Order Statistics : What's the deal with $n!$?

Question

Assuming I have $n$ IID Random Variables $X_1,...,X_n$ the Order Statistics of this set $:= (Y_1,...Y_n)$ has a distribution of $f_Y(y_1,...,y_n)=n!f_X(x_1,...,x_n)$ I have read a couple of derivations but I can't seem to understand what $n!$ is doing in the equation? I realize that there are $n!$ permutations of the random vectors and only one of them is the true Order Statistic for a given sample of the Random Variables but intuitively I feel we should be dividing by $n!$ and not multiplying. Can anyone explain?

Answer 1

This is not true in general, only when the probability for two independently drawn values to be equal is zero. If it is, then $n!$ different unordered tuples correspond to every ordered tuple. Since there are $n!$ times fewer ordered tuples, each of them is $n!$ more likely to be chosen than the corresponding unordered ones. You can think of the density of the unordered tuples "collapsed" onto their ordered representatives, with $n!$ unordered tuples each contributing the same density, leading to $n!$ times that density.

Answer 2

For any tuple $\{z_1, \dots, z_{10} \}, 0 < z_1 < \dots < z_{10} \leq 1$, we have  $$P( Y_1 \leq z_1, \dots, Y_{10} \leq z_{10} ) = \sum_{\sigma \in S_{10}} P(X_{\sigma (1)} \leq z_1, \dots,  X_{\sigma (10)} \leq z_{10}).$$ Taking the derivative, you get the claim using the other assumptions such as iid. So the intuition is the reverse (but I see what you mean).

Edit: the caveat of the other answer is obviously assumed (no point masses).