Deriving a distribution from the line generated from a point in a uniformly distributed circle and its origin

Let $a,r>0$ be two fixed numbers. A random point $(X,Y)$ is uniformly distributed over the circle {$(x,y) : x^2+(y-a)^2 = r^2$} with the centre of the circle at $(0,a)$. A line is drawn through $(X,Y)$ and $(0,a)$ and the line intersects the $x$ axis at some point $(\zeta, 0)$.

How would I find the distribution function and the density function of $\zeta$?

After thinking about it and sketching it out my intuitive guess is that $\zeta$ follows a normal distribution with $\mu = 0$ and $\sigma^2 = a$, although I'm not quite sure that this is correct. Would someone be able to give me a hint on how to start the problem if I am correct, and if I'm wrong would you be kind enough to give me a push in the right direction?

• It's bell-shaped, but there are many bell-shaped distributions that are not normal distributions, so even if this one were normal you would have to show that it is normal using exact mathematics (rather than merely looking at a graph) in order to make that conclusion--and in this case it isn't true. – David K Mar 29 '15 at 14:10

1 Answer

Let $\alpha \in [-\pi/2,\pi/2)$ the angle that $\frac{Y}{X}=\tan \alpha$. $\alpha$ is uniformly distributed and $\zeta=a \tan \alpha$ which lead us easily to compute the distribution of $\zeta$. Let $F$ the cumulative distribution function of $\zeta$. $$F(t)=Pr(a \tan \alpha < t)=Pr(\alpha < \arctan\frac{t}{a})=\frac{\arctan \frac{t}{a}+ \frac{\pi}{2}}{\pi}$$ because the $\tan$ is a monotone increasing function. $F$ is the cumulative distribution function of the famous Cauchy distribution. Its mean is indeed 0 but its variance isn't finite.

• Would you care to elaborate on your answer? I'm kind of stuck on what to do. I derived the marginal densities of $X$ and $Y$ but I don't know what to do next. – YellowPillow Mar 29 '15 at 17:59
• My hint shows that you doesn't need the marginal densities of X and Y. – Leonhardt von M Mar 29 '15 at 21:28
• I let $P(Z\le z) = P(Y \le \frac{xz}{a}) = P(Y \le \frac{xz}{a}) = \int_{-\sqrt{(r^2-(y-a)^2)}}^{\sqrt{(r^2-(y-a)^2)}}\int_{\sqrt{(r^2-x^2)+a}}^{ \frac {zX}{a}}\frac{1}{\pi r^2}dydx$ am I heading in the right direction? – YellowPillow Mar 29 '15 at 21:37
• Is Z the new name of $\zeta$? :) My hint was roughly: $P(Y<Xz/a)=P(Y/X<z/a)$ where we know $\arctan Y/X$ is uniformly distributed over the intervall $(-\pi/2,\pi/2)$. – Leonhardt von M Mar 29 '15 at 21:49
• Ahh I can't do it this is bugging me so much! And yes Z is $\zeta$ sorry for the change. – YellowPillow Mar 30 '15 at 3:12