Phase 13 - Lesson 13.3

Variance Reduction: Antithetic, Control Variates, Stratification, Importance Sampling

Same accuracy, far fewer paths - shrink \(\sigma\) instead of paying the \(1/\sqrt N\) tax.

⏱ 55 min● Advanced🔗 Prereqs: 13.2

↖ Phase 13 hub

Builds on: 13.2 fixed the \(1/\sqrt N\) rate; variance reduction attacks the \(\sigma\) in front of it.

Learning Objectives

Click a status chip to cycle: Not started → In progress → Studied → Practiced → Needs review → Mastered.

Derive the antithetic-variates estimator and prove it lowers variance when the payoff is monotone in the driver.
Construct a control-variate estimator and compute the optimal coefficient \(b^*\).
Explain stratified sampling and why proportional allocation cannot increase variance.
Derive the importance-sampling estimator and its likelihood-ratio weights, and identify the zero-variance ideal.
Empirically compare plain and antithetic variance on a concrete payoff.

Key Vocabulary

Antithetic variates: Pairing each draw \(Z\) with its mirror \(-Z\) so negatively-correlated payoffs average to a lower-variance estimate.

Control variate: A correlated quantity \(C\) with known mean \(\E[C]\) used to correct the estimator: \(Y-b(C-\E[C])\).

Optimal coefficient: The variance-minimizing \(b^*=\Cov(Y,C)/\Var(C)\) for a control variate.

Stratified sampling: Partitioning the sample space into strata and sampling each in fixed proportion to remove between-stratum variance.

Importance sampling: Sampling from a tilted density \(g\) and reweighting by the likelihood ratio \(f/g\) to concentrate effort where the payoff matters.

Likelihood ratio: The weight \(f(X)/g(X)\) that debiases an importance-sampling estimator drawn from \(g\) instead of \(f\).

Intuition & Motivation

Intuition

The Monte Carlo error is \(\sigma/\sqrt N\). Brute force attacks \(N\); variance reduction attacks \(\sigma\) - often far more cheaply. Four ideas recur. Antithetic: for each random path also run its mirror image, so lucky and unlucky draws cancel. Control variate: subtract off a correlated quantity whose mean you already know, removing its noise. Stratification: force the samples to fill the space evenly instead of clumping by chance. Importance sampling: sample where the action is (deep in the tail for a far out-of-the-money option) and reweight. Each keeps the estimator unbiased while shrinking its spread.

Antithetic variates

Write the payoff as \(Y=h(Z)\) for a symmetric driver \(Z\) (e.g. \(Z\sim\Normal(0,1)\), so \(-Z\) has the same law). The antithetic estimator averages the mirrored pair:

\[\hat\theta^{\text{anti}}=\frac1{N}\sum_{i=1}^{N}\frac{h(Z_i)+h(-Z_i)}2,\qquad \Var\!\Big(\tfrac{h(Z)+h(-Z)}2\Big)=\frac{\Var(h(Z))}2\big(1+\rho\big),\] (13.7)

where \(\rho=\mathrm{Corr}(h(Z),h(-Z))\). If \(h\) is monotone then \(\rho\lt 0\) and the paired variance beats two independent draws. Cost is identical (same total number of \(h\) evaluations), so any negative correlation is free variance reduction.

Proposition - When antithetics help

If \(h\) is monotone (nondecreasing or nonincreasing), then \(\Cov(h(Z),h(-Z))\le0\), so the antithetic estimator has variance no larger than the plain estimator with the same number of function evaluations. For non-monotone \(h\) the correlation can be positive and antithetics may hurt.

Control variates

Suppose alongside the target \(Y\) you can cheaply compute a correlated \(C\) with known mean \(\mu_C=\E[C]\). Form

\[\hat\theta^{\text{cv}}=\frac1N\sum_{i=1}^N\big(Y_i-b\,(C_i-\mu_C)\big),\qquad \E[\hat\theta^{\text{cv}}]=\theta\ \ \text{for any }b.\] (13.8)

It stays unbiased for every \(b\). Minimizing its variance over \(b\) gives the regression coefficient

\[b^*=\frac{\Cov(Y,C)}{\Var(C)},\qquad \Var(\hat\theta^{\text{cv}})=\frac{\Var(Y)}N\,(1-\rho_{YC}^2),\] (13.9)

so the variance drops by the factor \(1-\rho_{YC}^2\): a control 90% correlated with the payoff removes about 81% of the variance. In practice estimate \(b^*\) by least squares on the simulated pairs.

Worked Example - Optimal control-variate coefficient

Target \(Y\) (discounted option payoff); control \(C=S_T\) with known \(\E_{\mathbb{Q}}[S_T]=S_0e^{rT}\).

Estimate \(\widehat{\Cov}(Y,C)\) and \(\widehat{\Var}(C)\) from the simulated pairs.

Set \(b^*=\widehat{\Cov}(Y,C)/\widehat{\Var}(C)\).

Report \(\hat\theta^{\text{cv}}=\bar Y-b^*(\bar C-S_0e^{rT})\); its variance is \((1-\hat\rho^2)\) times the plain variance.

