Why the limiting distribution $M_n^*$ will be non-degenerate distribution?

Question

Let $X_1,\dots, X_n$ be iid data with common distribution $F$ and $M_n:=\max\{X_1,\dots, X_n\}$. The distribution of $M_n$: $$ P(M_n\le x)=F^n(x). $$

In the textbook, it says

we proceed by looking at the behaviour of $F^n$ as $n\to \infty$. But this alone is not enough: for any $z<z_+$ where $z_+$ is the upper end point of $F$ (the smallest value of $z$ so that $F(z)=1$), as $n\to \infty$, $$ F^n(z)\to 0 $$ so the distribution of $M_n$ degenerates to a point mass at $z_+$.

This difficulty is avoided by allowing a linear transformation of $M_n$: $$ M_n^*=\frac{M_n-b_n}{a_n} $$ for sequences of constants $\{a_n>0\}$ and $\{b_n\}$. Appropriate choices of the these sequences stabilize the location and scale of $M_n^*$.

Question: why $M_n^*$ will be fine? It seems that it still $$ P(M_n^*\le z)=F^n(a_nz+b_n)\to 0? $$ (because $0\le F(a_nz+b_n)\le 1$, then $[F(a_nz+b_n)]^n\to 0$ as $n\to \infty$, right?)

Answer 1

While it is true that for each $n\in\mathbb N$ we have $0\leq F(a_nx+b_n)\leq 1$, that is not enough to conclude that $F(a_nx+b_n)^n\to 0$ as $n\to\infty$, since the quantity $F(a_nx+b_n)$ can increase (i.e. go to $1$) as $n\to\infty$.

Consider the following example: Let $F$ be the distribution function given by $$F(x)=\begin{cases}1-x^{-1} & x\geq 1 \\ 0 & x<1 \end{cases}$$ then choose $b_n=0$ as well as $a_n=n$ to obtain for $x>0$ and sufficiently large $n$: $$F(a_nx+b_n)^n=F(nx)^n=(1-n^{-1}x^{-1})^n \to \exp(-x^{-1})$$ which you might recognize as the Frechet distribution. Note that while each $F(a_nx)$ is between $0$ and $1$, the value $F(a_nx)=1-n^{-1}x^{-1}$ is increasing to $1$ as $n\to\infty$.

Answer 2

I'd give another way to proof the assertion that the distribution of $M_n$ degenerates to a point mass at $z_+$.

Denote the limit distribution of $M_n$ by $F_0$, For any $n$, $P(M_n > z_+) = 1 - P(M_n \le z_+)=0$. Let $n$ tends to infinity, $F_0(z_+) = 1$ holds.

Given $\forall \epsilon > 0$, the probability $P(M_n < z_+ - \epsilon) = F^n(z_+ - \epsilon) \to 0$ since $F(z_+ - \epsilon) < 1$ strictly. Hence $F_0(z_+ - \epsilon)=0, \forall \epsilon > 0$.

Let $\epsilon \to 0$, we conclude that $F_0(z_+) = 0$ and $F_0(z)=0, \forall z < z_+$.