padgett-work
/
dissertation_work
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity
dissertation_work/Dissertation/Brownian_motion_monte_carlo...

486 lines
42 KiB
TeX
Raw Blame History

\chapter{Brownian Motion Monte Carlo}
\section{Brownian Motion Preliminaries}
We will present here some standard invariants of Brownian motions. The proofs are standard and can be found in for instance \cite{durrett2019probability} and \cite{karatzas1991brownian}.
\begin{lemma}[Markov property of Brownian motions]
Let $T \in \R$, $t \in [0,T]$, and $d \in \N$. Let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$ be a probability space. Let $\mathcal{W}_t: \lb 0, T \rb \times \Omega \rightarrow \R^d$ be a standard Brownian motion. Fix $s\in [0,\infty)$. Let $\mathfrak{W}_t = \mathcal{W}_{s+t}-\mathcal{W}_s$. Then $\mathfrak{W} = \left\{ \mathfrak{W}_t : t\in [0,\infty) \right\}$ is also a standard Brownian motion independent of $\mathcal{W}$.
\end{lemma}
\begin{proof}
We check against the Brownian motion axioms. First note that $\mathfrak{W}_0 = \mathcal{W}_{s+0} - \mathcal{W}_s = 0$ with $\mathbb{P}$-a.s.
Note that $t\mapsto \mathcal{W}_{s+t} - \mathcal{W}_s$ is $\mathbb{P}$-a.s. continuous as it is the difference of two functions that are also $\mathbb{P}$-a.s. continuous.
Note next that for $h\in \lp 0,\infty\rp$ it is the case that:
\begin{align}
\E\lb \mathfrak{W}_{t+h} -\mathfrak{W}_t\rb &= \E \lb \mathcal{W}_{s+t+h}-\mathcal{W}_{s+h} -\mathcal{W}_{s+t}+\mathcal{W}_s\rb \nonumber \\
&= \E \lb \mathcal{W}_{s+t+h}-\mathcal{W}_{s+t}\rb -\E \lb \mathcal{W}_{s+h}-\mathcal{W}_s\rb \nonumber \\
&=0-0 =0
\end{align}
We note finally that:
\begin{align}
\var \lb \mathfrak{W}_{t+h} -\mathfrak{M}_t\rb &= \var \lb \mathcal{W}_{s+t+h}-\mathcal{W}_{s} -\mathcal{W}_{s+t}+\mathcal{W}_s\rb \nonumber \\
&= \var \lb \mathcal{W}_{s+t+h}-\mathcal{W}_{s+t}\rb -\var \lb \mathcal{W}_{s}-\mathcal{W}_s\rb + \cancel{\cov \lp \mathcal{W}_{s+t+h}-\mathcal{W}_{s+t}, \mathcal{W}_{s+h}-\mathcal{W}_s\rp} \nonumber \\
&=h-0=h \nonumber
\end{align}
Finally note that two stochastic processes $\mathcal{W}$, $\mathcal{X}$ are independent whenever given a set of sample points $t_1,t_2,\hdots, t_n \in \lb 0,T\rb$ it is the case that the vectors $\lb \mathcal{W}_{t_1}, \mathcal{W}_{t_2},\hdots, \mathcal{W}_{t_n}\rb^\intercal$ and $\lb \mathcal{X}_{t_1},\mathcal{X}_{t_2},\hdots, \mathcal{X}_{t_n}\rb^\intercal$ are independent vectors.
That being the case note that the independent increments property of Brownian motions yields that, $\mathcal{W}_{s+t_1} - \mathcal{W}_s$, $\mathcal{W}_{s+t_2}-\mathcal{W}_s, \hdots, \mathcal{W}_{s+t_n}-\mathcal{W}_s$ is independent of $\mathcal{W}_{t_1},\mathcal{W}_{t_2},\hdots, \mathcal{W}_{t_n}$, i.e. $\mathfrak{W}$ and $\mathcal{W}$ are independent.
\end{proof}
\begin{lemma}[Independence of Brownian Motion]\label{iobm}
Let $T \in \lp 0,\infty\rp$. Let $\lp \Omega, \mathcal{F}, \mathbb{P}\rp$ be a probability space. Let $\mathcal{X}, \mathcal{Y}: \lb 0,T\rb \times \Omega \rightarrow \R^d$ be standard Brownian motions. It is then the case that they are independent of each other.
\end{lemma}
\begin{proof}
We say that two Brownian motions are independent of each of each other if given a sampling vector of times $\lp t_1,t_2,\hdots,t_n\rp$, the vectors $\lp \mathcal{X}_{t_1}, \mathcal{X}_{t_2},\hdots \mathcal{X}_{t_n}\rp$ and $\lp \mathcal{Y}_{t_1}, \mathcal{Y}_{t_2},\hdots, \mathcal{Y}_{t_n}\rp$ are independent. As such let $n\in \N$ and let $\lp t_1,t_2,\hdots t_n \rp$ be a vector or times with samples as given above. Consider now a new Brownian motion $\mathcal{X} - \mathcal{Y}$, wherein our samples are now $\lp \mathcal{X}_{t_1} - \mathcal{Y}_{t_1}, \mathcal{X}_{t_2}-\mathcal{Y}_{t_2}, \hdots, \mathcal{X}_{t_n} - \mathcal{Y}_{t_n} \rp$. By the independence property of Brownian motions, these increments must be independent of each other. Whence it is the case that the vectors $\lp \mathcal{X}_{t_1}, \mathcal{X}_{t_2},\hdots, \mathcal{X}_{t_n}\rp$ and $\lp \mathcal{Y}_{t_1}, \mathcal{Y}_{t_2},\hdots, \mathcal{Y}_{t_n}\rp$ are independent.
\end{proof}
\begin{lemma}[Scaling Invariance]
Let $T \in \R$, $t \in [0,T]$, and $d \in \N$. Let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$ be a probability space. Let $\mathcal{W}_t: \lb 0, T \rb \times \Omega \rightarrow \R^d$ be a standard Brownian motion. Let $a \in \R \setminus \{ 0\}$. It is then the case that $\mathcal{X}_t \coloneqq \frac{1}{a} \mathcal{W}_{a^2\cdot t}$ is also a standard Brownian motion.
\end{lemma}
\begin{proof}
We check against the Brownian motion axioms. Note for instance that the function $t \mapsto \mathcal{X}_t$ is a product of a constant with a function that is $\mathbb{P}$-a.s. continuous yielding a function that is also $\mathbb{P}$-a.s. continuous.
Note also for instance that $\mathcal{X}_0 = \frac{1}{a} \cdot \mathcal{W}_{a^2 \cdot 0} = 0$ with $\mathbb{P}$-a.s.
