My goal is to draw a random (3-dimensional) vector $X$ from a spherically symmetric exponential distribution
$$
X \sim f_X(x) = \frac{\lambda^3}{8\pi}e^{-\lambda |x|}
$$
with $x\in\mathbb{R}^3$.
My attempts:
From this question I got the idea of splitting the vector $X$ in norm and unit vector: Define a random variable $Y=RU$ with $R\in\mathbb{R}_+$ being a random variable for the radius of the vector $Y$ and $U\in\mathbb{R}^3$ with $|U|=1$ being a random unit vector, uniformly distributed on the surface of the unit 3-sphere.
How do I find the distribution $f_R$ of $R$, such that $Y=X$?
I tried $f_R(\rho) = \gamma e^{-\gamma \rho}$, with some $\gamma$ to adjust, but on comparison of the absolute moments of $X$ and $Y$, I find, that there is no $\gamma$, such that $Y$ and $X$ have the same absolute moments (which seems to me to be a neccessary condition for $Y=X$).
$$
M_{|X|}^{(k)} = \int f_X(x) |x|^k \text{d}x = \frac{(k+2)!}{2\lambda^k} \\
M_{|Y|}^{(k)} = \mathbb{E}[|Y|^k]=\mathbb{E}[R^k] = M_R^{(k)} = \frac{k!}{\gamma^k}
$$
Since I can calculate all the absolute moments of $|X|$ and these moments must be the same moments for $R$, I guess I run into the moment-problem.
My question is first of all, how to generate a random vector $X\sim f_X$. But I also would really like to know the radial distribution $f_R$ and how to get it.
Best Answer
It seems I solved the question myself.
The moments of $|X|$ and $R$ must be the same, and the moments of $|X|$ are known, so one can compute the moment generating function (MGF) of $R$: $$ M_R(t) = \sum_{k=0}^\infty \frac{t^k}{k!} M_R^{(k)} = \frac{1}{2}\sum_{k=0}^\infty \Big(\frac{t}{\lambda}\Big)^k (k+1)(k+2) $$ This Taylor series has the alternate form (via Wolfram alpha) $$ M_R(t) = \frac{\lambda^3}{(t-\lambda)^3} $$ From this post I got the information, that the MGF is linked to the PDF via inverse Laplace transform. And from the wikipedia article about the Laplace trafo I searched for the particular form of $M_R(t)$ to find its original function/inverse Laplace transformed. $$ f_R(\rho) = \mathcal{L}^{-1}[M_R(-\cdot)](\rho) = \frac{\lambda^3\rho^2 e^{-\lambda \rho}}{2} $$ This is the PDF of the gamma distribution $$ f_R(\rho) = \Gamma(\rho; 3, \lambda) = \text{Gamma}(\rho; 3, \lambda) $$ When checking the moments and MGF of $\text{Gamma}(3, \lambda)$ one gets the moments of $|X|$. So this is indeed the distribution I searched for.
Now to draw the random variable $R$, i.e. from the Gamma distribution, I follow this scheme (again from wikipedia).
Finally I get the exponentially distributed spherically symmetric random vector via $X=RU$ with $R\sim\Gamma(3,\lambda)$ and $U \sim \text{Unif}(S^{(3)}_0(1))$ (e.g. by the method posted here).
My "derivation" seems a little obscure to me. Could someone comment on this. Is there a easier, more straight forward way to get this result?