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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
- There is $\lambda >0$ such that
$$
\lambda^{-1} |\xi|^2 \le \xi^\top a(t, x) \xi \le \lambda |\xi|^2
\quad \forall (t, x, \xi) \in {\RR}_+ \times {\RR}^d \times {\RR}^d.
$$ - 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!
Best Answer
$\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand\la\lambda$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 \begin{equation} \frac{|\al|}{h^{-1+\eta/2} p_c (h,x,x')}\asymp h^{1/2-\eta/2}\to\infty \end{equation} as $h\downarrow0$ if $\eta>1$. $\quad\Box$