Note that for all $h \in \lp 0,\infty\rp$, and $t\in \lb 0,T\rb$ it is the case that:
\begin{align}
\E \lb \mathcal{X}_{t+h} - \mathcal{X}_t\rb &= \E \lb \frac{1}{a}\mathcal{W}_{a^2 \cdot \lp t+h\rp} - \frac{1}{a}\mathcal{W}_{a^2 \cdot t}\rb \nonumber \\
&=\frac{1}{a} \E \lb \mathcal{W}_{a^2\cdot \lp t+h\rp} - \mathcal{W}_{a^2\cdot t}\rb \nonumber \\
&=0\nonumber
\end{align}
Note that for all $h \in \lp 0,\infty\rp$, and $t\in \lb 0,T\rb$ it is the case that:
\begin{align}
\var\lb \mathcal{X}_{t+h } - \mathcal{X}_t\rb &= \var\lb \frac{1}{a}\mathcal{W}_{a^2\cdot \lp t+h\rp} - \frac{1}{a}\mathcal{W}_{a^2\cdot t}\rb \nonumber \\
&=\frac{1}{a^2}\var\lb\mathcal{W}_{a^2\cdot \lp t+h\rp} - \mathcal{W}_{a^2\cdot t}\rb \nonumber\\
&= \frac{1}{\cancel{a^2}}\cancel{a^2} \lp \cancel{t}+h-\cancel{t}\rp \nonumber\\
&=h
\end{align}
Finally note that for $t \in \lb 0,T\rb$ and $s \in \lb 0,t\rp$ it is the case that $\mathcal{W}_{a^2 \cdot t} - \mathcal{W}_{a^2 \cdot s}$ is independent of $\mathcal{W}_{a^2\cdot s}$. Whence it is also the case that $\mathcal{X}_t-\mathcal{X}_s$ is independent of $\mathcal{X}_s$.
\end{proof}
\begin{lemma}[Summation of Brownian Motions]
Let $T \in \R$, $t \in [0,T]$ and $d \in \N$. Let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$ be a probability space. Let $\mathcal{W}_t, \mathcal{X}_t: \lb 0,T \rb \times \Omega \rightarrow \R^d$ be a standard independent Brownian motions. It is then the case that the process $\mathcal{Y}_t$ defined as $\mathcal{Y}_t = \frac{1}{\sqrt{2}}\lp \mathcal{W}_t + \mathcal{X}_t \rp$ is also a standard Brownian motion.
\end{lemma}
\begin{proof}
Note that $t \mapsto \frac{1}{\sqrt{2}}\lp \mathcal{W}_t+\mathcal{X}_t\rp$ is $\mathbb{P}$-a.s. continuous as it is the linear combination of two functions that are also $\mathbb{R}$-a.s. continuous.
Note also that $\mathcal{Y}_0 = \frac{1}{\sqrt{2}}\lp \mathcal{W}_0+\mathcal{X}_0\rp = 0+0=0$ with $\mathbb{P}$-a.s.
Note that for all $h \in \lp 0,\infty\rp$ and $t \in \lb t,T\rb$ it is the case that:
\begin{align}
\E\lb \frac{1}{\sqrt{2}}\lp \mathcal{Y}_{t+h} - \mathcal{Y}_t\rp \rb &= \E \lb \frac{1}{\sqrt{2}} \lp\mathcal{W}_{t+h}+\mathcal{X}_{t+h} - \mathcal{W}_t-\mathcal{X}_t \rp\rb \nonumber \\
&= \frac{1}{\sqrt{2}}\E \lb \mathcal{W}_{t+h}-\mathcal{W}_t\rb + \frac{1}{\sqrt{2}}\E \lb \mathcal{X}_{t+h}-\mathcal{X}_t\rb \nonumber \\
&=0 \nonumber
\end{align}
Note that for all $h \in \lp 0,\infty\rp$, and $t\in \lb 0,T\rb$ it is the case that:
\begin{align}
\var \lb \frac{1}{\sqrt{2}}\lp\mathcal{Y}_{t+h} - \mathcal{Y}_{t}\rp\rb &= \var \lb \frac{1}{\sqrt{2}}\lp \mathcal{W}_{t+h}+\mathcal{X}_{t+h} - \mathcal{W}_t-\mathcal{X}_t\rp \rb \nonumber \\
&=\var \lb \frac{1}{\sqrt{2}}\lp \mathcal{W}_{t+h} - \mathcal{W}_t\rp + \frac{1}{\sqrt{2}}\lp \mathcal{X}_{t+h}-\mathcal{X}_t\rp\rb \nonumber\\
&= \frac{1}{2}\var \lb \mathcal{W}_{t+h}-\mathcal{W}_t\rb +\frac{1}{2}\var\lb \mathcal{X}_{t+h}-\mathcal{X}_t\rb + \cancel{\cov \lp \mathcal{W},\mathcal{X}\rp} \nonumber \\
&= h \nonumber
\end{align}
\end{proof}
\begin{definition}[Of $\mathfrak{k}$]\label{def:1.17}
Let $p \in [2,\infty)$. We denote by $\mathfrak{k}_p \in \R$ the real number given by $\mathfrak{k}:=\inf \{ c\in \R \}$ where it holds that for every probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and every random variable $\mathcal{X}: \Omega \rightarrow \R$ with $\E[|\mathcal{X}|] < \infty$ that $\lp \E \lb \lv \mathcal{X} - \E \lb \mathcal{X} \rb \rp^p \rb \rp ^{\frac{1}{p}} \leqslant c \lp \E \lb \lv \mathcal{X} \rv^p \rb \rp ^{\frac{1}{p}}.$
\end{definition}
\begin{definition}[Primary Setting]\label{primarysetting} Let $d,m \in \mathbb{N}$, $T, \mathfrak{L},p \in [0,\infty)$, $\mathfrak{p} \in [2,\infty)$ $\mathfrak{m} = \mathfrak{k}_{\mathfrak{p}}\sqrt{\mathfrak{p}-1}$, $\Theta = \mathbb{Z}$, $g \in C(\mathbb{R}^d,\mathbb{R})$, assume for all $t \in [0,T],x\in \mathbb{R}^d$ that:
\begin{align}\label{(2.1.2)}
\max\{|g(x)|\} \leqslant \mathfrak{L} \lp 1+\|x\|_E^p \rp
\end{align}
and let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space. Let $\mathcal{W}^{\theta}: [0,T] \times \Omega \rightarrow \mathbb{R}^d$, $\theta \in \Theta$ be independent standard Brownian motions, let $u \in C([0,T] \times \mathbb{R}^d,\mathbb{R})$ satisfy for all $t \in [0,T]$, $x\in \mathbb{R}^d$, that $\mathbb{E}[|g(x+\mathcal{W}^0_{T-t})|] < \infty$ and:
\begin{align}\label{(1.12)}
u(t,x) &= \mathbb{E} \lb g \lp x+\mathcal{W}^0_{T-t} \rp \rb
\end{align}
and let let $U^\theta:[0,T] \times \mathbb{R}^d \times \Omega \rightarrow \mathbb{R}$, $\theta \in \Theta$ satisfy, $\theta \in \Theta$, $t \in [0,T]$, $x\in \mathbb{R}^d$, that:
\begin{align}\label{(2.1.4)}
U^\theta_m(t,x) = \frac{1}{m}\left[\sum^{m}_{k=1}g\left(x+\mathcal{W}^{(\theta,0,-k)}_{T-t}\right)\right]
\end{align}
\end{definition}
\begin{lemma} \label{lemma1.1}
Assume Setting \ref{primarysetting} then:
\begin{enumerate}[label = (\roman*)]
\item it holds for all $n\in \N_0$, $\theta \in \Theta$ that $U^\theta:[0,T] \times \mathbb{R}^d\times \Omega \rightarrow \mathbb{R}$ is a continuous random field.
