# The answer is e, what was the question?!

February 11, 2016
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(This article was first published on R – Xi'an's Og, and kindly contributed to R-bloggers) A rather exotic question on X validated: since π can be approximated by random sampling over a unit square, is there an equivalent for approximating e? This is an interesting question, as, indeed, why not focus on e rather than π after all?! But very quickly the very artificiality of the problem comes back to hit one in one’s face… With no restriction, it is straightforward to think of a Monte Carlo average that converges to e as the number of simulations grows to infinity. However, such methods like Poisson and normal simulations require some complex functions like sine, cosine, or exponential… But then someone came up with a connection to the great Russian probabilist Gnedenko, who gave as an exercise that the average number of uniforms one needs to add to exceed 1 is exactly e, because it writes as $\sum_{n=0}^\infty\frac{1}{n!}=e$

(The result was later detailed in the American Statistician as an introductory simulation exercise akin to Buffon’s needle.) This is a brilliant solution as it does not involve anything but a standard uniform generator. I do not think it relates in any close way to the generation from a Poisson process with parameter λ=1 where the probability to exceed one in one step is e⁻¹, hence deriving  a Geometric variable from this process leads to an unbiased estimator of e as well. As an aside, W. Huber proposed the following elegantly concise line of R code to implement an approximation of e:

1/mean(n*diff(sort(runif(n+1))) > 1)

Hard to beat, isn’t it?! (Although it is more exactly a Monte Carlo approximation of $\left(1-\frac{1}{n}\right)^n$

which adds a further level of approximation to the solution….)

Filed under: Books, R, Statistics Tagged: Buffon’s needle, cross validated, Gnedenko, Monte Carlo integration, Poisson process, simulation  To leave a comment for the author, please follow the link and comment on their blog: R – Xi'an's Og.

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