Finding pdf of $Y=\min\{X_1,X_2,...,X_n\}$

Question

Let $X_1,X_2,...,X_n$ random indepedent variables that follow the same distribution with a pdf $f$ and possibility function $F$.Find the pdf of $Y=\min\{X_1,X_2,...,X_n\}$.

My solution: $F_Y(a)=P(Y<a)=P(\min\{X_1,X_2,...,X_n\}<a)$ ,if we let $X_k$ with $k \in \{1,2,...,n\}$ be the min then $F_Y(a)=P(X_k<a)=F_{X_k}(a)$ . So it follows that $f_X(y)=f_{X_k}(x)=f_{X_i}(x)$ for every $i \in \{1,2,...,n\}$ since $X_i$ follow the same pdf.

I dont really thing that answers the question but it is an obvious note.
Could someone help ?
Thank you in advance !

Answer 1

Let $X_{(1)}:=\min(X_k,k\leq n)$ $$\begin{aligned}P(X_{(1)}\leq x)&=P\bigg(\bigcup_{k\leq n}\{X_k\leq x\}\bigg)=\\ &=1-P\bigg(\bigcap_{k\leq n}\{X_k> x\}\bigg)=\\ &=1-(P(X_1>x))^n=\\ &=1-(1-P(X_1\leq x))^n \end{aligned}$$ So $$\frac{d}{dx}P(X_{(1)}\leq x)=n(1-P(X_1\leq x))^{n-1}f_{X_1}(x)$$