
Improving Efficiency
Owen, Chapters 8–10 and 15–17
August 24, 2026
Suppose
\[ \mu=\Ex[f(X)]=\Ex[g(X)], \qquad X\sim\Unif[0,1]^d \]
Both IID sample means are unbiased:
\[\begin{align*} \hmu_{f,n}&=\frac1n\sum_{i=0}^{n-1}f(X_i),& \hmu_{g,n}&=\frac1n\sum_{i=0}^{n-1}g(X_i) \end{align*}\]
Their mean squared errors are their variances:
\[\begin{align*} \var(\hmu_{f,n})&=\frac{\var\{f(X)\}}n,& \var(\hmu_{g,n})&=\frac{\var\{g(X)\}}n \end{align*}\]
An equivalent expectation can be computationally better when its integrand has smaller variance
Consider
\[ \mu=\int_{\reals^d}\cos(\norm{t})\exp(-\norm{t}^2)\,\dif t \]
Let \(x_j=\Phi(at_j)\), where \(\Phi\) is the standard normal CDF
Then \(t_j=\Phi^{-1}(x_j)/a\) and
\[\begin{multline*} \mu=\int_{[0,1]^d}\frac{(2\mpi)^{d/2}}{a^d} \cos\!\left(\frac{\norm{\Phi^{-1}(x)}}a\right)\\ \times\exp\!\left[-\frac{2-a^2}{2a^2}\norm{\Phi^{-1}(x)}^2\right]\,\dif x, \qquad a^2\le2 \end{multline*}\]
The parameter \(a\) creates a family of unbiased estimators with different variances
To approximate \(\mu=\int_{\mathcal T}g(t)\,\dif t\), choose an invertible map
\[ t=\psi(x), \qquad \psi:\mathcal X\to\mathcal T \]
Let \(J(x;\psi)=\det\{\partial\psi_i/\partial x_j\}_{i,j=1}^d\)
If \(X\) has density \(\varrho_X\), then
\[\begin{align*} \mu &=\int_{\mathcal X}g\{\psi(x)\}\abs{J(x;\psi)}\,\dif x\\ &=\Ex_{X\sim\varrho_X}\!\left[ \frac{g\{\psi(X)\}\abs{J(X;\psi)}}{\varrho_X(X)} \right] \end{align*}\]
For \(X\sim\Unif[0,1]^d\), the density in the denominator is one
Let \(T=\psi(X)\) for an invertible map \(\psi:\mathcal X\to\mathcal T\)
The densities satisfy
\[\begin{align*} \varrho_T(t) &=\varrho_X\{\psi^{-1}(t)\}\abs{J(t;\psi^{-1})},\\ \varrho_T\{\psi(x)\} &=\frac{\varrho_X(x)}{\abs{J(x;\psi)}} \end{align*}\]
Therefore
\[ \Ex_{T\sim\varrho_T}[h(T)] =\Ex_{X\sim\varrho_X}[h\{\psi(X)\}] \]
A change of variables in an integral and a transformation of a random variable are the same mechanism viewed from opposite directions
For a target density \(\varrho_T\), choose a proposal density \(\varrho_X\) with the same support:
\[\begin{align*} \mu &=\int_{\mathcal T}h(t)\varrho_T(t)\,\dif t\\ &=\Ex_{X\sim\varrho_X}\!\left[ h(X)\frac{\varrho_T(X)}{\varrho_X(X)} \right] \end{align*}\]
The ratio \(\varrho_T/\varrho_X\) is the importance weight
For \(X_0,\ldots,X_{n-1}\ \IIDsim\ \varrho_X\),
\[\begin{align*} \hmu_{\mathrm{IS},n} &=\frac1n\sum_{i=0}^{n-1}h(X_i)\frac{\varrho_T(X_i)}{\varrho_X(X_i)},\\ \widehat{\var}(\hmu_{\mathrm{IS},n}) &=\frac1{n(n-1)}\sum_{i=0}^{n-1} \left[h(X_i)\frac{\varrho_T(X_i)}{\varrho_X(X_i)}-\hmu_{\mathrm{IS},n}\right]^2 \end{align*}\]
Positive option payoffs may be rare under a discretized Brownian path
