In a path tracer, if we want to sample a ray direction in scattering medium, we can use phase function sampling, which actually use the direction of the ray before sampling as a sort of anchor to sample around (just like normal vector in surface scattering).

However, we could break this convention by devising new sampling approach. For example, if I want to point the ray towards the emitter. So maybe this sampling method does not sample around the incident (or, out-going, depends on what direction we are talking about) direction, as shown in the following figure. We sample $\beta$ instead of $\theta$. enter image description here

My question is: should we convert the PDF of $\cos\beta$ to $\cos\theta$? I think it is necessary when we want to use MIS. So how does this change of PDF work mathematically? Is this some kind of measure conversion like 'solid-angle to area product' problem? As I can tell, for example, if $\cos\beta$ is uniformly distributed, we won't have a uniformly distributed $\cos\theta$ so I think a conversion should be necessary.


1 Answer 1


I think I got the answer via extensive numerical simulation. The result of the simulation is not listed here, so if you need it then you can comment to let me know.

Ok, I think I've derived the mathematical form of this conversion, but whether the conversion is needed still remains unknown and I will leave it behind for now since I am contacting someone with this kind of area expertise to help, and the answer will be updated if I get a reasonable answer. But I do need some reassurance for the correctness of this derivation and I'd appreciate any help.

Ok so first we re-formulate the problem setup, as shown in the following figure: enter image description here

So we will always have $\theta = \alpha - \beta$ and note that $\alpha$ (its cosine term) is the value to sample for the new approach, $\theta$ is the angle for original phase function sampling and $\beta$ is a known constant (given the ray direction before sampling and the position of current vertex and target vertex). So: $$ \frac{dP}{d\cos\theta} =\underbrace{\frac{dP}{d\cos\alpha}}_{\text{known, should be converted}} \times |\frac{d\cos\alpha}{d\cos\theta}| = \frac{dP}{d\cos\alpha} |\frac{d\cos\alpha}{d\alpha}\left(\frac{d\cos(\alpha - \beta)}{d\alpha}\right)^{-1}|\\ =\frac{dP}{d\cos\alpha} |\frac{\sin \alpha}{\sin(\alpha - \beta)}| $$ The term: $$ |\frac{\sin \alpha}{\sin(\alpha - \beta)}| $$ should be able to do this conversion (if needed). Note that this is just a Jacobian conversion.

My point is, to apply MIS, we can't directly use PDF for sampling $\theta$ (cosine) and the PDF for sampling $\alpha$. The meaning of these two terms just doesn't match. We should convert both of them to the same meaning, like we are both talking about sampling $\alpha$ or $\theta$. I have done some numerical experiments, comparing hisogram of the cosine value with the analytical Jacobian I derived and I think the results match with each other.

So, this question now becomes:

  • Whether my derivation is correct?
    • The answer is yes. I did a numerical simulation, compared the histogram and the analytical curve and found they are matched.
  • Whether this conversion is needed for MIS? Or even, since Monte Carlo estimator is of form $f(x) / p(x)$, should I multiply the PDF in the denominator by this Jacobian?
    • I think the answer is no, it is not needed. A numerical simulation again showed that when I add this conversion, the result will not be correct (won't match).

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.