Finding the conditional distribution of single sample point given sample mean for $N(\mu, 1)$ 
Suppose that $X_1, \ldots, X_n$ are iid from $N(\mu, 1)$. Find the conditional distribution of $X_1$ given $\bar{X}_n = \frac{1}{n}\sum^n_{i=1} X_i$.

So I know that $\bar{X}_n$ is a sufficient statistic for $\mu$ and $X_1$ is an unbiased estimator of $\mu$, but I don't know how to proceed from here?
 A: Firstly, we need to find the joint distribution of $(X_1, \bar{X})$ (for simplicity, write $\bar{X}$ for $\bar{X}_n$). It is easily seen that 
\begin{equation}
\begin{bmatrix}
X_1 \\
\bar{X}
\end{bmatrix}
= \begin{bmatrix}
1 & 0 & \cdots & 0 \\
\frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n}
\end{bmatrix}
\begin{bmatrix}
X_1 \\ X_2 \\ \vdots \\ X_n
\end{bmatrix}\tag{1}
\end{equation}
In view of $(1)$, $(X_1, \bar{X})$ has jointly normal distribution, with the mean vector 
\begin{align}
\mu_0 = \begin{bmatrix}
1 & 0 & \cdots & 0 \\
\frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n}
\end{bmatrix}
\begin{bmatrix}
\mu \\ \mu \\ \vdots \\ \mu 
\end{bmatrix}
= \begin{bmatrix}
\mu \\ \mu
\end{bmatrix},
\end{align}
and the covariance matrix 
\begin{align}
\Sigma = \begin{bmatrix}
1 & 0 & \cdots & 0 \\
\frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n}
\end{bmatrix}
I
\begin{bmatrix}
1 & \frac{1}{n} \\
0 & \frac{1}{n} \\
\vdots & \vdots \\
0 & \frac{1}{n}
\end{bmatrix}
= \begin{bmatrix}
1 & \frac{1}{n} \\
\frac{1}{n} & \frac{1}{n}
\end{bmatrix}.
\end{align}
Now according to conditional distributions of a MVN distribution:
$X_1 | \bar{X} \sim N(\mu_1, \sigma_1^2),$
where 
\begin{align}
& \mu_1 = \mu + \frac{1}{n}\times n \times (\bar{X} - \mu) = \bar{X}, \\
& \sigma_1^2 = 1 - \frac{1}{n}\times n \times\frac{1}{n} = 1 - \frac{1}{n}.
\end{align}
