Show that the estimator $ \delta_n(X)=\Phi(\sqrt{\frac{n}{n-1}}(a-\bar{X}_n)) $ is an unbiased estimator [duplicate]

Question

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.

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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 $$ \sqrt{n}(\delta_n-g_a(\theta))\to^d \Phi'(a-\theta) N(0,1) $$

Answer 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.