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For iid data, the posterior on the parameter $$ p(\theta \mid x_{0:T}) = \prod_{t=0}^T p(x_t \mid \theta) p(\theta) $$ is known to become independent of the prior which is the Bernstein-von Mises theorem. As $T \rightarrow \infty$, the posterior also concentrates around the true parameter that generated data as a Gaussian density where the variance is determined by the Fisher matrix, i.e. $$ p(\theta \mid x_{0:T}) \approx \mathrm{exp} \left[-\frac{T}{2} \sum_{ij} (\theta_i - \theta^{\star}_i)\hat{J}_{ij} (\theta_j - \theta^{\star}_j) \right] $$ where $\hat{J}_{ij} = -\partial_i \partial_j \frac{1}{T} \sum_{\mu=1}^T \ln p(x_\mu \mid \theta^{\star})$. (reference, see eqn 5.4)

I am interested in a similar concentration argument for Markov models. $$ p(\theta \mid x_{0:T}) = \prod_{t=1}^T p(x_t \mid x_{t-1},\theta) p(\theta) $$ Experimentally (for identifiable models) I have observed the posterior to shrink to a Gaussian around the parameter with more data, but I am interested in whether a similar concentration density with respect to the Fisher information exist for the Markovian setting.

I would appreciate a derivation or at least the steps towards a derivation. I am an engineer grad student, so I would be happy if the recommended literature is accessible to non-stats people.

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  • $\begingroup$ The expression in the first line is not correct. Take $t=0$ and you should get Bayes theorem - you're missing a normalization constant. I suggest you figure that out (might be easy). That may help you understand what you really want. $\endgroup$
    – Yair Daon
    Commented May 20, 2016 at 22:00

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