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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!

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  • $\begingroup$ 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? $\endgroup$ – Dilip Sarwate Feb 4 '19 at 22:38
  • $\begingroup$ 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). $\endgroup$ – sblatt Feb 4 '19 at 22:48
  • $\begingroup$ Are the components independent? $\endgroup$ – kjetil b halvorsen Feb 4 '19 at 23:04
  • $\begingroup$ Yes. They are independent, but drawn from the same distribution. $\endgroup$ – sblatt Feb 4 '19 at 23:21
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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 $\frac{1}{2}$.

Unfortunately, there does not seem to be anything on the sum of iid Weibulls in general, and convolving even only two Weibulls with shape $\frac{1}{2}$ exceeds WolframAlpha's standard computing time.

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