Suppose that $X_1$, $X_2$, $X_3$ is random sample (iid) from a population with pdf
$f(x) = 3x^2$ if $0<x<1$
I would like to compute $g_2(y_2)$ which is the marginal pdf of order statistics $Y_2$ by using the joint pdf $g(y_1,y_2,y_3)$ of the order statistics $(Y_1,Y_2,Y_3)$. And also, I would like to find the marginal pdf of second order statistics, $Y_2$.
I could not do this question.
Maybe more a comment rather than an answer... Anyway without thinking too much and according to Wikipedia https://en.wikipedia.org/wiki/Order_statistic :
$F_{X(2)}(x)=\sum_{j=2}^3 {3\choose j} F^{j}_X(x)(1-F_X(x))^{n-j}$ [1]
So that :
$f_{X(2)}(x))=6F_X(x)(1-F_X(x))f_X(x)$
Now we can compute $F_X(x)=x^3$, so that:
$f_{X(2)}(x))=18x^5(1-x^3)$,
that you can check it integrates to 1 :
https://www.wolframalpha.com/input/?i=integrate_0%5E1+18x%5E5%281-x%5E3%29
( we consider all the domains $x \in [0,1]$ )
UPDATE. Proof of [1]: Since it looks that [1] is not so difficult to prove, I add for completeness an argument. Let's consider $P(X_{(2)}\le x)$. When $X_{(2)}\le x$ there are two possibilities. Either $X_1, X_2$ and $X_3$ are all $\le x$, an event happening with probability $(1-F_X(x))^3$, or two numbers are $\le x$ and one greater than $x$. This event happens with probability ${3 \choose 2}F_X(x)^2(1-F_X(x))$, since we have to choose two variables $\le x$ and the other greater $\ge x$. Summing up these mutually exclusive events we get [1]. A similar argument should lead also to the general result of Wikipedia, essentially we have to choose how many (and which) samples fall on the left of $x$, and how many on the right.