The cdf of $Z_n\equiv \max_{i=1,...,n} X_i$ is $(F(x))^n$

Question

Consider the random variables $X_1,...,X_n$ i.i.d. with cdf $F$. Consider $$ Z_n\equiv \max_{i=1,...,n} X_i $$ I've found in some sources that the cdf of $Z_n$ at $x\in \mathbb{R}$ is $(F(x))^n$.

Could you help to show it?

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If that result is true, what is its relation with the definition of a max-stable distribution? I'm confused as it seeems to me to implying that any distribution is max-stable.

[Def of max-stable distribution: Consider $\{X_i\}_{\forall i \in \{1,...,n\}}$ i.i.d. each with distribution $\mu$. $\mu$ is max-stable if there exists $\{a_n\}_{n\in \mathbb{N}}>0$ and $\{b_n\}_{n\in \mathbb{N}}\in \mathbb{R}$ such that, $\forall n \in \mathbb{N}$, $Z_n\equiv \frac{M_n-b_n}{a_n}\sim X_i$, where $M_n\equiv \max_{i\in \{1,...,n\}}X_i$. Equivalently, let $F$ be the cdf associated with $\mu$, i.e., $F(x)\equiv \mu((-\infty, x])$. $\mu$ is max-stable if there exists $\{a_n\}_{n\in \mathbb{N}}>0$ and $\{b_n\}_{n\in \mathbb{N}}\in \mathbb{R}$ such that, $\forall n \in \mathbb{N}$, $F^n(a_nx +b_n)=F(x)$.]

Answer 1

Consider the event $Z_n\leq z$ for some real number $z$. Because $Z_n=\max_{i=1,\ldots,n}X_i$, $Z_n\leq z$ implies that for each $i$, $X_i\leq z$. Conversely, if $X_i\leq z$ for each $i$, then $Z_n=\max_{i=1,\ldots,n}X_i\leq z$. Therefore, these 2 events are equivalent: $$ (Z_n\leq z)\iff (X_i\leq z,\forall i) $$ Thus, $$ \Pr(Z_n\leq z)=\Pr(X_i\leq z,\forall i)=\prod_{i=1}^n\Pr(X_i\leq z)=\prod_{i=1}^n F(z)=[F(z)]^n. $$ The 2nd equality uses the independence of $X_i$'s and the 3rd uses the fact that they share the same CDF $F$.

Answer 2

Consider that

\begin{align} \mathbb P(Z_n \leq x) &= \mathbb P\left(\max_{i=1,\dots,n} X_i \leq x\right) \\&= \mathbb P \left( \bigcap_{i=1}^n\, \left\{ X_i\leq x \right\}\right) \\&= \prod_{i=1}^n\,\mathbb P(X_i\leq x)\tag{$*$} \\&= \prod_{i=1}^n\,F(x) = {F(x)}^n \end{align}

where we've used independence to arrive at $(*)$.