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.
$\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}\newcommand{\e}{\operatorname{E}}$
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.)
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 $\mu$.
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
$$\Phi\left(\frac{u-\overline X}{\sqrt{1-\frac1n}}\right)$$
