Is $\sqrt{n}(\bar X_n - \bar X_m)$ tight? My question is a bit more specific than the title.
Suppose $\{X_i\}$ is iid with finite second moments, and suppose we select some random subset $D_n \subset \{1, \dots, n\}$ such that $n_d = |D_n|$, and suppose that $D_n$ is independent of $\{X_i\}_{i=1}^n$. Moreover, suppose that $n_d/n \xrightarrow{p} c$ for some constant $c \in (0,1)$.
Now, is it true that
$$\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n_d}\sum_{i \in D_n} X_i\right)= O_p(1)$$
I know that $\frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n_d}\sum_{i \in D_n} X_i = o_p(1)$, and I basically want to know if $\frac{1}{n}\sum_{i=1}^n X_i - \frac{1}{n_d}\sum_{i \in D_n} X_i = O_p(n^{-1/2})$
 A: Clarification I understand your big O notation as probabilistic big O notation 
Short Answer
Yes, you have the right speculation.
And, if my calculations are right, your speculation is still valid if $(n_d;n \ge 1)$ is a sequence of random variables such that :

*

*$n_d/n$ is bounded below by a real number $a>0$

*$n_d/n$ converges in law (as a sequence of real random variables)

Main idea
We'll prove the result the same way that we've done for the central limit theorem. That is, by characteristic functions and Levy's continuity theorem.
Long answer
WLOG: $\mathbb{E}(X_1)=0, \mathbb{E}(X^2)=1$
Let $\phi$ be the characteristic function of $X$, that is:
$$\phi(\lambda) = \mathbb{E}( e^{i\lambda X_1})$$
The finiteness of the second moment of $X$ implies that:
$$\phi(\lambda) = exp ( \lambda^2(-1+o_{\lambda}(1)))$$
where $|o_{\lambda}(1)| \le h( |\lambda|)$ for some increasing function $h$ bounded around $0$ and $h(\lambda) \xrightarrow{\lambda \rightarrow 0^+} 0$
($o_\lambda$  can be complex but doesn't matter)
Let

*

*$d^n_i:= 1_{ i \in \mathbb{D_n}}$ 
When there is no confusion, by abuse of notation, we'll let $d_i= d^n_i$

*$Y_n$ is the random quantity that we want to study, i.e
$$Y_n:= \sqrt{n}\left( \dfrac{1}{n}\left(\sum_{k=1}^n X_k\right) - \dfrac{1}{n_d}\left(\sum_{k \in D_n} X_k\right) \right)$$
So clearly, $Y_n$ can be written as
$$Y_n = \sqrt{n} \left(\sum_{k=1}^n \left(\dfrac{1}{n}-\dfrac{d_k}{n_d}\right)X_k\right)$$
So, by indepedence between $D_n$ and $(X_i)$, for all $\lambda$ real, we have:
$$\phi_{Y_n}(\lambda) = \mathbb{E}\big( \prod_{k=1}^n \phi\underbrace{\left( \lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)}_{ =o(1/\sqrt{n}) \text{regardless of the value of} d_i} \big)$$
Thus, by the remark we have to $\phi$, we imply that
$$\phi_{Y_n}(\lambda)  = \mathbb{E}\left( exp\left[ \sum_{k=1}^n \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)^2(-1+ o(1)) \right] \right)$$
where $|o(1)| \le h( \frac{b|\lambda|}{\sqrt{n}})$ for every $o(1)$ and some $b$ depending only on $a$.
As $ \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)^2 \ge 0$, we have:
$$ \sum_{k=1}^n \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_i}{n_d} \right)^2(-1+ o(1)) = (-1+o(1))\underbrace{\left[ \sum_{k=1}^n \left(\lambda \dfrac{1}{\sqrt{n}}-\lambda \sqrt{n}\dfrac{d_k}{n_d} \right)^2 \right]}_{ = \lambda^2(\dfrac{n}{n_d}-1)} $$
Thus
$$ \phi_{Y_n}(\lambda) = \mathbb{E}\left[ e^{ -(1+o(1))\lambda^2( n/n_d-1)} \right] \xrightarrow{n \rightarrow +\infty} e^{-\lambda^2(1/c-1) }$$
Therefore, by Levy's theorem, we have:
$$ \mathcal{L}( Y_n ) \longrightarrow \mathcal{N}( 0, 1/c-1)$$
Hence the conclusion.
$\square$
