# 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?

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 $$$$\begin{bmatrix} X_1 \\ \bar{X} \end{bmatrix} = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n} \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n \end{bmatrix}\tag{1}$$$$ In view of $$(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}