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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?

YellowPillow
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  • 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

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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.

Leonhardt von M
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  • 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 03:12
  • Wouldn't we be find the distribution of α, not $ \zeta $ itself? – YellowPillow Mar 30 '15 at 03:59
  • We computed the distribution of $\zeta$ from $\alpha$'s. Why is $\alpha$ uniformly distributed? Let $\varphi=(0,0)(0,a)(X,Y)\angle$. The distribution of $P=(X,Y)$ is uniform on the circle, it means that the probability that P is on a certain arc is proportional to the length of the arc. As we know from elementary geometry the length of arc corresponding to $\varphi$ is proportional to $\varphi$, hence $\varphi$ is uniformly distributed on the interval $[-\pi,\pi]$. $\alpha=\varphi$ if $|\varphi|<\pi/2$ and $\alpha=-\varphi$ if $|\varphi|\geq\pi/2$, so also $\alpha$ is uniformly distributed. – Leonhardt von M Mar 30 '15 at 09:35
  • And unfortunately I solved your problem instead give you a push into the right direction as you asked. But I hope it wasn't totally useless. If you would like I'll tell you a similar problem to practice. – Leonhardt von M Mar 30 '15 at 09:43
  • Yes please would you give me another problem? That would greatly help me in understanding this topic – YellowPillow Mar 31 '15 at 08:36
  • Let $(X,Y)$ point in the plane normally distributed such that its density is $\frac{1}{2\pi\sigma^2}\exp(-\frac{x^2+y^2}{2\sigma^2})$. Let $\theta$ the angle between the vector $(X,Y)$ and the $x$-axis. What is the distribution of $\theta$? – Leonhardt von M Mar 31 '15 at 23:13