Control Variate Monte Carlo

PyApprox Tutorial Library

Reducing Monte Carlo estimator variance using a correlated low-fidelity model with a known statistic.

Learning Objectives

After completing this tutorial, you will be able to:

  • Explain why a correlated low-fidelity model can reduce the variance of a Monte Carlo (MC) estimator
  • Write down the Control Variate Monte Carlo (CVMC) estimator and identify its free parameter \(\eta\)
  • State the optimal \(\eta\) and the resulting variance reduction in terms of model correlation
  • Identify when CVMC helps and when it does not

Prerequisites

Complete Monte Carlo Sampling and Estimator Accuracy and MSE before this tutorial.

Motivation

The Estimator Accuracy tutorial showed that the variance of the MC mean estimator is \(\sigma^2_\alpha / N\), where \(\sigma^2_\alpha\) is the variance of the high-fidelity model output \(f_\alpha\) and \(N\) is the number of samples. Reducing this variance requires either more samples — which is expensive — or a smarter estimator.

Control Variate Monte Carlo (CVMC) is the simplest example of a smarter estimator. The idea: if we have access to a cheap model \(f_\kappa\) that is correlated with \(f_\alpha\) and whose mean \(\mu_\kappa = \mathbb{E}_\theta[f_\kappa(\boldsymbol{\theta})]\) is known exactly, we can use the cheap model to cancel a large fraction of the MC error.

Figure 1 illustrates this with two models of a 2D input. The surface plots show that \(f_\alpha\) and \(f_\kappa\) have similar shape — when one is large, so is the other. The scatter plot on the right confirms that model outputs are tightly correlated (\(\rho \approx 0.9\)).

Figure 1: Left: overlaid response surfaces of \(f_\alpha\) (blue) and \(f_\kappa\) (orange) over \([-1,1]^2\), showing similar shape with a gap between them. Right: scatter plot of 100 random evaluations confirming tight output correlation \(\rho\).

The CVMC Estimator

The standard MC estimator of \(\mu_\alpha = \mathbb{E}_\theta[f_\alpha(\boldsymbol{\theta})]\) is

\[ \hat{\mu}_\alpha = \frac{1}{N} \sum_{k=1}^{N} f_\alpha(\boldsymbol{\theta}^{(k)}). \]

The CVMC estimator adds a correction term built from \(f_\kappa\):

\[ \hat{\mu}_\alpha^{\text{CV}} = \hat{\mu}_\alpha + \eta \left( \hat{\mu}_\kappa - \mu_\kappa \right) \tag{1}\]

where \(\hat{\mu}_\kappa = \frac{1}{N}\sum_{k=1}^N f_\kappa(\boldsymbol{\theta}^{(k)})\) is the MC mean of the low-fidelity model evaluated on the same \(N\) samples, and \(\eta\) is a scalar weight we are free to choose.

The correction term \(\hat{\mu}_\kappa - \mu_\kappa\) has mean zero — it is pure MC error in the low-fidelity estimate. Adding it to \(\hat{\mu}_\alpha\) does not introduce bias. But if the errors in \(\hat{\mu}_\alpha\) and \(\hat{\mu}_\kappa\) are correlated, choosing \(\eta\) with the right sign causes the correction to partially cancel the error in \(\hat{\mu}_\alpha\).

Variance Reduction Depends on Correlation

The key result (derived in Control Variate Analysis) is that with the optimal choice of \(\eta\), the CVMC estimator variance satisfies

\[ \mathbb{V}[\hat{\mu}_\alpha^{\text{CV}}] = \mathbb{V}[\hat{\mu}_\alpha] \left(1 - \rho^2_{\alpha\kappa}\right) \tag{2}\]

where \(\rho_{\alpha\kappa} = \mathrm{Corr}(f_\alpha, f_\kappa)\) is the correlation between the two model outputs. The factor \((1 - \rho^2_{\alpha\kappa})\) is always between 0 and 1, so CVMC always reduces (or at worst equals) the MC variance.

Figure 2 shows this relationship. Near-perfectly correlated models (\(|\rho| \approx 1\)) can reduce variance by orders of magnitude; uncorrelated models (\(\rho \approx 0\)) offer no benefit.

Figure 2: CVMC variance reduction factor \((1 - \rho^2)\) as a function of model correlation \(\rho\). A correlation of \(|\rho| = 0.9\) reduces variance by \(81\%\); \(|\rho| = 0.99\) reduces it by \(99\%\).

What CVMC Looks Like in Practice

Figure 3 shows the distribution of 1000 independent MC and CVMC mean estimates for a pair of models with \(\rho \approx 0.9\). Both estimators are centered on the true mean, confirming that CVMC is unbiased. But the CVMC histogram is dramatically narrower.

Figure 3: Distribution of 1000 independent MC and CVMC estimates of \(\mu_\alpha\). Both are centered on the true mean (black line), but CVMC has far smaller spread. The low-fidelity model has correlation \(\rho \approx 0.9\) with the high-fidelity model.

When Does CVMC Help?

CVMC requires two ingredients:

  1. A low-fidelity model \(f_\kappa\) with a known mean \(\mu_\kappa\). If \(\mu_\kappa\) is not known analytically, it must be estimated — introducing additional error. That case is handled by Approximate Control Variates.

  2. High correlation \(|\rho_{\alpha\kappa}|\) between the models. The variance reduction \((1 - \rho^2)\) is only substantial when \(|\rho| \gtrsim 0.5\). A weakly correlated low-fidelity model offers little benefit.

The cost of a single CVMC estimate is identical to plain MC: \(N\) evaluations of \(f_\alpha\) and \(N\) evaluations of \(f_\kappa\) on the same samples. The reduction in estimator variance is entirely due to cancellation of correlated errors, not a reduction in work.

Key Takeaways

  • CVMC adds a zero-mean correction \(\eta(\hat{\mu}_\kappa - \mu_\kappa)\) to the MC estimator; the correction is unbiased by construction
  • With the optimal \(\eta\), the variance reduction factor is \((1 - \rho^2_{\alpha\kappa})\), determined entirely by the correlation between models
  • \(|\rho| \approx 1\) gives near-perfect variance cancellation; \(\rho \approx 0\) gives no benefit
  • CVMC requires \(\mu_\kappa\) to be known; when it is not, use Approximate Control Variates
Tip

Ready to try this? See API Cookbook → CVEstimator.

Exercises

  1. If \(\rho = 0.7\), by what factor does CVMC reduce the variance compared to plain MC? How many fewer samples would you need to achieve the same standard error?

  2. Suppose \(f_\kappa\) is the true model \(f_\alpha\) itself (i.e., \(\rho = 1\)). What does \(\hat{\mu}^{\text{CV}}_\alpha\) reduce to? Is this useful in practice?

  3. The correction term is \(\eta(\hat{\mu}_\kappa - \mu_\kappa)\). Explain in words why a negative \(\eta\) is appropriate when the models are positively correlated (\(\rho > 0\)).

Next Steps