Exercise 23.1 - Sampling from a Cauchy

Answers

Recall that the p.d.f. of a standard Cauchy $𝒯 (x | 0, 1, 1)$ is:

𝒯 (x | 0, 1, 1) = \frac{1}{Z} \cdot \frac{1}{1 + x^{2}},

where:

Z = \int_{- \infty}^{\infty} \frac{1}{1 + x^{2}} d x = \arctan x |_{- \infty}^{\infty} = π .

Thus the c.d.f. is:

F (x) = \int_{- \infty}^{x} \frac{1}{π (1 + t^{2})} d t = \frac{\arctan x + \frac{π}{2}}{π},

whose inverse is:

\tan [π (p - \frac{1}{2})] .

So we can sample $p$ from $Uniform [- \frac{1}{2}, \frac{1}{2}]$ and transform it by $\tan (𝜋𝑝)$ , the result would follow a standard Cauchy.

solour_lfq

2021-03-24 13:42