There's good intuition for this. The key moral of the story is this: > The normal distribution provides uniform sampling on the > unit n-sphere. What does tha

There's good intuition for this. The key moral of the story is this: > The normal distribution provides uniform sampling on the > unit n-sphere. What does that mean? Assume one wants to sample a point uniformly from the unit circle. So we want to uniformly sample from the set { (x, y) : xx + yy = 1 }. The naive method of uniformly sampling x over [0, 1] and then uniformly sampling y in [0, 1 - xx] doesn't work. It doesn't work because for a given change in dx, the change in dy is not uniform. For example, if we are the point (x=0, y=1), in an infinitesimal region around this point, it looks like "flat land", so moving in dx does not change dy. That is, the tangent to the point (x=0, y=1) is horizontal. On the other hand, if we are at the point (x=1, y=0), in an infinitesimal region around this point, it looks like a cliff face: all we have is vertical. So the dy change is "infinity". That is, the tangent to the point (x=1, y=0) is vertical. Let's sample both x and y from the normal distribution. So the probability of getting x=r is e^{-rr}. The probability of getting y=s is e^{-ss}. The probability of getting some random point (x, y) = (r, s) is e^{-(rr+s*s)} = e^{-1} = constant. So if we sample two points using the normal distribution, the points are uniformly distributed on the unit circle. ### Box–Muller transform On the flip side, assume we have a uniformly distributed point on the circle. This means that the angle theta is uniformly distributed in [0, 2pi]. Given such a uniformly distributed point, we can generate two gaussian distributed numbers by taking their x and y component. The intuition for why this works comes from the above: - We saw that (x ~ normal, y ~ normal) => (theta ~ uniform) [as explained before]. - We can also show that (theta ~ uniform) => (x ~ normal, y ~ normal) which is what the box-muller transform does.