worked on slides

tex/slides.tex

@@ -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}
 

tex/src/mlmc.tex

-\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.
         \begin{equation}
             \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)
         \end{equation}
-        the cost for one sample can be bounded with $\gamma > 0$ by
+        The cost for one sample can be bounded with $\gamma > 0$ by
         \begin{equation}
             \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
         \end{equation}
-        and the variance of $\goal_l - \goal_{l-1}$ decays with $\beta > 0$
+        The variance of $\goal_l - \goal_{l-1}$ decays with $\beta > 0$
         \begin{equation}
             \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
-        \begin{equation}
-            \label{eq:model_problem}
-            - \div(\kappa(\omega,x) \nabla u(\omega,x)) = f(\omega,x)
-        \end{equation}
-        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}
+
+
+

tex/src/mlmc_algorithm.tex

 \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}