2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ "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? $\endgroup$ – Acccumulation Nov 16 '18 at 22:52
2
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.