# How to upper bound the difference between these two Gaussian-like densities?

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Let $$b:\RR_+ \times \RR^d \to \RR^d$$ and $$\sigma:\RR_+ \times \RR^d \to \mathcal M^{\text{sym}}_{d \times d} (\RR)$$ be measurable. Let $$a(t, x) := \sigma \sigma^\top (t, x)$$. We assume that

1. There is $$\lambda >0$$ such that $$\lambda^{-1} |\xi|^2 \le \xi^\top a(t, x) \xi \le \lambda |\xi|^2 \quad \forall (t, x, \xi) \in {\RR}_+ \times {\RR}^d \times {\RR}^d.$$
2. There is $$\eta, L>0$$ such that $$\sup_{(t, x) \in \RR_+ \times \RR^d} |b(t, x)| + \sup_{(t, x) \in \RR_+ \times \RR^d} |\sigma(t, x)| + \sup_{\substack{(x, y) \in \RR^d \times \RR^d \\ t \in \RR_+, x \neq y}} \frac{|a(t, x) - a(t, y)}{|x-y|^\eta} \le L.$$

We define Gaussian-like densities \begin{align} p_c (t, x, x') &:= \left ( \frac{c}{2 \pi t} \right )^{d/2} \exp \left ( - \frac{c |x'-x|^2}{2t} \right ) && \forall c, t >0, \forall x,x' \in {\RR}^d, \\ G(y) &:= (2\pi)^{-d/2} \exp (-|y|^2/2) && \forall y \in {\RR}^d. \end{align}

Let $$x,x' \in \RR^d$$ and $$h, t>0$$. We are interested in bounding the absolute value of the quantity $$\alpha := \frac{1}{h} \left ( \frac{G ((\sqrt h \sigma (t, x))^{-1}(x'-x-b(t, x)h)}{\sqrt{h^d \det a(t, x)}} - \frac{G ((\sqrt h \sigma (t, x'))^{-1}(x'-x-b(t, x')h)}{\sqrt{h^d \det a(t, x')}} \right ).$$

At the end of page 21 of the paper On Some non Asymptotic Bounds for the Euler Scheme, the authors said that

...tedious but elementary computations involving the mean value theorem yield that $$\exists c>0, \exists C \ge 1$$ s.t. $$|\alpha| \le C h^{-1 + \eta/2} p_c (h, x, x').$$

The authors did not provide these "elementary computations", so I'm trying to derive above inequality.

My attempt: Let \begin{align} y_1 &:= (\sqrt h \sigma (t, x))^{-1}(x'-x-b(t, x)h \\ y_2 &:= (\sqrt h \sigma (t, x'))^{-1}(x'-x-b(t, x')h \\ z_1 &:= \sqrt{h^d \det a(t, x)} \\ z_2 &:= \sqrt{h^d \det a(t, x')}. \end{align}

Then \begin{align} |\alpha| &= \frac{1}{h} \left | \frac{G(y_1)}{z_1} - \frac{G(y_2)}{z_2} \right | \\ &= \frac{1}{h} \frac{|z_2 G(y_1) - z_1 G(y_2)|}{z_1 z_2}. \end{align}

By assumption (1.), we can lower bound $$z_1 z_2$$.

Could you elaborate on how to upper bound $$|z_2 G(y_1) - z_1 G(y_2)|$$?

Thank you so much for your help!

• Is $h$ bounded? Commented May 11, 2023 at 15:12
• @IosifPinelis $h$ is a constant... Commented May 11, 2023 at 15:15
• If $h$ is a constant, why can't the factor $h^{-1 + \eta/2}$ be absorbed into the factor $C$ in $C h^{-1 + \eta/2}$? Commented May 11, 2023 at 16:16
• In other words, my question is the following: For what values of $h>0$ do you want the inequality $|\alpha| \le C h^{-1 + \eta/2} p_c (h, x, x')$ to hold? Commented May 11, 2023 at 16:18
• Yes, I know how to do this, but the derivation is rather long and tedious. You just group like terms at each step. Begin with $|z_2 G(y_1)-z_1 G(y_2)|\le|z_1-z_2|G(y_1)+z_1|G(y_1)-G(y_2)|$. Commented May 12, 2023 at 20:27

$$\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand\la\lambda$$We have $$\begin{equation*} |\al|=h^{-1-d/2}|g(1)-g(0)|\le h^{-1-d/2}\sup_{s\in[0,1]}|g'(s)|, \end{equation*}$$ where $$\begin{equation*} g(s):=\frac{G((\sqrt h\si(s))^{-1}y(s))}{\sqrt{\det a(s)}}, \end{equation*}$$ $$y(s):=u-hb(s)$$, $$u:=x'-x$$, $$\si(s):=(1-s)\si(t,x)+s\si(t,x')$$, $$b(s):=(1-s)b(t,x)+sb(t,x')$$, $$a(s):=(1-s)a(t,x)+sa(t,x')$$. For $$s\in[0,1]$$,
$$\begin{equation*} \frac{g'(s)}{g(s)}=g_1+g_2+g_3, \end{equation*}$$ where $$\begin{equation*} g_1:=-\frac{\det a(1)-\det a(0)}{2a(s)}, \quad g_2:=-(b(1)-b(0))^\top a(s)^{-1}y(s), \end{equation*}$$ $$\begin{equation*} g_3:=\frac1h\, y(s)^\top a(s)^{-1}(a(1)-a(0))a(s)^{-1}y(s). \end{equation*}$$

