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Let $a, b: \mathbb R_+ \to [0,1]$ be continuous functions. Let $k: \mathbb R_+\times\mathbb R \to [1,2]$ be $1-$Lipschitz. Set, for $0<s<t$ and $y>0$,

$$A(s,t,y):=\int_s^t\frac{k(u,y)}{1+a(u)}du \quad\mbox{and} \quad B(s,t,y):=\int_s^t\frac{k(u,y)}{1+b(u)}du.$$

Define further

$$f(s,x,t,y):=\frac{1}{\sqrt{2\pi A(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2A(s,t,y)}\right)\quad\mbox{and} \quad g(s,x,t,y):=\frac{1}{\sqrt{2\pi B(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2B(s,t,y)}\right).$$

Given a probability density $p$ on $(0,\infty)$ which can be assumed to be as good as possible, can we show the existence of some $C>0$ (depending only on $p$) s.t.

$$\left|\int_0^\infty p(x)dx \int_0^\infty f(s,x,t,y)dy - \int_0^\infty p(x)dx\int_0^\infty g(s,x,t,y)dy\right |\le C(t-s)^{1/2}\|a-b \|_t,$$

where $\|a-b \|_t:=\max_{0\le u\le t}|a(u)-b(u)|$.

PS : It is straightforward that $|A(s,t,y)-B(s,t,y)|\le C(t-s)\|a-b \|_t$ for some $C>0$. If $k$ is independent of $y$, i.e. $k(u,y)\equiv k(u)$, then $f,g$ are both Gaussian densities and the above inequality holds by computation.

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  • $\begingroup$ Do you want $C$ to be $<\infty$ and not to depend on $s,t,x,a,b,k$? All this should have been specified. $\endgroup$ Apr 5 at 19:46
  • $\begingroup$ @IosifPinelis I have corrected my desired inequality. I pursue a constant $C$ depending only on the density $p$ $\endgroup$
    – Philo18
    Apr 5 at 20:21

1 Answer 1

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$\newcommand{\si}{\sigma}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}\newcommand{\De}{\Delta}$Let us show that the desired bound holds if \begin{equation*} \int_0^\infty dx\,|p'(x)|<\infty, \tag{1}\label{1} \end{equation*} which in particular implies that there exists the limit \begin{equation*} p_0:=p(0+)\in[0,\infty). \tag{2}\label{2} \end{equation*} One may note that, for condition \eqref{1} to hold, it is enough that e.g. the density $p$ be continuously differentiable and bounded on $(0,\infty)$ with only finitely many modes.

Let \begin{equation*} A:=A(y):=A(s,t,y),\quad B:=B(y):=B(s,t,y), \end{equation*} \begin{equation*} h(x,y):=h(s,x,t,y):=f(s,x,t,y)-g(s,x,t,y), \end{equation*} \begin{equation*} I:=\int_0^\infty dx\,p(x) \int_0^\infty dy\,h(x,y). \tag{3}\label{3} \end{equation*}

We want to show that \begin{equation*} |I|\ll(t-s)^{1/2}\De a, \tag{$\clubsuit$}\label{*} \end{equation*} where $E\ll F$ means that $|E|\le cF$ for some real constant $c$ depending only on $p$ and \begin{equation*} \De a:=\|a-b\|_t. \end{equation*}

