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Ian Jauslin | 70e18cbb54 |
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@ -169,7 +169,7 @@ By the Central Limit Theorem, since $\varphi$ is bounded and continuous,
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\end{equation}
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where $\gamma(x)$ is a centered Gaussian probability density with variance $\sigma^2$.
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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 $c_n\sim n^{3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x)dx= \infty$.
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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 $c_n\sim n^{-3/2}$, $\sum_{n=1}^\infty c_n \int_{\mathbb R^d}|x| \star^n w(x)dx= \infty$.
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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 $R>0$,
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\begin{equation}\label{CLT2}
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