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.
Best Answer
Simply change your quantity of interest to $(f-\hat{\mu}(f(X))^2$ my dude. Here's the formulation I used:
$Q^\mathrm{CV}= \frac{1}{N}\sum_{i=1}^N\left[(f(x_i) - \hat{\mu}(f(X))^2 + \alpha\left((g(x_i) - \hat{\mu}(f(X))^2 - \mathbb{E}[(g(X) - \hat{\mu}(f(X))^2]\right)\right]$,