# Determining the posterior distribution for an Autoregressive or order 1 model

Question: For this question, note that the notation $y_{1:T} = (y_1, y_2, \cdots, y_T)$, ie, a vector of random variables.

Consider the following AR(1) model: \begin{align*} y_{t+1} = \phi y_t + \sigma \eta_t \ \ \cdots (a) \end{align*} for $t = 1, 2, \cdots, T$ where \begin{align*} \eta_t \stackrel{iid}{\sim} N(0,1) \ \ \cdots (b) \end{align*} with $\eta_1$ independent of $y_k$ for $k \le t$, and where \begin{align*} y_1 \sim N\left(0, \frac{\sigma^2}{1-\phi^2}\right) \ \ \cdots (c) \end{align*} Define $\theta = (\phi, \sigma)$ and consider a prior distribution given by: \begin{align*} p(\theta) \propto \frac{1}{\sigma} \ \text{for} \ \infty < \phi < \infty \ \text{and} \ 0<\sigma<\infty \ \ \cdots (1) \end{align*} Also define the conditional likelihood function to be: \begin{align*} p(y_{2:T} \mid y_1, \theta) = \left(2\pi \sigma^2 \right)^{-\frac{T-1}{2}}\exp\left[- \frac{1}{2\sigma^2}\sum_{t=2}^T \left(y_t - \phi y_{t-1} \right)^2\right] \ \ \cdots (2) \end{align*}

Show that the conditional posterior distribution, corresponding to the conditional likelihood function in $(2)$ and the prior density in $(1)$, may be obtained analytically, whereas the full posterior distribution, corresponding to the model in $(a)-(c)$ and the prior density in $(1)$, requires an alternative computational approach.

My Working: So firstly, I worked out the likelihood function corresponding to the AR(1) model described by $(a)-(c)$. My working is as follows:

We are given $y_1 \sim N\left(0, \frac{\sigma^2}{1-\phi^2}\right)$, so the pdf is given by: \begin{align*} p(y_1 \mid \theta) = \left(2\pi\left( \frac{\sigma^2}{1-\phi^2}\right) \right)^{-\frac{1}{2}}\exp\left[ - \frac{y_1^2}{2\left(\frac{\sigma^2}{1-\phi^2}\right)} \right] \end{align*} Conditionally, we have $y_2 \mid y_1 \sim N\left( \phi y_1, \sigma^2\right)$, so the pdf is given by: \begin{align*} p(y_2 \mid y_1, \theta) = \left(2\pi \sigma^2\right)^{-\frac{1}{2}} \exp\left[-\frac{1}{2\sigma^2} \left(y_2 - \phi y_1\right)^2 \right] \end{align*} Similarly, $y_3 \mid y_2, y_1 \equiv y_3 \mid y_2 \sim N\left(\phi y_2, \sigma^2\right)$, so the pdf is given by: \begin{align*} p(y_3 \mid y_2, y_1, \theta) = \left(2\pi \sigma^2\right)^{-\frac{1}{2}} \exp\left[-\frac{1}{2\sigma^2} \left(y_3 - \phi y_2\right)^2 \right] \end{align*} So in general, $y_t \mid y_{t-1}, y_{t-2}, \cdots, y_1 \equiv y_t \mid y_{t-1} \sim N\left(\phi y_{t-1}, \sigma^2\right)$, with pdf: \begin{align*} p(y_t \mid y_{t-1}, y_{t-2}, \cdots, y_1, \theta) = \left(2\pi \sigma^2\right)^{-\frac{1}{2}} \exp\left[-\frac{1}{2\sigma^2} \left(y_t - \phi y_{t-1}\right)^2 \right] \end{align*} Using the method of composition, we have: \begin{align*} p(y_{1:T} \mid \theta) & = p(y_1 \mid \theta)p(y_2 \mid y_1, \theta)p(y_3 \mid y_1, y_2, \theta) \cdots p(y_T \mid y_1, y_2, \cdots, y_{T-1}, \theta) \\ & = p(y_1 \mid \theta) \prod_{t=2}^T p(y_t \mid y_{t-1}, \theta) \end{align*} Thus the likelihood function computed for a given value of $\theta = \left(\phi, \sigma^2\right)$, is given by: \begin{align} L(\theta) & = \left\{ \left(2\pi\left( \frac{\sigma^2}{1-\phi^2}\right) \right)^{-\frac{1}{2}}\exp\left[ - \frac{y_1^2}{2\left(\frac{\sigma^2}{1-\phi^2}\right)} \right]\right\} \prod_{t=2}^T \left(2\pi \sigma^2\right)^{-\frac{1}{2}} \exp\left[-\frac{1}{2\sigma^2} \left(y_t - \phi y_{t-1}\right)^2 \right] \\ & \propto (1-\phi^2)^{\frac{1}{2}}\sigma^{-T} \exp\left[-\frac{\sum_{t=2}^T\left(y_t - \phi y_{t-1}\right)^2+y_1^2\left(1-\phi^2\right)}{2\sigma^2} \right] \ \ \cdots (W1) \end{align}

