What is the geometric standard deviation of a value, which is the result of dividing two independent values, each of which has its own geometric standard deviation ?

It is a frequent situation in science, that the the signal of the machines is in linear relationship to the logarithm of the analysed quantity, so it is appropriate to use geometric standard deviation to characterize the the error of the measured quantity. (I calculated geometric mean and geometric standard deviation from triplicate measurements for each value). Now I want to divide two such quantity values (one by another) to obtain a relative concentration. What will be the geometric standard deviation of the relative concentration ?

(I assume, that the two original values are independent - I do not want to deal with covariances etc.)

EDIT: In desperation, I came up with a theory, but would appreciate a confirmation from a matematician, because I am a molecular biologist.

I wanted to do x=a/b. I took my triplicate measurements and made all possible combinations of values, which are a1/b1, a1/b2, a1/b3, a2/b1, a2/b2, a2/b3, a3/b1, a3/b2, a3/b3. Then I calculated geometric mean and geometric standard deviation, as if these 9 results were 9 independent measurements.


Another approach: I assume, that "lognormal distribution, geometric means, multiplication/division of geometric means, logarithms of geometric standard deviations" are all in the same analogy to "normal distribution, arithmetic means, addition/substraction of arithmetic means, arithmetic standard deviations".

For my "x=a/b" , it leads me to these formulas: enter image description here

The resulting equation gives me the same results, as when I calculate from those 9 independent combinations described above. I calculate in Excel, which shows many decimal numbers, so it demonstrates very nicely, that the results of both calculations are identical.

Can someone confirm the formulas are correct, please ?


1 Answer 1


Yes, I believe your formula is correct. If $A$ is a random variable which takes positive values the geometric standard deviation of $A$ is by definition the exponential of the ordinary (arithmetic) standard deviation of $\log(A)$. Write $\sigma_g(A)$ for the geometric standard deviation and $\sigma(A)$ for the arithmatic standard deviation. Then: $$\log \sigma_g(A/B)= \sigma(\log(A/B)) = \sigma(\log(A)-\log(B))$$ but we are assuming that $\log(A)$ and $\log(B)$ are independent, since $A$ and $B$ are, and so their variances add. Therefore $$\sigma(\log(A)-\log(B))=\sqrt{(\sigma(\log(A)))^2 + (\sigma(\log(B)))^2} = \sqrt{(\log\sigma_g(A))^2 + (\log\sigma_g(B))^2}$$ which is what you found.

  • $\begingroup$ Thanks !!! Just out of curiosity - my/Your formulas and my manual calculating of geometric standard deviation from a1/b1, a1/b2, a1/b3, a2/b1, a2/b2, a2/b3, a3/b1, a3/b2, a3/b3 give the same result. Why does it not work that way for propagation of standard deviation (not geometric) in additive operations and for a1+b1, a1+b2, a1+b3, a2+b1, a2+b2, a2+b3, a3+b1, a3+b2 a3+b3 ? When I try that in Excel, stdevs are not the same, although similar. $\endgroup$
    – Barbara
    Sep 27, 2013 at 7:56
  • $\begingroup$ I am not sure but my guess is that it probably has something to do with using an $n$ or an $n-1$ in the standard deviation formula. Possibly Excel uses an $n-1$ by default, or something like that. $\endgroup$
    – Flounderer
    Sep 27, 2013 at 22:36
  • $\begingroup$ Nope, I used Stdevp in Excel, which does not use n-1. Btw, my questions about arithmetic stdev are not really necessary for my research, I just like to understand things... $\endgroup$
    – Barbara
    Sep 28, 2013 at 20:22

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