 24, 2013 at 20:25
  • $\begingroup$ 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. $\endgroup$
    – Michael Hardy
    Commented May 24, 2013 at 20:25

1 Answer 1

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You could bootstrap the entire process and use those results to create your confidence region for alpha and beta.

You could also use a Bayesian approach and place priors on alpha, beta, and lambda (and sigma) then work out a confidence region based on the posterior.

You could use simulations to see how often the regular confidence region (without adjusting for uncertainty in lambda) includes the true values and make adjustments based on that.

In any case, the purpose of Box-Cox transformations is not to find a "best" value of lambda and just use it blindly. The better approach is to use the information on lambda (including the confidence interval on lambda) to suggest possible values and combine that with your (and other experts) knowledge of the science that produced the data to choose a meaningful value of lambda. For example if the "best" value for lambda is estimated to be 0.4733268 and the 95% CI on lambda includes the value of 0.5 and there is a scientifically meaningful reason that a square root transformation would make sense then you should use the square root transformation instead of raising the data to a value like 0.4733268. Tools like the Box-Cox transformation are tools to be used along with scientific knowledge (and common sense) not to replace it.

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    $\begingroup$ +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. $\endgroup$
    – Michael Hardy
    Commented May 24, 2013 at 22:27
  • $\begingroup$ Related: stats.stackexchange.com/questions/122393/… $\endgroup$
    – kjetil b halvorsen ♦
    Commented Feb 25, 2017 at 13:58

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