Use Rao-Blackwell Theorem to find the UMVUE Suppose that $X_1,X_2,...,X_n$ is a random sample from a normal distribution, $X_i\sim N(\mu,9)$.
Find the UMVUE (uniformly minimum variance unbiased estimator) of $P(X\le c)$ where $c$ is a known constant.  Do this by finding the conditional distribution of $X_1$ given $\bar{X}={\bar{x}}$ and apply the Rao-Blackwell theorem with $T=u(X_1)$, where $u(x_1)=1$ if $x\le c$ and zero otherwise.
So clearly the parameter that we're trying to estimate given the problem description can be written as $\Phi\left( \frac{c-\mu}{3} \right)$ where $\Phi$ represents the CDF of the standard normal distribution.  However, I get stuck when I need to try and find the conditional distribution.  Without knowing the joint distribution or the other conditional distribution, how can I possibly find the conditional distribution of $X_1$ given $\bar{X}={\bar{x}}$?
 A: A complete sufficient statistic for $\mu$ is $\overline X$, or equivalently $S=\sum\limits_{i=1}^n X_i$. 
Now $(X_1,S)$ has a bivariate normal distribution (can be verified quickly using moment generating functions), more specifically $(X_1,S)\sim N_2\left(\mu\begin{pmatrix}1\\n\end{pmatrix},9\begin{pmatrix}1&1\\1& n\end{pmatrix}\right)$.
This gives the conditional distribution of $X_1$ given $S=s$ as a univariate normal $N\left(\frac sn,9\left(1-\frac1n\right)\right)$.
Hence the UMVUE of $P(X_1\le c)$ is 
\begin{align}
E_{\mu}\left[I_{X_1\le c}\mid S\right]&=P_{\mu}(X_1\le c\mid S)
\\&=\Phi\left(\frac{c-S/n}{3\sqrt{1-1/n}}\right)
\\&=\Phi\left(\frac{c-\overline X}{3\sqrt{1-1/n}}\right)
\end{align}
A: This is the same as the conditional distribution given the sum $S_n=\sum_{i=1}^n X_i.$ As with most conditional distributions where the conditioning seems 'backwards,' we can use Bayes: $$ f_{X_1\mid S_n}(x_1\mid S_n=s) = \frac{f_{S_n\mid X_1}(s\mid X_1=x_1)f_{X_1}(x_1)}{f_{S_n}(s)}.$$
The three things on the right are much clearer than the left. We have: $$ S_n\mid X_1=x_1 \sim N\left(x_1+\mu(n-1), 9(n-1)\right)\\X_1\sim N(\mu,9)\\S_n\sim N\left(\mu n,9n\right)$$
(Edit: I realize I may have been unclear when I said this was 'the same as the conditional distribution given then sum.' By this I meant they are related by a simple transformation: $$f_{X_1\mid \bar X }(x_1\mid \bar X=\bar x) = f_{X_1\mid S_n}(x_1\mid S_n=n\bar x)$$
