I have seen and read several similar questions, but mine pertains specifically to zero rich data.

I will be back transforming my data based on a first order Taylor series approximation. As outlined in


from page 7

$$\tilde{X} = exp(\hat{Y}) \\ \tilde{\sigma}_X = \tilde{X} \hat{\sigma}_Y$$

according to Wooldridge 2009 (p. 192)

"The percentage change interpretations are often closely preserved, except for changes beginning at y=0 (where the percentage change is not even defined). Generally, using log(1+y) and then interpreting the estimates as if the variable were log(y) is acceptable when the data contain relatively few zeros."

However, my data is zero rich so I came up with the following?

$$ \tilde{X} = exp(\hat{Y})-1 \\ \tilde{\sigma}_X = \tilde{X} \hat{\sigma}_Y $$

from my understanding because the formula for sample SD is:

$$\sigma_x = \sqrt{\frac{\sum{(X-\bar{X})^2}}{n-1}} $$

and relies on the difference of X and $\bar{X}$ shifting it by any constant c should not affect the standard deviation.

a simple test to show problem with zero rich data.

#log is natural log code is in R
test <- c(0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1 2)
mean(test) #mean transformed CASE1
#[1] 0.3333333
exp(mean(log(test + 1))) - 1 #CASE2 mean estimation adjusted for +1
#[1] 0.2300755
exp(mean(log(test + 1))) #CASE3 mean estimation based on exp(y)
#[1] 1.230076

gm_mean = function(a){prod(a)^(1/length(a))} # function from @doug
gm_mean(test) #CASE4 geometric mean
#[1] 0

CASE3 != CASE4 showing that transformation is affecting estimate of mean, as we would expect exp(y) to be equal to the geometric mean of y.

CASE4 and CASE2 are both close to CASE 1 with CASE2 being the best estimate.

This shows that for zero rich distributions there appears to be a discrepancy among different estimations of the mean.

If we perform the same analysis on the following data which does not need transformation

test2 <- c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3)
mean(test2) # CASE1 mean on untransformed 
#[1] 1.333333
exp(mean(log(test2 + 1))) - 1 # CASE2 log trans mean approx exp(y) -1
[1] 1.267067

exp(mean(log(test2))) #CASE3 mean exp(y)
[1] 1.230076
gm_mean(test2) #CASE4 geometric mean equals exp(y) as expected
[1] 1.230076

we find that CASE3 == CASE4 and both are roughly equal to CASE2. CASE 2-4 are all good approximations for the transformed mean.

TL;DR: are $$ \tilde{X} = exp(\hat{Y})-1 \\ \tilde{\sigma}_X = \tilde{X} \hat{\sigma}_Y $$ a good approximation for mean and standard deviation of zero rich ln + 1 transformed data?

edit: fixed sd formula

  • 1
    $\begingroup$ Why do you want to transform your data ? Why not fit a zero-inflated model ? $\endgroup$ Apr 24, 2019 at 3:58
  • $\begingroup$ Unfortunately, you haven't told us what you are actually trying to do do, so no one would be able to provide you with ANY guidance that would be useful at this stage. $\endgroup$ Apr 24, 2019 at 4:04
  • $\begingroup$ My data is not zero rich enough to justify and zero-inflated model (though I will look into this for anther project I am doing). Also, I transformed the data after looking at dist and using Box Cox procedure to justify my choice (Lambda was essentially zero). The question is more of a hypothetical. I merely want to know if adapting the back-transformation method from the pdf to adjust for log + 1 is correct. If it is not would it be correct for me to use the method from the included link for data that has been transformed with log+1 $\endgroup$
    – NicoFish
    Apr 24, 2019 at 4:12
  • $\begingroup$ Again, you haven't told us what it is you are trying to actually do. What is it you are trying to accomplish by taking these manipulations of your data? What is the context? Analysts typically don't just transform data for the fun of it. If we don't know your motivations it would be impossible for us to tell if any "back transformation" method is "correct." $\endgroup$ Apr 24, 2019 at 4:22
  • $\begingroup$ @StatsStudent. This is a question about back-transformation of data that has been transformed using log + 1. The analysis is irrelevant. You simply want to get the mean and sd of the transformed data NOT perform an analysis. You would like to report the mean and SE of the transformed data for the benefit of the reader. You cannot simply display mean(log(data + 1)). I would like to know if it is correct to represent the mean as exp(mean(log(data +1))) - 1 instead of exp(mean(log(data +1))). In this case back-transformation has nothing to do with any analysis or end goal. $\endgroup$
    – NicoFish
    Apr 24, 2019 at 5:03


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