I'm studying the Hough transform. Why is the Hough transform resistant to occlusion and noise in edge points?
The Hough Transform (HT) computes the lines passing trough two points, and then fills an accumulator (Hough space). The areas within the accumulator with the highest values are the potential lines.
If a line is cut in different points, the HT will statistically (if the number of iterations is big enough) use the remaining parts points during the process. As it is the same line that passes through these points, a peak will appear in the Hough space to represent the line. Remember, that the more points remaining in the line, the higher the chances that the points will be used for during the process.
That's also why the HT is robust to noise. When the process selects points from noise, then always vote for a different hypothetical line. However, the points belonging to the same line, always vote for that one. So in the case of a noisy images, the noise will fill the Hough space with sparse random small values, when the lines will generate a peak. You have a figures of the same image with/without noise here.