TITLE: Drawing natural numbers without replacement.
QUESTION [24 upvotes]: Suppose we start with an initial probability distribution on $\mathbb{N}$ that gives positive probability to each $n$. Let's call this random variable $X_1$ so we have $P(X_1=n)=p_{1,n}>0$ for all $n\in\mathbb{N}$. $X_1$ wil be the first draw from $\mathbb{N}$. For the next draw $X_2$ we define a new distribution on $\mathbb{N}\setminus\{ X_1 \}$ by rescaling the remaining probabilities so they add up to 1. So $p_{2,X_1}=0$ and $p_{2,n}=\frac{p_{1,n}}{1-p_{1,X_1}}$ for $n\neq X_1$. Continuing in this manner we get a stochastic process (certainly not Markov) that corresponds to drawing from $\mathbb{N}$ without replacement. My question is whether this process has ever been studied in the literature. In particular, I'm wondering if a clever choice of the initial distribution could result in tractable expressions for the distributions of $X_n$ for large $n$.

REPLY [4 votes]: Here are some preliminary computations. Assume the reference distribution is $(p(n))$. For every finite subset $I$ of $\mathbb N$, introduce the finite number $r(I)\ge1$ such that
$$
\frac1{r(I)}=1-\sum_{k\in I}p(k).
$$
Obviously, $P(X_1=n)=p(n)$ for every $n$. Likewise,
$P(X_2=n)=E(p(n)r(X_1);X_1\ne n)$ hence
$$
P(X_2=n)=p(n)(\alpha-p(n)r(n)),\qquad
\alpha=\sum\limits_kp(k)r(k).
$$
This shows that $X_1$ and $X_2$ are not equidistributed (if they were, $\alpha-p(n)r(n)$ would not depend on $n$, hence $p(n)$ would not either, but this is impossible since $(p(n))$ is a measure with finite mass on an infinite set).
One can also compute the joint distribution of $(X_1,X_2)$ as
$$
P(X_1=n,X_2=k)=p(n)r(n)p(k)[k\ne n],
$$
and this allows to expand 
$$
P(X_3=n)=E(p(n)r(X_1,X_2);X_1\ne n,X_2\ne n),
$$
as the double sum
$$
P(X_3=n)=p(n)\sum_{k\ne n}\sum_{i\ne n}[k\ne i]r(k,i)p(k)r(k)p(i),
$$
but no simpler or really illuminating expression seems to emerge.