In the Encyclopedia of Mathematics there is an inequality for Levy metric ($d_L$): $$d_L(E,F) \leq \{\beta_r(F)\}^{r/(r+1)},$$ where $E$ is a a distribution that is degenerate at zero, $\beta_r(F)$, $r>0$ is an absolute moment of the distribution $F$ (http://www.encyclopediaofmath.org/index.php/L%C3%A9vy_metric). I cannot prove this inequality. I can get only $$d_L(E,F) \leq \{\beta_r(F)\}^{1/(r+1)}.$$ Do you have any ideas how to get $r/(r+1)$?
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$\begingroup$ Did you look at the references cited there? $\endgroup$– Lutz MattnerCommented Oct 31, 2014 at 11:01
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$\begingroup$ Yes, I did. I found a proof for $r=2$ with estimate $1/3$. $\endgroup$– Yu XingCommented Oct 31, 2014 at 11:12
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1 Answer
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There is a confusion in the definition of $\beta_r(X)$. It does not match with the estimate. The estimate of the Encyclopedia should be written for $\gamma_r(X)=E(|X|^r)^{1/r}$ in place of $\beta_r(X)=E(|X|^r)$. The inequality that you can prove is $$d_L(E,F)\leq\gamma_r(F)^{\frac{r}{r+1}}.$$
On the other side, as we will see the same inequality is wrong for $\beta_r$. Let $F_\epsilon$ be uniform on $\{-\epsilon,+\epsilon\}$ and set $E$ the Dirac mass in $0$. You get $d_L(E,F_\epsilon)=\epsilon$ for every $\epsilon<1/2$ but also $\beta_r(F_\epsilon)=\epsilon^r$. Then $d_L\leq \beta_r^{r/(r+1)}$ simply writes $$\epsilon\leq \epsilon^{\frac{r^2}{r+1}}.$$ This is of course wrong, for instance for $r=2$.
Conclusion: use $\gamma_r$ as definition of the $r$-th moment, or use $\beta_r$ and replace the estimate of the Encyclopiedia by your estimate.
Nicolas
