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Chapter 5 • Theory & Derivations

Variance Reduction Techniques

Principles of variance reduction: efficiency improvements without increasing computational sample size; Antithetic Variates Method: negative covariance induction Cov(h(U), h(1-U)) < 0, monotonic transformations, and efficiency ratio; Control Variates Method: auxiliary variable Y with known expectation mu_Y, optimal correlation coefficient c* = Cov(X,Y)/Var(Y), and percentage variance reduction; Importance Sampling: change of measure, Radon-Nikodym derivative / likelihood ratio w(x) = f(x)/g(x), and optimal instrumental density g*(x) proportional to |h(x)|f(x); Stratified Sampling: proportional vs Neyman optimal allocation of sample points across strata, and variance decomposition; Latin Hypercube Sampling (LHS): multidimensional stratification, orthogonal arrays, and variance bounds.

§5.1 The Antithetic Variates Method

The Antithetic Variates Method, introduced by J. M. Hammersley and K. W. Morton in 1956, reduces variance by deliberately inducing negative covariance between pairs of simulation draws.

1. Mathematical Formulation

Suppose we desire to estimate $\theta = \mathbb{E}[h(U)]$, where $U \sim \mathcal{U}(0, 1)$. Because the standard uniform distribution is symmetric about $0.5$, if $U \sim \mathcal{U}(0, 1)$, its complement $1 - U$ is also identically distributed as $\mathcal{U}(0, 1)$.

Instead of drawing $2n$ independent uniform variates, the antithetic approach generates $n$ independent pairs $(U_i, 1 - U_i)$ and evaluates:

$$\hat{\theta}_{\text{anti}} = \frac{1}{2n} \sum_{i=1}^n \left[ h(U_i) + h(1 - U_i) \right] = \frac{1}{n} \sum_{i=1}^n \left( \frac{h(U_i) + h(1 - U_i)}{2} \right)$$

Let $W_i = \frac{1}{2} [h(U_i) + h(1 - U_i)]$. By linearity of expectation:

$$\mathbb{E}[\hat{\theta}_{\text{anti}}] = \mathbb{E}[W_i] = \frac{1}{2} (\mathbb{E}[h(U)] + \mathbb{E}[h(1 - U)]) = \frac{1}{2} (\theta + \theta) = \theta$$

The estimator is strictly unbiased.

2. Variance Analysis and Monotonicity

The variance of the antithetic average is:

$$\mathrm{Var}(W_i) = \mathrm{Var}\left( \frac{h(U) + h(1 - U)}{2} \right) = \frac{1}{4} \left[ \mathrm{Var}(h(U)) + \mathrm{Var}(h(1 - U)) + 2\,\mathrm{Cov}(h(U), h(1 - U)) \right]$$

Since $1 - U \stackrel{d}{=} U$, $\mathrm{Var}(h(1 - U)) = \mathrm{Var}(h(U)) = \sigma^2$. Therefore:

$$\mathrm{Var}(W_i) = \frac{\sigma^2}{2} + \frac{1}{2} \mathrm{Cov}(h(U), h(1 - U))$$

Comparing $\hat{\theta}_{\text{anti}}$ (which uses $2n$ function evaluations across $n$ pairs) with a Crude Monte Carlo estimator $\hat{\theta}_{\text{crude}}$ based on $2n$ independent draws:

$$\mathrm{Var}(\hat{\theta}_{\text{crude}}) = \frac{\sigma^2}{2n}$$
$$\mathrm{Var}(\hat{\theta}_{\text{anti}}) = \frac{\mathrm{Var}(W_i)}{n} = \frac{\sigma^2}{2n} + \frac{\mathrm{Cov}(h(U), h(1 - U))}{2n}$$

The antithetic estimator achieves strict variance reduction ($\mathrm{Var}(\hat{\theta}_{\text{anti}}) < \mathrm{Var}(\hat{\theta}_{\text{crude}})$) if and only if the covariance between paired evaluations is strictly negative:

$$\mathrm{Cov}(h(U), h(1 - U)) < 0$$

Theorem (Monotonic Functions Guarantee Variance Reduction): If $h(u)$ is a monotonically increasing (or monotonically decreasing) function on $[0, 1]$, then:

$$\mathrm{Cov}(h(U), h(1 - U)) \le 0$$

with strict inequality unless $h(u)$ is a constant almost everywhere.

Proof Intuition: If $h$ is monotonic increasing, when $U$ is larger than $0.5$, $h(U)$ is above average, while $1 - U$ is smaller than $0.5$, forcing $h(1 - U)$ below average. Deviations from the mean systematically cancel out, producing substantial variance suppression.

§5.2 The Control Variates Method

The Control Variates Method reduces variance by incorporating an auxiliary random variable $Y$ whose expectation $\mu_Y = \mathbb{E}[Y]$ is analytically known.

1. The Controlled Estimator

Suppose we want to estimate $\theta = \mathbb{E}[X]$, where $X = h(\mathbf{U})$. Let $Y = g(\mathbf{U})$ be an auxiliary variable simulated simultaneously with $X$, where $\mu_Y = \mathbb{E}[Y]$ is known exactly in closed form.

For any real scalar coefficient $c \in \mathbb{R}$, define the controlled estimator:

$$\hat{\theta}_c = X - c (Y - \mu_Y)$$

We evaluate its properties:

1. Unbiasedness:

$$\mathbb{E}[\hat{\theta}_c] = \mathbb{E}[X] - c (\mathbb{E}[Y] - \mu_Y) = \theta - c(0) = \theta$$

The controlled estimator is unconditionally unbiased for any arbitrary value of $c \in \mathbb{R}$!

2. Variance:

$$\mathrm{Var}(\hat{\theta}_c) = \mathrm{Var}(X - c Y) = \mathrm{Var}(X) - 2 c\,\mathrm{Cov}(X, Y) + c^2\,\mathrm{Var}(Y)$$

2. Derivation of the Optimal Control Coefficient $c^*$

To minimize $\mathrm{Var}(\hat{\theta}_c)$ with respect to $c$, differentiate and set to zero:

$$\frac{d}{dc} \mathrm{Var}(\hat{\theta}_c) = -2\,\mathrm{Cov}(X, Y) + 2 c\,\mathrm{Var}(Y) = 0$$

Solving for the optimal coefficient $c^*$:

$$c^* = \frac{\mathrm{Cov}(X, Y)}{\mathrm{Var}(Y)} = \rho_{XY} \frac{\sigma_X}{\sigma_Y}$$

Notice that $c^*$ is mathematically identical to the population regression slope of $X$ on $Y$ in ordinary least squares regression!

Substituting $c^*$ into the variance equation:

$$\mathrm{Var}(\hat{\theta}_{c^*}) = \sigma_X^2 - 2 \left( \frac{\mathrm{Cov}(X, Y)}{\sigma_Y^2} \right) \mathrm{Cov}(X, Y) + \left( \frac{\mathrm{Cov}(X, Y)}{\sigma_Y^2} \right)^2 \sigma_Y^2$$
$$\mathrm{Var}(\hat{\theta}_{c^*}) = \sigma_X^2 - \frac{[\mathrm{Cov}(X, Y)]^2}{\sigma_Y^2} = \sigma_X^2 \left( 1 - \frac{[\mathrm{Cov}(X, Y)]^2}{\sigma_X^2 \sigma_Y^2} \right)$$
$$\mathrm{Var}(\hat{\theta}_{c^*}) = \mathrm{Var}(X) \cdot (1 - \rho_{XY}^2)$$

where $\rho_{XY} = \frac{\mathrm{Cov}(X, Y)}{\sigma_X \sigma_Y}$ is the Pearson correlation coefficient between target $X$ and control $Y$.

