# Computing the limiting distribution of the Bayes estimator for exponential data with a Gamma prior (by using consistency?)

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?

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 $$\mathrm N(\theta, \mathcal I(\theta))$$.
• 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! Commented Apr 30 at 13:36