My question is related to this question: Is $\sqrt{n}(\bar X_n - \bar X_m)$ tight?
Suppose $\{(X_i, d_i)\}$ is iid with distribution $P$ with finite fourth moments. Suppose $d_i \in \{0,1\}$, and suppose $d_i$ is independent of $X_i$.
Let $n_d = \sum_{i=1}^n d_i$. In the previous question, I asked if $$\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n_d}\sum_{i}^n X_i\cdot d_i\right)= O_p(1)~,$$ which is true. The answer there shows it directly, using a characteristic function approach, but it is also to possible to show it using the multivariate CLT and the Delta method.
Now, I want to know if it is true that $$\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n X^*_{n,i} - \frac{1}{n^*_d}\sum_{i}^n X^*_{n,i}\cdot d^*_{n,i}\right)= O_p(1)~,$$ where, for each $n$, we draw $\{(X^*_{n,i}, d^*_{n,i})\}$ iid from $\hat P_n$, which is the empirical distribution based on $\{(X_{i}, d_{i})\}_{i=1}^n$ which is drawn iid from $P$. Basically, we use $\hat P_n$ as an approximation of $P$ (as you do in resampling, for example), and I ask if the claim still holds.
Here, it is easy to show that $\frac{1}{n}\sum_{i=1}^n X^*_{n,i} - \frac{1}{n^*_d}\sum_{i}^n X^*_{n,i}\cdot d^*_{n,i}= o_p(1)$ (using Slutsky for example), and I basically want to know if $\frac{1}{n}\sum_{i=1}^n X^*_{n,i} - \frac{1}{n^*_d}\sum_{i}^n X^*_{n,i}\cdot d^*_{n,i} = O_p(n^{-1/2})$.
It's not feasible to apply CLT directly here, since we are dealing with triangular arrays, but the finite fourth moment here does allow a multivariate version of the Lyapunov CLT to be used. The main challenge seems to be that the independence between $d_i$ and $X_i$ need not hold in the re-sampled samples that are drawn from $\hat P_n$.
