UMVUE for probability of cutoff Let $X_i \sim N(\mu,1)$, i.i.d. We aim to find UMVUE for $p(\mu) = P_{\mu}(X_1 \leq u)$ for some fixed $u$.
I have shown that $\bar{X}$ and $X_1 - \bar{X}$ are independent. ($\bar{X}$: sample mean).
Now I want to use this to compute $E_\mu[1\{X_1 < u\}\mid\bar{X}]$ and if I am not mistaken, by Lehmann-Scheffé, it is UMVUE.
I am quite new in statistics, maybe my arguments are wrong, or I do not know how to compute.
Any hint appreciated.
 A: $\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}\newcommand{\e}{\operatorname{E}}$

Lemma: Suppose $U,V$ are normally distributed, but also jointly normally distributed, i.e. so distributed that no matter which (non-random) scalars $a$ and $b$ are, the random variable $aU+bV$ is normally distributed. Suppose $\e(U) = \alpha,$ $\e(V) = \beta,$ $\v(U) = \sigma^2,$ $\v(V) = \tau^2,$ and $\c(U,V) = \rho\sigma\tau$ (so $\rho$ is the correlation). Then the conditional distribution of $V$ given $U$ is
  $$
V \mid U \sim N\left( \beta + \rho\tau\cdot\frac{U-\alpha} \sigma, \quad \tau^2(1-\rho^2) \right).
$$

Now suppose $V=X_1$ and $U= \overline X.$ Observe that these satisfy the hypotheses of the Lemma. And
\begin{align}
\alpha & = \mu, \\
\beta & = \mu, \\
\sigma^2 & = 1/n, \\
\tau^2 & = 1, \\
\rho & = 1/\sqrt n.
\end{align}
Therefore by the Lemma,
$$
X_1\mid \overline X \sim N\left( \overline X ,\quad \frac {n-1} n \right). \tag 1
$$
Therefore
$$
\Pr\left( X_1<u \mid \overline X \right) = \Phi\left( \frac{u - \overline X}{\sqrt{\frac{n-1} n}} \right).
$$
(The lack of dependence of line $(1)$ above upon $\mu$ is what it means to say that $\overline X$ is sufficient for this family of distributions.)
A: The parameter of interest
$$
p = P(X < u) = P(X-\mu \le u-\mu)=\Phi(u-\mu)
$$
so a possible estimator of $p$ is
$$
\hat p = \Phi(c(u-\bar X)).
$$
where $c$ is a suitably chosen constant.
This has expectation
\begin{align}
E\hat p 
  &=E \Phi(c(u-\bar X))
\\&=P(Z \le c(u- \bar X)) & \text{where $Z\sim N(0,1)$}
\\&=P(Z+c\bar X \le cu)
\\&=P(\frac{Z+c(\bar X -\mu)}{\sqrt{1+c^2/n}} \le \frac{c(u - \mu)}{\sqrt{1+c^2/n}})
\\&=\Phi(\frac{c(u - \mu)}{\sqrt{1+c^2/n}})
\\&=p
\end{align}
for
\begin{align}
\frac{c}{\sqrt{1+c^2/n}}&=1
\\ c^2&=1+c^2/n
\\ c&=1/\sqrt{1-1/n}.
\end{align}
So
$$
\hat p = \Phi(\frac{u-\bar X}{\sqrt{1-1/n}})
$$
is unbiased for $p$ and, by the Lehmann-Scheffé theorem, also UMVUE since $\bar X$ is sufficient for $\mu$.
A: I'll add another answer as this approach is another simple way to find the UMVUE.
Because $\overline X$ is a complete sufficient statistic for $\mu$ whenever $\mu\in \mathbb R$, the ancillary statistic $X_1-\overline X$ is independent of $\overline X$ by Basu's theorem. This independence can also be seen from the fact that $(X_1-\overline X,\overline X)$ is jointly normal and $\operatorname{Cov}(X_1-\overline X,\overline X)=0$.
This means
\begin{align}
P(X_1\le u\mid \overline X)&= P(X_1-\overline X\le u -\overline X\mid \overline X)
\\&=P(X_1-\overline X \le u-\overline X)
\end{align}
As $X_1-\overline X \sim N \left(0,1-\frac1n \right)$, the last probability equals
$$\Phi\left(\frac{u-\overline X}{\sqrt{1-\frac1n}}\right)$$
