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I'm attempting to use inversion sampling to generate points on a disk according to the following PDF:

$$ f(r) = \dfrac{2}{\pi(1+r^2)} $$

Here, the polar angle would just be a uniform random variable in $ [0,2\pi) $. My question is this: when I go to integrate the PDF to find the CDF, do I include the $r$ from the Jacobian in the integral:

$$ F(R) = \displaystyle\int_0^{R}f(r)r\hspace{0.1cm}dr $$

Or do I treat $R$ as any other random variable and just compute the following:

$$ F(R) = \displaystyle\int_0^{R}f(r)\hspace{0.1cm}dr $$

Or is there something else that I'm missing, as well? I'm relatively new to working with CDFs as my physics undergraduate education was lacking in terms of statistics, so I greatly appreciate any help with this!

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No, you should not include $r$. As a quick check, cdfs require that $\lim_{R \to \infty} F(R) = 1$. Without the $r$, you get

$$ F(R) = \int_0^R f(r) dr = \frac{2}{\pi} (\arctan(R) - \arctan(0)) = \frac{2}{\pi} \arctan(R) \to 1 \text{ as } R \to \infty, $$ which is what we expected. If you do include the $r$, you get

$$ F(R) = \int_0^R f(r) r dr = \frac{1}{\pi} \log(1 + R^2) \to \infty \text{ as } R \to \infty, $$ which is not a cdf.

Here, the $r$ term that you are considering is the determinant of the Jacobian of a change-of-coordinates transformation, namely $g(r, \theta) = (r \cos(\theta), r \sin(\theta))$, which maps polar coordinates to Cartesian coordinates. But you aren't actually changing coordinates here, as your pdf is already given as a pdf with respect to $r$.

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  • $\begingroup$ Ah, that limit check makes sense. Thanks for the help! $\endgroup$ – Ben A. Aug 1 at 2:33

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