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In my textbook, Identifiablity is defined as so:

For any $\theta_1, \theta_2 \in \Theta$ , if $\theta_1 \neq \theta_2 \Rightarrow \Bbb P_{\theta_1} \neq \Bbb P_{\theta_2}$ , where $\Bbb P_{\theta}$ is the probability distribution function.

Then it says that $X \sim N(\frac{\theta_1}{\theta_2}, \theta_3)$ where $\theta_1 \in \Bbb R$ , $\theta_2, \theta_3 \gt 0$ does not define an identifiable statistical model. Why?

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    $\begingroup$ Consider, for example, $(\theta_1,\theta_2,\theta_3)=(0,1,1)$ and $(\theta_1,\theta_2,\theta_3)=(0,2,1)$. Both triples correspond to the same $\text{N}(0,1)$ distribution. $\endgroup$ – Zen Jan 11 '20 at 16:58
  • $\begingroup$ How am I supposed to use the definition to prove this example? Since in the definition there are only 2 $\theta$'s. This is my main concern $\endgroup$ – The Poor Jew Jan 11 '20 at 17:04
  • $\begingroup$ Do we suppose that $\theta_1' = (\theta_1, \theta_2, \theta_3)$ and then if $\theta_1' = (0,1,1) \neq \theta_1''=(0,2,1) $ does not imply inequality of $\Bbb P_{\theta_1'}$ and $\Bbb P_{\theta_1"}$ ? Hence model is not identifiable? $\endgroup$ – The Poor Jew Jan 11 '20 at 17:15
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    $\begingroup$ Yes, also related is this question on identification $\endgroup$ – Jesper for President Jan 11 '20 at 17:59
  • $\begingroup$ Please add the self-study tag and read its wiki. $\endgroup$ – kjetil b halvorsen Jan 12 '20 at 14:43

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