(From Bayesian Essentials with R by Marin & Robert page 31)

We are given an iid sample $\mathfrak{D}_n = (x_1, \dots, x_n)$ from the normal distribution $\mathcal{N}(0, \sigma^2)$ and $\theta=(\mu, \sigma^2)$, prior distribution $\mathcal{N}(0, \sigma^2)$ on $\mu$, and exponential distribution $\mathcal{E}(1)$ for $\sigma^{-2}$.

Corresponding posterior density on $\theta$ is then given by \begin{align*} \pi((\mu, \sigma^2) | \mathfrak{D}_n) &\propto \pi(\sigma^2) \times \pi(\mu | \sigma^2) \times l((\mu, \sigma^2) | \mathfrak{D}_n)\\ &\propto (\sigma^{-2})^{(1/2 + 2)} \exp \big\{ -(\mu^2 + 2) / 2\sigma^2 \big\} \\ &\times (\sigma^{-2})^{n/2} \exp \big\{ -(n(\mu - \bar{x})^2 + s^2) / 2\sigma^2 \big\}\cdots (\textbf{eq1})\\ &\propto (\sigma^2)^{-(n+5)/2} \exp \big\{ -\big[ (n+1)(\mu-n\bar{x}/(n+1))^2 + (2+s^2) \big] /2\sigma^2\big\} \cdots (\textbf{eq2})\\ &\propto (\sigma^2)^{-1/2} \exp \big\{ -(n+1)\big[ \mu - n\bar{x}/(n+1)\big]^2 / 2\sigma^2 \big\} \times (\sigma^2)^{-(n+2)/2-1} \exp \big\{ -(2+s^2) / 2\sigma^2 \big\} \end{align*}

I am having trouble getting to eq2 from eq1. \begin{align} &(\sigma^{-2})^{(1/2 + 2)} \exp \big\{ -(\mu^2 + 2) / 2\sigma^2 \big\} \times (\sigma^{-2})^{n/2} \exp \big\{ -(n(\mu - \bar{x})^2 + s^2) / 2\sigma^2 \big\}\cdots (\textbf{eq1})\\ &=(\sigma^2)^{-(n+5)/2} \exp \big\{ (-\mu^2 -2 -n(\mu-\bar{x})^2 - s^2) / 2\sigma^2 \big\}\\ &=(\sigma^2)^{-(n+5)/2} \exp \big\{ (-\mu^2 -n(\mu-\bar{x})^2 -(2+ s^2) / 2\sigma^2 \big\} \end{align} We then try to complete the square in $-\mu^2 -n(\mu-\bar{x})^2$ \begin{align} (-\mu^2 -n(\mu-\bar{x})^2) &= -\mu^2 -n\mu^2 + 2n\mu\bar{x} - \bar{x}^2n \\ &= -(n+1) \mu^2 + 2n\mu\bar{x} - \bar{x}^2n\\ &= -(n+1)(\mu^2 - {2n\mu\bar{x} \over n+1} + {n^2\bar{x}^2 \over (n+1)^2}) + {n^2\bar{x}^2 \over n+1} - \bar{x}^2n \\ &= - (n+1)(\mu-n\bar{x}/(n+1))^2 + {n^2\bar{x}^2 \over n+1} - \bar{x}^2n \end{align}

At this point I get stuck since ${n^2\bar{x}^2 \over n+1} - \bar{x}^2n$ don't cancel out. Plus, they cannot be ignored since we divide that by $2\sigma^2$.


1 Answer 1


Apologies, you are correct, this is indeed a typo from hasty cut & paste.... It was pointed out on X validated a few years ago and I presented a corrected version there.

  • 1
    $\begingroup$ Hi Prof. Robert. I was not aware of the older post that addressed this problem. Thank you. $\endgroup$
    – MoneyBall
    Mar 2, 2019 at 8:29

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