4
$\begingroup$

I am struggling with the following sentence:

Using the fact that the cumulative distribution of the minimum of two i.i.d. random variables can be expressed as $1 - (1 - F(x))^2$....

Can anyone show me why this is the case?

$\endgroup$

1 Answer 1

16
$\begingroup$

It's a common trick. If $X = \min(Y_1,Y_2)$ and $F$, $F_X$ are the CDFs of the $Y_i$s and $X$, respectively, then \begin{align*} F_X(x) &= 1 - P(X > x) \\ &= 1- P(Y_1 > x, Y_2 > x) \\ &= 1 - P(Y_1 > x)P(Y_2 > x) \text{ independence}\\ &= 1 - [1-F(x)][1-F(x)] \text{ identicalness}. \end{align*}

$\endgroup$
2
  • 1
    $\begingroup$ :-) well that's easier than I thought. Thank you! $\endgroup$
    – Manuel R
    Jan 4, 2018 at 0:19
  • 1
    $\begingroup$ +1 for the answer, but why do you call it a "trick"? $\endgroup$ Jan 4, 2018 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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