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I know that the parameter space has to be convex (that I've proved using Jensen's inequality). Therefore, since it has to be a subset of $\mathbb R$, it follows that it has to be an interval or a point. Assuming that it is not a point, is it possible that the parameter space $\mathcal N$ for one-parameter exponential family distribution in canonical form, i.e. one with the following density: \begin{equation} f_\theta(x) = C(\theta)h(x)e^{\theta T(x)} \end{equation} is a closed interval?

Martund
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    $\mathbb{R}$ is a closed interval. – whuber Jan 12 '21 at 18:13
  • @whuber, $\mathbb R$ is both open and closed in $\mathbb R$ trivially. In this case, by closed interval, I mean some non-trivial case like $[a,b]$ where $-\infty<a<b<\infty$ – Martund Jan 12 '21 at 18:16
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    $[0,1]$ for the Binomial distribution. – Xi'an Jan 12 '21 at 18:39
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    If you want to rule out cases then you need to do so explicitly in your question. Otherwise the trivial answer(s) are valid. – whuber Jan 12 '21 at 18:53
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    @Xi'an For binomial distribution, $[0,1]$ is not correct, because the natural parameter is $\log\left[\frac{p}{1-p}\right]$, whose natural parameter space is $\mathbb{R}$. – Tan Jan 21 '21 at 19:12
  • The Binomial distribution is well defined for $p=0,1$. – Xi'an Jan 21 '21 at 19:14
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    True, but OP is about the natural parameter space of $\theta$, which is the natural parameter (because p.d.f is form of $e^{t\theta}$). $p$ is a not the natural parameter so $[0,1]$ is not the natural parameter space. – Tan Jan 21 '21 at 19:18

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