3 ms·
Oops, I misinterpreted how the graph was generated... but the complaint still holds. The averaging of a million trials will cut the amplitude of the fluctuation
by brilee 14y ago
Oops, I misinterpreted how the graph was generated... but the complaint still holds. The averaging of a million trials will cut the amplitude of the fluctuations by a factor of 1000, but the sqrt(N) effect should still be visible. (Note that if you multiply the Y-axis by sqrt(million) = 1000, then you get a scale of about 20, which is approx sqrt(500))
- yeahboats 14y agoI'm not very good at this stuff, but I don't see anything wrong with the graph. It hovers around 0. A Brownian motion starting which starts at 0, will, at time T be normally distributed with mean 0 and variance T. http://en.wikipedia.org/wiki/File:Random_Walk_example.svg http://en.wikipedia.org/wiki/File:Random_Walk_example.svg Some of them go up, some go down, but all will cross 0 infinitely many times, and all the paths averaged will equal 0.
- deleted 14y ago[deleted]