Let $\Phi$ and $\vpi$ denote the standard normal cdf and pdf, respectively. By \eqref{3} and \eqref{1},
\begin{equation*} \begin{aligned} I&=\int_0^\infty dx\,\Big(p_0+\int_0^x d\xi\,p'(\xi)\Big) \int_0^\infty dy\,h(x,y) \\ &=p_0 J+K, \end{aligned} \tag{4}\label{4} \end{equation*} where \begin{equation*} \begin{aligned} J&:=\int_0^\infty dy\,\int_0^\infty dx\, h(x,y), \\ K&:=\int_0^\infty dx\,\int_0^x d\xi\,p'(\xi) \int_0^\infty dy\,h(x,y) \\ &{\color{red}{\,\,=}}\int_0^\infty dy\,\int_0^\infty d\xi\,p'(\xi)\int_\xi^\infty dx\, h(x,y) \\ &=\int_0^\infty dy\,\int_0^\infty d\xi\,p'(\xi) \Big[\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big) -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big] \\ &=\int_0^\infty d\xi\,p'(\xi) \int_0^\infty dy\, \Big[\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big) -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big]. \end{aligned} \tag{5}\label{5} \end{equation*} The red equality in \eqref{5} holds by the Fubini theorem -- which is the crucial point of the entire proof, as it allows one to deal, instead of $\big|\vpi\big(\frac z\si\big)'_\si\big|$, with $\big|\Phi\big(\frac z\si\big)'_\si\big|$ as in \eqref{!} below and thus get the crucial additional factor $|z|$ in the numerators there, which alleviates the possible smallness of the denominator $t-s$ of the ratio $\frac{2|z|}{t-s}$ in \eqref{!}.

Note that $\{A,B\}\subset[\frac{t-s}2,2(t-s)]$, and hence for any real $z$ and any $\si$ between $\sqrt A$ and $\sqrt B$ we have \begin{equation*} \Big|\Phi\Big(\frac z\si\Big)'_\si\Big| =\frac{|z|}{\si^2}\,\vpi\Big(\frac z\si\Big) \le\frac{2|z|}{t-s}\,\vpi\Big(\frac z{\sqrt{2(t-s)}}\Big) \tag{$\heartsuit$}\label{!} \end{equation*} and \begin{equation*} |\sqrt A-\sqrt B|=\frac{|A-B|}{\sqrt A+\sqrt B}\ll (t-s)^{1/2}\De a, \end{equation*} so that (by, say, the mean value theorem) \begin{equation*} \Big|\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big) -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big| \ll\frac{|y-\xi|}{t-s}\,\vpi\Big(\frac{y-\xi}{\sqrt{2(t-s)}}\Big) (t-s)^{1/2}\De a \end{equation*} and
\begin{equation*} \begin{aligned} &\int_0^\infty dy\, \Big|\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big) -\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big| \\ &\ll\int_{-\infty}^\infty dy\, \frac{|y-\xi|}{t-s}\,\vpi\Big(\frac{y-\xi}{\sqrt{2(t-s)}}\Big) (t-s)^{1/2}\De a \\ &\ll(t-s)^{1/2}\De a. \end{aligned} \end{equation*}

So, by \eqref{5} and \eqref{1}, \begin{equation*} |K|\ll (t-s)^{1/2}\De a. \tag{6}\label{6} \end{equation*} Similarly and a bit easier, we get \begin{equation*} |J|\ll (t-s)^{1/2}\De a. \tag{7}\label{7} \end{equation*} Now \eqref{*} follows from \eqref{4}, \eqref{2}, \eqref{6}, and \eqref{7}.

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  • $\begingroup$ Thank you so much for the answer. Amazing analysis trick! I am thinking whether I can show $| \int_0^\infty f(s,0,t,y)dy -\int_0^\infty g(s,0,t,y)dy | \le C(t-s)^{1/2}\|a-b \|_t$ holds for some constant $C>0$. Do you believe it is true? If needed, I can post my further question independently $\endgroup$
    – Philo18
    Apr 6 at 14:38
  • $\begingroup$ Thank you for your appreciation. Concerning your question about $x=0$, it may indeed be better to post it separately -- just make sure it does not get perceived as a duplicate. $\endgroup$ Apr 6 at 14:53
  • $\begingroup$ Of course. I will post this question separately. Thanks again $\endgroup$
    – Philo18
    Apr 7 at 5:01
  • $\begingroup$ Hi Iosif, I have posted this question in a separate post mathoverflow.net/questions/420220/… which also includes the dependence of the integral of $\partial_s f$ on $(0,\infty)$ on its variance. Could you please take a look? $\endgroup$
    – Philo18
    Apr 12 at 13:55
  • $\begingroup$ @Philo18 : All right, I will look at it. $\endgroup$ Apr 13 at 3:01

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