Random points in sphere with probability p(r) How to pick random uniformly distributed points in a sphere has been asked before. The difference is that I don't want uniform distribution, rather I would like the number density to scale by $r^{-1}$.
The only method I can think of is to generate random x, y and z coordinates, reject those that are outside the sphere and then do another test. For simplicity I chose to distribute points between radius 1 and the sphere's radius R, with these values I calculated a probability function $p(r)=\frac{1}{\rm{ln}(R)r}$. The test was that I generated another random number and rejected the point if that number was larger than p(r). However I don't know if this is correct, and I'm hoping for a better idea since it was very slow for large R.
 A: If you look at the accepted answer for the post you linked, you see that the direction is generated by sampling from a multivariate normal distribution, and then the radius is chosen according to the desired distribution. In your case, you don't need to define a minimal radius $R_0 > 0$ because when you multiply the scaling factor $1/r$ for probability of the radius by the surface area of the sphere $\Theta(r^2)$ you get $\Theta(r)$, and so you have $P(r \leq R) \propto \int_{0}^R (1/r)r^2 dr = \Theta(R^2)$ where you have a maximum radius $R_{max}$. This gives you your cdf (cumulative distribution function ) of $f(R) = P(r \leq R) = R^2/R_{max}^2$ for sampling the radius. The inverse of the cumulative distribution function is $g(X) = R_{max} \sqrt{X}$. If you sample a uniform random variable $X$ from $[0,1]$ and compute $r = g(X)$, this gives you the desired distribution on the radius $r$ that has cumulative distribution function $f(R)$.  See http://en.wikipedia.org/wiki/Inverse_transform_sampling for an explanation on inverse transform sampling.