Deriving the conditional likelihood given in $(2)$ can be done as follows: \begin{align*} p(y_{2:T} \mid y_1, \theta) & = \frac{p(y_{1:T} \mid \theta)}{p(y_1 \mid \theta)} \\ & = \prod_{t=2}^T p(y_t \mid y_{t-1}, \theta) \\ & = \left(2\pi \sigma^2 \right)^{-\frac{T-1}{2}}\exp\left[- \frac{1}{2\sigma^2}\sum_{t=2}^T \left(y_t - \phi y_{t-1} \right)^2 \right] \end{align*}

My Query: I don't really get what the question is trying to ask me to do? What does it mean by conditional posterior? Full posterior? Any assistance will be appreciated!

EDIT 1 PROGRESS: Okay, so I've played around a bit more and made a bit of progress. I interpret 'conditional posterior' as follows:

Notice that when $T$ is large, then the factor $(1-\phi^2)$ in Eqn. $(W1)$ is small, so we can approximate the full likelihood with the conditional likelihood: \begin{align*} p(y_{2:T} \mid y_1, \theta) \propto \sigma^{-(T-1)}\exp\left[- \frac{1}{2\sigma^2}\sum_{t=2}^T \left(y_t - \phi y_{t-1} \right)^2 \right] \end{align*} So under the prior $p(\theta) \propto \frac{1}{\sigma}$, we have: \begin{align*} p(\theta \mid y_{1:T}) & \propto \sigma^{-T}\exp\left[- \frac{1}{2\sigma^2}\sum_{t=2}^T \left(y_t - \phi y_{t-1} \right)^2 \right] \\ & = \sigma^{-T}\exp\left[-\frac{1}{2\sigma^2}\sum_{t=2}^T \left(y_t^2 -2y_t \phi y_{t-1} + \phi^2 y_{t-1}^2 \right) \right] \\ & = \sigma^{-T}\exp\left[-\frac{1}{2\sigma^2}\left(\underbrace{\sum_{t=2}^Ty_t^2}_{C} - 2\phi\underbrace{\sum_{t=2}^Ty_t y_{t-1}}_{B} + \phi^2 \underbrace{\sum_{t=2}^T y_{t-1}^2}_{A} \right) \right] \end{align*} First, note that: \begin{align*} p(\phi \mid \sigma, y_{1:T}) & \propto \exp\left[-\frac{1}{2\sigma^2}\left(C-2\phi B + \phi^2A \right) \right] \\ & = \exp\left[-\frac{A}{2\sigma^2} \left(\phi^2 - 2\phi \frac{B}{A} + \frac{C}{A} \right) \right] \\ & = \exp\left[-\frac{A}{2\sigma^2}\left(\left(\phi - \frac{B}{A} \right)^2-\left(\frac{B}{A}\right)^2 + \frac{C}{A} \right) \right] \\ & \propto \exp\left[-\frac{A}{2\sigma^2}\left(\phi-\frac{B}{A}\right)^2 \right] \\ & = \exp\left[-\frac{1}{2\left(\frac{\sigma^2}{A} \right)}\left(\phi-\frac{B}{A}\right)^2 \right] \end{align*} So the distribution of $\phi \mid \sigma, y_{1:T}$ is given by: \begin{gather} \phi \mid \sigma, y_{1:T} \sim N\left(\frac{B}{A}, \frac{\sigma^2}{A} \right) \\ \implies \phi \mid \sigma, y_{1:T} \sim N\left(\frac{\sum_{t=2}^T y_t y_{t-1}}{\sum_{t=2}^T y_{t-1}^2}, \frac{\sigma^2}{\sum_{t=2}^T y_{t-1}^2} \right) \end{gather} Next, note