Statistics problem about computing UMVUE(Uniformly Minimum Variance Unbiased Estimator) [closed]

Question

Let $\underline{X}$ ~ Be(p) of size n. Obtain UMVUE for p.

My idea is the MLE of that problem is the UMVUE but i think its wrong. Can someone help me?

Answer 1

There are several ways to show that $T=\overline{X}_n$ is the UMVUE for $p$

This is one; I chose this because it allow us to do further considerations about that important estimator:

$$\mathbb{E}[\overline{X}_n]=p$$

Thus $T$ is unbiased and its variance is

$$\mathbb{V}[\overline{X}_n]=\frac{p(1-p)}{n}$$

Now let's calculate the lower bound for the variance of unbiased estimator for $p$ using Cramér-Rao inequality;

$$\mathbb{V}[T]\geq \frac{1}{-n\mathbb{E}\left\{\frac{\partial^2}{\partial p^2}\log f(x|p) \right\}}=\dots=\frac{p(1-p)}{n}$$

Thus $T=\overline{X}_n$ is UMVUE for $p$ AND its variance also attains the Cramér Rao lower bound....and this does not happen so frequently among UMVU Estimators

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here are the calculations for CR lower bound, as requested.

$$f=p^x(1-p)^{1-x}$$

$$\log f=x\log p+(1-x)\log(1-p)$$

$$\frac{\partial}{\partial p}\log f=\frac{x}{p}-\frac{1-x}{1-p}$$

$$\frac{\partial^2}{\partial p^2}\log f=-\frac{x}{p^2}+\frac{1-x}{(1-p)^2}$$

Now observe that $\mathbb{E}[X]=p$ and $\mathbb{E}[1-X]=1-p$ thus substitute and immediately conclude