Suppose a data generating process (DGP) is parameterized by some unknown parameter $\theta\_0$, say $P\_{\theta\_0}$, and we want to estimate the value of $\theta\_0$ using Bayesian method. Let $\pi(\theta)$ be the prior over the possible values of $\theta$, $\Theta$. 


I understand that for almost all priors, if the data observed are iid, then the Bayesian posterior will eventually concentrate on $\theta\_0$, as the number of observations ($n$) tends to infinity. (Correct me if I'm wrong.)


My question: **Are there any results that predict differential rates of convergence of the posterior distribution *based on different priors*?** 


For example, suppose $\Theta=\{\theta\_0,\theta\_1\}$ and the DGP is $P\_{\theta\_0}$. Consider two priors on $\Theta$: 
$$\pi\_1(\theta\_0)=0.2,\qquad \pi\_1(\theta\_1)=0.8$$ and $$\pi\_2(\theta\_0)=\pi\_2(\theta\_1)=0.5.$$ Are there any results that says the posterior based on $\pi\_2$ converges faster than that based on $\pi\_1$? (or the other way around?) 


Intuitively I would expect the posterior based on $\pi\_2$ to have faster convergence, as it is "closer" to $\theta\_0$. But in my experience, intuition is hardly reliable when it comes to probability theory.


Note: I've asked this quesiton on [Math.SE](https://math.stackexchange.com/q/1140024/70119) but received no answer. I thought I would try my luck here.



Rate of convergence of Bayesian posterior
One can measure the rate of convergence of the posterior distribution with density $p\_n$ to the Dirac probability distribution at $\theta\_0$ by how large the ratios $p\_n(\theta\_0)/p\_n(\theta)$ are for $\theta\ne\theta\_0$, where $\theta\_0$ is the "true" value of the parameter. One has 
$$
\frac{p\_n(\theta\_0)}{p\_n(\theta)}=
\frac{\pi(\theta\_0)}{\pi(\theta)}\,\frac{L\_n(\theta\_0)}{L\_n(\theta)}, 
$$
where $\pi$ is the prior density and $L\_n(\theta):=\prod\_1^n f\_\theta(X\_i)$ is the likelihood value based on the first $n$ of the iid observations $X\_1,X\_2,\dots$ with common density $f\_{\theta\_0}$. It is clear from the above display that, for the same "data" $X\_1,X\_2,\dots$, the greater is the prior ratio $\pi(\theta\_0)/\pi(\theta)$ the greater is 
the posterior "contrast" ratio $p\_n(\theta\_0)/p\_n(\theta)$. In particular, in the extreme case when (say) $\pi(\theta)$ is $1$ if $\theta=\theta\_0$ and is $0$ otherwise, the prior ratio is $\infty$ for any $\theta\ne\theta\_0$, and hence so is the "contrast" ratio $p\_n(\theta\_0)/p\_n(\theta)$. 


One may want to measure the posterior contrast by one number, say by averaging the posterior contrast over the data space and over the parameter space, to get something like 
$$\int\mu(d\theta)\,\mathsf{E}\ln\frac{p\_n(\theta\_0)}{p\_n(\theta)}=
\int\mu(d\theta)\,\ln\frac{\pi(\theta\_0)}{\pi(\theta)}
+n\int\mu(d\theta)\,\mathsf{E}\ln\frac{f\_{\theta\_0}(X\_1)}{f\_{\theta}(X\_1)},$$ 
where $\mu$ is a measure on the parameter space.  

So, the greater is the average prior contrast $\int\mu(d\theta)\,\ln\frac{\pi(\theta\_0)}{\pi(\theta)}$ the greater is 
the average posterior contrast $\int\mu(d\theta)\,\mathsf{E}\ln\frac{p\_n(\theta\_0)}{p\_n(\theta)}$. 


In particular, your prior $\pi\_2$ should result, by this logic, in a greater posterior contrast than $\pi\_1$. 


**Addendum:** The average prior contrast and hence the average posterior contrast may behave somewhat surprisingly, though. E.g., suppose that the parameter space is $\Theta=\mathbb{R}$, $\theta\_0=0$, $\pi$ is the normal density with mean $\theta\_1$ and variance $\tau^2$, $f\_\theta$ is the normal density with mean $\theta$ and variance $\sigma^2$, and $\mu$ is the normal distribution with mean $0$ and variance $\gamma^2$. Then for the average prior and posterior contrasts one has 
$$\int\mu(d\theta)\,\ln\frac{\pi(\theta\_0)}{\pi(\theta)}=\frac{\gamma^2}{2\tau^2}\quad\text{and}\quad 
\int\mu(d\theta)\,\mathsf{E}\ln\frac{p\_n(\theta\_0)}{p\_n(\theta)}=
\frac{\gamma^2}2\Big(\frac1{\tau^2}+\frac n{\sigma^2}\Big),$$
respectively. 
As could be expected, both these average contrasts are the greater the smaller is the variance $\tau^2$ of the prior distribution. However, neither of these contrasts depends on the prior mean $\theta\_1$ ! This will be so for any distribution $\mu$ symmetric about $0$ (or, more generally, about $\theta\_0$) -- because, for the normal prior $\pi$ as in this addendum, the symmetrized prior contrast 
$$\ln\frac{\pi(0)}{\pi(\theta)}+\ln\frac{\pi(0)}{\pi(-\theta)}
=\frac{\theta^2-2\theta \theta \_1}{2 \tau ^2}
+\frac{\theta^2+2\theta \theta \_1}{2 \tau ^2}
=\frac{\theta^2}{\tau^2}$$
does not depend on $\theta\_1$. 