The percentage variance reduction is exactly equal to the squared correlation coefficient:

$$\text{Percentage Variance Reduction} = \rho_{XY}^2 = R^2$$
  • If $|\rho_{XY}| = 0.90$, variance is reduced by $81\%$ (a 5.26-fold increase in efficiency).
  • If $|\rho_{XY}| = 0.99$, variance is reduced by $98\%$ (a 50-fold efficiency gain).

In practice, $c^*$ is estimated from the simulated sample using sample covariance and sample variance without inducing asymptotic bias.

§5.3 Importance Sampling: Change of Measure and Optimal Proposal

When simulating rare events (e.g., probability of default in financial stress testing $\theta < 10^{-5}$ or nuclear reactor breach probabilities), Crude Monte Carlo is useless because virtually all simulated paths fall in the uninformative normal operating region. Importance Sampling deliberately alters the sampling distribution to force frequent visitations to the critical rare-event region, adjusting each draw by a likelihood ratio weight to preserve strict unbiasedness.

1. Mathematical Formulation: The Radon-Nikodym Derivative

Let $\theta = \mathbb{E}_f[h(\mathbf{X})] = \int_{\mathcal{X}} h(\mathbf{x}) f(\mathbf{x})\,d\mathbf{x}$. Let $g(\mathbf{x})$ be an alternative probability density function (the importance proposal density) satisfying the absolute continuity condition:

$$h(\mathbf{x}) f(\mathbf{x}) \ne 0 \implies g(\mathbf{x}) > 0$$

We rewrite the integral by multiplying and dividing by $g(\mathbf{x})$:

$$\theta = \int_{\mathcal{X}} h(\mathbf{x}) \frac{f(\mathbf{x})}{g(\mathbf{x})} g(\mathbf{x})\,d\mathbf{x} = \int_{\mathcal{X}} \left[ h(\mathbf{x}) w(\mathbf{x}) \right] g(\mathbf{x})\,d\mathbf{x} = \mathbb{E}_g\left[ h(\mathbf{X}) w(\mathbf{X}) \right]$$

where:

$$w(\mathbf{x}) = \frac{f(\mathbf{x})}{g(\mathbf{x})}$$

is the likelihood ratio or importance weight (the Radon-Nikodym derivative $df / dg$).

The Importance Sampling Estimator based on $n$ independent draws $\mathbf{Y}_1, \dots, \mathbf{Y}_n \sim g(\mathbf{y})$ is:

$$\hat{\theta}_{\text{IS}} = \frac{1}{n} \sum_{i=1}^n h(\mathbf{Y}_i) w(\mathbf{Y}_i) = \frac{1}{n} \sum_{i=1}^n h(\mathbf{Y}_i) \frac{f(\mathbf{Y}_i)}{g(\mathbf{Y}_i)}$$

By construction, $\mathbb{E}_g[\hat{\theta}_{\text{IS}}] = \theta$, guaranteeing strict unbiasedness.

2. The Zero-Variance Optimal Proposal Density

The variance of the importance sampling estimator is:

$$\mathrm{Var}_g(\hat{\theta}_{\text{IS}}) = \frac{1}{n} \mathrm{Var}_g(h(\mathbf{Y}) w(\mathbf{Y})) = \frac{1}{n} \left[ \mathbb{E}_g\left( h(\mathbf{Y})^2 w(\mathbf{Y})^2 \right) - \theta^2 \right] = \frac{1}{n} \left[ \int_{\mathcal{X}} \frac{h(\mathbf{x})^2 f(\mathbf{x})^2}{g(\mathbf{x})}\,d\mathbf{x} - \theta^2 \right]$$

Theorem (Optimal Zero-Variance Importance Density): If $h(\mathbf{x}) \ge 0$, the importance proposal density $g^*(\mathbf{x})$ that minimizes the variance satisfies:

$$g^*(\mathbf{x}) = \frac{h(\mathbf{x}) f(\mathbf{x})}{\int_{\mathcal{X}} h(\mathbf{u}) f(\mathbf{u})\,d\mathbf{u}} = \frac{h(\mathbf{x}) f(\mathbf{x})}{\theta}$$

Under this optimal density, the variance is identically zero:

$$\mathrm{Var}_{g^*}(\hat{\theta}_{\text{IS}}) \equiv 0$$

Proof: Under $g^*(\mathbf{x})$, the importance weight is $w(\mathbf{x}) = \frac{f(\mathbf{x})}{g^*(\mathbf{x})} = \frac{\theta}{h(\mathbf{x})}$. Therefore, for every single candidate draw $\mathbf{Y}_i$:

$$h(\mathbf{Y}_i) w(\mathbf{Y}_i) = h(\mathbf{Y}_i) \cdot \frac{\theta}{h(\mathbf{Y}_i)} = \theta$$

Every simulated summand evaluates identically to the true integral $\theta$! A single draw ($n = 1$) yields the exact answer without any random variation.

Practical Reality: Because $g^*(\mathbf{x})$ requires knowing the normalizing constant $\theta$ (the very integral we wish to compute), $g^*$ cannot be implemented directly. However, the theorem provides the guiding principle: choose an importance density $g(\mathbf{x})$ that is shaped proportionally to $|h(\mathbf{x})| f(\mathbf{x})$, concentrating samples where the integrand is largest.

Cautionary Principle on Heavy Tails: The proposal $g(\mathbf{x})$ must have heavier tails than $f(\mathbf{x})$. If $g(\mathbf{x})$ decays faster than $f(\mathbf{x})$ in the tails, the likelihood ratio $w(\mathbf{x}) = f(\mathbf{x})/g(\mathbf{x}) \to \infty$, causing the variance $\mathbb{E}_g[h^2 w^2]$ to become infinite!

§5.4 Stratified Sampling and Latin Hypercube Sampling (LHS)

1. Stratified Sampling

Stratified sampling partitions the integration domain $\mathcal{D}$ into $K$ mutually disjoint strata $\mathcal{D}_1, \dots, \mathcal{D}_K$ such that $\bigcup_{k=1}^K \mathcal{D}_k = \mathcal{D}$ and $\mathcal{D}_j \cap \mathcal{D}_k = \emptyset$. Let $p_k = P(\mathbf{X} \in \mathcal{D}_k) = \int_{\mathcal{D}_k} f(\mathbf{x})\,d\mathbf{x}$ be the stratum probability weight, and let $\theta_k = \mathbb{E}[h(\mathbf{X}) \mid \mathbf{X} \in \mathcal{D}_k]$ be the conditional expectation.

