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 ... ... @@ -18,10 +18,7 @@ \section*{Monte Carlo methods} \input{src/mlmc} \input{src/experimental_setup} \section*{Intro} \input{src/numerical_results} % \input{src/experimental_setup} \section*{Stochastic Linear Transport problem} ... ...
 \begin{frame}{Multilevel Monte Carlo Method I} \begin{frame}{Introduction and Example I} \begin{itemize} \item \underline{Problem:} Let $u(\omega)$ be a random PDE solution on $\Omega$ and let $\goal(\omega)$ be some functional of $u(\omega)$. Estimate $\EE[\goal(\omega)]$. Estimate $\EE[\goal(\omega)]$ \item \underline{Assumptions:} Let $u_h(\omega, x) \in V_h$ be the corresponding FEM solution with convergence rate $\alpha > 0$, i.e. \item \underline{Example (1D Elliptic Model Problem):} Let $D \subset \RR^{d=1}$. Search for $u \in V$, such that \begin{align*} - (\kappa(\omega,x) u'(\omega,x))' = 0, \quad u'(0) = 1, \quad u(1) = 0 \end{align*} with $\kappa(\omega, x) = \log(g(\omega, x))$, where $g$ is a Gaussian field \item \end{itemize} \end{frame} \begin{frame}{Assumptions} \begin{itemize} \item \underline{FEM (Finite Element Method:)} Let $u_h(\omega, x) \in V_h$ be the corresponding FEM solution to $u(\omega, x)$ and $\goal_h(\omega)$ be the functional \item \underline{Assumptions:} The FEM method is convergent with convergence rate $\alpha > 0$, i.e. \label{eq:alpha-assumption} \abs{\EE[\goal_h - \goal]} \lesssim h^\alpha, \quad \abs{\EE[\goal_h - \goal]} \lesssim N^{-\alpha / d}, \quad N = \dim(V_h), N = \dim(V_h) the cost for one sample can be bounded with $\gamma > 0$ by The cost for one sample can be bounded with $\gamma > 0$ by \label{eq:gamma-assumption} \cost(\goal_h(\omega_m)) \lesssim h^{-\gamma}, \quad \cost(\goal_h(\omega_m)) \lesssim N^{\gamma / d}, \quad \omega_m \in \Omega and the variance of $\goal_l - \goal_{l-1}$ decays with $\beta > 0$ The variance of $\goal_l - \goal_{l-1}$ decays with $\beta > 0$ \label{eq:beta-assumption} \abs{\VV[\goal_l - \goal_{l-1}]} \lesssim h^\beta, \quad ... ... @@ -28,28 +46,13 @@ \end{itemize} \end{frame} \begin{frame}{Examples I} \begin{itemize} \item \underline{Elliptic Model Problem:} Let $D \subset \RR^d$. Search for $u \in V$, such that \label{eq:model_problem} - \div(\kappa(\omega,x) \nabla u(\omega,x)) = f(\omega,x) with Neumann and Dirichlet boundary conditions. \item Simple 1D Problem already with results? \item Vielleicht gemittelte Lösung mit unterschiedlichen epsilon \end{itemize} \end{frame} \begin{frame}{Multilevel Monte Carlo Method II} \begin{frame}{Monte Carlo Estimator} \begin{itemize} \item \underline{MC Estimator:} Draw $\omega_m \in \Omega$ and compute \item \underline{MC Estimator:} Draw $\omega_m \in\Omega$ and compute \begin{align*} \widehat{\goal}_{h,M}^{MC} = \frac{1}{M} \sum_{m=1}^M \goal_h(\omega_m) \end{align*} \item \underline{RMSE (Root mean square error):} \item \underline{RMSE (Root Mean Square Error):} \begin{align*} e(\widehat{\goal}^{MC}_{h,M})^2 = \EE \left[ (\widehat{\goal}^{MC}_{h,M} - \EE[\goal])^2 \right] = ... ... @@ -67,13 +70,20 @@ \end{itemize} \end{frame} \begin{frame}{Examples II} \begin{frame}{Example II} \begin{itemize} \item 2D Problem etwas irregulär \item \underline{Example (2D Elliptic Model Problem):} Let $D \subset \RR^{d=2}$. Search for $u \in V$, such that \begin{align*} - \div(\kappa(\omega,x) \nabla u(\omega,x)) = 0, \quad \nabla u(x) \cdot n = -1, \quad u(x) = 0 \end{align*} with $\kappa(\omega, x) = \log(g(\omega, x))$, where $g$ is a Gaussian field \end{itemize} \end{frame} \begin{frame}{Multilevel Monte Carlo Methods III} \begin{frame}{Multilevel Monte Carlo Methods I} \begin{itemize} \item \underline{Main idea:} Draw samples from several approximation levels and balance cost per level $\cost_l$ with total sample amount per level $M_l$ ... ... @@ -90,7 +100,7 @@ \widehat{\dgoal}^{MC}_{h,M_0} = \frac{1}{M_0} \sum_{m=1}^{M_0} \goal_0 (\omega_m) \end{align*} \item MLMC estimator: \item \underline{MLMC estimator:} \begin{align*} \widehat{\goal}^{MLMC}_{h,\{ M_l \}_{l=0}^L} = \sum_{l=0}^L \widehat{\dgoal}^{MC}_{h,M_l} = ... ... @@ -101,24 +111,81 @@ \begin{frame}{Multilevel Monte Carlo Methods II} \begin{itemize} \item \underline{RMSE (Root mean square error):} \begin{equation*} \item \underline{RMSE (Root Mean Square Error):} \begin{align*} e(\widehat{\goal}^{MLMC}_{h,\{ M_l \}_{l=0}^L})^2 = \underbrace{\sum_{l=0}^L \frac{1}{M_l} \VV[\dgoal_l]}_{\text{estimator error}} + \underbrace{\left( \EE[\goal_L - \goal] \right)^2}_{\text{FEM error}}. \underbrace{\left( \EE[\goal_L - \goal] \right)^2}_{\text{FEM error}} \end{align*} \item \underline{Total cost:} \begin{align*} \cost(\widehat{\goal}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) \lesssim \sum_{l=0}^L M_l \cost_l, \quad \cost_{\epsilon}(\widehat{\goal}^{MLMC}_{h,\{M_l \}_{l=0}^L}) \lesssim \epsilon^{-2-(\gamma - \beta)/\alpha} \end{align*} Here, we assumed $\beta < \gamma$. \item Arguments (...) \end{itemize} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Multilevel Monte Carlo method} \begin{theorem}[Multilevel Monte Carlo method] \label{MLMC_theorem} {\footnotesize Suppose that there are positive constants $\alpha, \beta, \gamma, c_1, c_2, c_3 > 0$ such that $\alpha \geq \frac{1}{2} \min(\beta, \gamma)$ and \begin{enumerate} {\footnotesize \item $\left| \E[Q_{h_l} - Q] \right| \leq c_1 h_l^\alpha$ \item $\V[Q_{h_l} - Q_{h_{l-1}}] \leq c_2 h_l^\beta$ \item $\mathcal{C}_l \leq c_3 h_l^{- \gamma}$.} \end{enumerate} Then, for any $0 < \epsilon < \frac{1}{e}$, there exists an $L$ and a sequence $\{ M_l \}_{l=0}^L$, such that \begin{equation*} e(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L})^2 = \E\left[( \widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L} - \E[Q])^2 \right] < \epsilon^2 \end{equation*} \item This leads leads to a better computational cost since: \begin{itemize} \item Assume $Q_h \rightarrow Q$, then $\VV[\left( Q_{h_l}(\omega_i) - Q_{h_{l-1}}(\omega_i) \right)] \rightarrow 