What’s the implication of $E(p)$ being fixed but $Var(p) \to 0$? Suppose that $X_1,\dots,X_n$ are iid sample from $Bernoulli(p)$: $f(x|p)=p^x(1-p)^{1-x}$ for $x=0,1$, $p\in(0,1)$. The prior distribution of $p \sim Beta(\alpha_0, \beta_0)$ with known $\alpha_0$ and $\beta_0$.
The Bayes estimator of $p$ is that
$$
\hat{p}=\frac{\alpha_0+\beta_0}{\alpha_0+\beta_0+n}\frac{\alpha_0}{\alpha_0+\beta_0}+\frac{n}{\alpha_0+\beta_0+n}\bar{X}_n
$$
which is the weighted of the prior mean and sample mean.

My question is that What’s the implication of $E(p)$ being fixed but $Var(p) \to 0$?


For example, when we fix the prior, $\hat{p}\to \bar{X}$ as $n\to \infty$. That means the Bayes estimator is dominated by the sample data and the role of the prior distribution disappears.

Assuming that your model is correctly specified, as $n \rightarrow \infty$ you will have $\bar{X}_n \rightarrow p$ and $|\hat{p}_n-\bar{X}_n| \rightarrow 0$ in probability, so your Bayes estimator of the parameter $p$ is consistent.  This can be demonstrated formally by showing that the posterior variance of your estimator converges to zero and the posterior mean of your estimator converges to the true parameter value; this implies convergence in mean-square which then implies convergence in probability.
