Let $X_1,X_2,\dots, X_N$ be i.i.d random variables with probability density function $f$ and distribution function $F$. Define thw following two random variables:
$X_{(1)}=\min\{X_1,X_2,\dots,X_n\}$ $X_{(n)}=\max\{X_1,X_2,\dots,X_n\}$
Problem: Calculate the probaility density function $f_{(n)}$ of $X_{(n)}$
Solution:
We first calculate the distribution function of $X_{(n)}$ and derive it to get $f_{(n)}$.
$F_{(n)}(t)=P[X_{(n)}\leq t] = P[X_1\leq t, \dots , X_n \leq t] = \prod _{k=1}^n P[X_k\leq t]=(F(t))^n$
Question: I'm not fully understanding the second equal sign. If we evaluate $X_{(n)}=\max\{X_1,X_2,\dots,X_n\}$ we get one $X_i$ (or several with the same value - which doesn't matter since all of them are distributed the same), so shouldn't we get:
$F_{(n)}(t)=P[X_{(n)}\leq t]=P[X_j \leq t]=...$? Whereas $\max\{X_1,X_2,\dots,X_n\} = X_j$?