In the context of ordered statistics, each of Y(1),Y(2),...,Y(n) a single observation or distributions that are I.I.D? In statistics one aspect of the I.I.D. concept that bothers is when I think about it in the context of ordered statistics. As most of you already know, $Y_1,Y_2,Y_3,...,Y_n$ are I.I.D. when the parameters are the same. 
Now, here are two things I'm confused about. 


*

*In the context of ordered statistics, each of $Y_{(1)},Y_{(2)},...,Y_{(n)}$ a single observation or distributions that are I.I.D?

*If they are distributions, how in the world is it possible to order distributions from the least to greatest??


If you have $Y_1,Y_2, \dotsc, Y_n$ and each are independent and follow some distribution $G$, then you could consider each $Y_i$ as a realization, or sample, taken from $G$. 
If you then order, then each $Y_{(i)}$ follows a new distribution. 
For example, say we have $Y_1, \dotsc, Y_n$, where each one is independent and  follows a $\text{unif}(0,1)$. Then, it should be clear, that each $Y_i\sim \text{unif}(0,1)$ (nothing changed). Once we order then, however, recall that for example, the first ordered statistic $Y_{(1)}\sim \text{Beta}(1,n)$ (Beta distribution). It is still an observation, but $Y_{(1)}$ follows a different distribution from $Y_1$. And in general, $Y_i\sim\text{unif}(0,1)$, but
$$Y_{(i)}\sim \text{Beta}(i, n+1 -i).$$

You don't order the distributions, you order the $Y_i$s.
For example, the distribution of $Y_{(1)}$ goes as follows
$$P(Y_{(1)}\leq z) = 1-P(Y_{(1)}>z) =1- (1-z)^n.$$
Then the pdf is $n(1-z)^{n-1}$. Notice that the interpretation is that
there is $\binom{n}{1}$ options for the smallest. Then the rest $\binom{n-1}{n-1} = 1$ have to fall in the interval $(1-z)$, hence $n(1-z)^{n-1}$. 

Let $Y_1,Y_2,Y_3$ iid exponential distributions with mean $1/\lambda$. Then, to find distribution of the minimum $M:=Y_{(1)}$, we must consider
$$P(M\in dm)$$
There are $3$ choices, for the minimum


*

*$Y_1$, or

*$Y_2$ or

*$Y_3$.


Once we have chosen the smallest, then there is only one way to choose the other two larger ones. So it must be the case that
$$P(M\in dm) = \binom{3}{1}f_Y(m)\binom{2}{2}(1-F_Y(m))^2 =3\lambda e^{-\lambda m}(e^{-\lambda m})^2 = 3\lambda e^{-3\lambda m}$$ 
Notice that the minimum $M$ (or $Y_{(1)}$) follows an exponential distribution with mean $\frac{1}{3\lambda}$.
