PDF of sum of two random variables

Question

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.

possible duplicate of Joint PDF of two random variables and their sum — sds, Jun 06 '14 at 22:54
@sds Also, here $U$ is uniformly distributed in the volume of a hypersphere. It is different from one dimensional pdf of the link. — M.X, Jun 07 '14 at 15:48
I wonder if this paper is the answer for this question http://www.jstor.org/discover/10.2307/2312184?uid=3738032&uid=2&uid=4&sid=21104270636923 — M.X, Jun 07 '14 at 17:32
@wordsthatendinGRY this problem says uniform distribution on $\mathbb{S}^n$ and Gaussian random variable in $\mathbb{R}^n$. — cactus314, Jun 10 '14 at 11:46
I'm pretty sure you're not going to get a closed form solution in terms of elementary functions. The answer can very likely be written in terms of a convolution of a Gaussian distribution with some type of function with finite support. Generally, the best you can probably do is to write the answer in terms of an integral. — Will Nelson, Jun 12 '14 at 06:30

Jack D'Aurizio · Answer 1 · 2014-06-15T01:34:02.297

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.

PDF of sum of two random variables

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