Limiting chance the smallest dips below 2/n

Question

Draw n independent numbers uniformly from [0,1] and let Ln be the smallest. As n grows large, what is the limiting probability that Ln falls below 	frac2n?

Accepted Answer

The minimum is above a level x when all n draws are above x, so P(Ln x) = (1-x)^n. With x = \frac2n: P\!\left(Ln 	frac2night) = \left(1 - 	frac2night)^n \longrightarrow e^{-2}. Therefore P\!\left(Ln < 	frac2night) \longrightarrow 1 - e^{-2} \approx 0.8647. Mirroring the maximum, the minimum of n uniforms lives within about \frac1n of the floor, and the rescaled value n Ln tends to an exponential distribution. Takeaway. The minimum hugs 0 at scale 1/n: P(Ln < c/n) 	o 1 - e^{-c}, the mirror image of the maximum near 1.

Limiting chance the smallest dips below 2/n

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