Distribution of $\sqrt{X^2+Y^2}$ when $X$ and $Y$ are Cauchy

Question

Suppose that $X$ and $Y$ are Cauchy-distributed with $\gamma=1$, i.e., with PDF $\frac 1 \pi \frac 1 {1+x^2}$. I tried to find the distribution of $R = \sqrt{X^2+Y^2}$. The PDF of $R$ should be given by integrating over an annulus $A: r^2 < x^2+y^2 < (r+dr)^2$ in the $x-y$ plane.

$P(r)dr = \int_A \frac 1 {\pi^2} \frac 1 {1 + x^2 + y^2 + (xy)^2} dxdy = dr \frac 1 {\pi^2} \int_0^{2\pi} \frac r {1 + r^2 + r^2 \sin\theta\cos\theta} d\theta$

The integral was evaluated as

$P(r)dr = \frac 2 \pi \frac r {\sqrt{1+2r^2+\frac 3 4 r^4}} dr$

I have checked the evaluation of the integral and I found it correct. But still, $\int_0^\infty P(r)dr=\infty$, while it should be 1 for a probability distribution. Where is the mistake?

Answer 1

The integral over $\theta$ should be

$$ \frac{1}{\pi^2} \int_0^{2\pi} \frac{r\; d\theta}{1+r^2 + r^4 \cos^2(\theta) \sin^2(\theta)} = \frac{4 r}{\pi (r^2+2)\sqrt{r^2+1}}$$

The integral of this from $r=0$ to $\infty$ is indeed $1$.