The probability that an order statistic is a particular sample member? Suppose I have $X_i, \ldots, X_n$ sample members that are all iid and follow some variable $Y$.
If I seek to find the probability that $Y_k = X_m$ (the $k$th order statistic where $1 \leq m \leq n$), my textbook claims that since all members are iid, the chance that any one of them equals some statistic $=\frac 1 n$.
I'm not convinced by this explanation. What about the independence of these sample members implies that they're equally to be assumed by some order statistic?
 A: To say that $X_1,X_2,X_3,X_4$ are i.i.d. and $X_1$ has a certain distribution is enough information to determine the joint distribution completely. In other words, if each of the random variables $W_1,W_2,W_3,W_4$ has the same distribution as $X_1$ and $W_1,W_2,W_3,W_4$ are independent then the whole sequence $(X_1,X_2,X_3,X_4)$ has the same distribution as $(W_1,W_2,W_3,W_4)$. That means, in particular, that permutations of $X_1,X_2,X_3,X_4$ have that same distribution, so that, for example, $(X_2,X_3,X_4,X_1)$ has the same distribution as $(X_1,X_2,X_3,X_4)$. Therefore, the probability that, for example, the third component of $(X_1,X_2,X_3,X_4)$ is in the second position after sorting into increasing order, is the same as the probability that the third component of $(X_2,X_3,X_4,X_1)$ is in the second position when sorted into increasing order. Thus $\Pr(X_3=X_{(2)}) = \Pr(X_4=X_{(2)}).$ And the same holds if some other pair than $3,4$ had been chosen as the indices and some other position than the second had been chosen. Consequently, every index among $1,2,3,4$ is equally likely to be the index of the random variable in the second position after sorting.  And similarly for other positions than the second.