\item it holds that for all $\theta \in \Theta$ that $\sigma \lp U^\theta \rp \subseteq \sigma \lp \lp \mathcal{W}^{(\theta, \mathcal{V})}\rp_{\mathcal{V} \in \Theta}\rp $.
\item it holds that $\lp U^\theta \rp_{\theta \in\Theta}$,$\lp \mathcal{W}^\theta \rp_{\theta \in \Theta}$, are independent.
\item it holds for all $n,m \in $, $i,k,\mathfrak{i},\mathfrak{k}\in \mathbb{Z}$, with $(i,k) \neq (\mathfrak{i},\mathfrak{k})$ that $(U^{(\theta,i,k)})_{\theta \in \Theta}$ and $\left(U^{(\theta,\mathfrak{i},\mathfrak{k})}\right)_{\theta \in \Theta}$ are independent and,
\item it holds that $\lp U^\theta \rp_{\theta \in \Theta}$ are identically distributed random variables.
\end{enumerate}
\end{lemma}
\begin{proof} For (i) Consider that $\mathcal{W}^{(\theta,0,-k)}_{T-t}$ are continuous random fields and that $g\in C(\mathbb{R}^d,\mathbb{R})$, we have that $U^\theta(t,x)$ is the composition of continuous functions with $m > 0$ by hypothesis, ensuring no singularities. Thus $U^\theta: [0,T] \times \mathbb{R}^d\times \Omega \rightarrow \mathbb{R}$.
\medskip
For (ii) observe that for all $\theta \in \Theta$ it holds that $\mathcal{W}^\theta$ is $\mathcal{B} \lp \lb 0, T \rb \otimes \sigma \lp W^\theta \rp \rp /\mathcal{B}\lp \mathbb{R}^d \rp$-measurable, this, and induction on prove item (ii).
\medskip
Moreover observe that item (ii) and the fact that for all $\theta \in \Theta$ it holds that $\lp\mathcal{W}^{\lp \theta, \vartheta\rp}_{\vartheta \in \Theta}\rp$, $\mathcal{W}^\theta$ are independently establish item (iii).
\medskip
Furthermore, note that (ii) and the fact that for all $i,k,\mathfrak{i},\mathfrak{k} \in \mathbb{Z}$, $\theta \in \Theta$, with $(i,k) \neq (\mathfrak{i},\mathfrak{k})$ it holds that $\lp\mathcal{W}^{\lp\theta, i,k,\vartheta\rp}\rp_{\vartheta \in \Theta}$ and $\lp\mathcal{W}^{\lp\theta,\mathfrak{i},\mathfrak{k},\vartheta\rp}\rp_{\vartheta \in \Theta}$ are independent establish item (iv).
\medskip
Hutzenhaler \cite[Corollary~2.5 ]{hutzenthaler_overcoming_2020} establish item (v). This completes the proof of Lemma 1.1.
\end{proof}
\begin{lemma}\label{lem:1.20} Assume Setting \ref{primarysetting}. Then it holds for $\theta \in \Theta$, $s \in [0,T]$, $t\in [s,T]$, $x\in \mathbb{R}^d$ that:
\begin{align}
\mathbb{E}\lb \lv U^\theta \lp t,x+\mathcal{W}^\theta_{t-s}\rp \rv \rb +\mathbb{E}\lb \lv g \lp x+\mathcal{W}^\theta_{t-s}\rp \rv \rb + \int^T_s \E \lb \lv U^\theta \lp r,x+\mathcal{W}^\theta_{r-s} \rp \rv \rb dr < \infty
\end{align}
\end{lemma}
\begin{proof}
Note that (\ref{(2.1.2)}), the fact that for all $r,a,b \in [0,\infty)$ it holds that $(a+b)^r \leqslant 2^{\max\{r-1,0\}}(a^r+b^r)$, and the fact that for all $\theta \in \Theta$ it holds that $\mathbb{E}\lb \|\mathcal{W}^\theta_T\|\rb < \infty$, assure that for all $s \in [0,T]$, $t\in[s,T]$, $\theta \in \Theta$ it holds that:
\begin{align}\label{(2.1.6)}
\mathbb{E}\lb \lv g(x+\mathcal{W}^\theta_{t-s})\rv \rb &\leqslant \mathbb{E}\lb\mathfrak{L}\lp 1+\|x+\mathcal{W}^\theta_{t-s}\|_E^p\rp\rb \nonumber\\
&\leqslant \mathfrak{L}\lb 1+2^{\max\{p-1,0\}}\lp \|x\|_E^p+\mathbb{E} \lb \left\|\mathcal{W}^\theta_T\right\|_E^p \rb \rp\rb<\infty
\end{align}
\label{eq:1.4}
We next claim that for all $s\in [0,T]$, $t\in[s,T]$, $\theta \in \Theta$ it holds that:
\begin{align}\label{(1.17)}
\mathbb{E}\lb \lv U^\theta \lp t,x+\mathcal{W}^\theta_{t-s} \rp \rv \rb+ \int^T_s \mathbb{E}\lb \lv U^\theta \lp r,x+\mathcal{W}^\theta_{r-s}\rp \rv \rb dr < \infty
\end{align}
To prove this claim observe the triangle inequality and (\ref{(2.1.4)}), demonstrate that for all $s\in[0,T]$, $t\in[s,T]$, $\theta \in \Theta$, it holds that:
\begin{align}\label{(1.18)}
\mathbb{E}\lb \lv U^\theta \lp t,x+\mathcal{W}^\theta_{t-s}\rp \rv \rb \leqslant \frac{1}{m}\left[ \sum^{m}_{i=1}\mathbb{E}\lb \lv g \lp x+\mathcal{W}^\theta_{t-s}+\mathcal{W}^{(\theta,0,-i)}_{T-t} \rp \rv \rb \rb
\end{align}
Now observe that (\ref{(2.1.6)}) and the fact that $(W^\theta)_{\theta \in \Theta}$ are independent imply that for all $s \in [0,T]$, $t\in [s,T]$, $\theta \in \Theta$, $i\in \mathbb{Z}$ it holds that:
\begin{align}\label{(1.19)}
\mathbb{E}\lb \lv g \lp x+\mathcal{W}^\theta_{t-s}+\mathcal{W}^{(\theta,0,i)}_{T-t} \rp \rv \rb = \mathbb{E}\lb \lv g \lp x+\mathcal{W}^\theta_{(t-s)+(T-t)}\rp \rv \rb = \mathbb{E}\lb \lv g \lp x+\mathcal{W}^\theta_{T-s}\rp \rv \rb <\infty
\end{align}
\medskip
Combining (\ref{(1.18)}) and (\ref{(1.19)}) demonstrate that for all $s \in [0,T]$, $t\in[s,T]$, $\theta \in \Theta$ it holds that:
\begin{align}\label{(1.20)}
\mathbb{E}\lb \lv U^\theta(t,x+\mathcal{W}^\theta_{t-s})\rv \rb < \infty
\end{align}
Finally observe that for all $s\in [0,T]$ $\theta \in \Theta$ it holds that:
\begin{align}\label{(1.21)}
\int^T_s \mathbb{E}\lb \lv U^\theta \lp r,x+\mathcal{W}^\theta_{r-s} \rp \rv \rb &\leqslant \lp T-s \rp \sup_{r\in [s,T]} \mathbb{E} \lb \lv U^\theta \lp r,x+\mathcal{W}^\theta_{r-s}\rp \rv \rb < \infty
\end{align}
Combining (\ref{(1.16)}), (\ref{(1.20)}), and (\ref{(1.21)}) completes the proof of Lemma \ref{lem:1.20}.
