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?
 A: 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.
A: I assume what you mean is something like this:

Let the $n > 1$ random variables $X_1, X_2, \dots, X_n$ over $\mathbb R$ be independent and identically distributed, and let $Y_i$ for $1 ≤ i ≤ n$ denote the $i$-th smallest value in $(X_1, X_2, \dots, X_n)$.  That is to say, $Y_1 = \min_{1 ≤ i ≤ n}(X_i)$ and $Y_n = \max_{1 ≤ i ≤ n}(X_i)$, and in general $i ≤ j \implies Y_i ≤ Y_j$.  Are the random variables $Y_1, Y_2, \dots, Y_n$ independent and identically distributed?

The answer is that the order statistics $Y_1, Y_2, \dots, Y_n$ are generally neither independent nor identically distributed.

*

*They are not independent because we know that e.g. $Y_1$ can never be larger than $Y_2$.


*They are not identically distributed because $Y_1$ is more likely to take smaller values than $Y_2$.

For example, let $n = 2$ and let us choose an arbitrary constant threshold $c$ such that $p = P(X_i ≤ c)$ is strictly between $0$ and $1$.  Then we have four possible cases for the values of $X_1$ and $X_2$, with their corresponding probabilities as follows:
$$\begin{array}{r|c|c|}
 & X_1 ≤ c & X_1 > c \\ \hline
X_2 ≤ c & p^2 & p(1-p) \\ \hline
X_2 > c & p(1-p) & (1-p)^2 \\ \hline
\end{array}$$
However, the corresponding table for $Y_1$ and $Y_2$ instead looks like this:
$$\begin{array}{r|c|c|}
 & Y_1 ≤ c & Y_1 > c \\ \hline
Y_2 ≤ c & p^2 & 0 \\ \hline
Y_2 > c & 2p(1-p) & (1-p)^2 \\ \hline
\end{array}$$
This is because the four cases in the two tables are not all in one-to-one correspondence:

*

*$Y_1 ≤ c \land Y_2 ≤ c \iff X_1 ≤ c \land X_2 ≤ c$, and

*$Y_1 > c \land Y_2 > c \iff X_1 > c \land X_2 > c$, but

*$Y_1 ≤ c \land Y_2 > c \iff (X_1 ≤ c \land X_2 > c) \lor (X_1 > c \land X_2 ≤ c)$, and

*$Y_1 > c \land Y_2 ≤ c$ cannot happen.

From the table above we can also calculate the marginal probabilities $P(Y_1 ≤ c) = p^2 + 2p(1-p) = 2p - p^2$ and $P(Y_2 ≤ c) = p^2$.  These can only be equal if $p = 0$ or $p = 1$, which was ruled out by the choice of $c$ above.  Thus,

*

*$Y_1$ and $Y_2$ cannot be identically distributed, since $P(Y_1 ≤ c) > P(Y_2 ≤ c)$, and

*$Y_1$ and $Y_2$ also cannot be independent, since $P(Y_1 ≤ c \mid Y_2 ≤ c) = 1 > P(Y_1 ≤ c)$.

The only case where the argument above does not work is if we cannot choose a threshold $c$ satisfying the criterion that $0 < P(X_i ≤ c) < 1$, i.e. if the distribution of the $X_i$ variables is a degenerate distribution concentrated at a single point $x$.  In that special case $Y_1$ and $Y_2$ are i.i.d., as $Y_1 = Y_2 = x$ almost surely.

The same argument can also be easily carried out, mutatis mutandis, for $n > 2$ and for any pair of indices $1 ≤ i < j ≤ n$; the only difference is that in that case the specific expressions in the second table become a bit more complicated.
Thus, for any $n > 1$, the only case where $Y_1, Y_2, \dots, Y_n$ are i.i.d. is if there is a constant $x$ such that $X_i = x$ almost surely.
A: 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 "order statistic". 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 :)