The overall expectation decomposes as:

$$\theta = \sum_{k=1}^K p_k \theta_k$$

In each stratum $k$, draw $n_k$ independent samples from the conditional density $f(\mathbf{x} \mid \mathbf{X} \in \mathcal{D}_k)$, and compute the stratified estimator:

$$\hat{\theta}_{\text{strat}} = \sum_{k=1}^K p_k \hat{\theta}_k, \quad \text{where } \hat{\theta}_k = \frac{1}{n_k} \sum_{i=1}^{n_k} h(\mathbf{X}_{ki})$$

The total variance decomposes via the Law of Total Variance:

$$\mathrm{Var}(\hat{\theta}_{\text{strat}}) = \sum_{k=1}^K \frac{p_k^2 \sigma_k^2}{n_k}$$

where $\sigma_k^2 = \mathrm{Var}(h(\mathbf{X}) \mid \mathbf{X} \in \mathcal{D}_k)$ is the within-stratum variance. Stratified sampling completely eliminates the between-stratum variance $\sum p_k (\theta_k - \theta)^2$ from the estimation error!

Under Neyman Optimal Allocation, the sample allocation per stratum that minimizes total variance for fixed total sample size $n = \sum n_k$ is:

$$n_k^* = n \frac{p_k \sigma_k}{\sum_{j=1}^K p_j \sigma_j}$$

2. Latin Hypercube Sampling (LHS)

Developed by McKay, Beckman, and Conover (1979), Latin Hypercube Sampling extends stratification to $d$-dimensional hypercubes. For $n$ simulation runs in $d$ dimensions:

  • The range of each coordinate variable $j \in \{1, \dots, d\}$ is partitioned into $n$ equiprobable intervals: $[0, 1/n), [1/n, 2/n), \dots, [(n-1)/n, 1]$.
  • In each interval along each dimension, exactly one point is placed.
  • The coordinates are matched across dimensions using random independent permutations $\pi_j \in S_n$:
$$\mathbf{U}_i = \left( \frac{\pi_1(i) - 1 + V_{1i}}{n}, \; \frac{\pi_2(i) - 1 + V_{2i}}{n}, \; \dots, \; \frac{\pi_d(i) - 1 + V_{di}}{n} \right)$$

where $V_{ji} \sim \mathcal{U}(0, 1)$. LHS guarantees that the marginal projections onto every single one-dimensional coordinate axis are completely uniform and non-overlapping, eliminating clustering and significantly reducing variance for additive integrands.

§5.9 Stratified Sampling and Optimal Neyman Allocation

In standard random sampling from a heterogeneous population or sample space $\Omega$, sample draws tend to cluster randomly in certain regions while under-representing others. Stratified Sampling partitions the sample space $\Omega$ into $K$ mutually exclusive and exhaustive strata:

$$\Omega = \Omega_1 \cup \Omega_2 \cup \dots \cup \Omega_K, \quad \Omega_j \cap \Omega_k = \emptyset \quad \forall j \ne k$$

Let $p_k = P(\mathbf{X} \in \Omega_k) = \int_{\Omega_k} f(\mathbf{x})\,d\mathbf{x}$ be the known probability weight of stratum $k$, such that $\sum_{k=1}^K p_k = 1$. The target quantity $\theta = \mathbb{E}[h(\mathbf{X})]$ can be decomposed by the law of total expectation:

$$\theta = \sum_{k=1}^K p_k \, \mathbb{E}[h(\mathbf{X}) \mid \mathbf{X} \in \Omega_k] = \sum_{k=1}^K p_k \, \mu_k$$

where $\mu_k = \mathbb{E}[h(\mathbf{X}) \mid \mathbf{X} \in \Omega_k]$ is the conditional stratum mean.

1. The Stratified Estimator

To estimate $\theta$, we allocate a total sample size $n = \sum_{k=1}^K n_k$ by drawing $n_k$ independent samples $\mathbf{X}_{k, 1}, \dots, \mathbf{X}_{k, n_k}$ strictly from the conditional density within stratum $k$:

$$f(\mathbf{x} \mid \mathbf{X} \in \Omega_k) = \frac{f(\mathbf{x}) \mathbb{I}(\mathbf{x} \in \Omega_k)}{p_k}$$

The stratified Monte Carlo estimator is:

$$\hat{\theta}_{\text{strat}} = \sum_{k=1}^K p_k \bar{Y}_k = \sum_{k=1}^K \frac{p_k}{n_k} \sum_{i=1}^{n_k} h(\mathbf{X}_{k, i})$$

Because each conditional sample average $\bar{Y}_k$ is an unbiased estimator of $\mu_k$, $\hat{\theta}_{\text{strat}}$ is strictly unbiased: $\mathbb{E}[\hat{\theta}_{\text{strat}}] = \sum_{k=1}^K p_k \mu_k = \theta$. Furthermore, because draws across distinct strata are generated independently, the variance is strictly additive:

$$\mathrm{Var}(\hat{\theta}_{\text{strat}}) = \sum_{k=1}^K p_k^2 \, \mathrm{Var}(\bar{Y}_k) = \sum_{k=1}^K \frac{p_k^2 \sigma_k^2}{n_k}$$

where $\sigma_k^2 = \mathrm{Var}(h(\mathbf{X}) \mid \mathbf{X} \in \Omega_k)$ is the within-stratum variance.

2. Proportional vs. Optimal Neyman Allocation

The variance of $\hat{\theta}_{\text{strat}}$ depends decisively on the choice of allocation $\{n_1, \dots, n_K\}$ subject to $\sum_{k=1}^K n_k = n$:

1. Proportional Allocation: Sample sizes are assigned in exact proportion to stratum probability weights: $n_k = n \cdot p_k$. Under proportional allocation:

$$\mathrm{Var}(\hat{\theta}_{\text{prop}}) = \sum_{k=1}^K \frac{p_k^2 \sigma_k^2}{n p_k} = \frac{1}{n} \sum_{k=1}^K p_k \sigma_k^2$$

Recall the Law of Total Variance:

$$\mathrm{Var}(h(\mathbf{X})) = \mathbb{E}[\mathrm{Var}(h(\mathbf{X}) \mid \text{stratum})] + \mathrm{Var}(\mathbb{E}[h(\mathbf{X}) \mid \text{stratum}]) = \sum_{k=1}^K p_k \sigma_k^2 + \sum_{k=1}^K p_k (\mu_k - \theta)^2$$

Comparing Crude Monte Carlo to proportional stratification:

$$\mathrm{Var}(\hat{\theta}_{\text{crude}}) - \mathrm{Var}(\hat{\theta}_{\text{prop}}) = \frac{1}{n} \sum_{k=1}^K p_k (\mu_k - \theta)^2 \ge 0$$

Proportional stratification is guaranteed to eliminate all between-stratum variation from the simulation error without any risk of increasing variance.

2. Optimal Neyman Allocation: Jerzy Neyman (1934) solved the constrained optimization problem of minimizing $\sum_{k=1}^K \frac{p_k^2 \sigma_k^2}{n_k}$ subject to $\sum_{k=1}^K n_k = n$ using Lagrange multipliers:

$$n_k^* = n \left( \frac{p_k \sigma_k}{\sum_{j=1}^K p_j \sigma_j} \right)$$

Under Neyman allocation, more computational effort is allocated to strata that are either more probable ($p_k$ is large) or more volatile ($\sigma_k$ is large). The resulting minimal variance is:

$$\mathrm{Var}(\hat{\theta}_{\text{Neyman}}) = \frac{1}{n} \left( \sum_{k=1}^K p_k \sigma_k \right)^2$$

By Cauchy-Schwarz, $\left( \sum p_k \sigma_k \right)^2 \le \sum p_k \sigma_k^2$, guaranteeing that Neyman allocation always achieves lower or equal variance compared to proportional allocation.

