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Consider the likelihood function for parameter vector $\boldsymbol{\theta}=(\theta_1,\theta_2)$:$L(\boldsymbol{\theta};\boldsymbol{x},\boldsymbol{y})=L_1(\boldsymbol{\theta};\boldsymbol{x})L_2(\theta_2;\boldsymbol{y})$; where $(\boldsymbol{x},\boldsymbol{y})$ are data vectors.

Now, assume the following conditions:

  1. The likelihood function $L_1(\boldsymbol{\theta};\boldsymbol{x})$ is non-identifiable, i.e. there exists a function $g(.)$ such that $\max_{\boldsymbol{\theta}}L_1(\boldsymbol{\theta};\boldsymbol{x})=k$ for all $\boldsymbol{\theta}$ satisfying $g(\boldsymbol{\theta})=0$, for some constant $k$.

  2. $\max_{\theta_1}L_1(\boldsymbol{\theta};\boldsymbol{x})$ is unique for all fixed values of $\theta_2$ in the parameter space.

  3. $\max_{\theta_2}L_2(\theta_2;\boldsymbol{y})$ is unique, i.e. $\theta_2$ is estimable from $L_2(\theta_2;\boldsymbol{y})$ alone.

It is intuitively clear that $L(\boldsymbol{\theta};\boldsymbol{x},\boldsymbol{y})$ is identifiable because $L_2(\theta_2;\boldsymbol{y})$ is identifiable. However, how can we prove identifiability of $L(\boldsymbol{\theta};\boldsymbol{x},\boldsymbol{y})$ mathematically?

Sean Easter
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NadeemK
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1 Answers1

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This gives the definition of identifiable. IIRC, the typical approach for a one-to-one proof is proof by contrapositive.

call-in-co
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  • Thanks for pointing this out. I think the situation here is somewhat different as the full likelihood function is a product of two sub-likelihood components, $L_1(.)$ and $L_2(.)$ , where one of them is non-identifiable. I am wondering about specific condition(s) (may be #2 in the question) is necessary for the identifiability of $L(.)$ as a whole. – NadeemK Jul 22 '15 at 16:00
  • I see what you mean. I don't have experience in this regard, but maybe [this](http://stats.stackexchange.com/questions/20608/what-is-model-identifiability?rq=1) post will help you, especially this line: "...in an identifiable model it is impossible for two distinct parameters (which could be vectors) to give rise to the same likelihood function." – call-in-co Jul 22 '15 at 16:25