Having come across this discussion I'm raising the question on the back-transformed confidence intervals conventions.

According to this article the nominal coverage back-transformed CI for the mean of a log-normal random variable is:

$\ UCL(X)= \exp\left(Y+\frac{\text{var}(Y)}{2}+z\sqrt{\frac{\text{var}(Y)}{n}+\frac{\text{var}(Y)^2}{2(n-1)}}\right)$ $\ LCL(X)= \exp\left(Y+\frac{\text{var}(Y)}{2}-z\sqrt{\frac{\text{var}(Y)}{n}+\frac{\text{var}(Y)^2}{2(n-1)}}\right)$

/and not the naive $\exp((Y)+z\sqrt{\text{var}(Y)})$/

Now, what are such CIs for the following transformations:

  1. $\sqrt{x}$ and $x^{1/3}$
  2. $\text{arcsin}(\sqrt{x})$
  3. $\log(\frac{x}{1-x})$
  4. $1/x$

How about the tolerance interval for the random variable itself (I mean a single sample value randomly drawn from the population)? Is there the same issue with the back-transformed intervals, or will they have the nominal coverage?

  • 2
    $\begingroup$ See Taylor expansion for moments of functions of rvs, and the Delta method. But care is needed. See for example discussion here and [here](stats.stackexchange.com/questions/41896/varx-is-known-how-to-calculate-var1-x/). Searching on taylor series will present several useful examples and discussions. $\endgroup$
    – Glen_b
    Commented Feb 17, 2014 at 0:22
  • $\begingroup$ I've made substantial edits to your formulas. Please check I didn't get any of them wrong. On my previous comment (sorry about the improperly formatted link there) - also see the precautionary comment under the answer here $\endgroup$
    – Glen_b
    Commented Feb 17, 2014 at 0:32
  • $\begingroup$ Thanks. Although I can hardly post a thing without being edited with those fancy expressions. $\endgroup$ Commented Feb 17, 2014 at 0:40

1 Answer 1


Why are you doing back transformations at all? That's critical to answering your question because in some cases the naive transform is the right answer. In fact, I think I'll argue that, if the naive back transform isn't the right answer then you shouldn't back transform at all.

I find the general issue of back transformation highly problematic and often filled with muddled thinking. Looking at the article you cited, what makes them think that it's a reasonable question that the back transformed CI doesn't capture the original mean? It's a mistaken interpretation of back transformed values. They think that the coverage should be for direct analysis in the back transformed space. And then they create a back transform to fix that mistake instead of their interpretation.

If you do your analyses on log values then your estimates and inferences apply to those log values. As long as you consider any back transform a representation of how that log analysis looks in the exponential space, and only as that, then you're fine with the naive approach. In fact, it's accurate. That's true of any transform.

Doing what they're doing solves the problem of trying to make the CI into something that it's not, a CI of the transformed values. This is fraught with problems. Consider the bind you're in now, the two possible CI's, one in transformed space where you do your analyses, and one back transformed, make very different statements about where the likely mu is in the other space. The recommended back transform creates more problems than it solves.

The best thing to take out of that paper is that when you decide to transform the data it has deeper impacts than expected on the meaning of your estimates and inferences.

  • $\begingroup$ Could you please explain that further? It seems to me the issue being the naive CI gives that of the geometric mean, rather than the arithmetic. Which is what would imply it to be strictly smaller, as they say, and hence the inconsistency and poor coverage. $\endgroup$ Commented Feb 17, 2014 at 9:18
  • $\begingroup$ Inconsistency with what? If you're going to analyze your exponential distribution directly and want to know the arithemtic mean then yes, it's poor coverage for that. But if you wanted to do that then you should have done that. If you're going to log transform your distribution and analyze the exponents then it's exactly the right coverage for that. $\endgroup$
    – John
    Commented Feb 17, 2014 at 16:25
  • $\begingroup$ I cant's see why do you object against the method in the article. The simulations show it performing fine, while the naive method is doing worse than the "Central-limit approach". $\endgroup$ Commented Feb 17, 2014 at 18:12
  • 1
    $\begingroup$ They show it doing better for what they want to make it be. The naive method works just fine for what it is. Look at the simulation in section 5. They set up a lnorm distribution mean 5, which has an exponent of 148.4. Then they go on to discuss coverage of the mean of 244.6!! That would only be important if you're going to model the mean of the original distribution, NOT the logs. They're trying to make it something it's not. The naive calculation has perfectly fine coverage over the mean of the log, 5. None of the other CI's are 95% CI's of that value and that's the one you're analyzing. $\endgroup$
    – John
    Commented Feb 17, 2014 at 19:07

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