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Please. I am trying to understand the proof, that cdf of minimum of $n$ random variables is $1-[1-F(x)]^n$

If I have $n$ independent random variables $X_1, \dots, X_n$, all of them have the same CDF $F(x)$, what is the point of searching for the $\min=\{X_1, \dots, X_n\}$? Shouldn't $X_1, \dots, X_n$ have the same value for every x? (In the meaning: $P (\min \{X_1, \dots, X_n\}≤x) = P (\bigcup_{i=1}^n X_i\}≤x) \cdots$)

Thanks.

Michael Hardy
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Lucy
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    "Shouldn't X1,…,Xn have the same value for every x?" It's not clear what you're saying here. Are you saying that each Xi has the same value as every other Xi? Or that the value of Xi doesn't depend on x? – Acccumulation Nov 16 '18 at 22:52

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Let's call the random variable $Y = \min\{X_1 \dots X_n\}$. One can notice that the complementary event of $Y<y$ is $\bigcap_{i=1}^n X_{i}>y$. Thus, the CDF of $Y$ can be computed as follows:

$\begin{array}{rcl} \text{CDF}_Y(y) & = & P(Y<y) \\ & = & 1 - P(X_1>y, X_2>y, \dots X_n>y)\\ & = & 1 - (1 - CDF_X(y))^n\\ \end{array}$

Carlos Campos
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