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Assume that you have two Poisson random variables, $y_{jk} ∼ Poi(\lambda_{jk} \psi_j)$ and $y_{kj}∼Poi(\lambda_{jk}\psi_k)$. I've read that this parameterization is not unique, but for me it is not trivial to see this. How to show that this parameterization is not unique? And how do I make it unique? So it is really a question about identifiability of the parameters given the parameterization.

EDIT: My way of viewing this is the following: The random variable $y=y_{jk}+y_{kj}$ is Poisson distributed with mean parameter $\lambda_{jk} \psi_j+\lambda_{jk} \psi_k = \lambda_{jk}(\psi_j+\psi_k)$. Denoting the parameter vector of interest as $\theta=(\lambda_{jk}, \psi_j, \psi_k)$, it is possible to find a $\theta^\prime\ne\theta$, such that $f_\theta=f_{\theta^\prime}$, which means that $\theta$ is not identified under this parameterization. Is this a correct reasoning?

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Let $\lambda_{jk} \psi_j = \alpha$ and $\lambda_{jk}\psi_k = \beta$.

You have 3 unknowns: $\lambda_{jk}, \psi_j, \psi_k$ and only 2 equations.

So when one unknowns are fixed at given values, you can resolved the second and third one based on the 2 equations. For example, let $\alpha =1$ and $\beta = 2$. If I fix $\lambda_{jk} = 1$, I get $ \psi_j = 1$ and $\psi_k = 2$. If I fix $\lambda_{jk} = 0.5$, i get $ \psi_j = 2$, and $\psi_k = 4$, etc....

To make them identifiable, need to add a condition, such as $ \psi_j = 1$.

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  • $\begingroup$ Great, thanks! But if the purpose still is to make inference about $\psi_j$, how would I "get rid of" $\lambda_{jk}$? $\endgroup$ – StatStud May 25 '17 at 14:10
  • $\begingroup$ Need to add a condition, otherwise, no inference is possible. $\endgroup$ – user158565 May 25 '17 at 14:19
  • $\begingroup$ So if I for instance add the condition $\lambda_{jk}=1$, I can derive the joint likelihood function of $y_{jk}$ and $y_{kj}$ and maximize it with respect to $\psi_j$ and $\psi_k$? $\endgroup$ – StatStud May 25 '17 at 14:31
  • $\begingroup$ Yes. (Only 'Yes' does not work, I need to type enough characters in order to Add Comment) $\endgroup$ – user158565 May 26 '17 at 3:22

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