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For $X_i \sim$ iid random variables:

For $1\le r_1 < ..<r_k \le n$ integers, I am trying to find the joint pdf of: $$ (X_{(r_1)},...,X_{(r_n)}) $$ where $X_{(r_1)}$ is the $r_1$th largest observation. I am wondering if anyone has seen the solution to this problem somewhere online? My current attempt:

Choosing $\epsilon$ small enough such that only one observation falls in an interval of width $\epsilon$

\begin{align*} &P(X_{(r_1)} \in (x_1 - \epsilon,x_1+\epsilon),.......,X_{(r_k)}\in (x_k - \epsilon,x_k+\epsilon))\\ &=P( n-k ~\text{of}~ X_1,....,X_n \in (-\infty, x_1-\epsilon),\\ &1 ~\text{of}~ X_1,....,X_n \in (x_1-\epsilon,x_1+\epsilon),\\ &....\\ &1 ~\text{of}~ X_1,....,X_n \in (x_k-\epsilon,x_k+\epsilon),)\\ &+\\ &P( n-k-1 ~\text{of}~ X_1,....,X_n \in (-\infty, x_1-\epsilon),\\ &1 ~\text{of}~ X_1,....,X_n \in (x_1-\epsilon,x_1+\epsilon),\\ &....\\ &1 ~\text{of}~ X_1,....,X_n \in (x_k-\epsilon,x_k+\epsilon),\\ &1 ~\text{of}~ X_1,....,X_n \in (x_k+\epsilon + \infty),\\ &+.... \end{align*} and so on and so forth accounting for all distributions of the remaining n-k observations between the area before $r_1$ and the area after $r_k$.

Each one of these is a multinomial, and has corresponding expressions in terms of the CDFs (dividing and taking epsilons to zero). After all the working, I get to the point:

$$ f_{(X_(r_1),....,X_(r_k)} (x_1,...,x_k)= $$ $$ n! \prod_{i=1}^{k} f(x_i) \left[ \frac{F(x_1)^{n-k}}{(n-k)!} +\frac{F(x_1)^{n-k-1} (1-F(x_k))}{(n-k-1)!} +......+\frac{ (1-F(x_k))^{n-k}}{(n-k)!} \right] $$

Not sure if I am on the right track, if anyone has seen this distribution before could you let me know if my attempt is so far correct?

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    $\begingroup$ i think you can refer to this question for (a) stats.stackexchange.com/questions/161145/… $\endgroup$
    – Deep North
    Aug 1, 2015 at 6:25
  • $\begingroup$ $=n!f(y_1)f(y_2)...f(y_r)\frac{[1-F(y_r)]^{n-r}}{(n-r)!}$, $y_1$ equal $X_{(r1)}$ for your case $\endgroup$
    – Deep North
    Aug 1, 2015 at 6:34
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    $\begingroup$ @DeepNorth the very last statement of this paper : google.com.au/… , has an expression for the joint pdf, I'm not sure if this agrees with the result you figured out in the question you linked ? $\endgroup$ Aug 1, 2015 at 11:10
  • $\begingroup$ @DeepNorth also, your answer in (a) assumes $r_1 = 1$, whereas what im trying to figure out doesn't necessarily have that? $\endgroup$ Aug 1, 2015 at 11:13
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    $\begingroup$ in this question two points are missing, first you did not tell us anything about $X_i,i=1,\ldots$ and their inter correlations and also type of variables. Under general situations that variables are iid, the answer is $f_{X_{(1)},\ldots,X_{(n)}}=n!\prod _1^n f_X(x_i)$ see here en.wikipedia.org/wiki/Order_statistic $\endgroup$
    – TPArrow
    Aug 1, 2015 at 11:51

1 Answer 1

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Since$$f_{X_{(1)},\ldots,X_{(n)}}(x_1,\ldots,x_n)=n!\prod_{i=1}^n f_X(x_i)\mathbb{I}_{x_1\le x_2\le\ldots\le x_n}$$the marginal of $(X_{(r_1)},\ldots,X_{(r_k)})$ is obtained by integration (with some abuses of notation, see e.g. the integral bounds): \begin{align} f_{X_{(r_1)},\ldots,X_{(r_k)}}(x_{r_1},\ldots,x_{r_k}) &=\int f_{X_{(1)},\ldots,X_{(n)}}(x_1,\ldots,x_n)\mathbb{I}_{x_{r_1}\le x_{r_2}\le\ldots\le x_{r_k}}\,\prod_{i\notin\{r_1,\ldots,r_k\}}\text{d}x_i\\ &=\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(x_{r_1})\prod_{i=r_1+1}^{r_2-1} f_X(x_i)\cdots \\ &\qquad\cdots f_X(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\\ &=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}}\\ &=n!\frac{F_X(x_{r_1})^{r_1-1}}{(r_1-1)!}\frac{[F_X(x_{r_2})-F_X(x_{r_1})]^{r_2-r_1-1}}{(r_2-r_1-1)!}\cdots\\ &\qquad\cdots\frac{[1-F_X(x_{r_k})]^{n-r_k-1}}{(n-r_k-1)!}\prod_{i=1}^k f_X(x_{r_i})\mathbb{I}_{x_{r_1}\le x_{r_2}\le\ldots\le x_{r_k}}\\ \end{align} which is the result produced in the reference (except for the use of $(x_1,...,x_k)$ as the argument of the density). The last integral follows from repeated integrations of $f_X(x)[F_X(x)-F_X(x_j)]^\alpha$; for instance, the first group of integrals leads to \begin{align*} \int_{x_{1}\le\ldots\le x_{r_1}}\prod_{i=1}^{r_1-1} f_X(x_i)\text{d}x_i &=\int_{x_2\le\ldots\le x_{r_1}}\prod_{i=2}^{r_1-1} f_X(x_i)\left\{\int_{-\infty}^{x_2} f_X(x_1)\right\}\text{d}x_1\prod_{i=2}^{r_1-1}\text{d}x_i\\ &=\int_{x_2\le\ldots\le x_{r_1}}\prod_{i=2}^{r_1-1}f_X(x_i)F_X(x_2)\text{d}x_i\\ &=\int_{x_3\le\ldots\le x_{r_1}}\prod_{i=2}^{r_1-1}f_X(x_i)\left\{\int_{-\infty}^{x_3} f_X(x_2)F_X(x_2)\text{d}x_2\right\}\prod_{i=3}^{r_1-1}\text{d}x_i\\ &=\int_{x_3\le\ldots\le x_{r_1}}\prod_{i=3}^{r_1-1}f_X(x_i)\frac{F_X(x_3)^2}{2!}\text{d}x_i\\ &=\ldots \end{align*}

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  • $\begingroup$ one last issue I am having, when evaluating one of the expressions like: $\prod_{i=1}^{r_1-1}\int_{x_{i-1}}^{x_{i+1}} f_X(x_i)\text{d}x_i\,f_X(x_{r_1})$, would this not evaluate out to be: \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} $\endgroup$ Aug 3, 2015 at 3:02
  • $\begingroup$ As indicated in my reply, the notation $\int_{x_{i-1}}^{x_{i+1}}f$ is abusive. The correct presentation is the one of my second set of equation. In your first equation, the dummy variable of the inner integral should be $x_1$ not $x_2$. And the lower bound of the inner integral of the last equation should be $-\infty$, not $x_1$ which has already been integrated out. $\endgroup$
    – Xi'an
    Aug 3, 2015 at 6:21

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