0
$\begingroup$

When summarizing normalized data (for example, percentage data), one must use the geometric mean instead of the arithmetic mean. Thus, instead of using the arithmetic standard deviation, one shall use the geometric standard deviation.

When using arithmetic 'descriptors', we can describe the results as mean $\pm$ standard deviation. However, when dealing with geometric 'descriptors', we must describe them as the range from (the geometric mean divided by the geometric standard deviation factor) to (the geometric mean multiplied by the geometric standard deviation factor), since one cannot add/subtract "geometric standard deviation factor" to/from geometric mean.


My question is: What is the correct notation for representing this operator?
I'm looking for the notation of what is equivalent to $\pm$ for the case of geometric mean and geometric standard deviation.

$\endgroup$
1
$\begingroup$

By definition $\mu_g=\left(\prod_{i=1}^nx_i\right)^{1/n}$, $\sigma_g=\exp\left(\sqrt{\frac{\sum_{i=1}^n(\log x_i-\log\mu_g)^2}{n}}\right)$, so if you take the logarithm of both you get arithmetic mean and standard deviation for $\log X$. If further $\log X$ is approximately normal you could have something like a confidence interval $\log\mu_g\pm z\times\log\sigma_g$. You can then exponentiate both side to get $\mu_g\times/\div\sigma_g^z$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I understand that. However, is there a compact notation for express geometric mean 'times div' geometric SD? $\endgroup$ – Iago Carvalho Apr 15 '19 at 8:13
  • $\begingroup$ @IagoCarvalho I am not sure. I don't think there is any... $\endgroup$ – Steven Xu Apr 16 '19 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.