This is a follow up to an earlier question Understanding step in derivation of joint distribution. In a derivation I am trying to understand, there is the following argument:

$$ n! \int \prod_{i=1}^n f_X(x_i) \mathbb{I}_{x_1\le x_2\le\ldots\le x_n}\prod_{i\notin\{r_1,\ldots,r_k\}}\text{d}x_i =n! \int \left(\prod_{i=1}^{r_1-1}f_X(x_i)\right)f_X(x_{r_1})\left(\prod_{i=r_1+1}^{r_2-1}f_X(x_i)\right)f_X(x_{r_2}) \cdots \left(\prod_{i=r_{k-1}+1}^{r_k-1}f_X(x_i)\right)f_X(x_{r_k})\left(\prod_{i=r_k+1}^{n}f_X(x_i)\right) \mathbb{I}_{x_1\le x_2\le\ldots\le x_n}\prod_{i\notin\{r_1,\ldots,r_k\}}\text{d}x_i $$ and on to the next step:

\begin{align} = & n!\prod_{i=1}^{r_1-1}\int_{x_{i-1}}^{x_{i+1}} f_X(x_i)\text{d}x_i\,f_X(x_{r_1})\prod_{i=r_1+1}^{r_2-1}\int_{x_{i-1}}^{x_{i+1}}\, f_X(x_i)\text{d}x_i\,f_X(x_{r_2})\cdots\\ &\quad\cdots f_X(x_{r_k})\,\prod_{i=r_k+1}^{n}\int_{x_{i-1}}^{x_{i+1}} f_X(x_i)\text{d}x_i\mathbb{I}_{x_{r_1}\le x_{r_2}\le\ldots\le x_{r_k}}\\ \end{align}

I think my confusion is due to the $\prod dx_i$ term and how this has been expanded to give smaller integrals, I also don't understand how the argument of the indicator function went from being $x_1,...,x_n$ to $x_{r_1}...x_{r_n}$


We have $k$ indices, such that, $0\leq r_1< r_2 <\cdots< r_k \leq n$ .

Then the product series' expansion is of the following form:

$$\prod_{i=1}^n G_i = \left(\prod_{i=1}^{r_1-1} G_i\right) \cdot G_{r_1} \cdot \left(\prod_{i=r_1+1}^{r_2-1} G_i\right) \cdot G_{r_2} \cdots \cdot \left(\prod_{i={r}_{(k-1)}+1}^{{r}_k-1} G_i\right)\cdot G_{r_k}\cdot \left(\prod_{i=r_k+1}^{n} G_i\right)$$

Where you have $G_i \mathop{:=} f_X(x_i)$

For the splitting of the integral you have

$$\begin{align} & \iiint \left(\prod_{i=1}^n G_i\right) \Bbb I_{x_1<x_2<\cdots < x_n} \left(\prod_{i\notin\{r_1\ldots r_k\}}\mathrm d x_i\right) \\[1ex] = & {\iiint \left(\prod_{i=1}^n G_i \Bbb I_{x_{(i-1)}<x_i<x_{(i+1)}} \right) \left(\prod_{i\notin\{r_1\ldots r_k\}}\mathrm d x_i\right)} \\[1ex] = & {\prod_{i=1}^{r_1-1} \left(\int G_i\Bbb I_{x_{(i-1)}<x_i<x_{(i+1)}} \mathrm d x_i \right) \cdot G_{r_1} \cdot \prod_{i=r_1+1}^{r_2-1} \left(\int G_i\Bbb I_{x_{(i-1)}<x_i<x_{(i+1)}} \mathrm d x_i \right) \cdots G_{r_k}\cdot \prod_{i=r_k+1}^{n} \left(\int G_i\Bbb I_{x_{(i-1)}<x_i<x_{(i+1)}} \mathrm d x_i \right)\cdot \Bbb I_{x_{r_1}<x_{r_2}<\ldots<x_{r_k}}} \end{align}$$

Where, again, $G_i \mathop{:=} f_X(x_i)$. Now since the $k$ indices ( $\{r_j\}_k$ ) do not occur as subscripts of the variables of integration, the various $G_{r_j}$ terms are effectively constants.

To help visualise, consider

$$\iiint f(x_i) \Bbb I_{x_1<x_2<x_3<x_4} \mathrm d x_1 \mathrm d x_3 \mathrm d x_4 = \int_{-\infty}^{x_2} f(x_1) \mathrm d x_1 \cdot f(x_2) \cdot \int_{x_2}^{\infty}\int_{x_2}^{x_4} f(x_3) f(x_4)\mathrm d x_3 \mathrm d x_4$$

  • $\begingroup$ So if I were to evaluate one of the expressions, would this make sense: \begin{align} \prod_{i=1}^{r_1-1} \int_{x_{i-1}}^{x_{i+1}}f(x_i)dx_i &=\prod_{i=2}^{r_1-1} \int_{x_{i-1}}^{x_{i+1}}f(x_i)dx_i \left (\int_{-\infty}^{x_2}f(x_2)dx_2 \right)\\ &=\prod_{i=2}^{r_1-1} \int_{x_{i-1}}^{x_{i+1}}f(x_i)F(x_2)dx_i\\ &=\prod_{i=3}^{r_1-1} \int_{x_{i-1}}^{x_{i+1}}f(x_i) dx_i\left(\int_{x_1}^{x_3}f(x_2)F(x_2)dx_2 \right)\\ &=\cdots \end{align} $\endgroup$ – dimebucker Aug 3 '15 at 2:56

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