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.

Let $m = \alpha_0 + \beta_0$ be the so-call sample size of the prior. We have $\mu := E(p) = \alpha_0 / m$ and
$$\text{var}(p) = \frac{\alpha_0 (m - \alpha_0)}{m^2(m+1)} = \frac{\mu(1-\mu)}{m+1}.$$
We see that for $\text{var}(p) \to 0$ given that $\mu$ is fixed, we need $m \to \infty$. Since the Bayes estimate is
$$
\hat p = \frac{m}{m+n} \mu + \frac{n}{m+n} \bar X_n,
$$
assuming that $n$ is fixed, letting $m \to \infty$ we have $\hat p \to \mu$, which is intuitively somewhat obvious.
