# How to perform MCMC integration when no prior over the integrated function is available? [closed]

As far as I can tell, MCMC integration (e.g. VEGAS) is performed by sampling from a distribution proportional to $$f(x)$$ using MCMC, then building a density estimator $$g(x)$$ using these samples (for example, in case of VEGAS it is a factorized histogram $$g(x_1, \dots, x_n) = g(x_1) \dots g(x_n)$$), and then using this estimator to perform importance sampling and reweighting of values of $$f(x)$$, like $$I(f) = \mathbb E_{x \sim g(x)} \frac{f(x)}{g(x)}.$$ Correct me if I am wrong. It is not exactly clear to me whether one can use values $$f_i$$ at points $$x_i \sim h(x) = \frac{f(x)}{\int f(x) dx}$$ to directly approximate the final integral without introducing a density model $$g(x)$$ that unavoidably introduces an error into the estimate, especially in higher dimensions where density estimation is hard?

Let me give an example of an error I am talking about. Consider that $$f(x)$$ is just a standard 1D Gaussian pdf so it sums up to one. We receive three $$(x_1, x_2, x_3)$$ points near zero and corresponding values $$(f_1, f_2, f_3)$$ and then estimate $$g_{K}(x)$$ from these $$x_i$$ points using a kernel density estimator. Depending on the choice of the kernel width $$K$$ the density estimate in these points $$g(x_i)$$ would fluctuate significantly, i.e. for small $$K$$ values of $$g(x_i)$$ would be high and therefore $$\sum_i \frac{f_i}{g(x_i)}$$ would be low.

Here $$K$$ corresponds to our assumption about the smoothness of $$f$$ and if it is wrong, we get wrong results. The approximation used in VEGAS assumes that the function $$f: R^n \to R$$ factors well into the product of scalar functions.

Is there a way to avoid these possibly wrong assumptions? Assume that we already have plenty of values $$(f_1, \dots , f_n)$$ measured at points $$(x_1, \dots , x_n)$$ and we know that $$x_i$$ follow the distribution $$x_i \sim h(x) \propto f(x)$$?

## closed as unclear what you're asking by Xi'an, mkt, kjetil b halvorsen, mdewey, Michael ChernickFeb 28 at 2:33

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• The title is disconnected from your question. There is no MCMC element there. – Xi'an Feb 27 at 7:15
• @Xi'an mcmc is a way to get plenty of sample from a distribution proportional to $f(x)$, but it is not exactly clear to me how one can use these samples to approximate the final integral without introducing a density model $q_{\theta}(x)$ that minimizes the variance of the fraction $\frac{f(x)}{q(x)}$. I guess, the estimate of this fraction is the value of the integral itself. – MInner Feb 27 at 17:26
• This idea is interesting but kills the main property of importance sampling, which is to produce an unbiased estimator of the integral. – Xi'an Feb 28 at 2:10
• @Xi'an better now? – MInner Mar 1 at 16:47
• Your interpretation is not correct: MCMC simulations from $f$ are used directly in estimating the integral without an intermediate non-parametric density estimation, by virtue of the ergodic theorem. – Xi'an Mar 2 at 2:39