**Note:** This question was asked in stats.stackexchange.com and [math.stackexchange.com](https://math.stackexchange.com/questions/33617/a-sequence-of-order-statistics-from-an-iid-sequence), with expired bounties on both sites. 


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).



A sequence of order statistics from an iid sequence
Another way to describe this sequence of random vectors is that you have an unordered set of $k$ points, initially sampled independently, and at each step you add a point and then with probability $p$ remove the minimum, and with probability $1-p$ remove the maximum.


Given $x \in (0,1)$, the number of points in the $n$th set lower than $x$ follows a random walk on the set $\{0,1,...,k\}$ whose initial distribution is binomial (and unimportant) and whose transitions occur with the following probabilities except at the edges:


* $-1$ with probability $(1-x)p$: Add a point greater than $x$ and delete the minimum.
* $+1$ with probability $x(1-p)$: Add a point smaller than $x$ and delete the maximum.
* $0$ with the complementary probability $(1-x)(1-p) + xp$: Add a point greater than $x$ and delete the maximum or add a point smaller than $x$ and delete the minimum.


The boundary cases are that when there are no points smaller than $x$, this will still increase to $1$ with probability $x(1-p)$, but the chance to stay $0$ is $1-x(1-p)$, and when all $k$ points are smaller than $x$, this will decrease to $k-1$ with probability $(1-x)p$ and stay $k$ with probability $1-(1-x)p$. 


The limiting distribution is the same as if you eliminate the chance to pick a new point lower than $x$ and then delete the minimum or to pick a point greater than $x$ and then delete the maximum.


The initial distribution doesn't matter. The limiting distribution of this random walk gives the limiting distributions for the coordinates, since the probability that the $\ell$th point is at most $x$ is the sum of the probabilities that there are exactly $\ell$, $\ell+1$, ... or $k$ points less than $x$. It does not say the joint distribution.