\end{proof}
\begin{corollary}\label{cor:1.20.1} Assume Setting \ref{primarysetting}, then we have:
\begin{enumerate}[label = (\roman*)]
\item it holds that $t \in [0,T],x\in \mathbb{R}^d$ that:
\begin{align}
\mathbb{E}\lb \lv U^0 \lp t,x \rp \rv \rb + \mathbb{E}\lb \lv g \lp x+\mathcal{W}^{(0,0,-1)}_{T-t} \rp \rv \rb < \infty
\end{align}
\item it holds that $t\in [0,T],x\in \mathbb{R}^d$ that:
\begin{align}
\mathbb{E}\lb U^0\lp t,x \rp \rb = \mathbb{E} \lb g \lp x+\mathcal{W}^{(0,0,-1)}_{T-t}\rp\rb
\end{align}
\end{enumerate}
\end{corollary}
\begin{proof}
(i) is a restatement of Lemma \ref{lem:1.20} in that for all $t\in [0,T]$:
\begin{align}
&\mathbb{E}\left[ \left| U^0\left( t,x \right) \right| \right] + \mathbb{E} \left[ \left|g \left(x+\mathcal{W}^{(0,0,-1)}_{T-t}\right)\right|\right] \nonumber\\
&<\mathbb{E} \left[ \left|U^\theta \lp t,x+\mathcal{W}^\theta_{t-s} \rp \right| \right] +\mathbb{E}\left[ \left|g \left(x+\mathcal{W}^\theta_{t-s}\right) \right| \right]+ \int^T_s \mathbb{E}\lb \lv U^\theta \lp r,x+\mathcal{W}^\theta_{r-s} \rp \rv \rb dr \nonumber\\
&< \infty
\end{align}
Furthermore (ii) is a restatement of (\ref{(1.14)}) with $\theta = 0$, $m=1$, and $k=1$. This completes the proof of Corollary \ref{cor:1.20.1}.
\end{proof}
\section{Monte Carlo Approximations}
\begin{lemma}\label{lem:1.21}Let $p \in (2,\infty)$, $n\in \mathbb{N}$, let $(\Omega, \mathcal{F}, \mathbb{P})$, be a probability space and let $\mathcal{X}_i: \Omega \rightarrow \mathbb{R}$, $i \in \{1,2,...,n\}$ be i.i.d. random variables with $\mathbb{E}[|\mathcal{X}_1|]<\infty$. Then it holds that:
\begin{align}
\lp\E \lb \lv \E \lb \mathcal{X}_1 \rb-\frac{1}{n} \lp \sum^n_{i=1} \mathcal{X}_i \rp \rv^p \rb \rp^{\frac{1}{p}} \leqslant \lb \frac{p-1}{n}\rb ^{\frac{1}{2}}\left(\E\lb \lv \mathcal{X}_1-\E \lb \mathcal{X}_1 \rb \rv^p \rp \rb^{\frac{1}{p}}
\end{align}
\end{lemma}
\begin{proof}
The hypothesis that for all $i \in \{1,2,...,n\}$ it holds that $\mathcal{X}_i:\Omega \rightarrow \mathbb{R}$ are i.i.d. random variables ensures that:
\begin{align}
\E \lb \lv \E \lb \mathcal{X}_1\rb - \frac{1}{n} \lp \sum^n_{i=1} \mathcal{X}_i \rp \rv ^p \rb
= \E \lb \lv \frac{1}{n} \lp \sum^n_{i=1} \lp \E \lb \mathcal{X}_1 \rb - \mathcal{X}_i \rp \rp \rv ^p \rb = \frac{1}{n^p} \E \lb \lv \sum^n_{i=1} \lp \E \lb \mathcal{X}_i \rb - \mathcal{X}_i \rp \rv^p \rb
\end{align}
This combined with the fact that for all $i \in \{1,2,...,n\}$ it is the case that $\mathcal{X}_i: \Omega \rightarrow \R$ are i.i.d. random variables and e.g. \cite[Theorem~2.1]{rio_moment_2009} (with $p \curvearrowleft p$, $ ( S_i )_{i \in \{0,1,...,n\}} \curvearrowleft ( \sum^i_{k=1} ( \E [ X_k ] - X_k))$, $( X_i )_{i \in \{1,2,...,n\}} \curvearrowleft ( \E [ X_i ] - X_i )_{i \in \{1,2,...,n\}}$ in the notation of \cite[Theorem~2.1]{rio_moment_2009} ensures that:
\begin{align}
\lp \E \lb \lv \E \lb \mathcal{X}_1 \rb - \frac{1}{n} \lp \sum^n_{i=1} \mathcal{X}_i \rp
\rv ^p \rb\rp ^{\frac{2}{p}} &= \frac{1}{n^2} \lp \E \lb \lv
\sum^n_{i=1} \lp \E \lb \mathcal{X}_i \rb - \mathcal{X}_i \rp \rv ^p \rb \rp ^{\frac{2}{p}}
\nonumber\\
&\leqslant \frac{p-1}{n^2} \lb \sum^n_{i=1} \lp \E \lb \lv \E \lb \mathcal{X}_i \rb -\mathcal{X}_i \rv^p \rb \rp ^{\frac{2}{p}} \rb \nonumber\\
&= \frac{p-1}{n^2} \lb n \lp \E \lb \lv \E \lb \mathcal{X}_1 \rb - \mathcal{X}_1 \rv ^p \rb \rp^{\frac{2}{p}} \rb \\
&= \frac{p-1}{n} \lp \E \lb \lv \E \lb \mathcal{X}_1 \rb -\mathcal{X}_1\rv ^p \rb \rp ^{\frac{2}{p}}
\end{align}
This completes the proof of the lemma.
\end{proof}
\begin{corollary}\label{corollary:1.11.1.}
Let $p\in [2,\infty)$, $n \in \N$, let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$ be a probability space, and let $\mathcal{X}_i: \Omega \rightarrow \R$, $i \in \{1,2,...,n\}$ be i.i.d random variables with $\E\lb \lv \mathcal{X}_1 \rv \rb < \infty$. Then it holds that:
\begin{align}\label{(1.26)}
\lp \E \lb \lv \E \lb \mathcal{X}_1 \rb - \frac{1}{n}\lp \sum^n_{i=1} \mathcal{X}_i \rp \rv ^p \rb \rp^{\frac{1}{p}} \leqslant \lb \frac{p-1}{n} \rb ^{\frac{1}{2}} \lp \E \lb \lv \mathcal{X}_1 - \E \lb \mathcal{X}_1 \rb \rv ^p \rb \rp ^{\frac{1}{p}}
\end{align}
\end{corollary}
\begin{proof}
Observe that e.g. \cite[Proposition~2.3]{grohsetal} and Lemma \ref{lem:1.21} establish (\ref{(1.26)}).