§5.10 Latin Hypercube Sampling (LHS) and Orthogonal Arrays

When simulating complex computer experiments or financial models in $d$ dimensions, classical stratification becomes infeasible: dividing each coordinate axis into $K$ intervals generates $K^d$ multidimensional strata. If $d = 20$ and $K = 10$, $K^d = 10^{20}$ cells, exceeding any feasible computational sample size $n$.

Michael D. McKay, Richard J. Beckman, and William J. Conover (1979) invented Latin Hypercube Sampling (LHS) to extend stratified sampling to arbitrary dimensions without combinatorial cell explosion.

1. Construction of a Latin Hypercube Sample

To generate an LHS design of size $n$ in $d$ dimensions:

  1. Divide the domain of each coordinate $j \in \{1, \dots, d\}$ into $n$ equiprobable intervals:
$$I_{j, k} = \left[ \frac{k-1}{n}, \frac{k}{n} \right], \quad k = 1, 2, \dots, n$$
  1. For each dimension $j$, generate an independent uniform random permutation of the integers $\{1, 2, \dots, n\}$, denoted $\pi_j = (\pi_j(1), \dots, \pi_j(n))$.
  2. Generate $n \times d$ independent uniform draws $U_{j, i} \sim \mathcal{U}(0, 1)$.
  3. Construct the $i$-th simulation point $\mathbf{X}_i = (X_{1, i}, \dots, X_{d, i})^T$ where:
$$X_{j, i} = \frac{\pi_j(i) - 1 + U_{j, i}}{n}$$

By construction, each interval $I_{j, k}$ of every coordinate axis contains exactly one sample point. When projected onto any 1-dimensional coordinate margin, the Latin Hypercube sample provides perfect 1D stratification.

2. Variance Reduction Properties of LHS

Michael Stein (1987) established the asymptotic distribution of Latin Hypercube estimators. Let $f: [0, 1]^d \to \mathbb{R}$ be square-integrable, and consider its additive main-effects approximation:

$$f_{\text{add}}(\mathbf{x}) = \mu + \sum_{j=1}^d [f_j(x_j) - \mu]$$

where $f_j(x_j) = \mathbb{E}[f(\mathbf{X}) \mid X_j = x_j]$. Let the residual non-additive component be:

$$r(\mathbf{x}) = f(\mathbf{x}) - f_{\text{add}}(\mathbf{x})$$

Theorem (Stein's Theorem): As $n \to \infty$, the Latin Hypercube estimator satisfies:

$$\sqrt{n}(\hat{\theta}_{\text{LHS}} - \theta) \xrightarrow{d} \mathcal{N}\left( 0, \int_{[0, 1]^d} r^2(\mathbf{x})\,d\mathbf{x} \right)$$

Whereas Crude Monte Carlo has asymptotic variance equal to the total variance $\mathrm{Var}(f(\mathbf{X})) = \mathrm{Var}(f_{\text{add}}) + \mathrm{Var}(r)$, Latin Hypercube Sampling asymptotically purges $100\%$ of the additive main-effects variance, leaving only non-linear interactions in the error term. If the target model is predominantly additive, LHS achieves order-of-magnitude variance reductions at zero additional computational cost.

§5.11 Adaptive Importance Sampling and the Cross-Entropy Method

Standard Importance Sampling requires specifying a proposal density $g(\mathbf{x})$ in advance. If $g(\mathbf{x})$ fails to cover the heavy tails of the integrand, the importance weights $w(\mathbf{x}) = f(\mathbf{x}) / g(\mathbf{x})$ can become unbounded, resulting in infinite variance.

Adaptive Importance Sampling (AIS) and the Cross-Entropy (CE) method, developed by Reuven Rubinstein (1997), iteratively optimize the proposal density $g(\mathbf{x}; \mathbf{v})$ within a parametric family to minimize the Kullback-Leibler (KL) divergence to the theoretical zero-variance optimal density $g^*(\mathbf{x}) \propto |h(\mathbf{x})| f(\mathbf{x})$.

1. Kullback-Leibler Divergence Minimization

The theoretical optimal density is $g^*(\mathbf{x}) = \frac{h(\mathbf{x}) f(\mathbf{x})}{\theta}$. The Cross-Entropy between $g^*$ and a parametric proposal $g(\mathbf{x}; \mathbf{v})$ is:

$$\mathcal{D}_{\text{KL}}(g^* \parallel g(\cdot; \mathbf{v})) = \int g^*(\mathbf{x}) \log \left( \frac{g^*(\mathbf{x})}{g(\mathbf{x}; \mathbf{v})} \right) d\mathbf{x} = \int g^*(\mathbf{x}) \log g^*(\mathbf{x})\,d\mathbf{x} - \int g^*(\mathbf{x}) \log g(\mathbf{x}; \mathbf{v})\,d\mathbf{x}$$

Minimizing this divergence is mathematically equivalent to maximizing the objective:

$$\max_{\mathbf{v}} D(\mathbf{v}) = \int \frac{h(\mathbf{x}) f(\mathbf{x})}{\theta} \log g(\mathbf{x}; \mathbf{v})\,d\mathbf{x} \iff \max_{\mathbf{v}} \mathbb{E}_{\mathbf{w}} \left[ h(\mathbf{X}) \frac{f(\mathbf{X})}{g(\mathbf{X}; \mathbf{w})} \log g(\mathbf{X}; \mathbf{v}) \right]$$

where $\mathbf{w}$ is an arbitrary reference parameter vector.

2. Multi-Stage Iterative Optimization

In rare-event simulation (e.g., estimating $P(S(\mathbf{X}) \ge \gamma)$ where $\gamma$ is a catastrophic risk threshold and the event occurs with probability $10^{-7}$), sampling directly from $f(\mathbf{x})$ produces zero hits. The CE method constructs a sequence of intermediate thresholds $\gamma_1 < \gamma_2 < \dots < \gamma_T = \gamma$ and parameter vectors $\mathbf{v}_1, \dots, \mathbf{v}_T$:

  1. Initialize $\mathbf{v}_0$ matching the prior distribution.
  2. At iteration $t$, generate $N$ draws from $g(\mathbf{x}; \mathbf{v}_{t-1})$.
  3. Set the intermediate threshold $\gamma_t$ as the $(1 - \rho)$-quantile of the sample scores (typically $\rho = 0.10$).
  4. Update the parameter vector $\mathbf{v}_t$ by solving the empirical maximization:
$$\mathbf{v}_t = \arg\max_{\mathbf{v}} \frac{1}{N} \sum_{i=1}^N \mathbb{I}(S(\mathbf{X}_i) \ge \gamma_t) \frac{f(\mathbf{X}_i)}{g(\mathbf{X}_i; \mathbf{v}_{t-1})} \log g(\mathbf{X}_i; \mathbf{v})$$

When $g(\mathbf{x}; \mathbf{v})$ belongs to a Natural Exponential Family (such as multivariate Gaussian), this update admits a closed-form analytical solution for the sample mean and covariance matrix.

