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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.

The inverse change of variables is $x=u$, $y=(v-u)/a$, with absolute Jacobian determinant $1/a$. Because the original random variables are independent, their joint density is $f(x)g(y)$. Thus the transformed joint density isThe support is important: $X,Y\ge 0$ implies $V\ge U\ge 0$.

For independent exponential random variables of rate $\lambda >0$ and $a=1/2$, this becomes $2{\lambda }^2{e}^{-2\lambda v+\lambda u}$ on $0\le u\le v$. Integrating out $u$ gives the density of the sum:It vanishes for $v<0$. On the other hand, the cumulative distribution function of the maximum isDifferentiating yields exactly the same density. The maximum and $X+Y/2$ have the same distribution, as described by maximum of two exponentials as a scaled sum; this does not assert that they are equal for the same pair of sampled values.