Excuse me if the question is ill-posed. I'll do my best to explain the problem.I have a vector $(x^{(n)}_1, x^{(n)}_2, \ldots x^{(n)}_n),$ whose individual components can be shown to be asymptotically normally distributed as $n \to \infty,$ with some suitable scaling (here $\sqrt{n}.$) That is to say, $\sqrt{n}x_i \xrightarrow{\mathcal{D}} \mathcal{N}.$ I do not much about the joint distribution; I assume they are independent. Under the assumption that they are independent, is it possible to say something about the asymptotic distribution of $\sum_{i=1}^{n}f(x_i),$ for some function $f.$ Because here I only have access to the asymptotic mean and variance of $x_i,$ so a direct application of the CLT doesn't seem warranted. I know we can not say much about the distribution of sums of random variables that individually converge in distribution(unless we have Skorohod embedding), but here I have an infinite number of them. Any input is welcome. Thanks a lot.

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    $\begingroup$ I believe that as long as you have independence, and the distributions are not becoming more and more crazy, there should be a version of the CLT. Some examples of this: dynamical central limit theorems (in these you don't have independence; just exponentially small dependence of $x_i$ and $x_j$ in $|i-j|$). $\endgroup$ – Anthony Quas Feb 11 '16 at 20:39
  • $\begingroup$ Anthony Quas Thanks for the response. My problem is I only know the asymptotic distribution of each of the components. Thanks for the suggestion, I'll look into them. $\endgroup$ – Arun Feb 11 '16 at 21:42
  • $\begingroup$ I am not sure what the connection between $x_i$ and $x_i^{(n)}$ is but just based on $\sqrt{n}x_i \xrightarrow{\mathcal{D}} \mathcal{N}$ you may be looking for the delta method. $\endgroup$ – user83457 Feb 12 '16 at 15:31
  • $\begingroup$ @michael Michael, they are the same variables; I added the superscript to denote dependence on n. Thing is I only know the asymptotic distribution of each x_i not even the variance or mean for finite n $\endgroup$ – Arun Mar 3 '16 at 18:15
  • $\begingroup$ @AnthonyQuas Thank you. I would greatly appreciate it if you expand a bit more or give some references. I tried, but couldn't. $\endgroup$ – Arun Mar 3 '16 at 18:16

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