Posterior derivation of normal model

Question

Working through the book Bayesian Essentials with R by Jean-Michel Marin & Christian Robert I am trying to work out the posterior for the model given on page 29 when the data is from a normal with unknown mean and variance.

The priors for $\mu$ and $\sigma^2$ are (p.29-30)

\begin{align} \pi(\mu | \sigma^2) &\sim N(0,\sigma^2)\\ \pi(\sigma^2) &\sim \exp(-\sigma^{-2})(\sigma^2)^{-2} \end{align}

and the likelihood (from top of page 29) as

$$l \propto \exp(-(n(\mu - \bar{x})^2 + s^2)/2\sigma^2)/\sigma^n$$

On the top of page 31 the posterior is given as

$$\pi((\mu, \sigma^2) | D) \propto \pi(\sigma^2) \times \pi(\mu | \sigma^2) \times l$$

Plugging the values in gives

\begin{align} \pi((\mu, \sigma^2) | D) \propto ~ & \exp(-\sigma^{-2})(\sigma^2)^{-2} \times \\ &\exp(-\mu^2 / 2\sigma^2) \times \exp(-(n(\mu - \bar{x})^2 + s^2)/2\sigma^2)/\sigma^n\tag{1} \end{align}

I cannot see how to move from this to what is given below.

I'd really appreciate any hints. Thank you.

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I can combined the exponent parts of the prior on $\sigma$ and $\mu$ to give $\exp(-\mu^2/2\sigma^2 - 1/\sigma^2)$ which is equal to $\exp(-(\mu^2 + 2)/2\sigma^2)$. I can see how the $1/\sigma^n$ from the likelihood is equal to $(\sigma^{-2})^{n/2}$ but I cannot see how $(\sigma^2)^{-2} = \sigma^{-4}$ is equal to $(\sigma^{-2})^{1/2 + 2}$ , which I think equals $\sigma^{-5}$.

Answer 1

There is a typo¹ in the exponential part if not in the power: Starting from \begin{align} \pi((\mu, \sigma^2) | D) \propto ~ & \overbrace{\exp(-\sigma^{-2})(\sigma^2)^{-2}}^{\pi(\sigma^2)} \times \overbrace{\sigma^{-1} \exp(-\mu^2 / 2\sigma^2)}^{\pi(\mu|\sigma^2)} \times \\ &\underbrace{\exp(-(n(\mu - \bar{x})^2 + s^2)/2\sigma^2)/\sigma^n}_{\ell(\mu,\sigma^2|D)} \end{align} one gets \begin{align} \pi((\mu, \sigma^2) | D) \propto ~ & (\sigma^{-2})^{\overbrace{\frac{1}{2}+2+\frac{n}{2}}^{(n+5)/2}}\times\\ &\exp\left\{-[(n+1)\mu^2-2n\mu\bar x+n\bar{x}^2+s^2+2]/2\sigma^2\right\}\\ \propto ~ & (\sigma^{-2})^{\frac{n+2}{2}+1+\frac{1}{2}}\times\\ &\exp\left\{-(n+1)(\mu-n\bar{x}/(n+1))^2\right\}\times\\ &\exp\left\{-[n\bar{x}^2/(n+1)+2+s^2]/2\sigma^2\right\}\\ = ~ & (\sigma^{-2})^{\frac{1}{2}} \exp\left\{-(n+1)(\mu-n\bar{x}/(n+1))^2\right\}\times\\ &(\sigma^{-2})^{\frac{n+2}{2}+1}\exp\left\{-[n\bar{x}^2/(n+1)+2+s^2]/2\sigma^2\right\} \end{align} Hence, compared with the derivation at the top

1. the powers of $\sigma^2$ are indeed $-1/2$ and $-(n+2)/2-1$: $$(\sigma^{-2})^{\frac{n+5}{2}}=(\sigma^2)^\frac{-1}{2}\times (\sigma^2)^{-\frac{n+2}{2}-1}$$
2. there is a missing $n\bar{x}^2/(n+1)$ term in the scale parameter of the posterior inverse Gamma distribution. (See Murphy's derivation for double-check.)

Apologies for the confusion!

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¹It was already pointed out in earlier X validated questions:

Computing posterior density of Normal with unknown $\mu$ and $\sigma^2$

Help with Bayesian derivation of normal model with conjugate prior