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Suppose I have a bivariate likelihood function, $L(\theta ,\lambda |\mathbf{x})$, where $\theta$ can take on continuous values, but $\lambda$ can only take 'count' values $(0,1,2,...)$, and $\mathbf{x}$ is the data. The MLE of $\lambda$ has 2 solutions, $\hat{\lambda}=11$ and $\hat{\lambda}=12$, since the the likelihood is equally greatest for those two $\lambda$ values, irrespective of the value of $\theta$.

To estimate the asymptotic standard error of $\theta$, I'm using the Fisher Information at the ML estimates (i.e. $\sqrt{I^{-1}(\hat{\theta} ,\hat{\lambda})^{11}}$ to estimate $se(\hat{\theta})$), but this is a function of $\hat{\lambda}$... which does not have a unique value!

Any ideas what to do in this case to find $se(\hat{\theta})$ using the Fisher information?

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  • $\begingroup$ Most of the proofs of maximum likelihood estimation assume that your parameter space is continuous and (at least) twice differentiable. The scenario you've described does not satisfy that case, why must the $\hat{\lambda}$ estimates be integer values? If this is a requirement you could try marginalizing the likelihood for the $\lambda$ and $\theta$ values if the correlation is not an object of interest. $\endgroup$ – Lucas Roberts Jun 9 '18 at 12:04
  • $\begingroup$ How did you find the Fisher info matrix $I(\theta, \lambda)$ *which is defined via differentiation, also with respect to the integer parameter $\lambda$? $\endgroup$ – kjetil b halvorsen Jan 16 at 20:33

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