2
$\begingroup$

I heard about an interesting technique for making ballpark guesses:

If you need a variable that you have no idea about (like the average number of pianos per household), you can try to come up with a range instead. So you can say that it must be larger than $min$ and smaller than $max$

Thenyou can take the geometric mean $\sqrt{max\cdot min}$ as an estimation.

Is there any justification for using here a geometric mean instead of the simpler arithmetic mean?

$\endgroup$
  • $\begingroup$ I am kind of confused here, do you only have the max and min for your data, or do you only want to use those two values? $\endgroup$ – Josh Jun 11 '17 at 15:42
  • $\begingroup$ @Josh: No, I don't have any data, they are also guesses... For example, i'm trying to guess how many piano tuners work in Chicago. So I'd need the number of pianos per household, and then I could say something like it must be larger than 1/25 pianos per household, but 1/10 pianos per household sounds too much... so instead of these extremes i'd take the geometric mean, 0.06 $\endgroup$ – AndrasG Jun 11 '17 at 16:01
  • $\begingroup$ So it is going to be hard to justify doing exactly what you are doing because it is fairly ad hoc and not based on data. There is the (arithmetic) midrange, but that is based on the idea of using max and min of the data, not guesses. I am also confused as to why you are using the geometric mean. You could try using a prior, and just not get a poterior, or something like that I guess. But doing statistics without data is not a trivial task. $\endgroup$ – Josh Jun 11 '17 at 16:30
2
$\begingroup$

Here's a couple of (somewhat related) reasons why:

For quantities that must be positive (consider that we know that at least one piano exists in at least one household) there's a tendency for errors on the high side to be further away than errors on the low side (you can be far more than 100% too big easily, but you can't be even 100% too small, because that would mean guessing $0$, a value we just said we know is not the case); indeed - at least to a rough approximation - guessing twice too high and guessing half too high would be nearer to equally likely. (Our relative uncertainties in such guesses are usually not absolute but relative.)

Or imagine that we had some kind of very noisy sense of how big it was, but there were several contributions to that uncertainty. We know that certain kinds of households are more likely to have pianos -- apartments, for example, will have trouble fitting even an upright in, while large houses might fit a grand piano, but we don't really have a good sense of a typical number of pianos per large house nor a typical proportion of large houses. Those effects tend to arise as products (proportion of dwellings big enough for a piano x pianos per larger dwelling).

When you have lots of contributions of small additive errors, means work well for estimating typical values. If you take logs, the product is a sum and the geometric mean is an arithmetic mean on the log scale.


I'm brought to mind of an essay in which Asimov explored the Drake equation to estimate the number of technological civilizations in the galaxy (I believe he also did a similar analysis in a book and a paper). At one point he treats one probability as "either being very small or large" (paraphrasing). He then chooses two representative values for those possibilities, and averages the final results. The outcome is very similar to just choosing a somewhat different large value (one merely half as big), so it gives almost no weight to the "small" option. When you have uncertainty in a positive quantity that spans orders of magnitude, the large one will completely dominate the outcome, even though it had by far the larger (absolute) uncertainty.

$\endgroup$

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.