# Posterior Distribution using Jeffreys prior

I'm trying to show that if $$X_1, \cdots, X_n \stackrel{iid}{\sim} N(\mu, \sigma^2)$$ with unknown $$\mu$$, $$\sigma$$ and the prior $$\pi(\mu, \sigma^2) \propto 1/\sigma^2$$ then the posterior distribution of $$\mu$$ is given by $$\frac{\sqrt{n} (\mu - \bar{x})}{s_{n-1}} \sim t_{n-1}$$ where $$\bar{x} = \frac{1}{n} \sum x_i$$ is the sample mean and $$s_{n-1} = \frac{1}{n-1} \sum (x_i - \bar{x})^2$$ is the sample variance. I know that $$\sum (x_i - \bar{x})^2 \sim \chi^2_{n-1}$$ by a standard result. So to show that $$\frac{\sqrt{n} (\mu - \bar{x})}{s_{n-1}} \sim t_{n-1} ,$$ I just need the numerator to be $$N(0, 1)$$ but I'm not too sure how. I note that $$\pi(\mu, \sigma^2|\mathbf{x}) \propto (\sigma^2)^{-(1 + n/2)} \text{exp} \left( \frac{-1}{2\sigma^2} \sum (x_i - \mu)^2 \right)$$ but I got stuck here. Anyone can give me suggestions from here?

• "Jeffreys", not "Jeffrey", after Sir Harold Jeffreys. Commented Mar 5 at 17:56
• A first step is rewriting the sum in the exponent as \begin{align} \sum(x_i-\mu)^2&=n\mu^2-2\mu\sum x_i+\sum x_i^2 \\&=n(\mu-\bar x)^2-n\bar x^2+\sum x_i^2 \\&=n(\mu-\bar x)^2+\sum(x_i-\bar x)^2 \end{align} by completing the square in $\mu$. It is then immediate that the posterior of $\mu$ conditional on $\sigma$ is normal and that marginal posterior of $\sigma^2$ is inverse gamma. Commented Mar 5 at 22:27
• Special case of stats.stackexchange.com/q/178199/7224 Commented Mar 7 at 10:45