Why is the approximation valid, in the formula provided by Brian Karis? I'm sorry I'm bringing this topic up again, but I need to expound some of the topics.

With reference to this question, I was wondering if someone can help me out in expanding some of the details.

In the answer to that question we have:


  However, since $L_i(l_k)$ is usually not a constant, this integral is
  not separable. Yet, in some cases (like when dealing with diffuse
  lighting or, more specifically, diffuse ambient lighting), the
  information that $f(l_k,v)$ carries is low-frequency enough that it
  can reasonably be approximated as a constant with respect to the
  distribution of $L_i(l_k)$. Experimentally, it can be shown that the
  final result is not too different from what you would expect.


I'm not sure I understand the specific bit "the information that $f(l_k,v)$ carries is low-frequency enough" and also the bit "with repsect to the distribution of $L_i(l_k)$".

I've been reading something about radiometry lately. The only clue I have is that since $f(l_k,v)$ is also function of the wavelength, which is inverse w.r.t. the frequency. Therefore low frequency means high wavelength, which in turn translates some specific assumption about the colour of both incoming and reflected light I guess. Experimental results are also mentioned and I'd like to find out those if they've been published.

Is my reasoning correct? If yes, I still have the feeling I'm missing something. If not what am I missing yet? Experimental results are also mentioned, can you point out any paper so I can read about those?

Thank you.
"the information that $f(l_k,v)$ carries is low-frequency enough": As IneQuation explains, low-frequency was used to refer to the detail of the brdf function. I did actually mean that $f_r$ was low frequency though (which is the case with diffuse lighting), not $L_i$.

"with respect to $L_i$": what this means it that, since there aren't any large peaks in $f_r$, no part of $L_i$ will be wheighted much more heavily than any other in the integral.

If each function was, say, reciprocal of the other (meaning in this case that the peaks in each one would cancel out when multiplied by the other) the correct integral would be very different to the separated integral where this cancelation does not occurr. but since the two functions are not correlated (or they shouldn't be) this difference is likely to be fairly small.

Also, if you use a really high frequency BRDF that takes into acount only incident light around the normal and you separate it, you will get something that's way off. in this example the estimate using separation is 4x larger that the correct result: https://www.desmos.com/calculator/c4zgypthrc

Experimental results can be generated easily. here is an example: https://www.desmos.com/calculator/poyswojjsa it shows a low frequency brdf $f_r$ and a high frequency incident light $L_i$ function being multiplied before and after integration. the ratio between the correct result and the one obtained using separation is of about 1.6

Addendum regarding your latest comment on my other answer:

$$\frac{1}{N}\sum^{N}_{k=1}f(L_k) = \sum^{N}_{k=1}(f(L_k)\cdot \frac{1}{N})$$

Here we have "moved" or distributed $\frac{1}{N}$ into the sum. This formulation is equivalent but looks a bit more similar to an integral. if we set $\Delta L = \frac{1}{N}$ then, we can get

$$\sum^{N}_{k=1}f(L_k)\Delta L\approx \int_{L}f(L)dL$$

keeping in mind that the separation that occurs is, conceptually, separation of integrals, and that every integral "needs" its own $dL$. it follows that every montecarlo estimator needs its own $1/N$

I hope this helps you understand where the extra$\frac{1}{N}$ comes from