Limiting chance the max of exponentials stays modest

Question

Draw n independent values from an exponential distribution with mean 1, and let Mn be the largest. As n grows large, what is the limiting probability that Mn stays below \ln n + \ln 3?

Accepted Answer

For the exponential with mean 1, a single value is below level x with probability 1 - e^{-x}, so the maximum is below x with probability P(Mn \le x) = \left(1 - e^{-x}\right)^n. Set x = \ln n + \ln 3, so e^{-x} = \frac{1}{3n}: P(Mn \le \ln n + \ln 3) = \left(1 - \frac{1}{3n}\right)^n \longrightarrow e^{-1/3} \approx 0.7165. This is the extreme-value (Gumbel) behavior: the maximum of n exponentials clusters around \ln n, and shifting the level by a constant t multiplies the limit's exponent, giving e^{-e^{-t}}. Takeaway. The maximum of n mean-1 exponentials sits near \ln n; the limiting law is…

Limiting chance the max of exponentials stays modest

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