I know perhaps this question is really obvious, but when I tried to prove it, I found it really confusing.
Suppose we have $X_{1},X_{2},\cdots$ are i.i.d random variables, and let $X_{1}',X_{2}',\cdots$ are the independent copies of the previous sequence. Now suppose that $n^{-1/2}\sum_{k=1}^{n}X_{k}\longrightarrow_{D}Z$ in distribution for some real-valued random variable, how could I prove that $$n^{-1/2}\sum_{k=1}^{n}X_{k}'\longrightarrow_{D}Z'\ \text{in distribution},$$ where $Z'$ and $Z$ are i.i.d?
I understand that since everything $\{X_{k}, X_{k}'\}$ are i.i.d, the sum $X_{1}+\cdots+X_{n}$ and $X_{1}'+\cdots+X_{n}'$ are i.i.d, and thus $n^{-1/2}(X_{1}+\cdots+X_{n})$ and $n^{-1/2}(X_{1}'+\cdots+X_{n}')$ are i.i.d
But how could I extend this result to the limit?
Thank you!
If $U_n$ and $V_n$ have the same distribution for each $n$ and $U_n \to U$ in distribution then $V_n \to U$ in distribution. This is immediate from the defintion of convergence in distribution.
In convergence in distribution (unlike convergence a.e. or in probability) the limiting random variable is not determined by the sequence. Only the distribution is determined. So we can say that your second sequence converges in distribution to $Z$ itself or to any other random variable with the same distribution. In particular we can take a random variable $Z'$ independent of $Z$ with the same distribution.