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Given a sequence of iid random variables $X_i$ (without loss of generality from $U(0,1)$), an integer $k \ge 1$ and some $p \in (0,1)$, construct the sequence of random vectors $Z^{(j)}$, $j=0,1,...$ in the following way. Let

$Z^{(0)}=(X_{(1)},...,X_{(k)}),$

where $X_{(l)}$ is the $l$-order statistic of sample $\{X_1,...,X_k\}$. Introduce notations

\begin{align} Z^{(j)}&=(Z_{j,1},...,Z_{j,k}),\\ m_j&=\min(Z_{j-1,1},...,Z_{j-1,k},X_{k+j}),\\ M_j&=\max(Z_{j-1,1},...,Z_{j-1,k},X_{k+j}) \end{align}

Then

$Z^{(j)}=(Y_{(1)},...,Y_{(k)})$

where $Y_{(l)}$ is the $l$-order statistic of the following set which is

  1. The set $\{Z_{j-1,1},...,Z_{j-1,k},X_{k+j}\}\backslash m_j$ with probability $p$
  2. The set $\{Z_{j-1,1},...,Z_{j-1,k},X_{k+j}\}\backslash M_j$ with probability $1-p$

The decision between cases 1. and 2. is made independently from the $X_i$ (and hence from the $Z^{(i)}$).

The $Z^{(j)}$ are supported on the $k$-dimensional simplex $S_k = \{(x_1, \dots, x_k) \in \mathbb{R}^k \, | \, 0 \le x_1 \le x_2 \le \dots \le x_k \le 1 \}$.

It appears that the $Z^{(j)}$ converge in distribution. Is this known? Is anything known about the limiting distribution?

For the case $k=1$, the answer is the following. Denote the cdf of $Z^{(j)}$ by $F_j$.

The cdf of $\min(X_{n+1},Z^{(n)})$ (for $U(0,1)$ case) is

$x+F_n(x)−xF_n(x)$ and the cdf of $\max(X_{n+1},Z^{(n)})$ is

$xF_n(x)$.

Hence

\begin{align} F_{n+1}(x)&=p(x+F_n(x)−xF_n(x))+(1−p)xF_n(x)\\ &=px+(p(1-x)+(1-p)x)F_n(x) \end{align}

Since $p(1-x)+(1-p)x\in(0,1)$ we have that

$\lim F_{n}(x)=\frac{px}{1-p(1-x)-(1-p)x}$

I am looking for general results (case k>1) either for the limiting distribution of the whole vector $Z^{(j)}$ or of some of its components (marginal distributions).

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    Cleaned up old, no-longer relevant comments. Said comments are still on display in the original posting at Stats.SE.2011-04-18

2 Answers 2

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I've asked this question in mathoverflow. It has one answer, which I do not find time to check, so I am giving the link as an answer here.

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    The answer given there is very useful, thank you for posting it there.2012-11-23
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EDIT I actually had a different answer for this question referring to markov chains, but then I saw a different and better way of solving the problem. So I changed my answer - can post the old one back if people want it

Now $Z^{(m)}$ will be a function of $X_1,\dots,X_{k+m-1}$. In particular it will be a function of the order statistics of this set.

For a uniformly distributed variable $X_i$, if we take a collection of $k+m-1$ of them, the rth order statistic has a beta distribution:

$X_{(r)}\sim Beta(r,k+m-r)$

Now what will the process have done by the mth step? It will have removed the top $n_t$ order statistics and the bottom $n_b$, with the constraint that $n_t+n_b=m-1$. Stated this way the probability of any such set occurring is given by the binomial distribution:

$Pr(n_t)={m-1 \choose n_t}p^{n_t}(1-p)^{m-1-n_t}$

Now given $n_t$ is the number chosen, we have $Z^{(m)}=(X_{(m-n_t)},\dots,X_{(k+m-1-n_t)})$ Now the joint distribution of $Z^{(m)}$ which looks like:

$P(Z^{(m)}|n_t)=\frac{(k+m-1)!X_{(m-n_t)}^{m-1-n_t}(1-X_{(k+m-1-n_t)})^{n_t}}{(m-1-n_t)!(n_t)!}dX_{(m-n_t)}\dots dX_{(k+m-1-n_t)}$

Which is a Dirichlet distribution with parameters $(m-n_t,1,1,\dots,1,n_t+1)$ with $k-2$ 1's in the middle (which is a general result for the joint distribution of order stats - not sure if its known).

Now we need to sum out $n_t$ with respect to its probability:

$P(Z^{(m)})=\sum_{n_t=0}^{m-1}P(n_t)P(Z^{(m)}|n_t)$ $=\frac{(k+m-1)!}{(m-1)!}\sum_{n_t=0}^{m-1}{m-1 \choose n_t}^{2}\left[(1-X_{(k+m-1-n_t)})p\right]^{n_t}\left[X_{(m-n_t)}(1-p)\right]^{m-1-n_t}$

Which looks tantalizingly close to a binomial distribution, if it wasn't for that annoying squared term in the summation. I don't know an exact solution for this summation. This is a far as I could get. You would likely require numerical work for this.

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    @probabilityislogic if you decide to go that route, let me know and I'll remove the downvote before you delete. Another option is to edit again and make clear that an inconsistency is present. Maybe someone can modify the approach and make it work. I tried a couple of different things, but couldn't find any traction.2011-04-18