Suppose we have the following weak convergence: $$ n^{-1/2}\cdot (X_n-n) \Rightarrow X \quad \text{as} \quad n\to\infty, $$ where $X_n$ and $X$ are random variables (note: $X$ is a non-degenerate r.v.), the "$\Rightarrow$" means convergence in distribution.
Can we prove that $$\mathbb{E}[X_n]=O(n) \quad \text{and} \quad \operatorname{Var}[X_n]=O(n) \quad \text{as} \quad n\to\infty ~?$$
If yes, please provide a detailed proof; if no, what conditions should we add to obtain the results?
Note that, I don't want to prove the convergence in mean, i.e., $\mathbb{E}[X_n]\to \mathbb{E}[X]$, instead I just want to show the orders of the mean and variance. Is it possible?
Thank you!
No it is not true. Assume $n^{-1/2} (Y_n - n) \Rightarrow Y$ for some nondegenerate $Y$. A counterexample is to choose $X_n$ such that $\mathbb{P}(X_n=2^n) = 1/n$ and $\mathbb{P}(X_n=Y_n)=1-1/n$. Then it holds that $n^{-1/2} (X_n - n) \Rightarrow Y$ but $\mathbb{E}[X_n]$ is of higher order than $n$.
I am not entirely sure what you need to require for your proposed statement to be true. Maybe something like uniform integrability of $(n^{-1/2} (X_n - n))_n$.