\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[Lemma~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