I thought I got problems like this already but looks like I'm stuck again.

Let $x_1, x_2$ be a random sample $\sim Ber(\theta)$

The objective is to find the UMVUE for $\theta^2$

A hint was provided in the form of a question, which is to show that $T=x_1x_2$ is unbiased for $\theta^2$ which I was able to do.

Next, I considered $x_1+x_2$ as a complete sufficient statistic for $\theta$. I worked this out using $x_1+x_2 \sim Binomial (2,\theta)$, the factorization theorem, and the theorem that says that any statistic with a distribution that is part of the exponential family is complete.

So, given this, I should get the UMVUE by evaluating $E(T|x_1+x_2)$, right?



Which is where I am stuck.

I fiddled around with $\sum x_i=x_1+x_2$ a bit and I think I stumbled on the UMVUE $\hat{\theta^2}=\frac{\sum x_i(\sum x_i-1)}{2}$. I think this is it because it is a function of the complete sufficient stat and is unbiased for the desired parameter. However, I don't know how to make the expression above appear this way.


Note that the product $x_1 x_2$ is Bernoulli$(\theta^2)$. However, by definiton of Bernoulli, we have $x_1 x_2 = \mathbf{1}_{[x_1=x_2=1]}$. Then, since $E[\mathbf{1}_{[x \in A}] = P(x \in A)$,

$$ E[x_1 x_2 \mid x_1 + x_2=t] = P( x_1 = x_2 =1 \mid x_1 + x_2 =t) = \dfrac{ P(x_1 = x_2 =1 \text{ and } x_1 + x_2 =t)}{P(x_1+x_2=t)} $$ Now check that this equals $\phi(t)=\begin{cases} 0 & t=0,1 \\ 1 & t=2 \end{cases}$ and $\phi(t)$ is your UMVUE


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