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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?

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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}

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