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Let $X_i$ be a sequence of iid random variables, $E [X] = 0$, $E [X^2] = 1$ and $E [|X|^k] < \infty$ for some $k \ge 3$. Classical local CLT says that the density function $f_n$ of $\frac1{\sqrt n}\sum_1^n X_i$ satisfies that $$ f_n(x) - \phi(x)\left(1 + \sum_{j=1}^{k-2} n^{-\frac j2}P_j(x)\right) = o\left(n^{-\frac{k-2}2}\right), \quad \phi(x) = \frac1{\sqrt{2\pi}}e^{-\frac{x^2}2} $$ where $P_j$ is some $(j + 2)$-order polynomial, and the RHS is uniformly small for $x \in R$.

This gives us very good estimate for constant $x$. My question is that can we get a similar expansion equation for $f_n(\sqrt nx)$? Since $\phi(\sqrt nx)$ decays faster than any polynomial order in $n$, we can not apply the local CLT directly.

Remark: I consider this in order to estimating the following expression for $x \not= 0$ and $y$: $$ \frac1{f_n(\sqrt nx)}f_{n-1}\left(\frac{nx + y}{\sqrt{n-1}}\right), \quad n \text{ sufficiently large}. $$ When $x = 0$ by local CLT this is bounded by $1 + C(1+y^2)/n + o(1/n)$. If $x \not= 0$, I expect the upper bound $$ \exp\left\{-\frac{x^2}2 - xy\right\}\left[1 + \frac{C(x^2 + y^2)}n + o\left(\frac1n\right)\right]. $$

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  • $\begingroup$ This will very much depend on the tail asymptotics of the density of $X$. $\endgroup$ Jun 24, 2019 at 17:23
  • $\begingroup$ We can do explicit calculation when X is Gaussian. So I want to know the case that X is subgaussian: $E [e^{sX}] < e^{c^2s^2/2}$ for any $s$ in $R$. $\endgroup$ Jun 24, 2019 at 22:54

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The asymptotics of the ratio $$r_n(x,y):=\frac1{f_n(\sqrt nx)}f_{n-1}\left(\frac{nx + y}{\sqrt{n-1}}\right) $$ will very much depend on the tail asymptotics of the density (say $f$) of $X$.

E.g., if $X$ is standard normal, then $r_n(x,y)\sim\exp\{-x^2/2-xy\}$ as $n\to\infty$.

If e.g. $f(x)\sim x^{-p}$ for some real $p>0$ as $x\to\infty$, then $f_n(\sqrt n x)\sim nf(\sqrt n x)$ for each real $x>0$ as $n\to\infty$ (cf. e.g. Vinogradov), so that $r_n(x,y)\to1$ as $n\to\infty$.

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  • $\begingroup$ Yes, I realized that it depends on the tail of X. I am typically interested in the case that X is sub-Gaussian: $Ee^{sX} < e^{cs^2}$ for all $s$. $\endgroup$ Jun 24, 2019 at 23:02
  • $\begingroup$ The subgaussian condition is not even mentioned in your question. Since your original question has been answered, I suggest you post a separate question involving the subgaussian condition. $\endgroup$ Jun 25, 2019 at 11:51

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