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?
$\begingroup$
$\endgroup$
3
-
6$\begingroup$ "Jeffreys", not "Jeffrey", after Sir Harold Jeffreys. $\endgroup$– jbowmanCommented Mar 5 at 17:56
-
7$\begingroup$ 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. $\endgroup$– Jarle TuftoCommented Mar 5 at 22:27
-
2$\begingroup$ Special case of stats.stackexchange.com/q/178199/7224 $\endgroup$– Xi'anCommented Mar 7 at 10:45
Add a comment
|