Suppose we have a sequence of random variable $X_n$, for example $X_n = \sum_{i=1}^{n} (Y_i - \mu)^2 $ for iid $Y_i$ with mean $\mu$ and sd $\sigma$, such that $X_n / n \rightarrow c$ for some constant $ c > 0$. Would it be possible to have a deterministic sequence $s_n$ that grows slower than $n$ so that there is a non-degenerate asymptotic distribution that $X_n / s_n$ converges to?

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    $\begingroup$ If $X_n/n\to c\ne0$ and $s_n/n\to0$ then $X_n/s_n=(X_n/n)(n/s_n)\to\infty$ hence the answer to your question is "No". $\endgroup$ – Did Feb 15 '17 at 7:17
  • $\begingroup$ Thank you! That's very helpful. $\endgroup$ – D.Williams Feb 15 '17 at 14:57

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