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The standard Bayesian parameter estimation deals with the problem of estimating $P(\theta|x_{1:k})$, where $\theta$ is a constant but unknown parameter and $x_{1:k}$ are observations/data. I know there exists some convergence results showing the point estimate, say MLE $\hat{\theta}$, converges to the true parameter value $\theta$.

My question is that, if $\theta$ is time variant and we know its deterministic dynamics, e.g. $\dot{\theta}=f(\theta)$, can we obtain some similar result that the error between the MLE $\hat{\theta}_t$ and the true value $\theta_t$, i.e. $e_t=\hat{\theta}_t-\theta_t$, converges to $0$ asymptotically?

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    $\begingroup$ Bayesians assume $\theta$ is random. Frequentists assume $\theta$ is non-random. Which framework are you using? $\endgroup$
    – Taylor
    Jul 30, 2016 at 3:43
  • $\begingroup$ @Taylor Sorry that I didn't make my point clear. I am actually thinking about Bayesian filtering, such as Kalman filter or particle filter, but in the application of estimating a non-random parameter. In this case, can I say the some point estimate, e.g. MAP estimate, converges to the "true" time-varying parameter? Or do I mess up some basic concepts here? Thanks! $\endgroup$
    – shionlau
    Aug 2, 2016 at 3:23
  • $\begingroup$ Even though filtering recursions (particle/kalman) can be justified with Bayes' rule, you may or may not be a Bayesian. If the parameter $\theta$ is fixed, you are not doing Bayesian filtering. There is no $p(\theta|x_{1:k})$ in this case. If the parameter is random with a prior $\theta \sim p(\theta)$, then you're doing Bayesian filtering. Then you do have $p(\theta|x_{1:k})$. When you say "estimating a non-random parameter," you are not doing Bayesian filtering. So it doesnt make sense to talk about a MAP estimate (there's no posterior distribution). You're flipping back and forth. $\endgroup$
    – Taylor
    Aug 2, 2016 at 3:43
  • $\begingroup$ @Taylor Thanks for clearing up these concepts! I think now my question is, if I want to estimate a non-random but time-varying parameter using Bayes' rule, it there some way that I can guarantee some result saying that the estimator is consistent that asymptotically converges to the "true value" of the parameter? Thanks! $\endgroup$
    – shionlau
    Aug 6, 2016 at 2:43

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