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

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: $$F_Y(a)=P(Y ,if we let $$X_k$$ with $$k \in \{1,2,...,n\}$$ be the min then $$F_Y(a)=P(X_k . 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 ?
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 $$\frac{d}{dx}P(X_{(1)}\leq x)=n(1-P(X_1\leq x))^{n-1}f_{X_1}(x)$$
• Great answer! Can you comment on the case when $X_i$ have a discrete distribution like Poisson? Jun 10, 2022 at 16:40
• 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. Jun 11, 2022 at 12:04