PDF of sum of two random variables Assume an $n$ dimensional random variable $U$ that is uniformly distributed in the volume of an $n$-sphere with radius $R$. Assume another $n$ dimensional random variable $N$ that is distributed according to Gaussian distribution with variance $\sigma^2$ per dimension and zero mean. 
Is it possible to derive the pdf (or cdf) of the sum of two random variables $U+N$? Note that $U$ and $N$ are independent.  
 A: You have two distributions supported on $\mathbb{R}^n$:
$$ f_1(\bar{x}) = A \exp({- B\|\bar{x}\|^2}),\quad f_2(\bar{x}) = C\cdot\mathbb{1}_{\|\bar{x}\|\leq R},$$
so the distribution of $U+N$ is given by:
$$\begin{eqnarray*} \left(f_1 * f_2\right)(\bar{z}) &=& \int_{\mathbb{R}^n}f_1(\bar{z}-\bar{x})\cdot f_2(\bar{x})\,d\bar{x}=AC\int_{\|\bar{x}\|\leq R}\exp\left(-B\|\bar{z}-\bar{x}\|^2\right)\,d\bar{x}\\
&=&AC\exp\left(-B\|\bar{z}\|^2\right)\int_{\|\bar{x}\|\leq R}\exp\left(2B<\bar{x},\bar{z}>-B^2\|\bar{x}\|^2\right)d\bar{x}.\tag{1}\end{eqnarray*}$$
By the isotropy of the sphere, the last integral is equivalent to:
$$I=\int_{\|\bar{x}\|\leq R}\exp\left(2B\|\bar{z}\|x_1-B^2\|\bar{x}\|^2\right)\,d\bar{x},$$
and by switching to spherical coordinates we get the following chain of identities, depending on the Bessel function $I_0$ and the incomplete $\Gamma$ function:
$$\begin{eqnarray*} I &=& (2\pi)^{n-1}\int_{0}^{R}\int_{0}^{\pi}\rho \exp\left(2B\|\bar{z}\|\rho\cos\theta-B^2\rho^2\right)d\theta d\rho\\ &=& 
\frac{1}{2B^2}(2\pi)^{n}\int_{0}^{RB}\rho e^{-\rho^2} I_0(2\|\bar{z}\|\rho)\,d\rho\\&=&\frac{(2\pi)^n}{2B^2}\int_{0}^{RB}\sum_{m=0}^{+\infty}\frac{\rho^{2m+1} e^{-\rho^2}\|\bar{z}\|^{2m}}{m!^2}\,d\rho\\
&=&\frac{(2\pi)^n}{4B^2}\sum_{m=0}^{+\infty}\frac{\|\bar{z}\|^{2m}}{m!^2}\left(\Gamma(m+1)-\Gamma(m+1,R^2 B^2)\right)\\
&=&\frac{(2\pi)^n}{4B^2}\sum_{m=0}^{+\infty}\frac{\|\bar{z}\|^{2m}}{m!}\left(1-\frac{\Gamma(m+1,R^2 B^2)}{\Gamma(m+1)}\right)\tag{2}
\end{eqnarray*}$$
hence we have a radial distribution that can be approximated by estimating the Mills ratios appearing in the last line. By taking small variances, the distribution of $U+N$ is almost uniform, since the convolution with the normal distribution just mollifies the uniform distribution.
