asymptotic distributon, posterior mean

by janekimpebe   Last Updated October 19, 2019 22:19 PM

Suppose $X \sim Bin(n, \theta)$, $\theta \in [0,1]$. Considering a prior distribution for $\theta$ from $Beta(a,b)$, we get a posterior distribution $\theta|X \sim Beta(a+X, b+n-X)$ and so we get the posetrior mean $\overline{\theta}_n(X)=\mathbb{E}[\theta|X]=\frac{a+X}{a+b+n}$. Now, assuming $X$ is sampled from $Bin(n, \theta_0)$, $\theta \in (0,1)$, I have to find the asymptotic distribution of $$\sqrt{n}(\overline{\theta}_n(X)-\theta_0)$$ as $n\rightarrow \infty$.

My solution:

As $\frac{\sqrt{n}a}{a+b+n} \rightarrow0$, it's equivalent to finding the distribution of $\sqrt{n}(\frac{X}{a+b+n}-\theta_0)$.

Now the MLE of $\theta$ for $X \sim Bin(n, \theta)$ is $\widehat{\theta}=\frac{X}{n}$.

So we can write $\sqrt{n}(\frac{X}{a+b+n}-\theta_0)=\sqrt{n}(\frac{X}{a+b+n}-\widehat{\theta})+\sqrt{n}(\widehat{\theta}-\theta_0).$

Now $\sqrt{n}(\frac{X}{a+b+n}-\widehat{\theta})=\sqrt{n}(\frac{X}{a+b+n}-\frac{X}{n})=\sqrt{n}X(\frac{-(a+b)}{n^2+na+nb})\rightarrow0$, as $X$ takes values in the range $\{0, 1, \dots n\}$.

So the distribution of $\sqrt{n}(\frac{X}{a+b+n}-\theta_0)$ is the same as the distribution of $\sqrt{n}(\widehat{\theta}-\theta_0)$, which by the asymptotic normality of the MLE is $N(0, \frac{\theta_0(1-\theta_0)}{n})$.

Is my solution correct? I am a bit unsure if we can assume the asymptotic normality of the MLE. Thank you!

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