EM for MAP variance of AR(1) this question is about MAP inference in an AR(1) model (exercise 1.6 from West, M., Time Series: Modeling, Computation and Inference). It's not a homework assignment.
Assume $n$ observations were generated from the model
$y_t = \phi y_{t-1} + \epsilon_t,\ \epsilon_t\sim\mathcal{N}(0, v)$
where $y_1$ (the first observation) is known, i.e. we can use the conditional likelihood $p(y_{2:n} | y_1, \phi, v)$ in the following.
The goal is to compute the mode of $p(v | y_{1:n})$
Priors and posteriors
We put a Gaussian and Inverse Gamma prior on $\phi$ and v, respectively and so
$\phi | v \sim \mathcal{N}(0,v)$
$v\sim IG(n_0/2, d_0/2)$
The posterior distributions are given by
$\phi | y_{1:n}, v \sim \mathcal{N}(m, vC)$ $\qquad(*)$
$v| y_{1:n} \sim IG(n^*/2, d^*/2)$
where $m=\frac{\sum_{t=2}^n y_{t}y_{t-1}}{\sum_{t=2}^n y_{t-1}+1},\ C=\frac{1}{\sum_{t=2}^n y_{t-1}+1},\ n^*=n+n_0-1,\ d^*=\sum_{t=2}^n y_{t}^2-m+d_0 $
Furthermore, the joint posterior $(\phi, v | y_{1:n})$ under the full likelihood $p(y_{1:n} | \phi, v)$ is proportional to
$v^{-n/2+1}(1-\phi^2)^{1/2}\exp\left(-\frac{\sum(y_t-\phi y_{t-1})}{2v}\right)$ $\qquad(**)$
EM
To find the MAP for $v$ we use the EM algorithm. The $m$-th E-step comprises of computing $E^{(m-1)}[\log(\phi,v| y_{1:n})]$ and so we need to compute the expression
$\int_{\mathbb{R}}\log p(\phi, v| y_{1:n})p(\phi| v^{m-1}, y_{1:n})d\phi$
Plugging in the expressions from $(*), (**)$ I find that this integral is very hard to compute. Is this really the way to go? How can I make progress?
Thanks for the help!
 A: Hm, maybe it's just algebra (correct errors if you see one). The expression from above becomes
$\int_{\mathbb{R}}\log p(\phi,v|y_{1:n})p(\phi| v^{m-1}, y_{1:n})d\phi=
\mathbb{H}^{m-1}+\underbrace{\int_{\mathbb{R}}\log p(v| y_{1:n})p(\phi| v^{m-1}, y_{1:n})d\phi}_{\log p(v| y_{1:n})\ \cdot\ 1}$
where $\mathbb{H}^{m-1}=\mathbb{E}^{m-1}[\log p(\phi| v, y_{1:n})]=\int_{\mathbb{R}}\log p(\phi| v, y_{1:n})p(\phi| v^{m-1}, y_{1:n})$ 
is an entropy term. The second term which simplifies to $\log p(v| y_{1:n})$ is the log-density of the Inverse Gamma and so
$\log p(v| y_{1:n})=\frac{n^*}{2}\log\frac{d^*}{2}-\Gamma(\frac{n^*}{2})-(\frac{n^*}{2}+1)\log v - \frac{d^*}{2v}$
Let's attack the entropy term
$\mathbb{H}^{m-1}=\int_{\mathbb{R}}\log p(\phi| v, y_{1:n})p(\phi| v^{m-1}, y_{1:n})= $
$= \underbrace{-\int_{\mathbb{R}}\frac{\log(2\pi vC)}{\sqrt{2\pi vC}}\exp(-\frac{(\phi - m)^2}{2vC})d\phi}_{-1/2 \log(2\pi vC)} - \underbrace{\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi vC}}\frac{(\phi - m)^2}{2vC}\exp(-\frac{(\phi - m)^2}{2vC})d\phi}_{=:\ A}$
With $f(\phi):=\exp(-\frac{(\phi - m)^2}{2vC}) and \eta:=\frac{1}{\sqrt{2\pi vC}}$, the last remaining term requires integration by parts
$A=\int_{\mathbb{R}}\eta\frac{(\phi-m)^2}{2vC}f(\phi)d\phi=-\int_{\mathbb{R}}\eta \frac{\phi-m}{2}\frac{df}{d\phi}d\phi=$
$=-(\int_{\mathbb{R}}\eta f(\phi)\frac{\phi-m}{2}-\underbrace{\int_{\mathbb{R}}\eta\frac{m}{2}f(\phi)d\phi)}_{m/2}$
$=-\frac{vC}{2}\int_{\mathbb{R}}\eta \frac{\phi-m}{vC}f(\phi)d\phi+\frac{m}{2} = \frac{m-vC}{2}$
