0
$\begingroup$

If you have data that are sampled from a normal distribution, what is the relationship between the arithmetic and geometric means? Would it ever make sense to report the geometric mean instead of the arithmetic mean? (Assume that all the values are positive; no zeros, no negative values)

$\endgroup$
  • $\begingroup$ Do you mean "arithmetic mean" rather than "arrhythmic mean"? If you really mean "arrhythmic mean", could you explain what one of those is? (A search shows the term occurring at least a few times, for example in a book with all French authors, but it's not a standard term in statistics written in English and would need explanation to be widely understood.) $\endgroup$ – Glen_b Feb 15 '16 at 23:14
  • $\begingroup$ Would you be happy to have a geometric mean $\sqrt{x_1x_2}$ have imaginary value when $x_1$ and $x_2$ have opposite sign? $\endgroup$ – Dilip Sarwate Feb 15 '16 at 23:38
  • $\begingroup$ Sorry I meant the arithmetic means $\endgroup$ – Bassam Feb 16 '16 at 10:02
  • 2
    $\begingroup$ Any normal distribution, no matter what $\mu$ & $\sigma^2$, will include $0$ & negative values. It is of course possible that your particular sample includes only strictly positive values, but the population must include negative values. $\endgroup$ – gung Feb 16 '16 at 15:11
  • $\begingroup$ Why do you want to use a geometric mean? $\endgroup$ – kjetil b halvorsen Feb 16 '16 at 18:12
1
$\begingroup$

Why do you want a geometric mean of normal distributed observations? I can see no good reason. https://www.math.toronto.edu/mathnet/questionCorner/geomean.html gives several good examples of use of geometric mean. A typical case is return on investment. Returns combine multiplicatively, so If you want one "typical" return that would result in same winning, if the return was held constant over years, you get that from the geometric mean of the yearly returns.

Common to all such examples is that the random variable in question cannot be negative, and, since every normal distributed variable is negative with some (maybe very small) positive probability, geometric means do not look natural to use. So, again, why do you want to use a geometric mean?

$\endgroup$

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.