# Can we use the control variate Monte Carlo variance reduction approach to estimate variance?

The control variate technique is a super useful was of fusing low- and high-fidelity models to reduce the variance in an estimate of an expected value. Consider an expensive, high-fidelity model $$f(x)$$ and inexpensive, low-fidelity model, $$g(x)$$. Say that $$X$$ is a random variable with probability distribution $$P$$, $$X\sim P(x)$$. Assume that the expected value of $$g(X)$$, $$\mathbb{E}[g(X)]$$, is known. Then, the control variate estimate is defined as

$$Q^\mathrm{CV}= \frac{1}{N}\sum_{i=1}^N\left[f(x_i) + \alpha\left(g(x_i) - \mathbb{E}[g(X)]\right)\right]$$,

where $$x_i\sim X$$.

Taking the variance of $$Q^\mathrm{CV}$$, we notice that, with inelligent choice of $$\alpha$$ and assuming some correlation between $$f(X)$$ and $$g(X)$$, the variance of $$Q^\mathrm{CV}$$ may be less than the the variance of the standard Monte Carlo approach:

$$\mathrm{Var}[Q^\mathrm{CV}] = \mathrm{Var}([f(X)] + \alpha^2 \mathrm{Var}[g(X)] + 2\alpha\ \mathrm{Cov}[f(X), g(X)]$$

In fact, it is not too hard to derive that the optimal value of $$\alpha$$ (the "control variate") is $$\alpha^* = -\frac{\mathrm{Cov}[f(X), g(X)]}{\mathrm{Var}[g(X)]}$$

This is all well and good if I want to estimate $$\mathrm{E}[f(X)]$$, the first moment of the uncertain function $$f(X)$$. What if I want to estimate $$\mathrm{Var}[f(X)]$$, the second moment of $$f(X)$$? Can the control variate approach me used to increase the accuracy of a Monte Carlo estimator by fusing $$f$$ with $$g$$?

I tried replacing the mean operator in the above formulation with variance and, so far, my numerical experiments indicate this is not a useful formulation for the accurate estimation of $$\mathrm{Var}[f(X)]$$.

For ease of notation, the control variate estimator can also be written as

$$Q^\mathrm{CV}= \mathrm{Mean}\left[\{f(x_i) + \alpha\left(g(x_i) - \mathbb{E}[g(X)]\right)\}_{i=1\dots N}\right]$$,

The formulation I thought would make sense to efficiently estimate the variance of $$F(X)$$ is the following:

$$\mathrm{Var}[f(X)] \approx \mathrm{Var}\left[\{f(x_i) + \alpha\left(g(x_i) - \mathrm{Var}[g(X)]\right)\}_{i=1\dots N}\right]$$

I believe this is not a useful formation, and hope there is a better one out there.

• I've never tried this before but, if you know $\mathbb E[g(X)^2]$, then you can use $g(X)^2$ as a control variate for $\mathbb E[f(X)^2]$? – mflopezabu Jan 19 at 0:35

## 1 Answer

Simply change your quantity of interest to $$(f-\hat{\mu}(f(X))^2$$ my dude. Here's the formulation I used:

$$Q^\mathrm{CV}= \frac{1}{N}\sum_{i=1}^N\left[(f(x_i) - \hat{\mu}(f(X))^2 + \alpha\left((g(x_i) - \hat{\mu}(f(X))^2 - \mathbb{E}[(g(X) - \hat{\mu}(f(X))^2]\right)\right]$$,