1
$\begingroup$

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 ?
$\endgroup$

1 Answer 1

2
$\begingroup$

This is a bit related to When do posteriors converge to a point mass?

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.

If the prior is also equal to the true distribution of the parameter (this doesn't need to be the case) then the convergence of the sample distribution of the estimate to the sample distribution of the true parameter will also be convergence to the prior distribution. (and the variances will equal)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.