Suppose $X_1, \dots, X_n$ are i.i.d samples from a normal distributoin with mean $\mu$ unknown and $\sigma^2$ known.
Then what's the value of $\mathbb{P}(X_1 < a | \bar{X}=b)$? where $a,b$ are known constants.
My thoughts on this. Apply Basu's theorem so that: $\mathbb{P}(X_1 < a | \bar{X}=b) = \mathbb{P}(X_1 / \bar{X} < a/\bar{X}\,|\,\bar{X}=b) = \mathbb{P} (X_1/\bar{X} < a/b)$ as we known $\bar{X}$ is complete and $X_1/\bar{X}$ is ancillary.
However, I don't think the distribution of $X_1/\bar{X}$ is easy to derive. Can someone help me on this? Thanks.
I think the easiest way is to find $\beta \in \mathbb{R}$ such that $X_1 - \beta \bar X$ is independent of $\bar X$. Then we can compute $$\mathbb{P}(X_1 < a | \bar X = b) = \mathbb{P}(X_1 - \beta \bar X < a - \beta b | \bar X = b) = \mathbb{P}(X_1 - \beta \bar X < a - \beta b)$$ using that $X_1 - \beta \bar X$ is normally distributed with a mean and variance we can compute.
To find such $\beta$, note that we can assume $\mathbb{E}[X_1] = \mathbb{E}[\bar X] = 0$ by replacing each $X_1$ with $X_1 - \mu$ and $\bar X$ with $\bar X - \mu$. Adding the same deterministic constant to both of them won't affect their independence, variance, or covariance. Then we must have $\mathbb{E}[\bar X (X_1 - \beta \bar X)] = 0$, and because uncorrelated jointly normal random variables are independent this is also enough to conclude $X_1 - \beta \bar X$ is independent of $\bar X$. We compute $\mathbb{E}[\bar X X_1] = \frac 1n \sigma^2$ and $\mathbb{E}[\bar X^2] = \frac 1n \sigma^2$, so we choose $\beta = 1$.
Plugging this into our first equation, we have $\mathbb{P}(X_1 < a | \bar X = b) = \mathbb{P}(X_1 - \bar X < a - b)$, so all we have left to do is find the mean and variance of $X_1 - \bar X$. Since $\mathbb{E}[\bar X] = \mathbb{E}[X_1] = \mu$, we have $\mathbb{E}[X_1 - \bar X] = 0$. Then we compute $$\operatorname{Var}(X_1 - \bar X) = \operatorname{Var}(X_1) + \operatorname{Var}(\bar X) - 2 \operatorname{Cov}(X_1,\bar X) = \sigma^2 + \frac 1n \sigma^2 - \frac 2n \sigma^2 = (1 - \frac 1n) \sigma^2,$$ so $X_1 - \bar X \sim N\left( 0, \left(1 - \frac 1n \right) \sigma^2 \right)$. Therefore we conclude $$\mathbb{P}(X_1 < a | \bar X = b) = \mathbb{P}(X_1 - \bar X < a - b) = \Phi_{0,\hat{\sigma}^2}(a-b)$$ where $\Phi_{0,\hat{\sigma}^2}$ is the CDF for a $N\left( 0, \left(1 - \frac 1n \right) \sigma^2 \right)$ random variable.