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 $\max\{X_1,X_2,\dots,X_n\} = X_j$?
Best Answer
if the maximum of $n$ variables is less than $t$ means that ALL the variables must be less than $t$
Same (speculate) story for the minimum..
If the minimum of $n$ variables is GREATER than $t$ that means ALL the variables are greater than $t$
...understood this, then using the i.i.d. fact, the solution is self evident