**Note:** This question was asked in stats.stackexchange.com and math.stackexchange.com, 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

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