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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 $\max\{X_1,X_2,\dots,X_n\} = X_j$?

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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

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Maximum of $n$ numbers is less than or equal to $t$ iff each one of them is less than or equal to $t$. This gives the second equality.

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