If $X$ is an observation from a $\text{Binom}(n, p)$, with $0 < p < 1$, how to find the UMVUE of $p^2$, and does it reach the Cramer-Rao lower bound?

If $X$ is an observation from a $\text{Binomial}(n, p)$, where $0 < p < 1$, how do I find the UMVUE of $p^2$, and does it reach the Cramer-Rao lower bound?

I was able to find UMVUE of $p$.

Since we know that $x$ is complete sufficient, as $E(X/n) = p$, the UMVUE of $p$ is then $X/n$. Then we have $\text{var}(X/n) = p(1-p)/n.$

To calculate the Cramer-Rao lower bound, $$L(p) = nCx\,*\,p^x(1-p)^{n-x},$$ $$l(p)= \log(nCx)+x\log(p)+(n-x)\log(1-p).$$

$$U(p) = x/p - (n-x)/(1-p) \,=\, x(1/p+1/(1-P)) - n/(1-p),$$

$$I(p) = \text{var}[U(p)] = np(1-p)(1/p+1/(1-P))^2 = n/p(1-p).$$

Cramer-Rao lower bound is $1/I(p) = p(1-p)/n,$ and we see that $X/n$ attains this lower bound.

I was wondering how can I do this with $p^2\,$?

It would be greatly appreciated for a solution.

Solution 1:

Consider $X=\sum_{i=1}^n Y_i$ where $Y_i \sim B(p)$, iid.

Let's set

$$T=\mathbb{1}_{\{Y_1=1;Y_2=1\}}$$

$T$ is evidently unbiased for $p^2$. To find its UMVUE we can use Rao Blackwell. The requested UMVUE will be

$$\begin{align} \mathbb{E}[T=1|X=x] & =\frac{p^2\mathbb{P}\left( \sum_{i=3}^{n}Y_i=x-2 \right)}{\mathbb{P}(X=x)}\\ & = \frac{p^2\binom{n-2}{x-2}p^{x-2}q^{n-x}}{\binom{n}{x}p^x q^{n-x}}\\ & = \frac{x(x-1)}{n(n-1)} \end{align}$$

Calculating its variance you can verify that it does not attain CR lower bound