Reverse BoxCox transformation with negative values Using the R package MASS, I transformed a variable, let's call it $V$, into another variable called $X$ with $\lambda = 1.25$.
Now, the BoxCox transformation has the following shape:
$X = (V^\lambda - 1) / \lambda$
So the reverse transformation is:
$V = (\lambda X + 1)^{1/\lambda}$
With $\lambda = 1.25$, $V < 1$ implies that $X < 0$. However, the reverse transformation only works for positive values with $\lambda = 1.25$. I therefore want to add a constant to $X$ before doing the reverse transformation. Let's call it $Z$ such that :
$Z = \begin{cases}1 - \mbox{min(X)} , & \mbox{if min(X) < 0}\\ 0 , & \mbox{otherwise}\end{cases}$
Then the reverse transformation becomes:
$V + E = (\lambda X + \lambda Z + 1)^{1/\lambda}$ 
Question: What transformation do I apply to $V + E$ in order to get back the true value $V$ only ?
 A: The Box-Cox family (I like the name a lot, but am not related to Sir David Cox) has the form it has so that powers and the logarithm belong neatly together and have similar behaviour around the origin. 
But while that is satisfying in principle, it need not govern what you do in practice. For example, if square roots ($\lambda = 0.5$ or $V^{0.5} = \root \of V$) are a good idea, I know of no reason whatsoever why $(\root \of V - 1) / 0.5$ is a better idea; indeed it's just an unnecessary linear re-expression of square roots and has the same statistical consequences. 
Similarly, if analysis points to $1.25$ as a power, I might use $V^{1.25}$ defined for positive and negative values alike as $\text{sign}(V) |V|^{1.25}$. Actually, I would need a strong reason not to use a variable as it comes for a power so close to $1$. 
A: There are two versions of the Box-Cox transformation: the one-parameter version (as above), and the two-parameter version, which is applied if some values of V are observed to be negative, or could be negative, in which case you transform V' using Box-Cox, where V'=V+ß. In most practical applications, it's usually OK to set ß=-MIN(V), but if you have some prior knowledge about the possible range of values of V, then it's even better.
