Editing Fermi problem (section)

== Justification ==
Fermi estimates generally work because the estimations of the individual terms are often close to correct, and overestimates and underestimates help cancel each other out. That is, if there is no consistent bias, a Fermi calculation that involves the multiplication of several estimated factors (such as the number of piano tuners in Chicago) will probably be more accurate than might be first supposed.

In detail, multiplying estimates corresponds to adding their logarithms; thus one obtains a sort of [[Wiener process]] or [[random walk]] on the [[logarithmic scale]], which diffuses as <math>\sqrt{n}</math> (in number of terms ''n''). In discrete terms, the number of overestimates minus underestimates will have a [[binomial distribution]]. In continuous terms, if one makes a Fermi estimate of ''n'' steps, with [[standard deviation]] ''σ'' units on the log scale from the actual value, then the overall estimate will have standard deviation <math>\sigma\sqrt{n}</math>, since the standard deviation of a sum scales as <math>\sqrt{n}</math> in the number of summands.

For instance, if one makes a 9-step Fermi estimate, at each step overestimating or underestimating the correct number by a factor of 2 (or with a standard deviation 2), then after 9 steps the standard error will have grown by a logarithmic factor of <math>\sqrt{9} = 3</math>, so 2<sup>3</sup> = 8. Thus one will expect to be within {{frac|8}} to 8 times the correct value – within an [[order of magnitude]], and much less than the worst case of erring by a factor of 2<sup>9</sup> = 512 (about 2.71 orders of magnitude). If one has a shorter chain or estimates more accurately, the overall estimate will be correspondingly better.