0$. \item The $Q_{h_0}(\omega_i)$ are not getting more expensive for more accuracy. \item The optimal choice for the sequence $M_l$ is given by \begin{equation*} M_l = \left\lceil 2 \epsilon^{-2} \sqrt{\frac{\VV[Y_l]}{\cost_l}} \left( \sum_{l=0}^L \sqrt{\VV[Y_l] \cost_l} \right) \right\rceil. \end{equation*} \end{itemize} \item This gives an overall cost of (given $\cost_{\epsilon}$ is best case) and \begin{equation*} \cost(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) = \sum_{l=0}^L M_l \cost_l, \quad \cost_{\epsilon}(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) \lesssim \epsilon^{-2}. \mathcal{C}_\epsilon(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) \lesssim \begin{cases} \epsilon^{-2}, &\text{if } \beta > \gamma \\ \epsilon^{-2}\log(\epsilon)^2, &\text{if } \beta = \gamma \\ \epsilon^{-2-(\gamma - \beta)/\alpha}, &\text{if } \beta < \gamma \end{cases}, \end{equation*} where the hidden constant depends on $c_1, c_2, c_3$.} \end{theorem} \end{frame} \begin{frame}{Multilevel Monte Carlo method - Algorithm} \begin{itemize} \item Challenge in using the MLMC method is showing that the assumptions hold \item The MLMC approach can be parallelized in an easy manner \end{itemize} \vspace{-0.4cm} \begin{algorithm}[H] \caption{Multilevel Monte Carlo method} \begin{algorithmic}[1] \label{MLMC algorithm} {\footnotesize \STATE Set $l_0 = 3$, $L_0 = 5$ and the initial number of samples $M_0 = \{ 200, 100, 50 \}$ \STATE Set range of levels $\{l_0, \dots, L_0 \}$ and the number of needed samples $\{ \Delta M_l = M_0 \}_{l = 0}^{L}$ \WHILE {$\Delta M_l > 0$ on any level} \FOR {levels with needed samples} \STATE Retrieve functionals and cost: $Y_l, \, \mathcal{C}_l \leftarrow \texttt{SubroutineEstimator}(\Delta M_l, l)$ \STATE Update statistics: $\mathcal{C}_l$, $|\E[Y_l]|$, $\V[Y_l]$ and set: $M_l = \Delta M_l$, $\Delta M_l = 0$ \ENDFOR \STATE Estimate exponents $\alpha$, $\beta$, $\gamma$ with the assumptions of the previous Theorem \STATE Estimate optimal $M_l$, $l = 0, \dots, L$ with $M_l = \left\lceil 2 \epsilon^{-2} \sqrt{\frac{\V[Y_l]}{\mathcal{C}_l}} \left( \sum_{l=0}^L \sqrt{\V[Y_l] \mathcal{C}_l} \right) \right\rceil$ \STATE Test for weak convergence with $|\E[Q_{h_L} - Q_{h_{L-1}}]| < (2^\alpha - 1) \frac{\epsilon}{\sqrt{2}}$ \STATE If not converged, increase range of levels by one level and initialize new $M_L$ \ENDWHILE} \end{algorithmic} \end{algorithm} \end{frame}
 \begin{frame}{Multilevel Monte Carlo method} \begin{itemize} \item Main idea MLMC: Sample from several approximation $Q_h$ on different fine triangulations. \item With the linearity of the expectation operator it holds \begin{equation*} \E[Q_h] = \E[Q_{h_0}] + \sum_{l=1}^L \E[Q_{h_l} - Q_{h_{l-1}}] = \sum_{l=1}^L \E[Y_l]. \end{equation*} \item Now estimate each $Y_l$ with the classical MC method, thus \begin{equation*} \widehat{Y}^{MC}_{h,M_l} = \frac{1}{M_l} \sum_{i=1}^{M_l} \left( Q_{h_l}(\omega_i) - Q_{h_{l-1}}(\omega_i) \right), \quad \widehat{Y}^{MC}_{h,M_0} = \frac{1}{M_0} \sum_{i=1}^{M_0} Q_{h_0}(\omega_i) \end{equation*} \item This gives the MLMC estimator \begin{equation*} \widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L} = \sum_{l=0}^L \widehat{Y}^{MC}_{h,M_l} = \sum_{l=0}^L \frac{1}{M_l} \sum_{i=1}^{M_l} Y_l(\omega_{i}). \end{equation*} \end{itemize} \begin{itemize} \item Main idea MLMC: Sample from several approximation $Q_h$ on different fine triangulations. \item With the linearity of the expectation operator it holds \begin{equation*} \E[Q_h] = \E[Q_{h_0}] + \sum_{l=1}^L \E[Q_{h_l} - Q_{h_{l-1}}] = \sum_{l=1}^L \E[Y_l]. \end{equation*} \item Now estimate each $Y_l$ with the classical MC method, thus \begin{equation*} \widehat{Y}^{MC}_{h,M_l} = \frac{1}{M_l} \sum_{i=1}^{M_l} \left( Q_{h_l}(\omega_i) - Q_{h_{l-1}}(\omega_i) \right), \quad \widehat{Y}^{MC}_{h,M_0} = \frac{1}{M_0} \sum_{i=1}^{M_0} Q_{h_0}(\omega_i) \end{equation*} \item This gives the MLMC estimator \begin{equation*} \widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L} = \sum_{l=0}^L \widehat{Y}^{MC}_{h,M_l} = \sum_{l=0}^L \frac{1}{M_l} \sum_{i=1}^{M_l} Y_l(\omega_{i}). \end{equation*} \end{itemize} \end{frame} \begin{frame}{Multilevel Monte Carlo method} \begin{itemize} \item The RMSE is then given by \begin{equation*} e(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L})^2 = \E\left[( \widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L} - \E[Q])^2 \right] = \underbrace{\sum_{l=0}^L \frac{1}{M_l} \V[Y_l]}_{\text{estimator error}} + \underbrace{\left( \E[Q_h - Q] \right)^2}_{\text{FEM error}}. \end{equation*} \item This leads leads to a better computational cost since: \begin{itemize} \item Assume $Q_h \rightarrow Q$, then $\V[\left( Q_{h_l}(\omega_i) - Q_{h_{l-1}}(\omega_i) \right)] \rightarrow 0$. \item Even with increasing required accuracy samples of $Q_{h_0}$ are not getting more expensive. \item The optimal choice for the sequence $M_l$ is given by \begin{equation*} M_l = \left\lceil 2 \epsilon^{-2} \sqrt{\frac{\V[Y_l]}{\mathcal{C}_l}} \left( \sum_{l=0}^L \sqrt{\V[Y_l] \mathcal{C}_l} \right) \right\rceil. \end{equation*} \end{itemize} \item This gives an overall cost of \begin{equation*} \mathcal{C}(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) = \sum_{l=0}^L M_l \mathcal{C}_l, \quad \mathcal{C}(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) \lesssim \sum_{l=0}^L \sqrt{\V[Y_l] \mathcal{C}_l}. \end{equation*} \begin{itemize} \item The RMSE is then given by \begin{equation*} e(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L})^2 = \E\left[( \widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L} - \E[Q])^2 \right] = \underbrace{\sum_{l=0}^L \frac{1}{M_l} \V[Y_l]}_{\text{estimator error}} + \underbrace{\left( \E[Q_h - Q] \right)^2}_{\text{FEM error}}. \end{equation*} \item This leads leads to a better computational cost since: \begin{itemize} \item Assume $Q_h \rightarrow Q$, then $\V[\left( Q_{h_l}(\omega_i) - Q_{h_{l-1}}(\omega_i) \right)] \rightarrow 0$. \item Even with increasing required accuracy samples of $Q_{h_0}$ are not getting more expensive. \item The optimal choice for the sequence $M_l$ is given by \begin{equation*} M_l = \left\lceil 2 \epsilon^{-2} \sqrt{\frac{\V[Y_l]}{\mathcal{C}_l}} \left( \sum_{l=0}^L \sqrt{\V[Y_l] \mathcal{C}_l} \right) \right\rceil. \end{equation*} \end{itemize} \item This gives an overall cost of \begin{equation*} \mathcal{C}(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) = \sum_{l=0}^L M_l \mathcal{C}_l, \quad \mathcal{C}(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) \lesssim \sum_{l=0}^L \sqrt{\V[Y_l] \mathcal{C}_l}. \end{equation*} \end{itemize} \end{itemize} \end{frame} \begin{frame}{Multilevel Monte Carlo method} \begin{theorem}[Multilevel Monte Carlo method] \label{MLMC_theorem} {\footnotesize Suppose that there are positive constants $\alpha, \beta, \gamma, c_1, c_2, c_3 > 0$ such that $\alpha \geq \frac{1}{2} \min(\beta, \gamma)$ and \begin{enumerate} {\footnotesize \item $\left| \E[Q_{h_l} - Q] \right| \leq c_1 h_l^\alpha$ \item $\V[Q_{h_l} - Q_{h_{l-1}}] \leq c_2 h_l^\beta$ \item $\mathcal{C}_l \leq c_3 h_l^{- \gamma}$.} \end{enumerate} Then, for any $0 < \epsilon < \frac{1}{e}$, there exists an $L$ and a sequence $\{ M_l \}_{l=0}^L$, such that \begin{equation*} e(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L})^2 = \E\left[( \widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L} - \E[Q])^2 \right] < \epsilon^2 \end{equation*} and \begin{equation*} \mathcal{C}_\epsilon(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) \lesssim \begin{cases} \epsilon^{-2}, &\text{if } \beta > \gamma \\ \epsilon^{-2}\log(\epsilon)^2, &\text{if } \beta = \gamma \\ \epsilon^{-2-(\gamma - \beta)/\alpha}, &\text{if } \beta < \gamma \end{cases}, \end{equation*} where the hidden constant depends on $c_1, c_2, c_3$.} \end{theorem} \begin{theorem}[Multilevel Monte Carlo method] \label{MLMC_theorem} {\footnotesize Suppose that there are positive constants $\alpha, \beta, \gamma, c_1, c_2, c_3 > 0$ such that $\alpha \geq \frac{1}{2} \min(\beta, \gamma)$ and \begin{enumerate} {\footnotesize \item $\left| \E[Q_{h_l} - Q] \right| \leq c_1 h_l^\alpha$ \item $\V[Q_{h_l} - Q_{h_{l-1}}] \leq c_2 h_l^\beta$ \item $\mathcal{C}_l \leq c_3 h_l^{- \gamma}$.} \end{enumerate} Then, for any $0 < \epsilon < \frac{1}{e}$, there exists an $L$ and a sequence $\{ M_l \}_{l=0}^L$, such that \begin{equation*} e(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L})^2 = \E\left[( \widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L} - \E[Q])^2 \right] < \epsilon^2 \end{equation*} and \begin{equation*} \mathcal{C}_\epsilon(\widehat{Q}^{MLMC}_{h,\{ M_l \}_{l=0}^L}) \lesssim \begin{cases} \epsilon^{-2}, &\text{if } \beta > \gamma \\ \epsilon^{-2}\log(\epsilon)^2, &\text{if } \beta = \gamma \\ \epsilon^{-2-(\gamma - \beta)/\alpha}, &\text{if } \beta < \gamma \end{cases}, \end{equation*} where the hidden constant depends on $c_1, c_2, c_3$.