Update to v1.1:

Fixed: Typo: c_n\sim n^{-3/2} not n^{3/2}
This commit is contained in:
Ian Jauslin 2021-05-19 19:55:02 -04:00
parent e8cb253fb2
commit 70e18cbb54
2 changed files with 6 additions and 1 deletions

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@ -169,7 +169,7 @@ By the Central Limit Theorem, since $\varphi$ is bounded and continuous,
\end{equation} \end{equation}
where $\gamma(x)$ is a centered Gaussian probability density with variance $\sigma^2$. where $\gamma(x)$ is a centered Gaussian probability density with variance $\sigma^2$.
This shows that there is a $\delta> 0$ such that for all sufficiently large $n$, $\int_{\mathbb R^d}|x| \star^n w(x)dx \geqslant \sqrt{n}\delta$, and then since $c_n\sim n^{3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x)dx= \infty$. This shows that there is a $\delta> 0$ such that for all sufficiently large $n$, $\int_{\mathbb R^d}|x| \star^n w(x)dx \geqslant \sqrt{n}\delta$, and then since $c_n\sim n^{-3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x)dx= \infty$.
To remove the hypothesis that $w$ has finite variance, note that if $w$ is a probability density with zero mean and infinite variance, $\star^n w(n^{1/2}x)n^{d/2}$ is ``trying'' to converge to a ``Gaussian of infinite variance''. In particular, one would expect that for all $R>0$, To remove the hypothesis that $w$ has finite variance, note that if $w$ is a probability density with zero mean and infinite variance, $\star^n w(n^{1/2}x)n^{d/2}$ is ``trying'' to converge to a ``Gaussian of infinite variance''. In particular, one would expect that for all $R>0$,
\begin{equation}\label{CLT2} \begin{equation}\label{CLT2}

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@ -1,3 +1,8 @@
v1.1:
* Fixed: Typo: c_n\sim n^{-3/2} not n^{3/2}
v1.0: v1.0:
* Fixed: Missing factor of 1/2 in expansion of f. * Fixed: Missing factor of 1/2 in expansion of f.