  1. Repeat until $\gamma_t \ge \gamma$, at which point the final proposal $g(\mathbf{x}; \mathbf{v}_T)$ samples rare events with high frequency, reducing the relative variance of rare-event estimators from thousands to near zero.

§5.12 Bridge Sampling and Path Sampling for Normalizing Constants

In Bayesian model selection and hypothesis testing, the fundamental challenge is computing the marginal likelihood (also called the evidence) of model $\mathcal{M}$:

$$Z = p(\mathbf{y} \mid \mathcal{M}) = \int_{\boldsymbol{\Theta}} p(\mathbf{y} \mid \boldsymbol{\theta}, \mathcal{M}) p(\boldsymbol{\theta} \mid \mathcal{M})\,d\boldsymbol{\theta}$$

Evaluating $Z$ is essential for calculating Bayes factors $B_{12} = Z_1 / Z_2$ and posterior model probabilities $P(\mathcal{M}_k \mid \mathbf{y})$. However, standard MCMC draws only from the unnormalized posterior $p(\boldsymbol{\theta} \mid \mathbf{y}) \propto q(\boldsymbol{\theta}) = p(\mathbf{y} \mid \boldsymbol{\theta}) p(\boldsymbol{\theta})$, where $Z$ cancels out of the Metropolis acceptance ratio.

Naive Monte Carlo integration using prior draws $\hat{Z}_{\text{prior}} = \frac{1}{N} \sum p(\mathbf{y} \mid \boldsymbol{\theta}^{(i)})$ has catastrophic variance because the posterior mass occupies a tiny fraction of the prior support. Conversely, the harmonic mean estimator of Newton and Raftery (1994) $\hat{Z}_{\text{HM}} = \left[ \frac{1}{N} \sum p(\mathbf{y} \mid \boldsymbol{\theta}^{(i)})^{-1} \right]^{-1}$ has infinite variance in virtually all non-trivial models.

Andrew Gelman and Xiao-Li Meng (1998) introduced Bridge Sampling and Path Sampling to provide robust, finite-variance estimators of normalizing constants.

1. Mathematical Formulation of Bridge Sampling

Let $p_1(\boldsymbol{\theta}) = q_1(\boldsymbol{\theta}) / Z_1$ be the target posterior and $p_0(\boldsymbol{\theta}) = q_0(\boldsymbol{\theta}) / Z_0$ be a tractable proposal distribution (such as a multivariate Gaussian or Student-$t$ distribution matching the posterior mean and covariance).

Let $\alpha(\boldsymbol{\theta})$ be an arbitrary bridge function. By multiplying and dividing by $\alpha$:

$$\int \alpha(\boldsymbol{\theta}) q_1(\boldsymbol{\theta}) p_0(\boldsymbol{\theta})\,d\boldsymbol{\theta} = \int \alpha(\boldsymbol{\theta}) \frac{q_1(\boldsymbol{\theta}) q_0(\boldsymbol{\theta})}{Z_0}\,d\boldsymbol{\theta}$$
$$\int \alpha(\boldsymbol{\theta}) q_0(\boldsymbol{\theta}) p_1(\boldsymbol{\theta})\,d\boldsymbol{\theta} = \int \alpha(\boldsymbol{\theta}) \frac{q_0(\boldsymbol{\theta}) q_1(\boldsymbol{\theta})}{Z_1}\,d\boldsymbol{\theta}$$

Taking the ratio of these two expectations yields the exact bridge identity:

$$\frac{Z_1}{Z_0} = \frac{\mathbb{E}_{p_0}\left[ \alpha(\boldsymbol{\theta}) q_1(\boldsymbol{\theta}) \right]}{\mathbb{E}_{p_1}\left[ \alpha(\boldsymbol{\theta}) q_0(\boldsymbol{\theta}) \right]}$$

Given $N_0$ independent draws from $p_0$ and $N_1$ MCMC draws from $p_1$:

$$\frac{\widehat{Z_1}}{Z_0} = \frac{\frac{1}{N_0} \sum_{i=1}^{N_0} \alpha(\boldsymbol{\theta}_{0, i}) q_1(\boldsymbol{\theta}_{0, i})}{\frac{1}{N_1} \sum_{j=1}^{N_1} \alpha(\boldsymbol{\theta}_{1, j}) q_0(\boldsymbol{\theta}_{1, j})}$$

2. The Optimal Bridge Function

Meng and Wong (1996) proved that the bridge function $\alpha^*(\boldsymbol{\theta})$ minimizing the asymptotic relative mean squared error is:

$$\alpha^*(\boldsymbol{\theta}) \propto \frac{1}{N_0 p_0(\boldsymbol{\theta}) + N_1 p_1(\boldsymbol{\theta})} = \frac{1}{N_0 q_0(\boldsymbol{\theta}) + N_1 \frac{Z_0}{Z_1} q_1(\boldsymbol{\theta})}$$

Because the optimal function depends on the unknown ratio $r = Z_1 / Z_0$, bridge sampling solves for $r$ via fixed-point iteration:

$$r^{(t+1)} = \frac{\frac{1}{N_0} \sum_{i=1}^{N_0} \frac{q_1(\boldsymbol{\theta}_{0, i})}{N_0 q_0(\boldsymbol{\theta}_{0, i}) + N_1 r^{(t)} q_1(\boldsymbol{\theta}_{0, i})}}{\frac{1}{N_1} \sum_{j=1}^{N_1} \frac{q_0(\boldsymbol{\theta}_{1, j})}{N_0 q_0(\boldsymbol{\theta}_{1, j}) + N_1 r^{(t)} q_1(\boldsymbol{\theta}_{1, j})}}$$

This fixed-point iteration converges in 5 to 10 steps, providing numerically stable estimates of marginal likelihoods with bounded variance.

§5.13 Regularized Multi-Response Control Variates

In large-scale risk simulations and derivative portfolio valuations, analysts frequently possess dozens or hundreds of candidate control variates $\mathbf{Y} = (Y_1, \dots, Y_P)^T$ with known expectations $\mathbb{E}[\mathbf{Y}] = \boldsymbol{\mu}_Y$. The multi-control estimator is:

$$\hat{\theta}_{\mathbf{c}} = \bar{X} - \mathbf{c}^T (\bar{\mathbf{Y}} - \boldsymbol{\mu}_Y)$$

When the number of controls $P$ is large relative to the simulation sample size $n$, estimating the optimal control vector $\mathbf{c}^* = \mathbf{\Sigma}_{YY}^{-1} \mathbf{\Sigma}_{YX}$ via ordinary least squares incurs severe sample estimation error, causing the variance of the controlled estimator to inflate due to loss of degrees of freedom.

1. Degree of Freedom Loss in OLS Control Variates

Lavenberg and Welch (1981) established that when $\mathbf{c}^*$ is estimated from the same simulation sample using sample covariance matrices, the variance of the controlled estimator satisfies:

$$\mathrm{Var}(\hat{\theta}_{\hat{\mathbf{c}}}) = \frac{\sigma^2 (1 - R^2)}{n} \left( \frac{n - 2}{n - P - 2} \right)$$

where $R^2 = \mathbf{\Sigma}_{YX}^T \mathbf{\Sigma}_{YY}^{-1} \mathbf{\Sigma}_{YX} / \sigma^2$ is the squared population multiple correlation coefficient.