We have $$\la^{-1}I\le a(s)\le\la I$$ and hence $$\begin{equation*} \la^{-1}\le |a(s)|\le\la, \quad \det a(s)\ge\la^{-d}, \end{equation*}$$ and, in view of this answer, $$|\det a(1)-\det a(0)|\le\la^{d-1}d\,|a(1)-a(0)|\le\la^{d-1}d\,L|u|^\eta$$. So, $$\begin{equation*} |g_1|\ll|u|^\eta; \end{equation*}$$ here and in what follows, $$A\ll B$$ and $$B\gg A$$ mean that $$A\le CB$$ for some real $$C>0$$ depending only on $$d,\la,L,T$$ -- given that $$h\in(0,T]$$. Next, $$\begin{equation*} |g_2|\le2L\la(|u|+hL)\ll|u|+h, \end{equation*}$$ $$\begin{equation*} |g_3|\le\frac1h\,\la^2 L|u|^\eta (|u|+hL)^2\ll\frac{|u|^{\eta+2}}h+|u|^\eta, \end{equation*}$$ since $$h\in(0,T]$$.

So, $$\begin{equation*} \frac{h^{1+d/2}|\al|}{g(s)}\ll |u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h. \end{equation*}$$

Next, $$\begin{equation*} g(s)\le\la^{d/2}\exp\Big(-\frac{|u|^2/2-(Lh)^2}{2\la h}\Big) \ll\exp\Big(-\frac{|u|^2}{4\la h}\Big) \ll h^{d/2}e^{-c|u|^2/h}p_c(h,x,x') \end{equation*}$$ for $$c:=\frac1{6\la}$$. So, $$\begin{equation*} \frac{|\al|}{h^{-1+\eta/2} p_c (h,x,x')} \ll \Big(|u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h\Big)h^{-\eta/2}e^{-c|u|^2/h}. \tag{10}\label{10} \end{equation*}$$ Further, $$\begin{equation*} |u|^\eta h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2}e^{-c|u|^2/h} \le\max_{w\ge0}w^{\eta/2}e^{-cw}\ll1; \tag{20}\label{20} \end{equation*}$$ $$\begin{equation*} |u| h^{-\eta/2}e^{-c|u|^2/h}=h^{(1-\eta)/2}(|u|^2/h)^{1/2}e^{-c|u|^2/h} \le T^{(1-\eta)/2}\max_{w\ge0}w^{1/2}e^{-cw}\ll1 \tag{30}\label{30} \end{equation*}$$ provided that $$\begin{equation*} \eta\le1; \tag{*}\label{*} \end{equation*}$$ $$\begin{equation*} h\,h^{-\eta/2}e^{-c|u|^2/h}\le h^{1-\eta/2}\ll1, \end{equation*}$$ again provided \eqref{*}; $$\begin{equation*} \frac{|u|^{\eta+2}}h h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2+1}e^{-c|u|^2/h} \le\max_{w\ge0}w^{\eta/2+1}e^{-cw}\ll1. \tag{40}\label{40} \end{equation*}$$

Now the desired inequality $$\begin{equation*} |\al|\ll h^{-1+\eta/2} p_c (h,x,x') \tag{50}\label{50} \end{equation*}$$ follows from \eqref{10}, \eqref{20}, \eqref{30}, and \eqref{40}, provided \eqref{*}. $$\quad\Box$$

Without condition \eqref{*}, inequality \eqref{50} will fail to hold in this generality.
E.g., suppose that for all $$t,x$$ we have $$\si(t,x)=I$$ and $$|b(t,x)|\in[0,1]$$, so that $$a(t,x)=I$$ and your conditions 1 and 2 hold with $$\la=1$$, $$L=2$$, and any real $$\eta>0$$. Suppose also that $$x=0$$, $$|x'|=\sqrt h$$, $$b(t,0)=0$$, and $$b(t,x')=-x'/|x'|$$. Then $$$$\frac{|\al|}{h^{-1+\eta/2} p_c (h,x,x')}\asymp h^{1/2-\eta/2}\to\infty$$$$ as $$h\downarrow0$$ if $$\eta>1$$. $$\quad\Box$$

• Thank you so much for your help! I really appreciate that you take time to write such an elaborate answer. Commented May 30, 2023 at 0:51