Interpretation of log(1 + x) transformed predictor

Interpretation of log transformed predictor neatly explains how to interpret a log transformed predictor in OLS. Does the interpretation change if there are 0s in the data and the transformation becomes log(1 + x) instead?

Some authors (e.g. Fox and Weisberg 2011) recommend adding a start (i.e. a positive constant) if a log transformation is necessary to correct skewness and improve symmetry, but the data contains zeros.

Consider a variation of the Ornstein example in CAR (p. 303):

require(car)
data(Ornstein)
boxplot(Ornstein$interlocks, horizontal = T)  The data is clearly right skewed, and contains 0s. summary(powerTransform(1 + Ornstein$interlocks))
## bcPower Transformation to Normality
##
##                         Est.Power Std.Err. Wald Lower Bound Wald Upper Bound
## 1 + Ornstein$interlocks 0.1248 0.053 0.0209 0.2287 ## ## Likelihood ratio tests about transformation parameters ## LRT df pval ## LR test, lambda = (0) 5.502335 1 0.0189911 ## LR test, lambda = (1) 262.431991 1 0.0000000  The powerTransform() function suggests that a log(1 + x) transformation here could be useful. boxplot(log(1 + Ornstein$interlocks), horizontal = T)


As you can see, symmetry is indeed improved.

Question: If this transformed variable were to be included in an OLS regression as an IV, would the coefficient estimates still have the usual interpretation of log transformed variables?

1 Answer

It depends. According to Wooldridge (2012) the percentage change interpretations are often closely preserved, except for changes beginning at $$y = 0$$ (where the percentage change is not defined). Strictly speaking, using $$\log(1+y)$$ and then interpreting the estimates as if the variable were $$\log(y)$$ is acceptable only if the data on $$y$$ contain relatively few zeros.

• Do you have a page number for Wooldridge? Jul 5, 2015 at 16:27
• @DimitriyV.Masterov In Wooldridge 2009 it's p.192 (Chapter 6.2 More on functional form). Jul 5, 2015 at 18:00
• In the 2012 EDT (US version), it is at the button of page 193 Jul 5, 2015 at 18:19
• This answer is not quite correct and might be misleading. What matters isn't whether $y$ includes "relatively few zeros," but the actual values of $y$ relative to $1.$ See stats.stackexchange.com/questions/576504 for more accurate answers.
– whuber
May 25 at 12:20