Let \(X\sim\Norm(\vzero,\mSigma)\) with \(\Sigma_{jk}=\min(t_j,t_k)\) and target density \(\varrho\)
Sample instead from \(\Norm(\va,\mSigma)\) with
\[ \va=\theta(t_1,\ldots,t_d)^\mathsf T \]
The likelihood ratio simplifies to
\[ \frac{\varrho(x)}{\varrho_{\mathrm{drift}}(x)} =\exp\!\left(-\theta x_d+\frac{\theta^2t_d}{2}\right) \]
Place more samples where the magnitude of the integrand times the target density is large, without creating extreme weights
Let \(\eta_1,\ldots,\eta_m\) have known zero means
For any coefficients \(\beta_1,\ldots,\beta_m\),
\[ \mu=\Ex[f(X)] =\Ex\!\left[f(X)-\sum_{j=1}^m\beta_j\eta_j(X)\right] \]
The estimator
\[ \hmu_{\mathrm{CV},n} =\frac1n\sum_{i=0}^{n-1}\left[f(X_i)-\sum_{j=1}^m\beta_j\eta_j(X_i)\right] \]
is unbiased for fixed \(\vbeta\)
Choose \(\vbeta\) to make
\[ \var\!\left[f(X)-\sum_{j=1}^m\beta_j\eta_j(X)\right] \]
as small as possible
Center the response and control values:
\[\begin{align*} y_i&=f(x_i)-\bar f,& H_{ij}&=\eta_j(x_i)-\bar\eta_j \end{align*}\]
Then
\[ \hvbeta=\argmin_{\vb}\norm{\vy-\mH\vb}^2 \]
When \(\mH^\mathsf T\mH\) is nonsingular,
\[ \hvbeta=(\mH^\mathsf T\mH)^{-1}\mH^\mathsf T\vy \]
The residual is orthogonal to the column space of \(\mH\), so least squares removes the component of \(f\) explained by the controls
A control variate turns known expectations into information about an unknown expectation
Recall
\[\begin{align*} \Ex(Y)&=\Ex_Z\{\Ex(Y\mid Z)\},\\ \var(Y)&=\Ex_Z\{\var(Y\mid Z)\}+\var_Z\{\Ex(Y\mid Z)\} \end{align*}\]
Therefore
\[ \var_Z\{\Ex(Y\mid Z)\}\le\var(Y) \]
If \(Y=f(X_1,\ldots,X_d)\) and
\[ g(X_{2:d})=\Ex(Y\mid X_{2:d}) \]
is analytic, estimate \(\mu\) by sampling \(g(X_{2:d})\) instead of \(Y\)
Analytically averaging over part of the randomness reduces variance
Let \(Y=f(X)\) with \(X\sim\Unif[0,1]^d\)
Suppose \(f\) is increasing in \(x_1\) and
\[ y=f(x_1,x_{2:d})iff x_1=g(y;x_{2:d}) \]
Then
\[\begin{align*} F_{Y\mid X_{2:d}}(y\mid x_{2:d}) &=\Prob\{X_1\le g(y;x_{2:d})\}\\ &=g(y;x_{2:d}),\\ \varrho_Y(y) &=\int_{[0,1]^{d-1}}\frac{\partial g}{\partial y}(y;x)\,\dif x \end{align*}\]
Thus a density value becomes an expectation of a conditional density
Let
\[ Y=\gamma_1X_1+\cdots+\gamma_dX_d, \qquad X\sim\Unif[0,1]^d, \qquad \gamma_j>0 \]
Solving for \(x_1\) gives
\[ g(y;x_{2:d})=\frac{y-\sum_{j=2}^d\gamma_jx_j}{\gamma_1} \]
when \(\sum_{j=2}^d\gamma_jx_j\le y\le\gamma_1+\sum_{j=2}^d\gamma_jx_j\)
Hence
\[ \varrho_Y(y)=\Ex\!\left[ \frac1{\gamma_1}\indic\!\left{ \sum_{j=2}^d\gamma_jX_j\le y\le \gamma_1+\sum_{j=2}^d\gamma_jX_j \right}\right] \]
If \(T(X)\) has the same distribution as \(X\), pair \(f(X_i)\) with \(f\{T(X_i)\}\):
