# What is the PDF for path tracing in the paper "Learning the light transport the reinforced way

It's the algorithm which combines path tracing and reinforcement learning. I can't understand what $$p_\omega$$ is.

The algorithm is clear. The actions are the directions and the states are the hit points. For updating the Q-values, the hit points are considered as parts of a discretized diagram like Voronoi diagram and the directions of rays are generated by a discretization of a hemisphere surface which divides it into equal areas. The Q-values are initialized as equal for each state.

First, we generate a camera ray, and when it hits a diffuse surface, it scatters according to a discrete PDF which is calculated by normalizing the Q-values of the state of hit point.

Second, The Q-values are updated along the path, we have the state $$s$$ and the next state $$s\prime$$ is generated from the direction generated by PDF of Q-values in $$s$$.

The area of each path on the hemispheres is $$2\pi/n$$, and integral in the update equation of Q-values is calculated like this:

$$\dfrac{2\pi}{n}\sum_{k=0}^{n-1}Q_k(y)f_s(\omega_k, y, -\omega)cos\theta_k$$

I understand all of the algorithm, but I don't know what $$p_\omega$$ is in the second part. Is it $$1/2\pi$$ or $$n/2\pi$$ ? Could someone prove what it really is? I need the proof.

I guess it is $$1/2\pi$$ because the path tracer here only use the PDF generated by Q-values. It doesn't use the Q-values at all. Am I correct? If I am correct, how can I prove it?

EDIT: For clarification, a ray generated through a patch uniformly and the Q-values are generated as I said. Because we only change the directions so more samples are generated inside the patches which receive more light.

I used $$n/2\pi$$ , but scene was so dark, so the PDF must be either $$1/2\pi$$ or a normalized combination of PDF's of patches. If the second is true, what it really is?

EDIT 2: Should I ask this question in the statistics community of stack exchange?

• I haven't looked into this beyond what you posted, but this seems like the probability with which you sampled your new ray. Since it mentions proportional to Q, I am assuming you do some inverse transform sampling, so you need to take it into account. – lightxbulb Nov 6 at 15:58

I have to answer my question. Consider that all patches have the same value. Then choosing a patch uniformly and selecting a random point uniformly on that patch is as the same as sampling the hemisphere uniformly. If $$p_s=Pr(Q_k \in Q)=1/n$$, which n is the number of patches, I can conclude that $$p_\omega=p_s*(n/2\pi)=1/2\pi$$ .
And if patches don't have the same value, then $$p_s=Pr(Q_k \in Q)=q_k/(q_1+...+q_n)$$ and $$p_\omega=p_s*(n/2\pi)=(q_k/(q_1+...+q_n))*(n/2\pi)$$