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Feb 7, 2022 at 21:00 history tweeted twitter.com/StackStats/status/1490792493520240642
Feb 7, 2022 at 16:02 answer added Sextus Empiricus timeline score: 2
Feb 7, 2022 at 14:43 history edited whuber
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Feb 7, 2022 at 14:41 comment added whuber @Scortchi Thank you -- I had only looked at the condition $\mu\in\mathbb{R}^p.$ Yes, indeed, this is a critical observation, for it reduces the question to two key aspects: (1) the smoothness of the map $S^{p-1}\times\mathbb{R}_+\to\mathbb{R}^p\setminus\{0\}$ defined by $(\mu,\kappa)\to\mu\kappa$ and (2) the interpretability of the resulting reparameterization.
Feb 7, 2022 at 8:38 comment added Scortchi @whuber: I don't follow the above: aren't $\kappa$ & $\mu$ just the length & direction, respectively, of the vector $\kappa\mu$, & so uniquely determined by it?
Feb 6, 2022 at 16:30 comment added whuber Because in the interior of the parameter space all the distributions are distinct, and within that space the product $\kappa\mu$ never determines the ordered pair $(\kappa,\mu),$ you can parameterize only a proper subset of all such distributions in this manner. This is exactly like asking why we might want to describe the shapes of rectangles using two non-negative numbers rather than employing only their areas.
Feb 6, 2022 at 2:59 comment added Henry A potentially more natural single parametrisation might be the expected value (which will lie in the interior of the sphere and have the same direction as $\mu$), and is related to a sufficient statistic of a sample from the distribution. Unfortunately the relationship with $\kappa$ is not simple
Feb 6, 2022 at 0:04 answer added Aksakal timeline score: 4
Feb 5, 2022 at 23:58 answer added jbowman timeline score: 5
Feb 5, 2022 at 23:06 history edited Rylan Schaeffer CC BY-SA 4.0
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Feb 5, 2022 at 23:00 history asked Rylan Schaeffer CC BY-SA 4.0