I already posted it on Physics SE, but maybe this is a better place:

I have a 5D integral being calculated with a Monte Carlo uniform random sampling. The issue is that the region of integration is very small and for 100000 points I get only around 20-30 points every time. The function has to be calculated several times for a certain parameter, which doesn't change the integration region but changes only the magnitude of the result (which is good, I suppose).

After several runs, I gathered some hundreds of points which are a valid match for the integration region. I was thinking about using those points to improve the integration result, but how can I do that, in a correct way?

EDIT: on the plane, the boundary is defined by an expression defining some kind of "little line" going from x = y = 150 up diagonally to infinity, so it's a thin domain. it is defined by all the five parameters in the function.

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    $\begingroup$ Can you provide some more detail? Why is the region of integration so small? What defines its boundaries? $\endgroup$ – Maxim Umansky Sep 18 '19 at 14:51
  • $\begingroup$ Maybe he is using rejection sampling. Most of the points are rejected. $\endgroup$ – R zu Sep 18 '19 at 18:22
  • $\begingroup$ the integration region is very sparse, that's it. As I wrote, out of 10^6 points I get only 35 usable points maximum. I collected all the points found with the various runs, and I want to know how can I use them. $\endgroup$ – LowFieldTheory Sep 18 '19 at 20:47
  • $\begingroup$ What's the shape of the integration region? You can't tweak the proposal distribution when moving from one point to the next to the shape of the region of interest. $\endgroup$ – Wolfgang Bangerth Sep 20 '19 at 16:10
  • $\begingroup$ Can you give more information about the region of integration? Is that given by some polynomial /bezier lines? As far as I know you are not chained to using a homogeneous random distribution as long as you use appropriate weights. If the integration domain is somehow geometrical in one of the 5 dimensions, you might make use of that by using a different random distribution for your points. $\endgroup$ – MPIchael Sep 27 '19 at 8:04

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