In Equation 3.6.3, there is a d!, it looks like a operator. It is used as d! cos theta d phi. However, I could not find the definition of this (friends and web.)

p.122 particle tracing Equation 4.32

In this Equation about

*alpha*, there is a mysterious (for me) term

*1/(q_{i+i})*.

*f*is projected solid angle,

*p_{i+1}*is approximation of BSDF, so no problem. But, what is this

*1/(q_{i+i})*?

Figure 1 shows theFigure 1: Sampling weight $\frac{1}{q_{i+1}}$. (1) terminate sampling by probability $p$, (2) bounce probability is $(1-p)$. Because, the sample value is better than nothing, the case (2) is respected by $\frac{1}{1-p}$, that is $\frac{1}{q_{i+1}}$.}

*alpha*update. The sampling is done by the Russian roulette method, then, the termination of a ray is decided by a probability. Intuitively, when you continue to sample, it is natural to respect the sampled result more. Because a sample has more information than no sample. Therefore, the sampled result has a weight of

*1/(sample probability)*.

For example, if the ray terminate probability 0.5, then the weight is 1/0.5 = 2.0. If the ray terminate probability 1/3, then, the weight is 1/(1-1/3) = 3/2. The following two examples are not Russian roulette anymore, but, If the ray didn't terminate all the time, then 1/(1-0) = 1, means using the sampled value. If the ray always terminate, there is no bounce, then, no weight defined since the weight only has meaning when the ray bounce.

So far, I told you ``intuitive'' or ``natural'' something. I confess, my intuition is not so good. The following is a proof of overview of why this is OK.

The issue is if this weight causes a bias, we are in trouble. I wrote what is unbias means in the other blog entry. An unbias algorithm has zero expectation of the sampled calculated error. Figure 1's expectation is (where the true answer is

*Q*),

But, sample value

*s_1*is zero because it is terminated. To be

*E*to

*Q*(or unbias means the error

*E-Q = 0*), sample value

*s_2*has a weight

*alpha*,

Therefore, we could compensate

*s_2*with alpha = 1/(1-p) and this lead us unbias.

Acknowledgements

Thanks to Leonhard G. who told me the overview of the proof.

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