The factor $\frac{n - 2}{n - P - 2}$ represents the Estimation Penalty: if $P$ is large (e.g., $P = 50$ controls with $n = 100$ draws), the penalty is $\frac{98}{46} \approx 2.13$, doubling the simulation variance and completely erasing the variance reduction gained from $R^2$.

2. Ridge and Lasso Penalized Control Variates

To solve the high-dimensional control variate problem, modern statistical simulation incorporates regularization from shrinkage regression:

1. Ridge Control Variates: Introduces an $L_2$ penalty on the coefficient vector:

$$\mathbf{c}_{\text{ridge}} = \left( \mathbf{S}_{YY} + \lambda \mathbf{I}_P \right)^{-1} \mathbf{S}_{YX}$$

where $\mathbf{S}_{YY}$ is the sample covariance matrix and $\lambda > 0$ is a tuning parameter selected via cross-validation. Ridge regularization stabilizes matrix inversion when candidate controls are collinear (such as option payoffs evaluated across adjacent strikes).

2. Lasso Control Variates: Solves the penalized optimization:

$$\mathbf{c}_{\text{lasso}} = \arg\min_{\mathbf{c}} \left\{ \frac{1}{n} \sum_{i=1}^n \left( X_i - \bar{X} - \mathbf{c}^T (\mathbf{Y}_i - \boldsymbol{\mu}_Y) \right)^2 + \lambda \sum_{j=1}^P |c_j| \right\}$$

The $L_1$ penalty enforces sparsity: the algorithm automatically selects the sparse subset of auxiliary variables that provide genuine explanatory power for $X$, setting non-contributing control weights to zero and preventing degree-of-freedom degradation.

§5.14 Control Variates with Unknown Mean and Sample Splitting

In complex multi-physics simulations, an auxiliary variable $Y$ may be strongly correlated with the output of interest $X$, but its theoretical expectation $\mu_Y$ is not known analytically. However, $Y$ may represent a low-fidelity, coarse-grid approximation that is computationally cheap to evaluate, whereas $X$ is an expensive high-fidelity simulation.

Sample Splitting Estimator

To exploit $Y$ without bias:

  1. Divide the total computational budget into a large sample of size $M$ for the cheap model $Y$, and a smaller sample of size $N \ll M$ where both $(X_i, Y_i)$ are evaluated simultaneously.
  2. Estimate the mean of $Y$ from the large independent sample:
$$\hat{\mu}_Y = \frac{1}{M} \sum_{k=1}^M Y_k^{\text{cheap}}$$
  1. Construct the controlled estimator on the paired sample:
$$\hat{\theta} = \bar{X}_N - c \left( \bar{Y}_N - \hat{\mu}_Y \right)$$

Because $\mathbb{E}[\bar{Y}_N - \hat{\mu}_Y] = 0$, the estimator is strictly unbiased. The variance is:

$$\mathrm{Var}(\hat{\theta}) = \frac{\sigma_X^2}{N} \left( 1 - \frac{M}{M + N} \rho^2 \right)$$

When $M \gg N$, the variance approaches $\frac{\sigma_X^2(1 - \rho^2)}{N}$, recovering the full variance reduction of a known-mean control variate at negligible extra computational cost.

§5.15 Computational Laboratory: Multi-Asset Variance Reduction Benchmarking

When pricing complex multi-asset options where no single control variate captures total payoff variation, combining multiple variance reduction techniques yields compounding efficiency gains.

Evaluating an out-of-the-money European basket call option:

  • Crude Monte Carlo: Baseline variance $\sigma^2 = 142.8$.
  • Antithetic Variates Alone: Reduces variance to $48.2$ (a $66.2\%$ variance reduction).
  • Control Variates Alone (Geometric Mean Option): Reduces variance to $3.15$ (a $97.8\%$ variance reduction).
  • Combined Antithetic and Control Variate Engine: Slashes variance to $0.42$, achieving a $99.7\%$ variance reduction relative to crude Monte Carlo.

The computational work required to achieve a specified pricing standard error is reduced by a factor of 340, demonstrating the immense value of variance reduction in quantitative finance.

§5.17 Unit 5 Review and Methodological Synthesis

Unit 5 investigated variance reduction techniques designed to minimize simulation estimation variance without introducing bias. We analyzed the Antithetic Variates method of Hammersley and Morton, proving that negative covariance between paired uniform draws $(U, 1-U)$ is guaranteed whenever the response function is monotonic. We formulated the Control Variates method, deriving the optimal coefficient vector $\mathbf{c}^* = \mathbf{\Sigma}_{YY}^{-1} \mathbf{\Sigma}_{YX}$ and demonstrating that variance is reduced by a factor of $1 - R^2$. We explored Stratified Sampling under proportional allocation and Jerzy Neyman's optimal allocation, showing that variance is reduced by shifting computational samples toward strata with larger population weights and internal variances. We introduced Latin Hypercube Sampling (LHS), proving via Stein's theorem that LHS asymptotically eliminates $100\%$ of additive main-effects variance. Finally, we developed Adaptive Importance Sampling, the Cross-Entropy method for rare events, and regularized multi-response control variates for high-dimensional financial derivative pricing.

§5.18 Diagnostic Guidelines for Variance Reduction Techniques

When deploying variance reduction techniques in complex simulation workflows, practitioners should observe four golden diagnostic principles:

1. Verify Covariance Direction in Antithetic Sampling: Always verify that the transformed response is monotonic with respect to each underlying uniform input. If a non-linear payoff is non-monotonic (such as a straddle or butterfly option), pairing $U$ with $1-U$ can yield positive covariance $\mathrm{Cov}(h(U), h(1-U)) > 0$, inadvertently *increasing* estimator variance above crude Monte Carlo.

2. Monitor Condition Numbers in Multi-Control Inversion: When deploying dozens of control variates, compute the condition number $\kappa(\mathbf{S}_{YY})$ of the empirical sample covariance matrix. If $\kappa(\mathbf{S}_{YY}) > 10^6$, multicollinearity among controls will cause the estimated weights $\hat{\mathbf{c}}$ to oscillate erratically, requiring Tikhonov ridge regularization.

3. Check for Finite Variance in Importance Sampling: Never trust an importance sampling estimate without examining the empirical distribution of the unnormalized weights $w_i = f(\mathbf{x}_i) / g(\mathbf{x}_i)$. If the maximum observed weight constitutes more than $10\%$ of the sum of all weights, or if a Pareto tail fit to the weights yields shape parameter $\alpha \le 2$, the importance estimator has infinite variance and will yield deceptively tight, yet invalid, confidence intervals.

4. Evaluate Wall-Clock Efficiency: The true metric of variance reduction is the Efficiency Ratio:

$$\text{Eff} = \frac{\mathrm{Var}(\hat{\theta}_{\text{crude}}) \times \text{CPU Time}_{\text{crude}}}{\mathrm{Var}(\hat{\theta}_{\text{reduced}}) \times \text{CPU Time}_{\text{reduced}}}$$

A technique that slashes variance by $50\%$ but requires four times the CPU time to compute auxiliary transformations is inefficient ($\text{Eff} = 0.5$). Only deploy variance reduction methods that yield net efficiency ratios strictly greater than $1.0$.

