let$ ~X_1, ...X_n ~$be iid Bernoulli$(p)$ let $W_n$ is the MVUE of $p^2$
Find $W_n$ and determine it is asymptotically normally distributed in the sense that $ \sqrt{ n}(W_n − \mu)$
converges in distribution to a normal distribution, for some constant $\mu$.
I try:
This family of distribution belongs to the exponential family and the complete and sufficient statistics is $Y_n = ∑X_i $ binomial$(n,p)$
I have found the MVUE of $ p^2$ is $(Y_n ^2$-$Y_n$)/$n(n-1) $ by Lehmann and Scheffe Theorem
Now, I know the MLE of $p$ is $Y_n/n $
Then by CLT we have, $\sqrt n(Y_n/n - p) $ converges in distribution to $N(0, p(1-p))$
from here I tried to use the delta method but I cannot proceed because I got $n$ on the right-hand side. Is that enough to say it is not convergent?
\begin{align} \sqrt n \left(W_n - p^2\right) & = \sqrt n \left(\frac {Y_n^2 - Y_n}{n(n-1)} - p^2\right) \\ &= \sqrt n \left(\left(\frac{Y_n}n\right)^2 - p^2\right) + \sqrt n Y_n^2 \left(\frac 1{n(n-1)} - \frac1{n^2}\right) - \frac {Y_n}{\sqrt n (n-1)}\\ &= \sqrt n \left(\left(\frac{Y_n}n\right)^2 - p^2\right) + \frac{Y_n^2}{n\sqrt n (n-1) } - \frac {Y_n}{\sqrt n (n-1)} \end{align}
Since $\sqrt n \left(\frac {Y_n}n - p \right) \to_{d} \mathcal N\left(0, p(1-p)\right) $ then by using delta method with $g(p) = p^2$ $$\sqrt n \left(\left(\frac {Y_n}n\right)^2 - p^2 \right) \to_{d} \mathcal N\left(0, 2p\times p(1-p)\right)$$
On the other hand $\frac {Y_n}n \to_{a.s.} p$ $$\frac{Y_n^2}{n\sqrt n (n-1) } - \frac {Y_n}{\sqrt n (n-1)} \to_{a.s.} 0$$ this proves that $$\sqrt n \left(W_n - p^2 \right) \to_{d} \mathcal N\left(0, 2p\times p(1-p)\right)$$