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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.

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  • $\begingroup$ 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 $\mathbb E[f(X)^2]$? $\endgroup$
    – foreignvol
    Commented Jan 19, 2019 at 0:35

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

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

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  • $\begingroup$ Does this proposal miss an opportunity to further reduce variance, e.g., by perhaps using: $\hat{\mu} := \tfrac{1}{N}\sum_{i=1}^N \left( f(x_i) + \alpha_2(g(x_i) - \mathbb{E}[g(X)])\right)$ where $\alpha_2$ is a separately tuned variance reducing parameter? $\endgroup$
    – jvdillon
    Commented Jun 19, 2020 at 19:39

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