MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose, for example, you are simulating samples from a (multivariate) Gaussian with mean zero and covariance $\Gamma=BB^T$. If you had generated a sample $x$, you could generate more (dependent) samples by some unitary transformation $U$, I think via $\tilde{x}=BUB^{-1}x$ in this case. If you knew something about the statistic you were trying to estimate (for example, maybe we estimate $\xi=\mathbb{E}(f(x))$ where $f$ is smooth and is //not// invariant to these transformations) you could try to use the correlated samples to do some sort of variance reduction.

Does this kind of technique where you use a symmetry of the distribution have a name? Is it even a technique ... maybe it is fatally flawed somehow? Maybe it would just fall under the Control Variate or CRN techniques? Uses of "Gauge invariance" feels like the right area to look in but I don't have much experience in physics.

share|cite|improve this question
I think this is a question of sufficiency. Depending on what you are estimating, there is only so much information in the initial sample $x$ whether you extract it via simulation or otherwise. Certainly in cases of symmetry you can find that a given statistic is sufficient where, absent that symmetry, it wouldn't be. Take a uniform random variable with unknown support $[a,b]$; then $\max(|x_{1:n}|)$ is not sufficient for $a$. But if you knew that $a=-b$, then it would be. – R Hahn Nov 27 '12 at 15:30
Good point. In my case (at least as I meant to describe it), the problem lies in the fact the probability distribution has a symmetry but the function that we are estimating breaks this symmetry. Basically, only part of the integrand has the symmetry. – mathtick Nov 28 '12 at 17:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.