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Given $$X_i ∼ Uniform[0, 1] $$for $$i = 1, .., n.$$ What is the distribution of $$M := min(X_1, ...., X_n)?$$ You may assume that $$X_1,X_2, ... ,X_n are independent.$$ My solution: $$X_1 ∼U [0,1]$$ $$PDF f(x)=1, 0≤ x≤ 1$$ $$ F(x)= \int_0^{x} f_x(x)\,dx$$ $$=\int_0^{x} 1\,dx$$ $$x, 0≤ x≤ 1$$ CDF of min $$= 1-[1-x]^n $$ PDF of min $$=n(1-x)^{n-1}$$

This is my solutions, and just wanted to check my answers with you guys

user29
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    @DavidG.Stork Please be nicer to user29. They are new to Math StackExchange. – MathMagician Mar 05 '22 at 17:29
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – William Mar 05 '22 at 17:38
  • @William I want to know if my solutions for cdf and pdf are correct – user29 Mar 05 '22 at 17:39
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    This is a correct answer but as a solution I would want to see a presentation of the main idea that the min is bigger than x if and only if all are bigger than x. So a few more lines of development would be useful. – Michael Mar 05 '22 at 18:27

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