Assume that we have objects (e.g. particles) whose properties (e.g. diameter) follow a log-normal distribution that can be described by a geometric mean $\mu_g$ and a geometric standard deviation $\sigma_g$.

Now consider that we want to measure $\mu_g$ and $\sigma_g$, by sampling $n$ samples from the distribution (e.g. measuring the diameters of $n$ individual particles). Consider further that the utilized measurement technique is imperfect, i.e. we don't receive the true value $A_i$ from a measurement, but instead we receive $A_i'=A_i+\Delta A_i$.

If we define that the errors are random, then a reasonable assumption for $A_i'$ could be that it is normally distributed around $A_i$ with a certain standard deviation $\sigma_{measurement}$.

We then calculate the measured properties of the initial log-normal distribution, based on the imperfect measurements, so we receive

$\mu_g'=\mu_g+\Delta \mu_g= \sqrt[n]{(A_1+\Delta A_1) (A_2+\Delta A_2) \cdots (A_n+\Delta A_n)}$


$\sigma_g'=\sigma_g+\Delta \sigma_g=\exp{\sqrt{ \sum_{i=1}^n \left(\ln { {A_i+\Delta A_i} \over \mu_g' } \right)^2 \over n }}$.


  1. Can we make general statements about $\Delta \mu_g$ and $\Delta \sigma_g$? E.g. $\Delta \mu_g<0$ or $\Delta \sigma_g>0$, or something along these lines?
  2. How does the sample size $n$ play into this? Often, we would assume that errors will approach 0 for $n\rightarrow\infty$. Is this also the case in this scenario?

Note: I simulated the process, and obviously I'm willing to share the results, but I'd like to have an unbiased discussion first.

Thanks a lot!

  • $\begingroup$ Why do you assume that the measurement error is normal and not lognormal? Isn't it more likely that the error is (on average) a certain fraction of the diameter, so larger errors (usually) with larger particles? $\endgroup$ May 3 at 15:35
  • $\begingroup$ @HarveyMotulsky Thanks for tuning in. Great point! I thought about that as well, and for many measurement techniques you might be right. However, I'm concerned with manual image-based measurements, where you typically click on opposing particle borders, while having zoomed in. IMHO, the error comes from the image having a little blur. So I guess that for both small and large particles, the clicks will be a few pixels off. Sometimes a little more, sometimes a little less. That being said, I think one could also argue for a log-normal error. Would that make the mathematical problem easier? $\endgroup$
    – Nos
    May 3 at 20:20
  • $\begingroup$ Your post asks about observations of the form $e^X+Y$, with additive normal errors. If the observations have the form $e^X+e^Y-1$ instead, with additive lognormal errors, the math is slightly more complicated. But if the observations have the form $e^Xe^Y$, with multiplicative lognormal errors as Harvey Motulsky suggests, then the math is easier. $\endgroup$
    – Matt F.
    May 3 at 20:41

1 Answer 1


I find this question easier after replacing $A$ by $e^X$ and $\Delta A$ by $Y$:

If $X$ and $Y$ are independent normals and $Y$ has mean $0$, are the moments of $\ln(e^X+Y)$ bigger or smaller than the moments of $X$?

This only makes sense when $Y$ is small by comparison with $e^X$, so that the cases where $e^X+Y$ is negative are negligible -- which is the same as requiring that the errors are small by comparison with the measurements, so that the geometric mean mentioned in the question is meaningful.

With that mild assumption, the answer is:

  • $E[\ln(e^X+Y)]<E[X]$
  • $Var[\ln(e^X+Y)]>Var[X]$

The claim about means follows from the concavity of $\log$, or in more detail:

\begin{align} (e^x+y)(e^x-y)&<(e^x)^2\\ \frac12[\ln(e^x+y)+\ln(e^x-y)]&<x\\ E[\ln(e^x+Y)]&<x\\ E[\ln(e^X+Y)]&<E[X] \end{align}

The claim about variances follows, at least approximately, from

$$Var[\ln(e^X+Y)]-Var[X]\simeq Var[e^{-X}Y]$$

which is accurate to second order in $Y$.

  • $\begingroup$ Thanks a ton for the awesome answer! Is there a way to quantify the errors of the resulting geometric standard deviation and the geometric mean? $\endgroup$
    – Nos
    May 4 at 7:09
  • 1
    $\begingroup$ The Taylor series for the error of an observation is $\ln(e^X+Y)-X\simeq e^{-X}Y-\frac12 e^{-2X}Y^2$, so each observation has a mean error of approximately $-\frac12E[e^{-2X}]Var[Y]$, and the overall estimate of the mean has the same expected error. You can also ask: “If each observation has an error of $aY+bY^2$, what are the means and variances of its sample mean and sample variance”, but the questions with variance are more involved than I can answer here. $\endgroup$
    – Matt F.
    May 4 at 9:56

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