\[ \hmu_{\mathrm{Anti},n} =\frac1{2n}\sum_{i=0}^{n-1}\left[f(X_i)+f\{T(X_i)\}\right] \]
Examples:
\[\begin{align*} T(X)&=1-X,& X&\sim\Unif[0,1]^d,\\ T(X)&=-X,& X&\sim\Norm(\vzero,\mSigma) \end{align*}\]
Although \(X_i\) and \(T(X_i)\) are dependent, the pairs are independent across \(i\)
The variance is
\[ \var(\hmu_{\mathrm{Anti},n}) =\frac{\var\{f(X)\}}{2n} \left[1+\corr\{f(X),f(T(X))\}\right] \]
Antithetic sampling helps when the paired function values are negatively correlated
If \(X\) has continuous CDF \(F\), then
\[\begin{gather*} U=F(X)\sim\Unif[0,1],\\ 1-U\sim\Unif[0,1],\\ T(X)=F^{-1}\{1-F(X)\}\sim F \end{gather*}\]
For a symmetric distribution, this often reduces to reflection about the center
\(\exstar\) For which monotone functions \(f\) does \(f(X)\) become negatively correlated with \(f\{T(X)\}\)?
Place one independent uniform point in each of \(n\) equal strata:
\[ X_i=\frac{i+U_i}{n}, \qquad U_i\ \IIDsim\ \Unif[0,1], \qquad i=0,\ldots,n-1 \]
Then \(X_i\sim\Unif[i/n,(i+1)/n]\) and
\[\begin{align*} \Ex(\hmu_{\mathrm{LHS},n}) &=\frac1n\sum_{i=0}^{n-1}\Ex\{f(X_i)\}\\ &=\sum_{i=0}^{n-1}\int_{i/n}^{(i+1)/n}f(x)\,\dif x\\ &=\int_0^1f(x)\,\dif x \end{align*}\]
The \(X_i\) are independent but not identically distributed
Because the strata are independent,
\[ \var(\hmu_{\mathrm{LHS},n}) =\frac1{n^2}\sum_{i=0}^{n-1}\var\{f(X_i)\} \]
For differentiable \(f\), the mean value and Taylor theorems imply
\[\begin{align*} \var(\hmu_{\mathrm{LHS},n}) &\le\frac{\norm{f'}_\infty^2}{n^3},\\ \rmse(\hmu_{\mathrm{LHS},n}) &\le\frac{\norm{f'}_\infty}{n^{3/2}} \end{align*}\]
In one dimension, stratification can improve the IID \(n^{-1/2}\) RMSE rate to \(n^{-3/2}\) for smooth \(f\)
Let \(\pi_1,\ldots,\pi_d\) be independent random permutations of \(0,\ldots,n-1\):
\[ X_{i\ell}=\frac{\pi_\ell(i)+U_{i\ell}}n, \qquad U_{i\ell}\ \IIDsim\ \Unif[0,1] \]
Each coordinate uses every stratum exactly once
For \(R\) independent Latin hypercubes,
\[\begin{align*} \hmu_{\mathrm{LHS},n}^{(r)} &=\frac1n\sum_{i=0}^{n-1}f(X_i^{(r)}),\\ \hmu_{\mathrm{LHS},n,R} &=\frac1R\sum_{r=1}^R\hmu_{\mathrm{LHS},n}^{(r)},\\ \widehat{\var}(\hmu_{\mathrm{LHS},n,R}) &=\frac1{R(R-1)}\sum_{r=1}^R \left(\hmu_{\mathrm{LHS},n}^{(r)}-\hmu_{\mathrm{LHS},n,R}\right)^2 \end{align*}\]
Each design has \(n=64\) points in \([0,1]^2\)

Low discrepancy sampling, or quasi-Monte Carlo, improves space filling beyond one-dimensional stratification
For nodes \(x_0,\ldots,x_{n-1}\) and target distribution \(F\),
\[\begin{align*} D^2(\{x_i\}_{i=0}^{n-1},F;K) &=\int_{\mathcal X^2}K(x,x')\,\dif F(x)\,\dif F(x')\\ &\quad-\frac2n\sum_{i=0}^{n-1}\int_{\mathcal X}K(x,x_i)\,\dif F(x)\\ &\quad+\frac1{n^2}\sum_{i,j=0}^{n-1}K(x_i,x_j) \end{align*}\]
