It would help me if someone can verify the following two proofs I made for the statement below (it's regarding only the second proof).
Let us have a unit ball centered at $(0,1,0)$. And let $(X,Y,Z)$ be a multivariate random variable uniformly distributed in that ball. Then I want to prove/disprove the statement that the distribution of $(X,Y,Z)/\sqrt{X^2+Y^2+Z^2}$ is cosine distributed over the unit hemisphere, that is: $p(\theta) = \frac{\cos\theta\sin\theta}{\pi}$.
First let's consider the $2D$ case. For each point on the unit hemisphere $(x,y) : x = r\sin\theta, y = r\cos\theta, \theta \in [-\pi/2,\pi/2], r = 1$, I find the intersection of the ray starting at $(0,0)$ with direction $(x,y)$ with the unit circle centered at $(0,1)$, with canonical equation $x^2+(y-1)^2=1$. Plugging in x and y I solve for r: $r^2\sin\theta+r^2\cos\theta – 2r\cos\theta + 1 – 1 = 0$, $r(r-2\cos\theta)=0$. Obviously one of the roots is $0$ and the other is $r = 2\cos\theta$. Then the first intersection point is $(0,0)$ and the second is $(2\cos\theta\sin\theta,2\cos^2\theta)$. Considering that, I argue that the probability density function induced by the normalization transformation is $p(\theta)=\frac{\cos\theta}{2}$ over the unit hemicircle. Here's the part I am not so sure about: since we have uniformly distributed points in the unit ball, then the probability for picking a specific direction $\theta$ is $p(\theta) = C|(2\cos\theta\sin\theta,2\cos^2\theta)|$. I am not certain that I can do this – precisely using the distance as the corresponding probability, since I do not provide justification about this except for the fact that it 'follows' from the fact that the distribution inside the ball is uniform, so I can integrate it along each ray to get the corresponding probability for picking a point onto the unit hemisphere in that direction $\theta$. Expanding $p(\theta) = C\sqrt{4\cos^2\theta\sin^2\theta + 4\cos^4\theta} = 2C|\cos\theta|$. After integrating: $\int_{-\pi/2}^{\pi/2}{2C|\cos\theta|d\theta} = 1$, one gets $C = 1/4$, and $p(\theta) = \frac{\cos\theta}{2}$.
The $3D$ case is similar. Once again we generate the ray: $x=r\sin\theta\cos\phi, y=r\cos\theta, z=r\sin\theta\sin\phi$, and intersect it with $x^2 + (y-1)^2 + z^2 = 1$. Plugging in the ray equation into the canonical equation once again I get the solutions: $r(r-2\cos\theta)=0$. Then $p(\theta) = C|(2\cos\theta\sin\theta\cos\phi, 2\cos^2\theta, 2\cos\theta\sin\theta\sin\phi)|\sin\theta$. After a few transformations: $p(\theta) = 2C\cos\theta\sin\theta$. Integrating $\int_{0}^{2\pi}{d\phi}\int_{-\pi/2}^{\pi/2}{2C\cos\theta\sin\theta d\theta}$ yields $C = 1/2\pi$. Ultimately I get $p(\theta) = \frac{\cos\theta\sin\theta}{\pi}$.
Best Answer
My proof seems to be incorrect as I didn't take into account the $r$ and $r^2$ factors when integrating, yielding $C_1\cos^2\theta$ and $C_2\cos^3\theta$ for the pdfs.