In Veach's thesis the MIS weights (e.g. using the balance heuristic) for two strategies with densities $p_1, p_2$ are given as $w_1(x) = \frac{p_1(x)}{p_1(x)+p_2(x)}$ and $w_2(x) = \frac{p_2(x)}{p_1(x)+p_2(x)}$. Veach uses his path integral formulation, which essentially treats all densities as defined w.r.t. the area measure instead of the solid angle measure. I don't see anything stopping me from rewriting those w.r.t. the solid angle measure, or even keeping $p_1$ w.r.t. the solid angle measure, and $p_2$ w.r.t. the area measure. I have not seen the latter anywhere, however, and I assume there is a good reason for that.

I mention that there is no constraint on the measure w.r.t. which the densities are defined in the weighting function, since the requirement here is that $w_1(x)+w_2(x) = 1$, as we use that to split the integral in two parts. It is clear that this holds regardless how I pick $p_1$ and $p_2$ (provided that $p_1+p_2\ne 0$). So is there a formal mathematical justification for the densities being considered exclusively w.r.t. the area measure or the solid angle measure? Would I necessarily get a higher variance by mixing densities w.r.t. different measures?

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    $\begingroup$ As a start, I would try adapting the proof of Theorem 9.2 (page 288) with two different measures, for the simple case of two strategies. If you express both in a common measure (e.g. solid angle), this will introduce a Jacobian term in the denominator somewhere. You would then have to produce similar bounds for each term of Eq. 9.16. Intuitively I would expect the proof to break somewhere so that balance heuristic prevails as the best minimizer, regardless of the measure. $\endgroup$
    – Hubble
    Commented Jun 14, 2023 at 13:27


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