\end{proof}
\begin{corollary}\label{cor:1.22.2}
Let $p \in [2,\infty)$, $n\in \N$, let $(\Omega, \mathcal{F}, \mathbb{P})$, be a probability space, and let $\mathcal{X}_i: \Omega \rightarrow \R$, $i \in \{1,2,...,n\}$, be i.i.d. random variables with $\E[|\mathcal{X}_1|] < \infty$, then:
\begin{align}
\lp \E \lb \lv \E \lb \mathcal{X}_1\rb - \frac{1}{n} \lp \sum^n_{i=1} \mathcal{X}_i \rp \rv ^p \rb \rp ^{\frac{1}{p}} \leqslant \frac{\mathfrak{k}_p \sqrt{p-1}}{n^{\frac{1}{2}}} \lp \E \lb \lv \mathcal{X}_1 \rv^p \rb \rp ^{\frac{1}{p}}
\end{align}
\end{corollary}
\begin{proof}
This a direct consequence of Definition \ref{def:1.17} and Corollary \ref{corollary:1.11.1.}.
\end{proof}
\section{Bounds and Covnvergence}
\begin{lemma}\label{lem:1.21} Assume Setting \ref{def:1.18}. Then it holds for all $t\in [0,T]$, $x\in \mathbb{R}^d$
\begin{align}
&\left(\E\left[\left|U^0(t,x+\mathcal{W}^0_t)-\E \left[U^0 \left(t,x+\mathcal{W}^0_t \right)\right]\right|^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}} \nonumber\\
&\leqslant \frac{\mathfrak{m}}{m^{\frac{1}{2}}} \left[\left(\E\left[ \lv g \lp x+\mathcal{W}^0_T \rp \rv^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}}\right]
\end{align}
\end{lemma}
\begin{proof} For notational simplicity, let $G_k: [0,T] \times \mathbb{R}^d \times \Omega \rightarrow \mathbb{R}$, $k\in \mathbb{Z}$, satisfy for all $k\in \mathbb{Z}$, $t\in[0,T]$, $x\in \mathbb{R}^d$ that:
\begin{align}
G_k(t,x) = g\left(x+\mathcal{W}^{(0,0,-k)}_{T-t}\right)
\end{align}
\medskip
Observe that the hypothesis that $(\mathcal{W}^\theta)_{\theta \in \Theta}$ are independent Brownian motions and the hypothesis that $g \in C(\mathbb{R}^d,\mathbb{R})$ assure that for all $t \in [0,T]$,$x\in \mathbb{R}^d$ it holds that $(G_k(t,x))_{k\in \mathbb{Z}}$ are i.i.d. random variables. This and Corollary \ref{cor:1.22.2} (applied for every $t\in [0,T]$, $x\in \mathbb{R}^d$ with $p \curvearrowleft \mathfrak{p}$, $n \curvearrowleft m$, $(X_k)_{k\in \{1,2,...,m\}} \curvearrowleft (G_k(t,x))_{k\in \{1,2,...,m\}}$), with the notation of Corollary \ref{cor:1.22.2} ensure that for all $t\in [0,T]$, $x \in \mathbb{R}^d$, it holds that:
\begin{align}
\left( \E \left[ \left| \frac{1}{m} \left[ \sum^{m}_{k=1} G_k(t,x) \right] - \E \left[ G_1(t,x) \right] \right| ^\mathfrak{p} \right] \right)^{\frac{1}{\mathfrak{p}}} \leqslant \frac{\mathfrak{m}}{m^{\frac{1}{2}}}\left(\E \left[|G_1(t,x)|^\mathfrak{p} \right] \right)^{\frac{1}{\mathfrak{p}}}
\end{align}
\medskip
Combining this, with (1.16), (1.17), and item (ii) of Corollary \ref{cor:1.20.1} yields that:
\begin{align}
&\left(\E\left[\left|U^0(t,x) - \E \left[U^0(t,x)\right]\right|^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}} \nonumber\\
&= \left(\E \left[\left|\frac{1}{m}\left[\sum^{m}_{k=1}G_k(t,x)\right]- \E \left[G_1(t,x)\right]\right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}} \\
&\leqslant \frac{\mathfrak{m}}{m^{\frac{1}{2}}}\left(\E \left[\left| G_1(t,x)\right| ^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}} \\
&= \frac{\mathfrak{m}}{m^{\frac{1}{2}}} \left[\left(\E \left[\left|g\left(x+\mathcal{W}^1_{T-t}\right)\right|^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}}\right]
\end{align}
This and the fact that $\mathcal{W}^0$ has independent increments ensure that for all $n\in $, $t\in [0,T]$, $x\in \mathbb{R}^d$ it holds that:
\begin{align}
\left(\E \left[\left| U^0 \left(t,x+\mathcal{W}^0_t\right) - \E \left[U^0 \left(t,x+\mathcal{W}^0_t\right)\right]\right|^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}} \leqslant \frac{\mathfrak{m}}{m^{\frac{1}{2}}} \left[\left(\E \left[\left| g \left(x+\mathcal{W}^0_T\right)\right|^\p\right]\right)^{\frac{1}{\mathfrak{p}}} \right]
\end{align}
This completes the proof of Lemma \ref{lem:1.21}.
