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 !
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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 $$\frac{d}{dx}P(X_{(1)}\leq x)=n(1-P(X_1\leq x))^{n-1}f_{X_1}(x)$$
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$\begingroup$ Great answer! Can you comment on the case when $X_i$ have a discrete distribution like Poisson? $\endgroup$ Jun 10, 2022 at 16:40
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1$\begingroup$ @orangeskid thanks! In the case of discrete distributions we don't have a pdf in the usual sense, so I would stop at the minimum distribution step $\endgroup$– SnoopJun 10, 2022 at 16:58
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$\begingroup$ Of course, the probability mass function will be $\mathsf P(X_{(1)} =x)= \mathsf P(X_{(1)} \leq x)-\mathsf P(X_{(1)} \leq x-1)$ ... if the variables are integer-valued. $\endgroup$ Jun 11, 2022 at 12:04