that: \begin{align*} p(\sigma \mid y_{1:T}) & \propto \int_{\phi} \sigma^{-T} \exp\left[-\frac{1}{2\sigma^2} \sum_{t=2}^T \left(y_t - \phi y_{t-1}\right)^2 \right]d\phi \\ & = \sigma^{-T} \int_{\phi} \exp\left[-\frac{A}{2\sigma^2} \left(\left(\phi - \frac{B}{A} \right)^2 - \left(\frac{B}{A}\right)^2 + \frac{C}{A} \right) \right]d\phi \\ & = \sigma^{-T} \int_{\phi} \exp\left[-\frac{A}{2\sigma^2} \left(\phi - \frac{B}{A} \right)^2 + \left(\frac{A}{2\sigma^2}\right) \left(\frac{B}{A}\right)^2 - \left(\frac{A}{2\sigma^2}\right)\left(\frac{C}{A}\right) \right]d\phi \\ & = \sigma^{-T} \exp\left[\frac{B^2/A - C}{2\sigma^2} \right] \int_{\phi} \exp\left[-\frac{A}{2\sigma^2} \left(\phi - \frac{B}{A} \right)^2\right] d\phi \\ & = \sigma^{-T} \exp\left[\frac{B^2/A - C}{2\sigma^2} \right] \left(2\pi\left(\frac{\sigma^2}{A} \right) \right)^{\frac{1}{2}} \\ & \propto \frac{1}{\sigma^{(T-2)+1}}\exp\left[\frac{B^2/A - C}{2\sigma^2} \right] \end{align*} Now define $b = \frac{B}{A}$, notice: \begin{align*} Q(y_{2:T}, b) & = \sum_{t=2}^T \left(y_t - b y_{t-1}\right)^2 \\ & = \sum_{t=2}^T \left(y_t^2 - 2y_t b y_{t-1} + b^2 y_{t-1}^2 \right) \\ & = \sum_{t=2}^T y_t^2 - 2b\sum_{t=2}^T y_t y_{t-1} + b^2 \sum_{t=2}^T y_{t-1}^2 \\ & = C - 2bB+b^2A \end{align*} Then clearly, \begin{align*} -Q(y_{2:T}, b) = B^2/A-C \end{align*} So the distribution of $\sigma \mid y_{1:T}$ is given by: \begin{gather} \sigma \mid y_{1:T} \sim IG\left(v = T-2, \widehat{\sigma^2} = \frac{1}{T-2} \left(B^2/A - C \right) \right) \\ \implies \sigma \mid y_{1:T} \sim IG\left(v = T-2, \widehat{\sigma}^2 = \frac{1}{T-2} \left(\frac{\left(\sum_{t=2}^T y_t y_{t-1} \right)^2}{\sum_{t=2}^T y_{t-1}^2} - \sum_{t=2}^T y_t^2 \right) \right) \end{gather} Or equivalently: \begin{gather*} \sigma \mid y_{1:T} \sim IG\left(v = T-2, \widehat{\sigma^2} = -\frac{1}{T-2}Q(y_{2:T}, b) \right) \end{gather*} Thus, we can derive the conditional posterior distribution analytically as: \begin{align*} p(\phi, \sigma \mid y_{1:T}) = p(\phi \mid \sigma, y_{1:T})p(\sigma \mid y_{1:T}) \end{align*} where $p(\phi \mid \sigma, y_{1:T})$ and $p(\sigma \mid y_{1:T})$ are derived above.

However, if we use the full likelihood (which is what the second part of the question is asking), how can we derive the joint posterior? I do not see any obvious ways to find the appropriate integrating constants. I'm assuming I need to use Gibbs/M-H/or some other kind of MCMC sampling scheme?