\end{proof}
\begin{lemma}\label{lem:1.22} Assume Setting \ref{primarysetting}. Then it holds for all, $t\in [0,T]$, $x\in \mathbb{R}^d$ that:
\begin{align}
\left(\E \left[ \left| U^0 \left(t,x+\mathcal{W}^0_t\right) - u \left(t,x+\mathcal{W}^0_t\right) \right|^\mathfrak{p} \right]\right)^\frac{1}{\mathfrak{p}} \leqslant \left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right) \left( \E \left[\left| g \left(x+\mathcal{W}^0_T\right) \right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}}
\end{align}
\end{lemma}
\begin{proof}
Observe that from Corollary \ref{cor:1.20.1} item (ii) we have:
\begin{align}
\E\left[U^0(t,x)\right] = \E \left[ g \left(x+\mathcal{W}^{(0,0,-1)}_{T-t}\right) \right]
\end{align}
This and (\ref{(1.12)}) ensure that:
\begin{align}
u(t,x) - \E \left[U^0(t,x) \right] &= 0 \nonumber \\
\E \lb U^0(t,x) \rb - u \lp t,x \rp &= 0
\end{align}
This, and the fact that $\mathcal{W}^0$ has independent increments, assure that for all, $t\in [0,T]$, $x\in \mathbb{R}^d$, it holds that:
\begin{align}
\left(\E \left[\left| \E \lb U^0 \lp t,x+\mathcal{W}^0_t \rp\right] - u \lp t,x+\mathcal{W}^0_t \rp\right| ^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}} = 0 \leqslant \left(\E \left[ \lv u \lp t,x+\mathcal{W}^0_t \rp \rv^\p\right]\right)
\end{align}
This along with (\ref{(1.12)}) ensure that:
\begin{align}
\left(\E \left[\left| \E \left[U^0 \lp t,x+\mathcal{W}^0_t \rp \right] - u \lp t,x+\mathcal{W}^0_t \rp \right| ^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}} = 0 \leqslant \left( \E \left[\left| g \left(x+\mathcal{W}^0_T\right) \right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}}
\end{align}
Notice that the triangle inequality gives us:
\begin{align}
\left(\E \left[ \left| U^0 \left(t,x+\mathcal{W}^0_t\right) - u \left(t,x+\mathcal{W}^0_t\right) \right|^\mathfrak{p} \right]\right)^\frac{1}{\mathfrak{p}} \leqslant \left(\E \left[\left| U^0(t,x+W^0_t) - \E \left[U^0(t,x+\mathcal{W}^0_t)\right]\right|^\p\right]\right)^{\frac{1}{\p}} \nonumber \\
+\left(\E \left[\left| \E \left[U^0 \lp t,x+\mathcal{W}^0_t \rp \right]-u \lp t,x+\mathcal{W}^0_t \rp\right|^\p\right]\right)^{\frac{1}{\p}}
\end{align}
This, combined with (1.26), (1.21), the independence of Brownian motions, gives us:
\begin{align}
\left(\E \left[ \left| U^0 \left(t,x+\mathcal{W}^0_t\right) - u \left(t,x+\mathcal{W}^0_t\right) \right|^\mathfrak{p} \right]\right)^\frac{1}{\mathfrak{p}} &\leqslant \left(\frac{\mathfrak{m}}{m^{\frac{1}{2}}}\right) \left( \E \left[\left| g \left(x+\mathcal{W}^0_T\right) \right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}} \nonumber \\
&= \left(\frac{\mathfrak{m}}{m^{\frac{1}{2}}}\right) \left( \E \left[\left| g \left(x+\mathcal{W}^0_T\right) \right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}}
\end{align}
This completes the proof of Lemma \ref{lem:1.22}.
\end{proof}
\begin{lemma}\label{lem:1.25} Assume Setting \ref{primarysetting}. Then it holds for all $t\in [0,T]$, $x\in \mathbb{R}^d$ that:
\begin{align}
\left( \E \left[ \left| U^0 \left(t,x+\mathcal{W}^0_t \right) - u \left( t, x+\mathcal{W}^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} \leqslant \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_s \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
\end{lemma}
\medskip
\begin{proof} Observe that Lemma \ref{lem:1.22} ensures that:
\begin{align}\label{(1.46)}
\left( \E \left[ \left| U^0 \left(t,x+\mathcal{W}^0_t \right) - u \left( t, x+\mathcal{W}^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} \leqslant \left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\E \left[\left| g \left(x+\mathcal{W}^0_T\right)\right|^\p\right]\right)^\frac{1}{\p}
\end{align}
Observe next that (\ref{(1.12)}) ensures that:
\begin{align}\label{(1.47)}
\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\E \left[\left| g \left(x+\mathcal{W}^0_T\right)\right|^\p\right]\right)^\frac{1}{\p} \leqslant \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_T \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
Which in turn yields that:
\begin{align}\label{(1.48)}
\mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_T \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p} \leqslant \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_s \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
Combining \ref{(1.46)}, \ref{(1.47)}, and \ref{(1.48)} yields that:
\begin{align}
\left( \E \left[ \left| U^0 \left(t,x+\mathcal{W}^0_t \right) - u \left( t, x+\mathcal{W}^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} &\leqslant \left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\E \left[\left| g \left(x+\mathcal{W}^0_T\right)\right|^\p\right]\right)^\frac{1}{\p} \nonumber\\
&\les\mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\sup_{s\in[0,T]}\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_s \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
This completes the proof of Lemma \ref{lem:1.25}.
\end{proof}
\begin{corollary}\label{cor:1.25.1} Assume Setting \ref{primarysetting}. Then it holds for all $t\in[0,T]$, $x\in \R^d$ that:
\begin{align}
\left( \E \left[ \left| U^0 \lp t,x \rp -u(t,x) \rv ^\p \right] \right) ^{\frac{1}{\p}} \leqslant \mathfrak{L} \left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}} \right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \right\| x+\mathcal{W}^0_s \left\|_E^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
\end{corollary}
\begin{proof} Observe that for all $t \in [0,T-\mft]$ and $\mft \in [0,T]$, and the fact that $W^0$ has independent increments it is the case that:
\begin{align}
u(t+\mft,x) = \E \left[g \left(x+\mathcal{W}^0_{T-(t+\mft)}\right)\right] = \E \left[g \left(x+\mathcal{W}^0_{(T-\mft)-t)}\right)\right]
\end{align}
It is also the case that:
\begin{align*}
U^\theta(t+\mft,x) = \frac{1}{m} \left[\sum^{m}_{k=1} g \left(x+\mathcal{W}^{(\theta,0,-k)}_{T-(t+\mft)}\right)\right] = \frac{1}{m} \left[\sum^{m}_{k=1} g \left(x+\mathcal{W}^{(\theta,0,-k)}_{(T-\mft)-t}\right)\right]
\end{align*}
\medskip
Then, applying Lemma \ref{lem:1.25}, applied for all $\mft \in [0,T]$, with $\mathfrak{L} \curvearrowleft \mathfrak{L}$, $p \curvearrowleft p$, $\mathfrak{p} \curvearrowleft \mathfrak{p}$, $T \curvearrowleft (T-\mft)$ is such that for all $\mft \in [0,T]$, $t \in [0,T-\mft]$, $x \in \R^d$ we have:
\begin{align}
&\left( \E \left[ \left| U^0 \left(t+\mft,x+\mathcal{W}^0_t \right) - u \left( t+\mft, x+\mathcal{W}^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} \nonumber \\
&\leqslant \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\sup_{s \in [0,T-\mft]}\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_s \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p} \nonumber \\
&\leqslant \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_s \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
Thus we get for all $\mft \in [0,T]$, $x\in \R^d$, $n \in $:
\begin{align}
\left( \E \left[ \left| U^0 \left(\mft,x \right) - u \left(\mft, x \right) \right|^\p \right] \right)^{\frac{1}{\p}} &= \left( \E \left[ \left| U^0 \left(\mft,x+\mathcal{W}^0_0 \right) - u \left(\mft, x+\mathcal{W}^0_0 \right) \right|^\p \right] \right)^{\frac{1}{\p}}\nonumber\\
&\leqslant \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{1}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left\| x+\mathcal{W}^0_s \right\|_E^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
This completes the proof of Corollary \ref{cor:1.25.1}.
