Show that the estimator $ \delta_n(X)=\Phi(\sqrt{\frac{n}{n-1}}(a-\bar{X}_n)) $ is an unbiased estimator Suppose that $X_1,\dots,X_n$ are iid from the normal location model $N(\theta,1)$ and we wish to estimate the critical or cutoff value $g_{a}(\theta)=P(X_1\le a)$, where $a\in R$ is some fixed number.
(a) Let $\Phi$ denote the CDF of the standard normal distribution. Show that the estimator
$$
\delta_n(X)=\Phi(\sqrt{\frac{n}{n-1}}(a-\bar{X}_n))
$$
is an unbiased estimator of $g_a(\theta)$. Prove that $\sqrt{n}(\delta_n-g_a(\theta))\to^d N(0,\phi^2(a-\theta))$ where $\phi$ is the PDF of the standard normal.

For (a), we want to show that $E\delta_n(X)=g_a(\theta)$, then it is enough to show that
$$
E\delta_n(X)=E\left[P(Z\le \sqrt{\frac{n}{n-1}}(a-\bar{X}_n)|X_1,\dots, X_n)\right]
$$
where $Z$ is standard normal and independent of $X_i$. But how to go the next step?
For the second part,
$$\sqrt{n}(\delta_n-g_a(\theta))=\sqrt{n}(\Phi(\sqrt{\frac{n}{n-1}}(a-\bar{X}_n))-\Phi(a-\theta))$$
Note that
$$
\sqrt{n}(\sqrt{\frac{n}{n-1}}(a-\bar{X}_n)-(a-\theta))\to N(0,1)
$$
Then from the Delta method
$$
\sqrt{n}(\delta_n-g_a(\theta))\to^d \Phi'(a-\theta) N(0,1)
$$

*

*

Sometimes the iterated expectation is easier to solve, but in this case, it is not.
\begin{align*}
E\delta_n 
&= E_{\bar{X}_n}\left[P(Z\le \sqrt{\frac{n}{n-1}}(a-\bar{x}_n)|\bar{X}_n =\bar{x}_n)\right]\\
&= P(Z \le \sqrt{\frac{n}{n-1}}(a-\bar{X}_n)) \\
&= P(Z - \sqrt{\frac{n}{n-1}}(a-\bar{X}_n) \le 0)
\end{align*}
2.
Use the (first-order) Delta-Method.