We usually take \(\mathcal X=[0,1]^d\) and seek nodes with small discrepancy
For the uniform target on \([0,1]^d\), discrepancy is
worst-case error
\[ \left\lvert\int f(x)\,\dif x-\frac1n\sum_{i=0}^{n-1}f(x_i)\right\rvert \le D\norm{f}_{\mathcal H_K} \]
average-case root mean squared error for \(f\) drawn from a Gaussian process with covariance kernel \(K\)
distance between the target distribution and the empirical distribution
The kernel specifies the function class and the distributional features being measured
Write a nonnegative integer in base \(b\):
\[ i=(\cdots i_2i_1i_0)_b \]
Reverse the digits across the radix point:
\[ \phi_b(i)=(0.i_0i_1i_2\cdots)_b =\frac{i_0}b+\frac{i_1}{b^2}+\frac{i_2}{b^3}+\cdots \]
The van der Corput construction turns a fixed-size grid into an extensible sequence
For \(i=0,1,\ldots\):
| Family | Construction | Typical preferred size |
|---|---|---|
| Kronecker | \(x_i=i\vzeta\bmod1\) | none |
| Lattice | \(x_i=\phi_b(i)\vzeta\bmod1\) | \(n=b^m\) |
| Halton | \(x_i=\{\phi_2(i),\phi_3(i),\phi_5(i),\ldots\}\) | none |
| Digital | \(x_i=i_0c_0\oplus i_1c_1\oplus\cdots\) | \(n=b^m\) |
Here \(\oplus\) denotes digitwise addition modulo \(b\)
Different families balance extensibility, projection quality, and fast construction
Generators for lattice rules and digital nets may be chosen by number theory or computer search
The component-by-component strategy:
A high-dimensional design is built by optimizing one coordinate at a time
Randomization aims for
\[ X_{\mathrm{rand},i}\sim\Unif[0,1]^d \]
for every \(i\), while preserving low discrepancy jointly
Independent randomizations provide replications without reverting to IID point placement
The estimator
\[ \hmu_n=\frac1n\sum_{i=0}^{n-1}f(X_{\mathrm{rand},i}) \]
is unbiased for \(\Ex[f(X)]\), \(X\sim\Unif[0,1]^d\)
Randomized nodes almost surely avoid the boundary, so Gaussian quantile transforms remain finite
Independent randomizations allow variance and error estimation
For smooth integrands, scrambled digital sequences can attain nearly \(\Order(n^{-3/2})\) RMSE
Randomization combines QMC space filling with Monte Carlo uncertainty quantification
Two practical questions are
\[\begin{align*} \abs{\mu-\hmu_n}&\le\text{what?}\\ \abs{\mu-\hmu_n}&\le\varepsilon \quad\text{for what $n$?} \end{align*}\]
We want answers based only on the computed data \(f(x_0),f(x_1),\ldots\)
An efficient method needs both fast error decay and a reliable stopping rule
© 2026 Fred J. Hickernell · Illinois Tech · assisted by ChatGPT and Codex · Improving Efficiency · MATH 565 — Fall 2026 Website · \(\exstar\) = exercise