Stable local limit theorems - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T03:01:33Zhttp://mathoverflow.net/feeds/question/52945http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/52945/stable-local-limit-theoremsStable local limit theoremsIgor Kortchemski2011-01-23T11:44:35Z2011-01-23T11:44:35Z
<p>Consider a sequence of integer valued indentically distributed <em>centered</em> independent random variables $X_1, X_2, \ldots$ with the additional condition that the support of $X_1$ is aperiodic. Suppose that the distribution of $X_1$ is in the domain of attraction of a stable law of index $\alpha \in (1,2]$. This means that there exists a sequence $(B_n)_n$ such that $(X_1+X_2+\cdots+X_n)/B_n$ converges in distribution to a stable law of index $\alpha$.</p>
<p>Then the following convergence, known as the Local Limit Theorem (see e.g. I.A.Ibragimov, Yu.V.Linnik, <em>Independent and stationary sequences of random variables</em>, Theorem 4.2.1), holds:
$$\lim_{n \rightarrow \infty} \sup_{k} \left | B_n \mathbb{P}[X_1+\cdots+X_n=k]-g(\frac{k}{B_n})\right| =0,$$
where $g(x)$ is the density of some $\alpha$-stable distribution.</p>
<p>When $X_1$ has a finite second moment (which implies $\alpha=2$), a refinement of this theorem is known, which states that:
$$\lim_{n \rightarrow \infty} \sup_{k} (1 \vee \frac{k^2}{n}) \left| \sqrt{n} \mathbb{P}[X_1+\cdots+X_n=k]-p_{\sigma^2}(\frac{k}{\sqrt{n}})\right|=0,$$
where $\sigma^2$ is the variance of $X_1$ and $p_t(y)=\exp(-\frac{y^2}{2t})/\sqrt{2 \pi t}$.</p>
<p>Is there an analog of this strong local limit theorem in the case $1 < \alpha < 2$? For instance, is it true that:
$$\lim_{n \rightarrow \infty} \sup_{k} (1 \vee \frac{k^\alpha}{B_n^\alpha}) \left | B_n \mathbb{P}[X_1+\cdots+X_n=k]-g(\frac{k}{B_n})\right| =0 ?$$</p>