# wigner semi-circle distribution random numbers generation

I am trying to generate random numbers in Wigner semi-circle distribution. Since this one does not have the analytical solution for the inverse function of the pdf. I wonder if anyone familiar with a standard way to generation random numbers (RNs) follow this distribution and what are the pros and cons for each method.

My initial guess from what I have researched is, I can use the rejection method or sample method on the uniform distributive random numbers to get the Wigner semi-circle one.

Previously, I have generated normally distributive RNs from uniformly distributive RNs using the Box-Muller transformation. Even I am not a statistics major, it was pretty straight forward process. However, I am having hard time grasp my head around other distributions, specifically this Wigner semi-circle.

Any instructions or source recommendations will be highly appreciated. Thank you.

• This is a shifted, scaled Beta$(3/2,3/2)$ distribution. Its CDF does have an explicit analytic inverse called the inverse regularized incomplete beta function, but you scarcely need that to generate random values: just look at the graph of the PDF--what kind of geometric figure is it? Hint: the answer lies in the name of the distribution. – whuber Apr 15 at 20:24
• @whuber thank you for your answer. The graph of the Wigner semi circle pdf is a semicircle or semi ellipse. Are you saying that I just need to use this pdf and sample the uniform random variables to get this Wigner dist? – Lac Apr 15 at 23:50
• Given the shape of the subgraph of the density, there is no need for rejection. – Xi'an Apr 26 at 11:50

[Following whuber's comments:] Since $$f(x)=\frac{2}{\pi R^2}\sqrt{R^2-x^2}$$the sub-graph of $$f$$ $$\mathcal S_R=\{(x,y);0\le y\le f(x)\}$$ is the half-disk of radius $$R$$. Thus, by the fundamental lemma of simulation, simulating $$X\sim f$$ is equivalent to simulating $$(X,Y)$$ uniformly on $$\mathcal S$$, which corresponds in spherical coordinates to simulating $$(\rho,\theta)\sim \frac{2}{\pi R}\rho \Bbb I_{(0,R)}(\rho) \Bbb I_{(0,\pi)}(\theta)$$ which is obvious:
1. simulate $$U_r,U_a\sim\mathcal U(0,1)$$
2. compute $$X=R\, \sqrt{U_r} \cos (\pi U_a)$$ [and do not compute $$Y=R\, \sqrt{U_r} \sin (\pi U_a)$$!]
3. return $$X$$
[and shows proximity with the Box-Mueller algorithm, although for the latter $$\rho$$ is distributed as an Exponential $$\mathcal E(1/2)$$].
• It may be worth noting that $Y$ need not be calculated at all. This suggests rejection actually might be more efficient because it requires generating only $4/\pi$ URNs per realization rather than $2$ per realization. – whuber Apr 30 at 12:26