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Let data be $X_i \sim \text{Exp}(\theta)$ iid, $i=1,...,n$. Let the prior be $\text{Gamma}(\alpha, \beta)$. The posterior is then of course $\text{Gamma}(\alpha + n, \beta + \sum X_i)$. The Bayes estimator $\pi_n$ for $\theta$ is then the posterior mean, $\frac{\alpha + n}{\beta + \sum X_i}$.

I would like to compute the limiting distribution of $\sqrt{n}(\pi_n - \theta)$ as $n \rightarrow \infty$.

Note that I have already shown (using law of large numbers) that the Bayes estimator is consistent ($\pi_n \overset{P}{\longrightarrow} \theta$). Could this help me calculate this limiting distribution, if so, how?

I have been trying to do some algebraic manipulation to get $\sqrt{n}(\frac{\alpha + n}{\beta + \sum X_i} - \theta)$ into a form where I can start applying Slutsky/continuous mapping/etc, but this hasn't worked. This leads me to believe I need to use the consistency result that I previously proved. However, I am unsure how to make use of my consistency result in order to get the limiting distribution.

How can I calculate this limiting distribution?

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As $n \rightarrow \infty$ you can ignore the effect of $\alpha,\beta$ in the posterior (the data swamp the prior), so your estimator is just the usual MLE for the rate of an exponential, $\bar{X}^{-1}$.

This converges in distribution to $\mathrm N(\theta, \mathcal I(\theta))$.

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  • $\begingroup$ Dear Doctor Milt, thank you for your answer. Yes, I understand this intuitively (and indeed this is what my hypothesis was for what the distribution would converge to). Is there a simple way to formally argue that you can ignore the effect of $\alpha$, $\beta$ in the posterior? I have been a bit confused on how to lay out doing that formally. I will accept your answer if you can further elaborate on that point. Thank you! $\endgroup$ Commented Apr 30 at 13:36
  • $\begingroup$ The Bernstein-von Mises theorem and proof are here: adamnsmith.com/files/notes/bayes-asymptotics.pdf. $\endgroup$
    – jbowman
    Commented Apr 30 at 15:27

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