Stratified sampling

Partition \([0,1)\) (or the sample space) into \(m\) equal strata and draw exactly \(N/m\) points inside each. Total variance splits as within-plus-between; stratification zeroes out the between-strata part:

\[\Var(Y)=\underbrace{\E[\Var(Y\mid \text{stratum})]}_{\text{within}}+\underbrace{\Var(\E[Y\mid \text{stratum}])}_{\text{between}}.\] (13.10)

With proportional allocation the stratified estimator’s variance equals the within-strata term alone - never larger than plain Monte Carlo, and strictly smaller whenever \(\E[Y\mid\text{stratum}]\) varies across strata.

Importance sampling

To estimate \(\theta=\E_f[h(X)]=\int h(x)f(x)\,dx\) when the important region is rare under \(f\), sample from a tilted \(g\) and reweight:

\[\theta=\int h(x)\frac{f(x)}{g(x)}\,g(x)\,dx=\E_g\!\Big[h(X)\frac{f(X)}{g(X)}\Big],\qquad \hat\theta=\frac1N\sum_{i=1}^N h(X_i)\frac{f(X_i)}{g(X_i)},\ X_i\sim g.\] (13.11)

The estimator is unbiased for any \(g\) with \(g\gt 0\) where \(hf\ne0\). The ideal \(g^*\propto|h|f\) gives zero variance (unreachable, since it needs \(\theta\)), but a \(g\) that mimics \(|h|f\) - e.g. shifting the normal’s mean toward a deep strike - can cut variance by orders of magnitude for rare-event payoffs.

Interactive: antithetic vs plain variance

Common Mistakes to Avoid

Applying antithetics to a strongly non-monotone payoff (e.g. a straddle around the mean), where \(\rho\gt 0\) and variance can increase.
Using a control variate whose mean you only estimate; the method requires \(\E[C]\) known exactly, or the correction is itself biased.
Choosing an importance density \(g\) with lighter tails than \(f\), making the weights \(f/g\) blow up and the variance infinite.
Double-counting the evaluation budget: compare methods at equal cost (same number of payoff evaluations), not equal number of ‘samples’.

Quant Practitioner Tips

Antithetics are almost free to code and help whenever the payoff is monotone in the Gaussian driver - make them your default.
The delta-hedged or underlying \(S_T\) (with known \(\E_{\mathbb{Q}}[S_T]=S_0e^{rT}\)) is a superb control variate for vanilla options.
For deep out-of-the-money or rare-event pricing, importance sampling is often the only method that turns an impossible run into a feasible one.
Estimate \(b^*\) by regressing \(Y\) on \(C\); you get the coefficient and the achieved \(R^2=\rho^2\) (the variance reduction) for free.

Knowledge Check

Q1 Medium

Antithetic variates reduce variance provided the payoff \(h(Z)\) is:

Symmetric about zero

Monotone in the driver \(Z\) (so \(h(Z),h(-Z)\) are negatively correlated)

Discontinuous

Always positive

Q2 Medium

The optimal control-variate coefficient is:

\(b^*=\Var(C)/\Cov(Y,C)\)

\(b^*=\Cov(Y,C)/\Var(C)\)

\(b^*=\rho_{YC}\)

\(b^*=1\)

Q3 Hard

An importance-sampling density \(g\) gives an infinite-variance estimator when:

It has the same mean as \(f\)

Its tails are lighter than those of \(h f\), so the weights \(f/g\) explode

It equals \(f\)

It is uniform

Practical Exercise

You simulate a call payoff \(Y\) and use the underlying \(C=S_T\) (known mean \(\mu_C=S_0e^{rT}\)) as a control. From \(10^5\) paths you estimate \(\widehat{\Cov}(Y,C)=120\) and \(\widehat{\Var}(C)=400\), with sample correlation \(\hat\rho_{YC}=0.7\). (a) Give \(b^*\). (b) By what factor does the control variate reduce the estimator variance?

▶ Show full solution

(a) \(b^*=\widehat{\Cov}(Y,C)/\widehat{\Var}(C)=120/400=0.30\).

(b) Variance is multiplied by \(1-\hat\rho_{YC}^2=1-0.7^2=1-0.49=0.51\), i.e. a 49% reduction. Equivalently you would need roughly \(1/0.51\approx1.96\) times as many plain paths to match this accuracy.

After the reveal, answer for yourself: If you had a second control variate correlated with the first, would stacking them help, and how would you choose the two coefficients?

Lesson Summary

Variance reduction shrinks the \(\sigma\) in \(\sigma/\sqrt N\) while keeping the estimator unbiased. Antithetic variates mirror the driver (help when the payoff is monotone); control variates subtract a correlated known-mean quantity with optimal \(b^*=\Cov(Y,C)/\Var(C)\) and factor \(1-\rho^2\); stratification removes between-stratum variance; importance sampling tilts toward rare but important regions with likelihood-ratio weights.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: Give the optimal control-variate coefficient and the resulting variance factor.
A: \(b^*=\Cov(Y,C)/\Var(C)\); the estimator variance is multiplied by \(1-\rho_{YC}^2\), so a highly correlated control removes most of the noise.

Q: Why can an importance-sampling estimator have infinite variance?
A: If the sampling density \(g\) has lighter tails than \(|h|f\), the likelihood ratio \(f/g\) is unbounded and \(\E_g[(hf/g)^2]\) diverges; \(g\) must not under-weight the tails.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

← 13.2 Next: 13.4 →