Consider pseudo Gaussian densities for $0<s<t$ and $x,y\in\mathbb R$

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

where $A(s,t,y):=\int_s^t k(u,y)/(1+a(u))du$ and $B(s,t,y):=\int_s^tk(u,y)/(1+b(u))du$. I'm interested in the dependence of the difference $f-g$ on the parameters $a,b$. Assume $a, b: \mathbb R_+ \to [0,1]$ are continuous, $k: \mathbb R_+\times\mathbb R \to [1,2]$ is $1-$Lipschitz. Does there exist $C>0$ depending only on $T>0$ s.t.

\begin{eqnarray} \left|\int_0^\infty f(0,0,t,y)dy - \int_0^\infty g(0,0,t,y)dy\right | &\le& Ct^{1/2}\|a-b \|_t,\quad \forall 0<t\le T\quad (\ast) \\ \int_0^t \left|\int_0^\infty \partial_sf(s,0,t,y)dy - \int_0^\infty \partial_s g(s,0,t,y)dy\right |ds &\le& C(t-s)^{1/2}\|a-b \|_t,\quad \forall 0<s<t\le T\quad (\star), \end{eqnarray}

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

PS : $(\star)$ can be shown if

\begin{eqnarray} \left|\int_0^\infty \partial_sf(s,0,t,y)dy - \int_0^\infty \partial_s g(s,0,t,y)dy\right | \le C(t-s)^{-1/2}\|a-b \|_t. \end{eqnarray}

Iosif Pinelis has shown in How does the integral of pseudo Gaussian kernel on $(0,\infty)$ depend on its variance? that

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

where $p$ is some probability density on $(0,\infty)$.


1 Answer 1


$\newcommand{\De}{\Delta}\newcommand{\vpi}{\varphi}$Let \begin{equation*} A:=A(y):=A(0,t,y),\quad B:=B(y):=B(0,t,y), \end{equation*} so that \begin{equation*} f(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{A(y)}(y), \quad g(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi_{B(y)}(y), \end{equation*} where \begin{equation*} \vpi_a(u):=\frac1{\sqrt a}\,\exp\Big(-\frac{u^2}{2a}\Big). \end{equation*} Note that \begin{equation*} \frac{\partial}{\partial a}\vpi_a(u) =\frac12\Big(\frac{u^2}{a^{5/2}}-\frac1{a^{3/2}}\Big)\exp\Big(-\frac{u^2}{2a}\Big). \end{equation*}

So, \begin{equation*} 2\sqrt{2\pi}\,\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)] \tag{1}\label{1} =\int_0^1 dv\,I(v), \end{equation*} where \begin{equation*} I(v):=I_1(v)-I_2(v), \tag{2}\label{2} \end{equation*} \begin{equation*} I_1(v):=\int_0^\infty dy\,H(y) \frac{y^2}{c_v(y)^{5/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big), \end{equation*} \begin{equation*} I_2(v):=\int_0^\infty dy\,H(y) \frac1{c_v(y)^{3/2}}\exp\Big(-\frac{y^2}{2c_v(y)}\Big), \end{equation*} \begin{equation*} H:=B-A, \end{equation*} \begin{equation*} c_v:=A+v(B-A). \end{equation*}

Next is the crucial step:
\begin{equation*} I_1(v)=I_{11}(v)+I_{12}(v), \tag{3}\label{3} \end{equation*} where \begin{equation*} I_{11}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big) \Big(\frac{y}{c(y)}-\frac{y^2c'(y)}{c(y)^2}\Big) \frac{y}{c(y)^{3/2}}H(y), \end{equation*} \begin{equation*} I_{12}(v):=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big) \frac{y^3c'(y)}{c(y)^{7/2}} H(y), \end{equation*} \begin{equation*} c(y):=c_v(y), \end{equation*} and $c'$ is the derivative of the Lipschitz function $c=c_v$; this derivative exists almost everywhere (a.e.), since $k$ is $1$-Lipschitz. Moreover, \begin{equation*} |c'|\le t \end{equation*} a.e. Note also that for $y\ge0$ and $t>0$ \begin{equation*} t/2\le c(y)\le2t,\quad |H(y)|\ll t\,\De a,\quad |H'(y)|\ll t\,\De a, \end{equation*} where \begin{equation*} \De a:=\|a-b\|_t \end{equation*} and $E\ll F$ means that $|E|\le CF$ for some universal real constant $C$.

