Let $(\epsilon\_t)\_t$ be a sequence of iid random variables, distributed according to the density $f:\mathbb{R}\to (0,\infty)$ and
$$
 x\_t = q( \theta^\star, x\_1,x\_2, \ldots, x\_{t-1}) + \epsilon\_t \,.
$$
Assume that the derivative of $q$ with respect to $\theta$ is bounded from below, i.e. for all $\theta, x$
$$
\frac{\partial q(\theta, (x\_r)\_{r<t})}{\partial \theta} \geq c > 0\,.
$$
The posterior likelihood is defined as
$$
 L\_t(\theta) = \sum\_{s=1}^t \log f \Big(x\_s-q(\theta, (x\_r)\_{r<t})\Big) + \log \pi\_0(\theta)\,,
$$
where the prior density $\pi\_0:\mathbb{R} \to (0,\infty)$ is strictly positive.


**Question:** Are those assumptions sufficient to conclude that the posterior 
$$
\pi\_t(\theta) = \frac{e^{L\_t(\theta)}}{\int e^{L\_t(z)} dz}
$$
converges almost surely to a dirac measure on $\theta^\star$? How could I show this? What would be a good reference?


I know that Bernstein von-Mises Theorem yields this type of result, but I could not find a version which is general enough to cover this case.


Thanks a lot!



convergence of Bayesian posterior with non iid data
Probably not true just with the assumptions you have there. It seems to accommodate a model in which $\theta$ is the starting point of a random walk. That would only require $q = \theta + x\_1 + ... + x\_t$. It is known and easy to believe that in this case you cannot estimate $\theta $ consistently, and I would think that therefore the posterior should not collapse to a point mass. You can calculate the posterior explicitly in case $\theta, \epsilon\_t$ are i.i.d. normal see that that is true.


