# Estimating with the geometric mean

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?

• 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?
– Josh
Jun 11, 2017 at 15:42
• @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 Jun 11, 2017 at 16:01
• 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.
– Josh
Jun 11, 2017 at 16:30
• For reference, here is a possible source for this technique Nov 16, 2022 at 15:32

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.)
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