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Let $n\ge 1$ be an integer and let $X_1,\dots ,X_n$ be independent and identically distributed random variables with an exponential distribution of rate $\lambda >0$. Their sample maximum $M_n$ is nonnegative. Substituting the exponential distribution's cumulative distribution function into the distribution functions of independent minima and maxima gives the exact lawFor example, at $x=(\log 2)/\lambda$ each observation is below the threshold with probability $1/2$, but all $n$ observations are below it with probability $2^{-n}$. Thus an individual observation's median becomes an increasingly unlikely upper bound for the whole batch. The general order statistic density in Siegrist's treatment, with $k=n$, also gives this law on integration. ^[ref].

Solving $(1-e^{-\lambda x}{)}^n=p$ gives the exact quantile function, for $0<p<1$,In a toy model with $\lambda =1{\mathrm{s}}^{-1}$, the median maximum is $0.693147$ seconds for $n=1$, $2.703555$ seconds for $n=10$, and $4.975147$ seconds for $n=100$. All three use the exact distribution function; the increase comes from taking a maximum over more observations. The $n=2$ case connects to the existing maximum of two exponentials as a scaled sum.

The Gumbel distribution explains the logarithmic shift visible here: after subtracting $(\log n)/\lambda$ and measuring deviations in units of $1/\lambda$, the distribution functions approach one fixed shape. The next derivation makes that convergence in distribution explicit.

For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.