# PDF for length of Laplace-distributed vectors

I am interested in finding an analytic expression for the length of a 3-vector whose components are distributed according to a Laplace distribution with zero mean and the same scale parameter.

I have been unable to find anything useful so far in the literature, but it seems from my numeric tests that the answer should be close to $$r^2 e^{-r/\lambda} dr$$, where $$r = \sqrt{x_1^2 + x_2^2 + x_3^2}$$ and the $$x_i$$ are Laplace variables with zero mean and exponential scale parameter $$\lambda$$. My heuristic result makes some sense to me from a transformation to spherical coordinates.

It would already help if I knew something more about the distribution of $$r^2$$ (for normally-distributed $$x_i$$, I know that I would get a $$\chi^2$$ distribution).

I would appreciate any hints or literature pointers you might have.

Thanks!

• Length of a vector is generally considered to be a nonnegative quantity and so why are you modeling it as a Laplacian radio variable rather than an exponential random variable? – Dilip Sarwate Feb 4 '19 at 22:38
• Exponential distributions are perfectly fine, I would appreciate any input in that regard as well. It is just that my physical problem starts out with a Laplace distribution (these are real-space vectors pointing in all directions). – sblatt Feb 4 '19 at 22:48
• Are the components independent? – kjetil b halvorsen Feb 4 '19 at 23:04
• Yes. They are independent, but drawn from the same distribution. – sblatt Feb 4 '19 at 23:21

Consider $$X_1, X_2, X_3\sim\text{Laplace}(0,b)$$, iid for a parameter $$b$$.
Then $$|X_i|\sim\text{Exp}\big(\frac{1}{b}\big)$$.
Then $$X_i^2=|X_i|^2\sim\text{Weibull}\big(b^2,\frac{1}{2}\big)$$.
So your question is answered by the sum of three iid Weibull variables with common scale $$b^2$$ and the specific shape $$\frac{1}{2}$$.