Thanks for this awesome series! A question: Wouldn't the direct lighting example here not really be "importance sampling", as it seems we're just changing the domain of integration rather than sampling one portion of the domain with higher probability?
When you say "terminate 50% of all contributions with luminance less than 0.25" in the Roussian roulette example, how do you do that? In the Roussian roulette explanation you only showed how to terminate paths with a certain probability at each bounce, but how can you make it luminance-dependent?
or is the idea that there is some relation between the russian roulette probability p_rr and the percentage of paths below a certain luminance that you discard? Also why talk about luminance rather than radiance here?
compute radiance. generate a random number. if random number is smaller than some predecided value and computed radiance is smaller than 0.25 kill that ray
The russion roulette section feels very incoherent: 1) 1:10:00 Is the "tentative contribution" being tested, COMMULATIVE across the entire path up-to-end-including the current event/vertex, or is it just the contribution of the CURRENT event by itself that is being tested? 2) In either of the above cases, should it also include the direct illumination component (NEE) or exclude it? 3) 1:11:00 Anything multiplied by 0 is 0, and the expected value of a constant is that constant, so the right part of the equation would be "E[(something)*0]" which is "E[0]" whichi is just 0 so the whole thing would always be 0. So - what's the point? What is that notation meant to convey?
the calulation at 1:11:00 is meant to show that the *new* random variable obtained by introducing the russian roulette procedure, has the same expected value as the original random variable. So it's valid for our Monte Carlo estimation
100 out of 100 for this course. Thank you so much @KeenanCrane.
Thanks for this awesome series! A question: Wouldn't the direct lighting example here not really be "importance sampling", as it seems we're just changing the domain of integration rather than sampling one portion of the domain with higher probability?
The variance of sum of random variables is equal to the sum of variances (22:10) if they are independent no?
in 51:30, the expression for d omega should say cos theta'
@@GuillermoValleCosmos yes to all
yes, if and only iff cov(x, y) = 0. And cov(x, y) = 0 if x and y are independent.
When you say "terminate 50% of all contributions with luminance less than 0.25" in the Roussian roulette example, how do you do that?
In the Roussian roulette explanation you only showed how to terminate paths with a certain probability at each bounce, but how can you make it luminance-dependent?
or is the idea that there is some relation between the russian roulette probability p_rr and the percentage of paths below a certain luminance that you discard? Also why talk about luminance rather than radiance here?
compute radiance. generate a random number. if random number is smaller than some predecided value and computed radiance is smaller than 0.25 kill that ray
@@GuillermoValleCosmos I think most of the time luminance and radiance refer to same thing
Great Lecture!
A question: at 52:00, shouldn't dw = dA' * cos(theta')/... instead of dA' * cos(theta)/... ?
you are right. it should be dA'cosθ' / |p - p'|^2
The russion roulette section feels very incoherent:
1) 1:10:00 Is the "tentative contribution" being tested, COMMULATIVE across the entire path up-to-end-including the current event/vertex, or is it just the contribution of the CURRENT event by itself that is being tested?
2) In either of the above cases, should it also include the direct illumination component (NEE) or exclude it?
3) 1:11:00 Anything multiplied by 0 is 0, and the expected value of a constant is that constant, so the right part of the equation would be "E[(something)*0]" which is "E[0]" whichi is just 0 so the whole thing would always be 0.
So - what's the point? What is that notation meant to convey?
the calulation at 1:11:00 is meant to show that the *new* random variable obtained by introducing the russian roulette procedure, has the same expected value as the original random variable. So it's valid for our Monte Carlo estimation