I am back-transforming y=log(x+1) data that represent individuals/m^2 of area. Data are in the form of X ± SE. I back-transformed the means, but I am wondering if it's possible to directly back-transform the standard errors and, if so, how? I need to back-transform these data because I am comparing to non-transformed data. Obtaining the raw data is not possible.

  • $\begingroup$ Is this a matter of confidence interval for the mean of a skewed variable with zeros? Or do you have a model for which that was the response or outcome? Why not use a generalised linear model with log link, either way? $\endgroup$
    – Nick Cox
    Jul 4, 2018 at 14:35
  • $\begingroup$ Hi Nick -- thanks for your comment. I am conducting a systematic review, and these data are from a paper that meets the review criteria. I am looking to back-transform the means and SE so that I can use the untransformed values to generate effect sizes to be used in meta-analysis. This is why I'm not analysing the data with a generalised linear model with log link -- I need the untransformed values to compare to other studies in the review. The authors originally transformed the data to accommodate for a variable with zeros. $\endgroup$
    – pvichm
    Jul 4, 2018 at 14:51
  • $\begingroup$ Could you please zoom-in on the exact issue. Why doesn't backtracking each step do the trick? $\endgroup$
    – Jim
    Jul 4, 2018 at 15:06
  • $\begingroup$ H Jim -- The paper discusses invertebrate response to fire. Sampling occurred in burnt and unburnt plots at six different times. Data were presented in a plot, wherein each point represented the mean number of individuals per m^2 of leaves with SE bars included, but these are log(x+1) transformed data. I lifted the means and SE off the plot using Plot Digitizer. I then back-transformed the means via y = 10^(x)-1, wherein y = untransformed mean and x = transformed mean. However, I am unsure how to back-transform the SE since they are dependent on variation among individual datapoints. $\endgroup$
    – pvichm
    Jul 4, 2018 at 15:16
  • $\begingroup$ was my answer clear? did it help? $\endgroup$
    – Ben Bolker
    Jul 9, 2018 at 1:12

1 Answer 1


In general, if you use a smooth nonlinear transformation $f(x)$ to transform a random variable $x$, its standard deviation will be approximately scaled (multiplied) by a factor $\phi=\left. \frac{\partial f}{\partial x} \right|_{x=\bar x}$, i.e. $\sigma(f(x)) = \phi \cdot \sigma(x)$, or $\sigma(x) = \sigma(f(x))/\phi$ (you can derive this using the first term in a Taylor expansion). In this case $\partial[\log(x+1)]/\partial x = 1/x$. So you can back-transform the value you have to get $x$, then multiply the standard error you have by that $x$ value.

This is a rather crude approach, but I think it's what I would recommend in the absence of any more information.

I might actually recommend that you transform your other (non-log-transformed) values in the other direction, to work on the log scale; if you are implicitly assuming Normally distributed values in your analysis, that's more likely to be true for abundances on the log scale than on the original (abundance/m$^2$) scale.

PS. If you are using $\log_{10}$ rather than natural log in your transformation, you should be aware that $\partial \log_{10}(x)/\partial x$ is $1/(\ln(10) \cdot x)$ (where $\ln$ is the natural logarithm; same as $\log$ above).


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