I have encountered a problem in my textbook 'The Bayesian Choice' by Christian P. Robert. It goes something like this:

$"$For a particular case of AR(1) model, $(x_t)_{1\leq t\leq T}$. Where $x_t = \rho x_{t-1}+\epsilon_t$, $\epsilon_t$ i.i.d. $\mathcal{N}(0,\sigma^2)$. Given a sequence of observations till time $T-1$, $x_{1:(T-1)} = (x_1,\dots,x_{(T-1)})$.

The predictive distribution of $x_T$ is then $$x_T\mid x_{1:(T-1)}\sim \int\dfrac{1}{\sqrt{2\pi}}\sigma^{-1}\exp(-(x_T-\rho x_{(T-1)})^2/2\sigma^2)\pi(\rho,\sigma\mid x_{1:(T-1)})d\rho d\sigma.$$

Show that, in this scenario, the joint distribution $\pi(\rho,\sigma^2\mid x_{1:(T-1)})$ is explicit under the conjugate prior, $$\rho\sim\mathcal{N}(0,k\sigma^2),\quad \sigma^2\sim\mathcal{IG}(\alpha,\beta)."$$

Now my problem here is how do I show this? Can I show it without calculationg the joint posterior? I'm assuming this is not possible, so how do I calculate the joint posterior here? (Can I just take the product of the two posteriors? And on that note, how do I know when I can and cannot do this?)

From what I've understood, one can only calculate the joint posterior by the product of each posterior if we know that they are not dependent on each other. In this case it seems like this is what is meant to be done. However, I have not been successful in finding more information on wheter or not this is how you would go about this calculation. Nor have I been able to establish that my interpretation of the requirements for deriving the joint posterior is correct.

  • $\begingroup$ We treat the AR(p) model in more details in our book, Bayesian Core.. Please add a self-study tag and detail why you cannot derive the joint posterior. $\endgroup$
    – Xi'an
    Apr 7, 2019 at 15:27
  • $\begingroup$ @Xi'an Thank you for responding, I have not yet read 'Bayesian Core', will definately pick it up! I have added the tag you suggested and tried to further explain what I'm having trouble understanding. $\endgroup$
    – L200123
    Apr 7, 2019 at 16:26

1 Answer 1


The joint prior on $𝜌,𝜎^2$ is the product$$(\sigma^2)^{-1/2}\exp\{-\rho^2/2k\sigma^2\}\times(\sigma^2)^{-\alpha-1}\exp\{-\beta/\sigma^2\}$$ The likelihood is$$\prod_{t=2}^T (\sigma^2)^{-1/2}\exp\{-(x_t-\rho x_{t-1})^2/2k\sigma^2\}$$ Therefore the joint posterior on $𝜌,𝜎^2$ is proportional to $$(\sigma^2)^{-1/2}\exp\{-\rho^2/2k\sigma^2\}\times(\sigma^2)^{-\alpha-1}\exp\{-\beta/\sigma^2\} \prod_{t=2}^T (\sigma^2)^{-1/2}\exp\{-(x_t-\rho x_{t-1})^2/2\sigma^2\}$$ which factorises as $$(\sigma^2)^{\alpha-n/2-3/2}\exp\{-[s^0_{1:T}+\beta]/\sigma^2\} \exp\{-\rho^2[k^{-1}+s^1_{1:T}]/2\sigma^2+2\rho s^2_{1:T}]/2\sigma^2\}$$ where \begin{align} s^0_{1:T} &= \sum_{t=2}^T x_t^2\\ s^1_{1:T} &= \sum_{t=1}^{T-1} x_t^2\\ s^2_{1:T} &= \sum_{t=2}^T x_tx_{t-1} \end{align} From there I presume you can derive the joint posterior distribution of $(\rho,\sigma^2)$.

  • 1
    $\begingroup$ Thank you so much! I had completely misunderstood, this was really helpful! $\endgroup$
    – L200123
    Apr 8, 2019 at 13:48

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