# Probability Theory relating distance and the Cauchy Distribution

This is quite a detailed probability theory question and any hints or a place to start would be greatly appreciated. The question is as follows:

A radioactive source emits particles with all directions being equally likely. The source is held at distance 1 from a vertical infinite plane photographic plate. Consider the nearest point of that plane to the source as the origin. Show that, given the particle hits the plate, the horizontal coordinate of the point of its impact has the Cauchy distribution, i.e., the density function is $$f_X(x)= \frac{1}{π(x^2+1)}$$

• Start with a s.v. $\theta$ decribing the angle in which the particle is emitted. Then $\theta\sim Uniform[0,\pi]$.
– Marc
Mar 17, 2014 at 21:30
• Thank you for helping me start the problem...sorry could you tell me what s.v. represents I assume the v is for variable? Mar 17, 2014 at 22:25
• the horizontal coordinate of the point of its impact >> the vertical coordinate of the point of its impact.
– Did
Mar 17, 2014 at 23:00
• stochastic variable
– Marc
Mar 18, 2014 at 7:31

Note that $X=\tan\Theta$ where the random variable $\Theta$ is uniform on $(-\frac\pi2,\frac\pi2)$ hence, for every bounded measurable function $u$, $$E(u(X))=E(u(\tan\Theta))=\frac1\pi\int_{-\pi/2}^{\pi/2}u(\tan t)\mathrm dt.$$ The change of variable $x=\tan t$ has Jacobian $\mathrm dx=\mathrm dt/\cos^2t=(1+x^2)\mathrm dt$ hence $$E(u(X))=\frac1\pi\int_{-\infty}^{\infty}u(x)\frac{\mathrm dx}{1+x^2}.$$ This identity holds for every bounded measurable function $u$ hence $X$ has a density $f_X$ such that, for every $x$, $$f_X(x)=\frac1{\pi(1+x^2)}.$$