 Solution

Let $M_n={\max }_{1\le j\le n}{W}_j$ for independent and identically distributed random variables with distribution function $F$. Then $P(M_n\le u)=F(u{)}^n$. Membership in the maximum domain of attraction of a nondegenerate distribution function $G$ means that some deterministic $a_n>0$ and $b_n\in \mathbb{R}$ satisfyor equivalently $F(a_{n}x+b_n{)}^n\to G(x)$ at every continuity point of $G$.

Under the stated convergence, put $Z_n=(M_n-b_n)/a_n$, $r_n={\alpha }_n/a_n\to a>0$, and $s_n=({\beta }_n-b_n)/a_n\to b$. The Slutsky theorem says that combining a sequence with convergence in distribution with sequences converging in probability to constants preserves the corresponding sum, product and ratio limits, provided a limiting denominator is nonzero. Applying it here givesThe limiting distribution function is $P((Z-b)/a\le x)=G(ax+b)$, continuous for every real $x$ because $G$ is continuous and $a>0$. Consequently,