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Every non-degenerate limit in convergence in distribution of positively scaled and centered sample maxima is a max-stable distribution. Up to positive affine transformations its law is the Gumbel distribution, a Fréchet distribution with positive shape parameter, or a negative Weibull distribution with positive shape parameter. This classifies possible limits, without asserting that every distribution function admits a limiting normalization.

For a nondegenerate distribution function $G$, membership $F\in D(G)$ in its maximum domain of attraction means that there exist $a_n>0,b_n\in \mathbb{R}$ such thatat every continuity point of $G$. A max-stable distribution satisfies: for each positive integer $k$, there exist $c_k>0,d_k\in \mathbb{R}$ such that $G(c_{k}x+d_k{)}^k=G(x)$. Thus a maximum of $k$ independent copies has the same probability distribution after a change of location and scale.

Suppose first that $F\in D(G)$. For fixed $k$, the distribution functions $H_n=F^{kn}$ have the two nondegenerate limitsThe given convergence-of-types result supplies constants $a>0,b$ such that $G(x{)}^k=G(ax+b)$. Equality extends from continuity points to all points by right continuity. Substitute $x=(z-b)/a$ to obtain $G((z-b)/a{)}^k=G(z)$, which proves max-stability with $c_k=1/a,d_k=-b/a$. Conversely, if $G$ is max-stable, take $F=G$ and its max-stability constants for $k=n$. Then $F(c_{n}x+d_n{)}^n=G(x)$ exactly for every $n$, proving $G\in D(G)$. Therefore the maximum domain of attraction is nonempty exactly when $G$ is max-stable.

For the particular logarithmic tail, complete the distribution function by taking $F(x)=0$ for $x\le x_0$. Let $a_n>1$ be the unique solution of $a_n\log a_n=n$, and take $b_n=0$. Equivalently $F(a_n)=1-1/n$, including $n=1$ with $a_1=x_0$. For each fixed $x>0$, eventually $a_{n}x>x_0$, andHere $a_n\to \infty$ follows directly from its defining equation. Since $\log (1-z)=-z+O(z^2)$ as $z\to 0$, it follows that $n\log F(a_{n}x)\to -1/x$. For $x\le 0$, the probability is zero. ConsequentlyThis is the shape-one Fréchet distribution, and the calculation proves the limit at every real $x$ without needing a regular variation theorem.

The function $t\log t$ is strictly increasing on $t>1$. For every integer $n\ge 3$, $n\log n>n=a_n\log a_n$, so $a_n<n$. The defining equation now gives $a_n=n/\log a_n>n/\log n$. Taking logarithms yields $\log a_n>\log n-\log \log n$, whose right side is positive, soMultiplying by $\log n/n$ and squeezing gives $a_n\log n/n\to 1$. Slutsky theorem applied to the normalized sample maximum therefore gives the simpler logarithmic-tail Fréchet normalization

A nondegenerate distribution function $G$ is a max-stable distribution if, for every integer $r\ge 1$, there are $a_r>0$ and $b_r\in \mathbb{R}$ such thatEquivalently, the suitably centered and positively scaled sample maximum of $r$ independent and identically distributed random variables with law $G$ again has law $G$. The extremal types theorem states that every nondegenerate limiting law of normalized sample maxima is a max-stable distribution. Thus this property identifies all possible limiting types, although a particular starting distribution function need not admit any such limiting normalization.

The three standard types are the Gumbel distributionthe Fréchet distribution, with $\alpha >0$,and the negative Weibull distribution, with $\alpha >0$,They are max-stable distributions, as follows respectively by taking $(a_r,b_r)=(1,\log r)$, $(r^{1/\alpha },0)$, and $(r^{-1/\alpha },0)$. The necessary and sufficient classification is: $G$ is max-stable exactly when $G(x)=H((x-b)/a)$ for $a>0$, $b\in \mathbb{R}$, and one of these three types $H$. A positive affine map sends a max-stable distribution to another max-stable distribution.

For the requested threshold equivalence, set $p_n=1-F(u_n)$. By independence,If $n{p}_n\to \tau <\infty$, then $p_n\to 0$ and the expansion $\log (1-p)=-p+O(p^2)$ gives $n\log (1-p_n)\to -\tau$, since $n{p}_n^2=(n{p}_n)p_n\to 0$. Exponentiation proves the claimed limit. Conversely, if $(1-p_n{)}^n\to e^{-\tau }>0$, then $p_n\to 0$: any subsequence with $p_n\ge \delta >0$ would have its powers tending to zero. Taking logarithms now gives $-n\log (1-p_n)\to \tau$. Since $-\log (1-p)/p\to 1$, including the continuous extension at $p=0$, it follows that $n{p}_n\to \tau$. Hence

The exceedance count $S_n$ has the binomial distribution with parameters $n,p_n$. For $\tau >0$ and each fixed nonnegative integer $s$, its probability mass function satisfieswhere $(n{)}_s=n(n-1)\cdots (n-s+1)$ and $(n{)}_0=1$. Indeed $(n{)}_s/n^s\to 1$, $(n{p}_n{)}^s\to {\tau }^s$, and $(1-p_n{)}^{n-s}\to e^{-\tau }$. If $\tau =0$, the bound $\mathbb{P}(S_n\ge 1)\le \mathbb{E}{S}_n=n{p}_n\to 0$ gives the degenerate Poisson distribution at zero. Summing the finitely many terms proves, in both cases,This derives the needed instance of the Poisson limit theorem directly.

Finally fix $x$ with $G(x)>0$ and put $u_n=a_{n}x+b_n$. The assumed limit for the normalized sample maximum and the proved equivalence give $n{p}_n\to -\log G(x)$. The second-largest order statistic is at most $u_n$ exactly when at most one observation is strictly greater than $u_n$. This event identity holds even if observations have ties. Apply the preceding Poisson limit theorem with $k=1$ to obtainThe endpoint $G(x)=1$ corresponds to $\tau =0$ and is already included. This is the case $r=2$ of the exceedance count limit for extreme order statistics.