\end{proof}
\begin{theorem}\label{tentpole_1} Let $T,L,p,q, \mathfrak{d} \in [0,\infty), m \in \mathbb{N}$, $\Theta = \bigcup_{n\in \mathbb{N}} \Z^n$, let $g_d\in C(\R^d,\R)$, and assume that $d\in \N$, $t \in [0,T]$, $x = (x_1,x_2,...,x_d)\in \R^d$, $v,w \in \R$ and that $\max \{ |g_d(x)|\} \leqslant Ld^p \left(1+\Sigma^d_{k=1}\left|x_k \right|\right)$, let $\left(\Omega, \mathcal{F}, \mathbb{P}\right)$ be a probability space, let $\mathcal{W}^{d,\theta}: [0,T] \times \Omega \rightarrow \R^d$, $d\in \N$, $\theta \in \Theta$, be independent standard Brownian motions, assume for every $d\in \N$ that $\left(\mathcal{W}^{d,\theta}\right)_{\theta \in \Theta}$ are independent, let $u_d \in C([0,T] \times \R^d,\R)$, $d \in \N$, satisfy for all $d\in \N$, $t\in [0,T]$, $x \in \R^d$ that $\E \left[g_x \left(x+\mathcal{W}^{d,0}_{T-t} \right)\right] < \infty$ and:
\begin{align}
u_d\left(t,x\right) = \E \left[g_d \left(x + \mathcal{W}^{d,0}_{T-t}\right)\right]
\end{align}
Let $U^{d,\theta}_m: [0,T] \times \R^d \times \Omega \rightarrow \R$, $d \in \N$, $m\in \Z$, $\theta \in \Theta$, satisfy for all, $d\in \N$, $m \in \Z$, $\theta \in \Theta$, $t\in [0,T]$, $x\in \R^d$ that:
\begin{align}
U^{d,\theta}_m(t,x) = \frac{1}{m} \left[\sum^{m}_{k=1} g_d \left(x + \mathcal{W}^{d,(\theta, 0,-k)}_{T-t}\right)\right]
\end{align}
and for every $d,n,m \in \N$ let $\mathfrak{C}_{d,n,m} \in \Z$ be the number of function evaluations of $u_d(0,\cdot)$ and the number of realizations of scalar random variables which are used to compute one realization of $U^{d,0}_m(T,0): \Omega \rightarrow \R$.
There then exists $c \in \R$, and $\mathfrak{N}:\N \times (0,1] \rightarrow \N$ such that for all $d \in \N$, $\varepsilon \in (0,1]$ it holds that:
\begin{align}\label{(2.48)}
\sup_{t\in[0,T]} \sup_{x \in [-L,L]^d} \left(\E \left[\left| u_d(t,x) - U^{d,0}_{\mathfrak{N}(d,\epsilon)}\right|^\p\right]\right)^\frac{1}{\p} \leqslant \epsilon
\end{align}
and:
\begin{align}\label{2.3.27}
\mathfrak{C}_{d,\mathfrak{N}(d,\varepsilon), \mathfrak{N}(d,\varepsilon)} \leqslant cd^c\varepsilon^{-(2+\delta)}
\end{align}
\end{theorem}
\begin{proof} Throughout the proof let $\mathfrak{m}_\mathfrak{p} = \sqrt{\mathfrak{p} -1}$, $\mathfrak{p} \in [2,\infty)$, let $\mathbb{F}^d_t \subseteq \mathcal{F}$, $d\in \N$, $t\in [0,T]$ satisfy for all $d \in \N$, $t\in [0,T]$ that:
\begin{align}\label{2.3.29}
\mathbb{F}^d_t = \begin{cases}
\bigcap_{s\in[t,T]} \sigma \left(\sigma \left(W^{d,0}_r: r \in [0,s]\right) \cup \{A\in \mathcal{F}: \mathbb{P}(A)=0\}\right) & :t<T \\
\sigma \left(\sigma \left(W^{d,0}_s: s\in [0,T]\right) \cup \{ A \in \mathcal{F}: \mathbb{P}(A)=0\}\right) & :t=T
\end{cases}
\end{align}
Observe that (\ref{2.3.29}) guarantees that $\mathbb{F}^d_t \subseteq \mathcal{F}$, $d\in \N$, $t\in [0,T]$ satisfies that:
\begin{enumerate}[label = (\Roman*)]
\item it holds for all $d\in \N$ that $\{ A \in \mathcal{F}: \mathbb{P}(A) = 0 \} \subseteq \mathbb{F}^d_0$
\item it holds for all $d \in \N$, $t\in [0,T]$, that $\mathbb{F}^d_t = \bigcap_{s \in (t,T]}\mathbb{F}^d_s$.
\end{enumerate}
Combining item (I), item (II), (\ref{2.3.29}) and \cite[Lemma 2.17]{hjw2020} assures us that for all $d \in \N$ it holds that $W^{d,0}:[0, T] \times \Omega \rightarrow \R^d$ is a standard $\left(\Omega, \mathcal{F}, \mathbb{P}, \left(\mathbb{F}^d_t\right)_{t\in [0, T]}\right)$-Brownian Brownian motion. In addition $(58)$ ensures that it is the case that for all $d\in N$, $x\in \R^d$ it holds that $[0,T] \times \Omega \ni (t,\omega) \mapsto x + W^{d,0}_t(\omega) \in \R^d$ is an $\left(\mathbb{F}^d_t\right)_{t\in [0,T]}/\mathcal{B}\left(\R^d\right)$-adapted stochastic process with continuous sample paths.
\medskip
This and the fact that for all $d\in \N$, $t\in [0,T]$, $x\in \R^d$ it holds that $a_d(t,x) = 0$, and the fact that for all $d\in \N$, $t \in [0,T]$, $x$,$v\in \R^d$ it holds that $b_d(t,x)v = v$ yield that for all $d \in \N$, $x\in \R^d$ it holds that $[0,T] \times \Omega \ni (t,\omega) \mapsto x+W^{d,0}_t(\omega) \in \R^d$ satisfies for all $t\in [0,T]$ it holds $\mathbb{P}$-a.s. that:
\begin{align}
x+W^{d,0}_t = x + \int^t_0 0 ds + \int^t_0 dW^{d,0}_s = x + \int^t_0 a_d(s,x+W^{d,0}_s) ds + \int^t_0 b_d(s,x+W^{d,0}_s) dW^{d,0}_s
\end{align}
\medskip
This and \cite[Lemma 2.6]{hjw2020} (applied for every $d \in \N$, $x \in \R^d$ with $d \curvearrowleft d$, $m \curvearrowleft d$, $T \curvearrowleft T$, $C_1 \curvearrowleft d$, $ C_2 \curvearrowleft 0$, $\mathbb{F} \curvearrowleft \mathbb{F}^d$, $ \xi \curvearrowleft x, \mu \curvearrowleft a_d, \sigma \curvearrowleft b_d, W \curvearrowleft W^{d,0}, X \curvearrowleft \left(\left[0,T\right] \times \Omega \ni (t, \omega) \mapsto x+W^{d,0}_t(\omega) \in \R^d\right)$ in the notation of \cite[Lemma 2.6]{hjw2020} ensures that for all $r\in [0,\infty)$, $d\in \N$, $x\in \R^d$, $t \in [0,T]$ it holds that
\begin{align}
\E \left[\left\| x+W^{d,0}_t\right\|^r\right] \leqslant \max\{T,1\} \left(\left(1+\left\| x\right\|^2\right)^{\frac{r}{2}} + (r+1)d^{\frac{r}{2}}\right) \exp \left(\frac{r(r+3)T}{2}\right) < \infty
\end{align}
This, the triangle inequality, and the fact that for all $v$,$w\in [0,\infty)$, $r\in (0,1]$, it holds that $(v+w)^r \leqslant v^r + w^r$ assure that for all $\p \in [2,\infty)$, $d\in \N$, $x \in \R^d$ it holds that:
\begin{align}
&\sup_{s\in [0,T]} \left(\E \left[\left(1+ \left\| x+W^{d,0}_s \right\|_E^q\right)^\p\right]\right)^\frac{1}{\p} \nonumber
