How find MVUE for $g( \theta)$?

Question

let $X=(x_1,...,x_n)$ ,$n>1$ random sample of $B(1, \theta)$ find MVUE of $g( \theta)=P_ \theta(x_1\geq x_2 )$

i did something like that:

$\prod_{i=1}^{n}f_x(x;\theta )=e^{\sum_{i=1}^{n}x_iln(\theta )+\sum_{i=1}^{n}(1-x_i)ln(1-\theta )}$

and that tells me that there exist a unique MVUE but i dont know how to find $T(X)$ that such that $E(T(X))=g(\theta)$

i am a bit confused can someone explain how to do it?

Answer 1

One easy unbiased estimator is the following:

$$T=\mathbb{1}_{\{X_1\geq X_2\}}$$

You surely know that $S=\sum_{i=1}^n X_i$ is complete and sufficient, thus you can use Rao - Blackwell and Lehmann - Scheffé together finding that the UMVUE is

$$\mathbb{E}[T|S=s]=\mathbb{P}[T=1|S=s]=\frac{ \binom{n-2}{s} + \binom{n-2}{s-1}+ \binom{n-2}{s-2} }{\binom{n}{s}}=1-\frac{\binom{n-2}{s-1}}{\binom{n}{s}}=1-\frac{s}{n}\cdot\frac{n-s}{n-1}$$

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The unbiased estimator I showed is not the only one you can use..I leave to you as an exercise to find the other obvious and more suitable unbiased estimator for $g(\theta)$