So, \begin{equation*} |I_{12}(v)|\ll\int_0^\infty dy\,\exp\Big(-\frac{y^2}{4t}\Big) \frac{y^3\,t}{t^{7/2}}\, t\,\De a \asymp\sqrt t\,\De a. \tag{4}\label{4} \end{equation*}

Integrating by parts, we have \begin{equation*} \begin{aligned} I_{11}(v)&=-\int_0^\infty dy\,\Big[\frac d{dy}\exp\Big(-\frac{y^2}{2c_v(y)}\Big)\Big] \frac{y}{c(y)^{3/2}}H(y) \\ &=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big) \frac d{dy}\Big[\frac{y}{c(y)^{3/2}}H(y)\Big] \\ &=I_2(v)+I_{111}(v)-\frac32\,I_{112}(v), \end{aligned} \tag{5}\label{5} \end{equation*} where \begin{equation*} \begin{aligned} I_{111}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big) \frac{y}{c(y)^{3/2}}H'(y), \end{aligned} \end{equation*} \begin{equation*} \begin{aligned} I_{112}(v)&:=\int_0^\infty dy\,\exp\Big(-\frac{y^2}{2c_v(y)}\Big) \frac{yc'(y)}{c(y)^{5/2}}H(y). \end{aligned} \end{equation*} Similarly to \eqref{4}, we get \begin{equation} |I_{111}(v)|+|I_{112}(v)|\ll \sqrt t\,\De a. \tag{6}\label{6} \end{equation}

Collecting \eqref{1}, \eqref{2}, \eqref{3}, \eqref{4}, \eqref{5}, and \eqref{6}, we conclude that \begin{equation*} \Big|\int_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)]\Big|\ll \sqrt t\,\De a, \end{equation*} which proves the first inequality of the two ones in question.

The other inequality can apparently be proved similarly. Since this answer is already rather long and complicated, I will leave the other inequality as an exercise (or to be posted elsewhere).

  • $\begingroup$ Thank you very much Iosif for the elegant answer. Very appealing idea to treat the difference of the pseudo Gaussian densities as the integral of some derivative w.r.t. the variance. I will try to adopt this idea to estimate the integral of derivative $\endgroup$
    – user478492
    Apr 14, 2022 at 7:38
  • $\begingroup$ Indeed, the aim is to estimate $\int_0^\infty \partial_s p\otimes H^{(r)}(s,0,t,y)dy-\int_0^\infty \partial_s p_{\epsilon}\otimes H_{\epsilon}^{(r)}(s,0,t,y)dy$, see the paper hal.archives-ouvertes.fr/hal-01169685v3/document for the definitions of $p\otimes H$ and $p_{\epsilon}\otimes H_{\epsilon}^{(r)}$. In this paper, the authors provide an estimation of $ p\otimes H^{(r)}(s,x,t,y)-p_{\epsilon}\otimes H_{\epsilon}^{(r)}(s,x,t,y)$, while it is not enough for my purpose. I will use your idea to estimate and I will post this separately if I cannot succeed. Thanks again $\endgroup$
    – user478492
    Apr 14, 2022 at 7:45
  • $\begingroup$ Thank you for your appreciation of this answer. However, I think the main idea here is, not just to differentiate w.r. to the variance, but then mostly remove this badly behaving derivative by the integration by parts. To do this integration by parts, we need the Lipschitz condition on $k$. This Lipschitz condition was not needed or used in the previous answer, at mathoverflow.net/a/419721/36721 . So, here one may say the reasoning is a bit subtler. Hopefully, this comment will be of help in your further work. $\endgroup$ Apr 14, 2022 at 13:50
  • $\begingroup$ Dear Iosif, please allow me to come towards you again for the continuation question on $(\star)$. After around one week's efforts, I still cannot obtain the desired result (I can only show the inequality when $k\equiv k(t)$). I have formulated my question separately in a generic version and given its origin. Could you please take a look (mathoverflow.net/questions/420644/…)? Thank you very kindly for your consideration. $\endgroup$
    – user478492
    Apr 19, 2022 at 14:16
  • $\begingroup$ All right, I will look at that question. $\endgroup$ Apr 19, 2022 at 16:27

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