Timeline for Convex support of an exponential family and its mean parameter space $\mathcal{M}$
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
replaced http://stats.stackexchange.com/ with https://stats.stackexchange.com/
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| Apr 11, 2017 at 12:03 | history | edited | Henry.L | CC BY-SA 3.0 |
added 180 characters in body
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| Mar 23, 2017 at 23:40 | comment | added | Henry.L | @passerby51 A more detail explanation is available now on stats.stackexchange.com/questions/266691/…, hope helps! | |
| Feb 21, 2017 at 20:00 | history | edited | Henry.L | CC BY-SA 3.0 |
corrected.
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| Feb 21, 2017 at 19:59 | comment | added | Henry.L | @passerby51 And by saying that $\mathcal{M}$ is closed by Helly's selection theorem because any limit of BV functions will yet still converge to a BV function. Note that I emphasize $X$ is compact in order to make my argument hold, your example is on $\mathbb{R}^d$,which is not compact. My comment in the bracket is to say that you can choose $\mathbb{R}^{d}$ as the underlying imbedded space, which means you want to inherit Euclidean structure. I do not mean you can choose $X=\mathbb{R}^d$ and still hope the interpretation to be valid. I have make it clear now. | |
| Feb 21, 2017 at 19:53 | comment | added | Henry.L | @passerby51 A curved exponential family is still an exponential family, what you raised in the example is just $N(\mu,\mu)\in\partial\mathcal{M}$, a variance zero degenerate distribution is the apex of the $\mathcal{M}$. | |
| Feb 21, 2017 at 16:58 | comment | added | passerby51 | thanks for the detailed response. Not sure if I follow everything. I have added an example to my post for which I believe $\mathcal M$ is open. Am I missing something? | |
| Feb 21, 2017 at 12:08 | history | edited | Henry.L | CC BY-SA 3.0 |
complement
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| Feb 21, 2017 at 11:59 | history | edited | Henry.L | CC BY-SA 3.0 |
add more about boundary case
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| Feb 20, 2017 at 23:23 | history | answered | Henry.L | CC BY-SA 3.0 |