# Understanding step in derivation of joint distribution

In a derivation I am trying to understand, there is the following argument: \begin{align} &=\int n!\prod_{i=1}^n f_X(x_i)\mathbb{I}_{x_1\le x_2\le\ldots\le x_n}\prod_{i\notin\{r_1,\ldots,r_k\}}\text{d}x_i\\ &=n!\int \prod_{i=1}^{r_1-1} f_X(x_i)f(x_{r_1})\prod_{i=r_1+1}^{r_2-1} f_X(x_i)\cdots \\ &\qquad\cdots f(x_{r_k})\prod_{i=r_k+1}^{n}f_X(x_i)\mathbb{I}_{x_1\le x_2\le\ldots\le x_n}\prod_{i\notin\{r_1,\ldots,r_k\}}\text{d}x_i\\ \end{align}

Which doesn't make sense to me, possibly due to the abuse of notation, I'm not quite sure how the first product is broken up into smaller products

• I see that $f_X(x_i)$ is the marginal of the $i$th random variable. I presume you use the same index $X$ for all $i$ since they have the same distribution. What do the indices $r_j$ and the function $f$ (with no subscript) stand for? – muaddib Aug 2 '15 at 15:03
• $x_{r_i}$ is the $r_i$ th order statistic, and i was assuming that the function $f = f_X$ – dimebucker Aug 2 '15 at 15:11
• @muaddib that's actually something im also confused about though so I could be wrong... – dimebucker Aug 2 '15 at 15:13

$$\prod_{i=1}^n f_X(x_i) = \left(\prod_{i=1}^{r_1-1}f_X(x_i)\right)f_X(x_{r_1})\left(\prod_{i=r_1+1}^{r_2-1}f_X(x_i)\right)f_X(x_{r_2}) \cdots \left(\prod_{i=r_{k-1}+1}^{r_k-1}f_X(x_i)\right)f_X(x_{r_k})\left(\prod_{i=r_k+1}^{n}f_X(x_i)\right)$$
with the usual convention that an empty product (which occurs whenever two of the $r$'s are adjacent: $r_{j+1} = r_j+1$) has value $1$.