Let $P=(p(1),p(2),\ldots,p(i),\ldots)$ be a discrete probability distribution on $\mathbb Z_+$. Thus, $p(i) \ge 0$ for all $i$ and $\sum_i p(i) = 1$. Moreover, assume for simplicity that $p(1) \ge p(2) \ge \ldots$.
Let $i_1,\ldots,i_T$ be an iid sample from $P$, and for any $i \in \mathbb Z_+$, let $n_T(i) = \#\{t\, \mid\, i_t = i\}$ be the number of times $i$ occurs in the sample. Also define $N_T := \#\{i \, | \,n_T(i) \ge 1\}$, the number of distinct $i$ which occur at least once. Finally, define a probability distribution $P_T=(p_T(1),p_T(2),\ldots)$ on $\mathbb Z_+$ by
$$ p_T(i) = \frac{1}{N_T}\begin{cases} 1,&\mbox{ if }n_T(i) \ge 1,\\ 0,&\mbox{ else.} \end{cases} $$
Question. When $T \to \infty$, does $P_T$ have a limit in some sense ? If yes, how is this limit related to the base distribution $P$ ?
For illustrative purposes, consider the case where $p(i) = i^{-(1+\alpha)}$ for all $i$, where $\alpha \gt 0$ is a constant.
Observations
- $\mathbb E\,1[n_T(i) \ge 1] = \mathbb P(n_T(i) \ge 1) = 1 - (1-p(i))^T$.
- $n_T(i)/T \to p(i)$.
- $1 \le N_T \le T$.
Update: Case of finite support
Suppose $P$ has finite support with $N$ atoms. In this case, $P_T$ converges to the uniform distribution on this support.
Informal argument: If one conditions on $n_T(i) = k$ for some $i \in [N]$ and $k \in [0,T]$, then $1[n_T(j) \ge 1] = Bern(1-(1-p(j))^{T-k})$ for all $j \ne i$, while $1[n_T(i) \ge 1] = 1$. Thus, $N_T = 1 + \sum_{j=1}^N Bern(1-(1-p(j))^{T-k})$. Now, in the limit when $T \to \infty$, one has $ (1-p(j))^{T-k} \to 0$ for all $j \ne i$. Thus, $N_T \to 1 + N-1 = N$ and $n_T(j) \to 1$. We deduce that $P_T \to U([N])$.