Are sorted i.i.d still i.i.d? I need help with that question.
We have i.i.d. Let us sort the values in ascending order.Is new series is i.i.d?
From my opinion the answers is yes because function y that sort the x is only mapping from x, but im not sure. Can anybody correct my understanding?
No, unless $X$ is a constant (non-random). First of all, the $y_i$ are clearly dependent: you already know that $y_2\geq y_1$. Furthermore, they are not identically distributed: the later values are more likely to be large.
It depends whether you know the values of the $\{y_i\}$ or not. The knowledge of the values makes the random variables not IID anymore. If you are curious about the case when you do know the values at ordering, check out the &quot;order statistic&quot;. Deriving the laws is a bit cumbersome in my opinion but completly doable by any students knowing probability.
https://en.wikipedia.org/wiki/Order_statistic
However, if you do not know the values at ordering time, then yes you can rearrange the IID random variables as you wish yourself :)