THREE WORKED QUESTIONS

Step-by-Step Statistical Simulation and Monte Carlo Solutions

Three original questions connect calculations, definitions and interpretation. Open each solution to follow the reasoning.

Solved Problem Example 5.1: Antithetic Variates for Exponential Decay Integration

An actuary computes the expected discounted payout of a continuous perpetuity:

$$\theta = \int_0^1 e^{-2x}\,dx$$
  1. Evaluate the exact analytical value of $\theta$.
  2. Compute the theoretical variance of the Crude Monte Carlo estimator $\hat{\theta}_{\text{crude}}$ based on $2n$ independent uniform draws from $\mathcal{U}(0, 1)$.
  3. Compute the theoretical variance of the Antithetic Variates estimator $\hat{\theta}_{\text{anti}}$ based on $n$ antithetic pairs $(U_i, 1 - U_i)$, and calculate the percentage variance reduction and efficiency gain.
Exact Analytical Integration

The integrand is $h(x) = e^{-2x}$ on $[0, 1]$:

$$\theta = \int_0^1 e^{-2x}\,dx = \left[ -\frac{1}{2} e^{-2x} \right]_0^1 = -\frac{1}{2} e^{-2} - \left(-\frac{1}{2}\right) = \frac{1 - e^{-2}}{2}$$

Evaluating numerically:

$$e^{-2} = \frac{1}{e^2} \approx \frac{1}{7.389056} \approx 0.1353353$$
$$\theta = \frac{1 - 0.1353353}{2} = \frac{0.8646647}{2} \approx 0.432332$$
Variance of Crude Monte Carlo

Let $U \sim \mathcal{U}(0, 1)$. The second moment is:

$$\mathbb{E}[h(U)^2] = \int_0^1 (e^{-2u})^2 du = \int_0^1 e^{-4u}\,du = \left[ -\frac{1}{4} e^{-4u} \right]_0^1 = \frac{1 - e^{-4}}{4}$$
$$e^{-4} \approx 0.0183156 \implies \mathbb{E}[h(U)^2] = \frac{1 - 0.0183156}{4} = \frac{0.9816844}{4} \approx 0.2454211$$

The population variance is:

$$\sigma^2 = \mathrm{Var}(h(U)) = \mathbb{E}[h(U)^2] - \theta^2 = 0.2454211 - (0.432332)^2$$
$$0.432332^2 \approx 0.186911 \implies \sigma^2 = 0.2454211 - 0.1869110 = 0.058510$$

For $2n$ independent evaluations, the Crude Monte Carlo variance is:

$$\mathrm{Var}(\hat{\theta}_{\text{crude}}) = \frac{\sigma^2}{2n} = \frac{0.058510}{2n} = \frac{0.029255}{n}$$
Variance of Antithetic Estimator and Efficiency Gain

For an antithetic pair $(U, 1 - U)$, the cross-product expectation is:

$$\mathbb{E}[h(U) h(1 - U)] = \int_0^1 e^{-2u} e^{-2(1 - u)}\,du = \int_0^1 e^{-2u - 2 + 2u}\,du = \int_0^1 e^{-2}\,du = e^{-2} \approx 0.1353353$$

The covariance between antithetic evaluations is:

$$\mathrm{Cov}(h(U), h(1 - U)) = \mathbb{E}[h(U) h(1 - U)] - \theta^2 = 0.1353353 - 0.1869110 = -0.0515757$$

Notice that the covariance is strongly negative!

The variance of the antithetic pair average $W = \frac{1}{2}[h(U) + h(1 - U)]$ is:

$$\mathrm{Var}(W) = \frac{\sigma^2}{2} + \frac{1}{2} \mathrm{Cov}(h(U), h(1 - U)) = \frac{0.058510}{2} + \frac{-0.0515757}{2} = 0.029255 - 0.025788 = 0.003467$$

The variance of the antithetic estimator based on $n$ pairs ($2n$ total evaluations) is:

$$\mathrm{Var}(\hat{\theta}_{\text{anti}}) = \frac{\mathrm{Var}(W)}{n} = \frac{0.003467}{n}$$

Comparing the variances for the exact same computational cost ($2n$ evaluations):

$$\text{Variance Ratio} = \frac{\mathrm{Var}(\hat{\theta}_{\text{anti}})}{\mathrm{Var}(\hat{\theta}_{\text{crude}})} = \frac{0.003467 / n}{0.029255 / n} = \frac{0.003467}{0.029255} \approx 0.1185$$
$$\text{Percentage Variance Reduction} = (1 - 0.1185) \times 100\% = 88.15\%$$
$$\text{Efficiency Gain} = \frac{\mathrm{Var}(\hat{\theta}_{\text{crude}})}{\mathrm{Var}(\hat{\theta}_{\text{anti}})} = \frac{0.029255}{0.003467} \approx 8.44$$

Conclusion: The antithetic estimator achieves an 8.44-fold computational speedup, delivering the same precision with less than $12\%$ of the computational evaluations.

Solved Problem Example 5.2: Control Variates for Non-Linear Function Integration

A financial engineer estimates the integral:

$$\theta = \int_0^1 \frac{1}{1 + x}\,dx = \ln(2) \approx 0.693147$$

using Monte Carlo simulation with $U \sim \mathcal{U}(0, 1)$. To reduce variance, the engineer chooses the linear Taylor control variate:

$$Y = 1 - \frac{1}{2} U$$

whose expectation is $\mu_Y = \mathbb{E}[Y] = 1 - \frac{1}{2}(0.5) = 0.750$.

  1. Compute the analytical variance of the target variable $X = \frac{1}{1 + U}$.
  2. Compute the covariance $\mathrm{Cov}(X, Y)$ and variance $\mathrm{Var}(Y)$.
  3. Determine the optimal control coefficient $c^*$, the resulting variance of the controlled estimator, and the percentage variance reduction.
Variance of Target Variable $X$

The target is $X = h(U) = \frac{1}{1 + U}$ for $U \sim \mathcal{U}(0, 1)$:

$$\theta = \mathbb{E}[X] = \int_0^1 \frac{1}{1 + u}\,du = [\ln(1 + u)]_0^1 = \ln(2) \approx 0.693147$$

The second moment is:

$$\mathbb{E}[X^2] = \int_0^1 \left(\frac{1}{1 + u}\right)^2 du = \left[ -\frac{1}{1 + u} \right]_0^1 = -\frac{1}{2} - (-1) = \frac{1}{2} = 0.500$$

The variance is:

$$\mathrm{Var}(X) = \mathbb{E}[X^2] - \theta^2 = 0.500 - (0.693147)^2 = 0.500 - 0.480453 = 0.019547$$
Covariance and Control Variance

The control variate is $Y = 1 - \frac{1}{2}U$:

$$\mathrm{Var}(Y) = \mathrm{Var}\left( 1 - \frac{1}{2}U \right) = \left(-\frac{1}{2}\right)^2 \mathrm{Var}(U) = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48} \approx 0.020833$$

Now compute $\mathbb{E}[X Y]$:

$$X Y = \left(\frac{1}{1 + U}\right) \left( 1 - \frac{1}{2}U \right)$$

Notice that:

$$\frac{1 - \frac{1}{2}u}{1 + u} = \frac{\frac{3}{2} - \frac{1}{2}(1 + u)}{1 + u} = \frac{3/2}{1 + u} - \frac{1}{2}$$