} \end{theorem} \end{frame} \begin{frame}{Multilevel Monte Carlo method - Algorithm} \begin{itemize} \item Challenge in using the MLMC method is showing that the assumptions hold \item The MLMC approach can be parallelized in an easy manner \end{itemize} \begin{itemize} \item Challenge in using the MLMC method is showing that the assumptions hold \item The MLMC approach can be parallelized in an easy manner \end{itemize} \vspace{-0.4cm} \vspace{-0.4cm} \begin{algorithm}[H] \caption{Multilevel Monte Carlo method} \begin{algorithmic}[1] \label{MLMC algorithm} {\footnotesize \STATE Set $l_0 = 3$, $L_0 = 5$ and the initial number of samples $M_0 = \{ 200, 100, 50 \}$ \STATE Set range of levels $\{l_0, \dots, L_0 \}$ and the number of needed samples $\{ \Delta M_l = M_0 \}_{l = 0}^{L}$ \WHILE {$\Delta M_l > 0$ on any level} \FOR {levels with needed samples} \STATE Retrieve functionals and cost: $Y_l, \, \mathcal{C}_l \leftarrow \texttt{SubroutineEstimator}(\Delta M_l, l)$ \STATE Update statistics: $\mathcal{C}_l$, $|\E[Y_l]|$, $\V[Y_l]$ and set: $M_l = \Delta M_l$, $\Delta M_l = 0$ \ENDFOR \STATE Estimate exponents $\alpha$, $\beta$, $\gamma$ with the assumptions of the previous Theorem \STATE Estimate optimal $M_l$, $l = 0, \dots, L$ with $M_l = \left\lceil 2 \epsilon^{-2} \sqrt{\frac{\V[Y_l]}{\mathcal{C}_l}} \left( \sum_{l=0}^L \sqrt{\V[Y_l] \mathcal{C}_l} \right) \right\rceil$ \STATE Test for weak convergence with $|\E[Q_{h_L} - Q_{h_{L-1}}]| < (2^\alpha - 1) \frac{\epsilon}{\sqrt{2}}$ \STATE If not converged, increase range of levels by one level and initialize new $M_L$ \ENDWHILE} \end{algorithmic} \end{algorithm} \begin{algorithm}[H] \caption{Multilevel Monte Carlo method} \begin{algorithmic}[1] \label{MLMC algorithm} {\footnotesize \STATE Set $l_0 = 3$, $L_0 = 5$ and the initial number of samples $M_0 = \{ 200, 100, 50 \}$ \STATE Set range of levels $\{l_0, \dots, L_0 \}$ and the number of needed samples $\{ \Delta M_l = M_0 \}_{l = 0}^{L}$ \WHILE {$\Delta M_l > 0$ on any level} \FOR {levels with needed samples} \STATE Retrieve functionals and cost: $Y_l, \, \mathcal{C}_l \leftarrow \texttt{SubroutineEstimator}(\Delta M_l, l)$ \STATE Update statistics: $\mathcal{C}_l$, $|\E[Y_l]|$, $\V[Y_l]$ and set: $M_l = \Delta M_l$, $\Delta M_l = 0$ \ENDFOR \STATE Estimate exponents $\alpha$, $\beta$, $\gamma$ with the assumptions of the previous Theorem \STATE Estimate optimal $M_l$, $l = 0, \dots, L$ with $M_l = \left\lceil 2 \epsilon^{-2} \sqrt{\frac{\V[Y_l]}{\mathcal{C}_l}} \left( \sum_{l=0}^L \sqrt{\V[Y_l] \mathcal{C}_l} \right) \right\rceil$ \STATE Test for weak convergence with $|\E[Q_{h_L} - Q_{h_{L-1}}]| < (2^\alpha - 1) \frac{\epsilon}{\sqrt{2}}$ \STATE If not converged, increase range of levels by one level and initialize new $M_L$ \ENDWHILE} \end{algorithmic} \end{algorithm} \end{frame} ... ...
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