Assume a continuous random variable $X$ that is uniformly distributed $\underline{\text{on}}$ a $k$-sphere. For simplicity, lets assume a simple circle with radius $R$ in 2 dimension. Therefore $$f(r,\theta)=\frac{1}{2\pi R}, \;\;\;r=R$$ Assume $Y$ is another random variable identical to $X$
Can someone please help me to derive the probability density function (pdf) of $Z=X+Y$?
In dimension two, one considers some i.i.d. random variables $X$ and $Y$ and it seems that each is uniformly distributed on the circle $\{(u,v)\mid u^2+v^2=1\}$ (this assumes without loss of generality that $R$ in the post is $R=1$). (Note in particular that $X$ and $Y$ have no density.) The distribution of $Z=X+Y$ is invariant by the rotations centered at the origin and is concentrated on the disk $\{(u,v)\mid u^2+v^2\leqslant2\}$. Thus, for every $r$, $$ P(\|Z\|\geqslant r)=P(\|Z\|^2\geqslant r^2)=P((1+\cos\alpha)^2+(\sin\alpha)^2\geqslant r^2), $$ where $\alpha$ is uniformly distributed on $(0,2\pi)$ or, equivalently, on $(0,\pi)$. Furthermore, $(1+\cos\alpha)^2+(\sin\alpha)^2=2(1+\cos\alpha)$ hence $$ P(\|Z\|\geqslant r)=P\left(\cos\alpha\geqslant\frac12r^2-1\right)=\frac1\pi\arccos\left(\frac12r^2-1\right). $$ Differentiating this and using the fact that $(\arccos)'(x)=1/\sqrt{1-x^2}$, one gets the density of $\|Z\|$ and finally that $Z=(R\cos\Theta,R\sin\Theta)$ where $(R,\Theta)$ has density $$ f_{R,\Theta}(r,\theta)=\frac{2\,\mathbf 1_{0\lt r\lt2}}{\pi\sqrt{4-r^2}}\cdot\frac{\mathbf 1_{(0,2\pi)}(\theta)}{2\pi}. $$