# asymptotic distributon, posterior mean

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

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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