Integrating over $[0, 1]$:

$$\mathbb{E}[X Y] = \int_0^1 \left( \frac{3/2}{1 + u} - \frac{1}{2} \right) du = \frac{3}{2} \ln(2) - \frac{1}{2} = 1.5 \times 0.693147 - 0.500 = 1.039721 - 0.500 = 0.539721$$

The covariance is:

$$\mathrm{Cov}(X, Y) = \mathbb{E}[X Y] - \mathbb{E}[X] \mathbb{E}[Y] = 0.539721 - (0.693147 \times 0.750) = 0.539721 - 0.519860 = 0.019861$$
Optimal Coefficient $c^*$ and Variance Reduction

The optimal coefficient is:

$$c^* = \frac{\mathrm{Cov}(X, Y)}{\mathrm{Var}(Y)} = \frac{0.019861}{1 / 48} = 0.019861 \times 48 \approx 0.953328$$

The squared correlation coefficient is:

$$\rho_{XY}^2 = \frac{[\mathrm{Cov}(X, Y)]^2}{\mathrm{Var}(X) \mathrm{Var}(Y)} = \frac{(0.019861)^2}{0.019547 \times 0.020833} = \frac{0.00039446}{0.00040722} \approx 0.96866 \quad (96.87\%)$$

The variance of the controlled estimator is:

$$\mathrm{Var}(\hat{\theta}_{c^*}) = \mathrm{Var}(X) \cdot (1 - \rho_{XY}^2) = 0.019547 \times (1 - 0.96866) = 0.019547 \times 0.03134 \approx 0.0006126$$

Comparing with the original variance $\mathrm{Var}(X) = 0.019547$:

$$\text{Efficiency Gain} = \frac{0.019547}{0.0006126} \approx 31.9$$

Conclusion: The control variate achieves an astounding 96.87% variance reduction, delivering a 31.9-fold computational efficiency gain using a simple linear approximation!

Solved Problem Example 5.3: Importance Sampling for Extreme Gaussian Tail Estimation

An extreme value analyst estimates the deep tail probability $\theta = P(Z > 4.5)$ for $Z \sim \mathcal{N}(0, 1)$, whose true value is approximately $3.39767 \times 10^{-6}$.

  1. Explain why Crude Monte Carlo with $n = 10^5$ draws will fail to provide any meaningful estimate.
  2. Formulate an Importance Sampling estimator using a shifted Exponential distribution proposal:
$$g(y) = \lambda e^{-\lambda (y - 4.5)} \quad \text{for } y \ge 4.5$$
  1. Determine the optimal parameter $\lambda^*$ and compute the variance of the resulting importance sampling estimator.
Failure of Crude Monte Carlo

Under Crude Monte Carlo:

$$\hat{\theta}_{\text{crude}} = \frac{1}{n} \sum_{i=1}^n I(Z_i > 4.5)$$

The probability of observing a single success in $n = 100,000$ draws is:

$$\mathbb{E}[\text{Number of successes}] = n \theta = 100,000 \times 3.39767 \times 10^{-6} \approx 0.34$$

In roughly $71\%$ of all simulation runs, zero points will fall in the target region ($P(\text{Binomial}(10^5, \theta) = 0) = (1 - \theta)^{10^5} \approx e^{-0.34} \approx 0.712$). The Crude Monte Carlo estimator will output $\hat{\theta} = 0$, reporting that the probability is zero!

Formulation of the Importance Sampling Estimator

The target probability is:

$$\theta = \int_{4.5}^\infty \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,dx$$

We choose a shifted exponential proposal starting at $x_0 = 4.5$:

$$g(x) = \lambda e^{-\lambda(x - 4.5)} \quad \text{for } x \ge 4.5$$

To sample $X \sim g(x)$: draw $U \sim \mathcal{U}(0, 1)$ and set $X = 4.5 - \frac{\ln U}{\lambda}$. The importance weight is:

$$w(x) = \frac{f(x)}{g(x)} = \frac{\frac{1}{\sqrt{2\pi}} e^{-x^2/2}}{\lambda e^{-\lambda(x - 4.5)}} = \frac{1}{\lambda \sqrt{2\pi}} \exp\left( -\frac{x^2}{2} + \lambda(x - 4.5) \right)$$

The Importance Sampling estimator is:

$$\hat{\theta}_{\text{IS}} = \frac{1}{n} \sum_{i=1}^n w(X_i)$$

Because all generated draws $X_i \ge 4.5$ automatically satisfy the condition, every single draw contributes a non-zero, informative weight!

Optimal Parameter $\lambda^*$ and Variance

To make the importance weight $w(x)$ as close to constant as possible, we choose $\lambda$ to match the rate of decay of the normal density at the threshold $x_0 = 4.5$:

$$\frac{d}{dx} \left( -\frac{x^2}{2} \right) \Bigg|_{x = 4.5} = -4.5 \implies \lambda^* = 4.5$$

Substituting $\lambda = 4.5$:

$$w(x) = \frac{1}{4.5 \sqrt{2\pi}} \exp\left( -\frac{x^2}{2} + 4.5 x - 4.5^2 \right) = \frac{1}{4.5 \sqrt{2\pi}} \exp\left( -\frac{(x - 4.5)^2}{2} - \frac{4.5^2}{2} \right)$$
$$w(x) = \frac{e^{-4.5^2 / 2}}{4.5 \sqrt{2\pi}} \exp\left( -\frac{(x - 4.5)^2}{2} \right)$$

Notice that for all $x \ge 4.5$, the term $-\frac{(x - 4.5)^2}{2} \le 0$, which guarantees that the weight is bounded above by:

$$w(x) \le \frac{e^{-10.125}}{4.5 \sqrt{2\pi}} \approx \frac{3.996 \times 10^{-5}}{11.2798} \approx 3.54 \times 10^{-6}$$

Evaluating the second moment:

$$\mathbb{E}_g[w(X)^2] \le (3.54 \times 10^{-6}) \times \theta \approx (3.54 \times 10^{-6}) \times (3.398 \times 10^{-6}) \approx 1.20 \times 10^{-11}$$
$$\mathrm{Var}_g(w(X)) = \mathbb{E}_g[w(X)^2] - \theta^2 \approx 1.20 \times 10^{-11} - (3.398 \times 10^{-6})^2 \approx 1.20 \times 10^{-11} - 1.15 \times 10^{-11} = 5.0 \times 10^{-13}$$

For $n = 10,000$ draws, the standard error of the Importance Sampling estimate is:

$$\text{SE}(\hat{\theta}_{\text{IS}}) = \sqrt{\frac{5.0 \times 10^{-13}}{10,000}} = \sqrt{5.0 \times 10^{-17}} \approx 7.07 \times 10^{-9}$$

The relative error is:

$$\frac{\text{SE}}{\theta} \approx \frac{7.07 \times 10^{-9}}{3.398 \times 10^{-6}} \approx 0.00208 \quad (0.21\%)$$

Conclusion: Importance Sampling achieves an astonishing 0.21% relative error with just 10,000 draws on a $10^{-6}$ rare event where Crude Monte Carlo failed completely!