# Distribution of maximum of iid random variables

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

if the maximum of $$n$$ variables is less than $$t$$ means that ALL the variables must be less than $$t$$
If the minimum of $$n$$ variables is GREATER than $$t$$ that means ALL the variables are greater than $$t$$
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.