Solved – Box Cox transformation in R

data transformationrregression

I understand how to use the box cox transformation in R and how to get the graph and lambda.

These are the things that are confusing me. For simplicity assume this example:

 Weight = Gender + Height + Age + Income

gender = categorical variable 1 = male, 0 = female

continuous variables – weight, height, age, income.

I've done:

model = lm(weight ~ gender + height + age + income)

and applied box-cox to this model which is about 1 so no transformation is needed for weight.

The questions are:

How do I apply Box-Cox to the 'x' variables to see if they need to be transformed (I read they can be applied to all x variables, the only disadvantage is its time consuming, but this isn't an issue for me).
How do I know if they need a transformation for sure e.g. if lambda is 0.4 should I use a square root transformation or does it 'have' to be 0.5? what is lambda is 2?
If one or two or all variables need a transformation then how do I adjust the main model formula?

Best Answer

When referring to Box-Cox transformations there are really 2 concepts that look like they are being mixed up. The first is what the original paper was about, the methodology of finding a transformation within a family of transformations that gives the "best" transformation assuming the truth results in normal residuals with equal variance and a linear relationship. This is what you already did with the response (dependent) variable.

In the paper Box and Cox presented a family of transformations (actually the paper has more than 1, but variations on the main one are what people mainly refer to), this family of transformations (or variations on it) is also often called the Box-Cox transformations. If you want to apply one of this family of transformations to a predictor variable, then just plug the variable into the formula.

To determine what types of transformations to apply to predictor variables, the most important thing to use is knowledge about your data and the science behind it. How do you expect weight to change with age, height, or income? (I would be surprised if a Box-Cox transform is the best for any of these).

There are tools like spline fits, ACE, AVAS, and others that can suggest transformations, but you need to use knowledge and common sense to convert these into meaningful transformations.

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Solved – Box Cox Transformation with swift

By reading documentation from the R geoR package we can find that the boxcox transform with the extra parameter $\lambda_2$ is defined as $$ Y' =\begin{cases} \log(Y+\lambda_2) \text{if $\lambda=0$} \\ \frac{(Y+\lambda_2)^\lambda -1}{\lambda} \text{otherwise} \end{cases} $$ so if $\lambda_2=0$ this is the usual boxcox transform, and the boxcoxfit function will estimate the two parameters $\lambda, \lambda_2$ by maximum likelihood. From the examples on the help page it seems like the boxcoxfit function can take as first argument either a vector of data values or a model object, then presumably using the residuals from the fit. From your code it seems like you have just given a data vector, thus finding the boxcox transformation parameters based on the marginal distribution of the response variable. That is usually less useful than using the residuals from the model fit, so you should reanalyze doing that!

You asks then why the $\lambda_2=\text{3.116280e+04}$ is so large, if that is reasonable? Well, we cannot judge that, since you did'nt tell us about the marginal distribution of your $Y$ variable! But, you told us that $Y=\text{annual foreign sales of companies (in US\$ thousands)}$ and I would guess that contains many large values, so in that context maybe the estimated $\lambda_2$ is not so large.

Solved – How to get the true mean forecast using the Arima package with a Box-Cox transformation

The approach adopted by the forecast package is to fit a model to the transformed data and to backtransform both the point forecasts and prediction intervals. This is not "wrong". It is a legitimate choice. This preserves the coverage of the prediction intervals, and the back-transformed point forecast can be considered the median of the forecast densities (assuming the forecast densities on the transformed scale are symmetric). For many purposes, this is acceptable, but occasionally the mean forecast is required. For example, with hierarchical forecasting the forecasts need to be aggregated, and medians do not aggregate but means do.

It is easy enough to derive the mean forecast using a Taylor series expansion. Suppose $f(x)$ represents the back-transformation function, $\mu$ is the mean on the transformed scale and $\sigma^2$ is the variance on the transformed scale. Then using the first three terms of a Taylor expansion around $\mu$, the mean on the original scale is given by $$f(\mu) + \frac{1}{2}\sigma^2f''(\mu).$$

For a Box-Cox transformation, $$f(x) = \left\{\begin{array}{ll} (\lambda x+1)^{1/\lambda} & \text{if $\lambda\ne0$;}\\ e^x & \text{if $\lambda=0$.}\end{array}\right. $$ So $$f''(x) = \left\{\begin{array}{ll} (1-\lambda)(\lambda x+1)^{1/\lambda-2} & \text{if $\lambda\ne0$;}\\ e^x & \text{if $\lambda=0$.}\end{array}\right. $$ and the backtransformed mean is given by $$\left\{\begin{array}{ll} (\lambda \mu+1)^{1/\lambda}\left[1 + \frac{\sigma^2(1-\lambda)}{2(\lambda \mu+1)^{2}}\right] & \text{if $\lambda\ne0$;}\\ e^\mu\left[1 + \frac{\sigma^2}{2}\right] & \text{if $\lambda=0$.}\end{array}\right. $$ Therefore, to adjust the back-transformed mean obtained by R, the following code can be used.

library(forecast)

fit <- auto.arima(AirPassengers, lambda=0)
fc <- forecast(fit, h=50, level=95)
fvar <- ((BoxCox(fc$upper,fit$lambda)-BoxCox(fc$lower,fit$lambda))/qnorm(0.975)/2)^2
plot(fc)
fc$mean <- fc$mean * (1 + 0.5*fvar)
lines(fc$mean,col='red')

fit <- auto.arima(AirPassengers, lambda=0.2)
fc <- forecast(fit, h=50, level=95)
fvar <- ((BoxCox(fc$upper,fit$lambda)-BoxCox(fc$lower,fit$lambda))/qnorm(0.975)/2)^2
plot(fc)
fc$mean <- fc$mean * (1 + 0.5*fvar*(1-fit$lambda)/(fc$mean)^(2*fit$lambda))
lines(fc$mean,col='red')

An extended version of this answer is given on my blog.

Best Answer

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Solved – How to get the true mean forecast using the Arima package with a Box-Cox transformation

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