\leqslant 1 + \sup_{s\in [0,T]} \left(\E \left[\left\| x+W^{d,0}_{s}\right\|_E^{q\p}\right]\right)^{\frac{1}{\p}} \nonumber \\
&\leqslant 1 + \sup_{s \in [0,T]} \left(\max\{T,1\} \left(\left(1+\left\| x\right\|_E^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{q\p(q\p+3)T}{2}\right) \right)^\frac{1}{\p} \nonumber \\
&\leqslant 1 + \max\{T^\frac{1}{\p},1\} \left(\left(1+\left\| x\right\|_E^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{q(q\p+3)T}{2}\right) \nonumber \\
&\leqslant 2\left(\left(1+\left\| x\right\|_E^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{q(q\p+3)T}{2}+\frac{T}{\p}\right) \nonumber \\
&\leqslant 2\left(\left(1+\left\| x\right\|_E^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right)
\end{align}
Given that for all $d\in \N$, $x \in [-L,L]^d$ it holds that $\left\| x \right\|_E \leqslant Ld^{\frac{1}{2}}$, this demonstrates for all $\p \in [2,\infty)$, $d\in \N$, it holds that:
\begin{align}
&L\left( \frac{ \mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\sup_{x \in [-L,L]^d} \sup_{s \in [0,T]} \left( \E \left[\left(1+\left\| x + W^{d,0}_s \right\|_E ^q\right)^\p\right]\right)^{\frac{1}{\p}}\right) \nonumber\\
&\leqslant L\left( \frac{ \mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\sup_{x \in [-L,L]^d} \left[\left(\left(1+\left\| x\right\|_E^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right)\right] \right)\nonumber\\
&\leqslant L\left( \frac{ \mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right)
\end{align}
Combining this with Corollary \ref{cor:1.25.1} tells us that:
\begin{align}\label{(2.3.33)}
&\left( \E \left[ \left| u_d(t,x)-U^{d,0}_m\lp t,x \rp \right| ^\p \right] \right) ^{\frac{1}{\p}} \nonumber\\
&\leqslant L\left(\frac{\mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\sup_{x \in [-L,L]^d} \sup_{s \in [0,T]} \left( \E \left[\left(1+\left\| x + W^{d,0}_s \right\|_E^q\right)^\p\right]\right)^{\frac{1}{\p}}\right) \nonumber\\
&\leqslant L\left( \frac{\mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right)
\end{align}
This and the fact that for all $d \in \N$ and $\ve \in (0,\infty)$ and the fact that $\mathfrak{m}_\p \leqslant 2$, it holds that for fixed $L,q, \mathfrak{p}, d,T$ there exists an $\mathfrak{M}_{L,q,\mathfrak{p},d,T} \in \R$ such that $\mathfrak{N}_{d,\epsilon} \geqslant \mathfrak{M}_{L,q,\mathfrak{p},d,T}$ forces:
\begin{align}\label{2.3.34}
L\left[ \frac{\mathfrak{m}_\p}{\mathfrak{N}_{d,\epsilon}^\frac{1}{2}}\right]\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \leqslant \ve
\end{align}
Thus (\ref{(2.3.33)}) and (\ref{2.3.34}) together proves (\ref{(2.48)}).
Note that $\mathfrak{C}_{d,\mathfrak{N}_{d,\epsilon},\mathfrak{N}_{d,\epsilon}}$ is the number of function evaluations of $u_d(0,\cdot)$ and the number of realizations of scalar random variables which are used to compute one realization of $U^{d,0}_{\mathfrak{N}_{d,\epsilon}}(T,0):\Omega \rightarrow \R$. Let $\widetilde{\mathfrak{N}_{d,\ve}}$ be the value of $\mathfrak{N}_{d,\ve}$ that causes equality in $(\ref{2.3.34})$. In such a situation the number of evaluations of $u_d(0,\cdot)$ do not exceed $\widetilde{\mathfrak{N}_{d,\ve}}$. Each evaluation of $u_d(0,\cdot)$ requires at most one realization of scalar random variables. Thus we do not exceed $2\widetilde{\mathfrak{N}_{d,\epsilon}}$. Thus note that:
\begin{align}\label{(2.3.35)}
\mathfrak{C}_{d,\mathfrak{N}_{d,\ve},\mathfrak{N}_{d,\ve}} \leqslant 2\lb L\mathfrak{m}_\p\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \rb \ve^{-1}
\end{align}
Note that other than $d$ and $\ve$ everything on the right-hand side is constant or fixed. Hence (\ref{(2.3.35)}) can be rendered as:
\begin{align}
\mathfrak{C}_{d,\mathfrak{N}_{d,\ve},\mathfrak{N}_{d,\ve}} \leq cd^k\ve^{-1}
\end{align}
Where both $c$ and $k$ are dependent on $L,\mathfrak{p,m},L$
%Next observe that:
%\begin{align}
% \lb L\left( \frac{\mathfrak{m}_\p}{\mathfrak{N}_{d,\epsilon}^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \rb^{-1} \geqslant \frac{1}{\ve} \\
% \mathfrak{N}_{d,\epsilon}\lb L\left(\mathfrak{m}_\p\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \rb^{-1} \geqslant \frac{1}{\ve} \\
%\end{align}
%
%Note that $\mathfrak{C}_{d,\mathfrak{N}_{d,\epsilon},\mathfrak{N}_{d,\epsilon}}$ is the number of function evaluations of $u_d(0,\cdot)$ and the number of realizations of scalar random variables which are used to compute one realization of $U^{d,0}_{\mathfrak{N}_{d,\epsilon}}(T,0):\Omega \rightarrow \R$. In such a situation the number of evaluations of $u_d(0,\cdot)$ do not exceed $\mathfrak{N}_{d,\ve}$. In other words, for a given precision $\ve$ in dimension $d$, the number computations required for evaluating $u_d(0,\cdot)$ is at most:
% \begin{align}
% \lb L\mathfrak{m}_\p\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \rb \ve^{-1}
% \end{align}
% Furthermore note that for each evaluation of $u_d$ is associated with an evaluation of $\mathcal{W}_t^{d,(\theta, 0,-k)}$ which requires at most one realization of scalar random variables and thus in turn is also bounded by:
% \begin{align}
% \lb L\left( \frac{\mathfrak{m}_\p}{\mathfrak{N}_{d,\epsilon}^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \rb \ve^{-1}
% \end{align}
% And thus we have that:
% \begin{align}
% \mathfrak{C}_{d,\mathfrak{N}_{d,\ve},\mathfrak{N}_{d,\ve}} \leqslant 2 \lb L\left( \frac{\mathfrak{m}_\p}{\mathfrak{N}_{d,\epsilon}^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \rb \ve^{-1}
% \end{align}
\end{proof}
Reference in New Issue View Git Blame Copy Permalink