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This question arises from the study of the k-parameter exponential family of distributions.

When the canonical form is specified i.e. the family $\mathcal{P}_\theta$ is reparametrized by $\eta_i = \eta_i(\theta), i=\{1,\dotsc,k\}$ with $\theta \in \Theta \subset \mathbb{R}^k$, there is an implicit assumption that the dimension of $\Theta$ and the dimension of the image under the map $(\theta_1,\dotsc,\theta_k)\rightarrow (\eta_1(\theta),\dotsc,\eta_k(\theta)) $ are equal to k. Can we conclude that the mapping is one-to-one, which is crucial for identifiability of $\theta$?

I did not see this in the notes I find so far. Can someone offer his/her wisdom on the matter?

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For arbitrary continuous functions, no such implication holds. The map $e^{i\theta}$ is continuous and surjective from the real line onto a circle. Both have dimension $1$, but the map rather dramatically fails to be one to one, as the inverse image of every point is infinite.

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  • $\begingroup$ Thanks Matt, I totally agree with you. I suppose identifiability is too elementary a topic that the statisticians hardly brush on it if ever. $\endgroup$ – David KWH Sep 6 '16 at 18:16
  • $\begingroup$ @DKWH No problem. $\endgroup$ – Matt Samuel Sep 6 '16 at 18:19

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