# Convergence of the variance of the posterior expectation

Consider a classical Bayesian model : $$\begin{array}{cc} \theta \sim \pi \\ X = (X_1, ..., X_n) \overset{i.i.d.}{\sim} \pi(.\mid\theta) \end{array}$$ where the prior does $$\pi$$ and the likelihood $$\pi(.\mid \theta)$$ do not change with $$n$$ and data $$X$$ are i.i.d. conditionnaly on $$\theta$$ (no weird stuff here).

I'm wondering wether the variance of the expectation a posteriori should coincide asymptotically with the variance of $$\theta$$. That is, wether the following property is true: $$\mathrm{Var}\left(\mathbb{E}[\theta \mid X]\right) \underset{n\to\infty}{\longrightarrow}\mathrm{Var}[\theta] .$$ I expect it to be true from an informal reasonning based on variance decomposition, but I would like to have that intuition confirmed (or unvalidated) by a more formal argument.

So my questions are :

• Is this result true ? If yes, under what hypothesis ?
• How would one prove this rigorously ?
• Are there cases (not totally degenerated like a point mass prior) where this result does not hold?
• What about the posterior variance ? Shouldn't it provide information on the variance of the posterior expectation ?

If the prior is covering the entire range then $$\hat{\theta} \xrightarrow{P} \theta$$, the parameter estimate will converge to the true parameter. So a Bayesian type of estimate will converge to the true parameter and the sample distribution of the estimate will converge to the distribution of the true parameter.