Limiting chance the max stays below 1 minus 1/n

Question

Draw n independent numbers uniformly from [0,1] and let Mn be the largest. As n grows large, what is the limiting probability that Mn stays below 1 - 	frac1n?

Accepted Answer

The maximum is below a level t when all n numbers are below t, so P(Mn \le t) = t^n. Here t = 1 - \frac1n: P\!\left(Mn \le 1 - 	frac1night) = \left(1 - 	frac1night)^n. This is the classic limit \left(1 - 	frac1night)^n \longrightarrow e^{-1} = \frac1e \approx 0.3679. The natural scale for the gap below 1 is \frac1n: the maximum of n uniforms typically falls within about \frac1n of the ceiling, and the rescaled shortfall n(1 - Mn) settles into an exponential distribution. Takeaway. Near the top of a uniform sample, distances shrink like 1/n; setting t = 1 - c/n gives P(Mn \le t) 	o e…

Limiting chance the max stays below 1 minus 1/n

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