
Selected Topics
October 9, 2026
Share the work across processors, samples, and levels
Modern CPUs and GPUs can evaluate many operations concurrently
Parallelism changes computational cost
It does not repair biased estimators or poor sampling designs
For
\[ \vtheta_* = \argmin_{\vtheta\in\reals^m}\operatorname{Loss}(\vtheta), \]
gradient descent iterates
\[ \vtheta_{k+1}=\vtheta_k-\eta_k\nabla\operatorname{Loss}(\vtheta_k) \]
Move downhill using all available data
Suppose
\[ \operatorname{Loss}(\vtheta) =\sum_{i=0}^{N-1}\operatorname{loss}_i(\vtheta) \]
Computing the full gradient costs \(N\) component gradients
Stochastic gradient descent uses an inexpensive random component:
\[ \vtheta_{k+1}=\vtheta_k-\eta_k\nabla\operatorname{loss}_{I_k}(\vtheta_k), \qquad I_k\sim\Unif\{1,\ldots,N\} \]
Common variants include
The stochastic gradient is cheap but noisy
Averaging a mini-batch reduces its variance
Both methods minimize the same quadratic loss
Stochastic gradients use one randomly selected component at each step

Suppose
\[ Y=Y_1+Y_2, \qquad \mu=\mu_1+\mu_2, \qquad \mu_\ell=\Ex(Y_\ell) \]
Estimate the two means independently:
\[\begin{align*} \hmu&=\hmu_1+\hmu_2,& \hmu_\ell&=\frac1{n_\ell}\sum_{i=0}^{n_\ell-1}Y_{i\ell} \end{align*}\]
If one sample of \(Y_\ell\) costs \(\$_\ell\), then
\[\begin{align*} \mse(\hmu) &=\frac{\var(Y_1)}{n_1}+\frac{\var(Y_2)}{n_2},\\ \cost(\hmu)&=n_1\$_1+n_2\$_2 \end{align*}\]
The sample counts should reflect both variance and cost per sample
For a fixed total cost \(C\), minimize
\[ \frac{V_1}{n_1}+\frac{V_2}{n_2} \quad\text{subject to}\quad n_1\$_1+n_2\$_2=C, \qquad V_\ell=\var(Y_\ell) \]
The optimal ratio is
\[ \frac{n_1}{n_2} =\sqrt{\frac{V_1\$_2}{V_2\$_1}} \]
and the optimal mean squared error is
\[ \mse(\hmu) =\frac1C\left(\sqrt{V_1\$_1}+\sqrt{V_2\$_2}\right)^2 \]
Use more samples at a level with larger variance and fewer samples at a level with larger cost
The decomposition is useful when the expensive correction has small variance
Let
\[ Y=Y_1+\cdots+Y_L, \qquad \mu=\sum_{\ell=1}^L\mu_\ell, \qquad \mu_\ell=\Ex(Y_\ell) \]
With independent samples at each level,
\[\begin{align*} \hmu&=\sum_{\ell=1}^L\hmu_\ell,& \hmu_\ell&=\frac1{n_\ell}\sum_{i=0}^{n_\ell-1}Y_{i\ell},\\ \mse(\hmu)&=\sum_{\ell=1}^L\frac{V_\ell}{n_\ell},& \cost(\hmu)&=\sum_{\ell=1}^Ln_\ell\$_\ell \end{align*}\]
The challenge is to choose both the level decomposition and the sample allocation
For total cost \(C\), the optimal allocation is proportional to
\[ n_\ell\propto\sqrt{\frac{V_\ell}{\$_\ell}} \]
More explicitly,
\[ n_\ell =\frac{C\sqrt{V_\ell/\$_\ell}} {\sum_{k=1}^L\sqrt{V_k\$_k}} \]
The resulting mean squared error is
\[ \mse(\hmu) =\frac1C\left(\sum_{\ell=1}^L\sqrt{V_\ell\$_\ell}\right)^2 \]
Efficiency depends on keeping \(V_\ell\$_\ell\) small across levels
Suppose
\[ \mu=\lim_{d\to\infty}\Ex\{f_d(\vX_{1:d})\} \]
as in an option payoff based on an increasingly fine time discretization
Choose \(d_1<\cdots<d_L\) and define coupled differences
\[\begin{align*} Y_1&=f_{d_1}(\vX_{1:d_1}),\\ Y_\ell&=f_{d_\ell}(\vX_{1:d_\ell}) -f_{d_{\ell-1}}(\vX_{1:d_{\ell-1}}), \qquad \ell\ge2 \end{align*}\]
Then
\[ \Ex\{f_{d_L}(\vX_{1:d_L})\} =\sum_{\ell=1}^L\Ex(Y_\ell) \]
with a remaining truncation bias \(\mu-\Ex\{f_{d_L}(\vX_{1:d_L})\}\)
If \(d_\ell=m^\ell\), \(V_\ell\lesssim\beta^\ell\), and \(\$_\ell\lesssim m^\ell\), then
\[ \sum_{\ell=1}^L\sqrt{V_\ell\$_\ell} \lesssim\sum_{\ell=1}^L(\beta m)^{\ell/2} \]
Multilevel Monte Carlo is effective when correction variance decays faster than correction cost grows
From Monte Carlo estimates to computational design
Good simulation balances accuracy, computational cost, and evidence for stopping
Bring the course methods together in a concrete model or project:
The next step is a defensible end-to-end simulation
© 2026 Fred J. Hickernell · Illinois Tech · assisted by ChatGPT and Codex · Selected Topics · MATH 565 — Fall 2026 Website · \(\exstar\) = exercise