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Let $c>0$ and $T>0$ be fixed. Denote by $F$ the Gaussian CDF, i.e. $F:\mathbb R\to\mathbb R$ is defined by

$$F(x):=\int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz.$$

For every $a\in [0,1)$, does there exist $C\equiv C_a>0$ s.t.

$$\left|\int_0^\infty \frac{1}{s-t}\left(F\left(\frac{y}{\sqrt{c(s-t)}}\right)-F\left(\frac{y-x}{\sqrt{c(s-t)}}\right)\right) dy\right| \le \frac{C|x|^a}{(s-t)^{(1+a)/2}}$$

holds for all $0\le t<s\le T$ and $x\in\mathbb R$?

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  • $\begingroup$ Just to clarify, may $C$ depend on $T$ and $c$ as well? $\endgroup$
    – Steve
    May 18, 2022 at 14:55
  • $\begingroup$ Yes, $C$ must depend on $c$. In this question, $C$ can also depend on $T$ (while I believe $C$ does not depend on $T$ but I didn't get it by calculus) $\endgroup$
    – GJC20
    May 18, 2022 at 14:56

1 Answer 1

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No. Indeed, let $$u:=\frac x{\sqrt{c(s-t)}}.$$ Then, with the substitution $z:=\dfrac y{\sqrt{c(s-t)}}$, the inequality in question can be rewritten as $$|I(u)|\le C|u|^a \tag{1}\label{1}$$ for all real $u$, where $$I(u):=\int_0^\infty dz\,(F(z)-F(z-u)).$$

For $u>0$, $$I(u)=\int_0^\infty dz\,\int_{z-u}^z dF(t) =\int_{-u}^\infty dF(t)\int_0^\infty dz\,1(t<z<t+u) \\ \ge\int_0^\infty dF(t)\,u=\frac12\,u. $$ Similarly, $|I(u)|\le|\int_{-\infty}^\infty dz\,(F(z)-F(z-u))|=|u|$ for all real $u$.

So, \eqref{1} holds for all real $u$ if and only if $a=1$. (This conclusion actually holds for any cdf $F$ such that $F(0-)<1$.)

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  • $\begingroup$ Nice arguments. Many thanks Iosif! $\endgroup$
    – GJC20
    May 19, 2022 at 4:17
  • $\begingroup$ You are very welcome! $\endgroup$ May 19, 2022 at 18:02

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