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There are two convergent in probability sequences of random variables $f_n \to f$ and $g_n \to g$, defined over a metric space $X$. From this post we know that under uniform continuity of $f$, the composition of sequences is convergent in probability as well, i.e. $f_n(g_n) \to f(g)$.

Is it true that

$$ \sqrt{n}(f_n(g_n) - f(g)) \to N\left(0, \mathbb{E}\left(f(g) - \mathbb{E}f(g)\right)^2\right) $$

In particular, should there be no impact from a possible dependence between $f_n$ and $g_n$ on variance?

wlq
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  • What are your assumptions? It is false that, if $h_n \rightarrow h$, then $\sqrt{n}(h_n-h) \sim N(0,\sigma^2)$. – madprob Oct 30 '17 at 11:51
  • I guess this is the link that I am missing between convergence in probability and CLT. What assumptions are needed for the CLT to hold in this case? – wlq Oct 30 '17 at 12:15
  • For example, let $f_{n} \equiv 0$ and $g_n \equiv 0$. In this case, $\sqrt{n}(f_n(g_n)-f(g)) \equiv 0$. I think that there is no general answer. However, the usual conditions in order to obtain a CLT, is to impose that $g_{n}$ is some kind of sample average. Or you can simply impose that $g$ has normal distribution. – madprob Oct 30 '17 at 12:22
  • Can I ask what line of argument you would use with $g$ having a normal distribution? Indeed, $g_n$ is the sample average in my setup. Actually, both $f_n$ and $g_n$ are empirical measures. – wlq Nov 02 '17 at 08:47
  • In this case, check the delta method: https://en.wikipedia.org/wiki/Delta_method – madprob Nov 06 '17 at 11:54

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