Log-concave posterior density function adaptive rejection sampling

[numbersense]

Let $x_1,x_2,\ldots,x_n$ be binary observations which have independent and identical Bernoulli distributions with parameter $\theta$, so that $f(x_i | \theta) = \theta^{x_i} (1 - \theta)^{1-x_i}$. Suppose that the prior density for $\theta$ is
\begin{equation*}
f(\theta) \propto \theta^{-1} (1 - \theta)^{-1} \exp\left(- \frac{1}{2\nu} \log \left(\frac{\theta}{1-\theta}\right)^2\right)
\end{equation*}
Show that the posterior density for $\theta$ is not generally log-concave, but that the posterior density for $\displaystyle \phi = \frac{\theta}{1-\theta}$ is and hence can be sampled by adaptive rejection sampling. Write down the acceptance probability for a random walk Metropolis-Hastings algorithm where the proposal $\phi^*$ is generated from a normal distribution with mean $\phi^i$ and variance $\sigma^2$.
Likelihood function is
\begin{equation*}
f(\boldsymbol{x} | \theta) = \theta^{\sum x_i} (1- \theta)^{n - \sum x_i}
\end{equation*}
Posterior density function is
\begin{equation*}
g(\theta) = f(\theta | \boldsymbol{x}) \propto f(\boldsymbol{x} | \theta) f(\theta) = \theta^{(\sum x_i) - 1} (1-\theta)^{n-1-\sum x_i} \exp\left(-\frac{1}{2\nu} \log\left(\frac{\theta}{1-\theta}\right)^2\right)
\end{equation*}
\begin{equation*}
\ln g(\theta) = \left(\left(\sum x_i\right) - 1\right) \ln \theta + \left(n-1-\sum x_i\right) \ln (1-\theta) - \frac{1}{2\nu} \log\left(\frac{\theta}{1-\theta}\right)^2
\end{equation*}
\begin{align*}
\frac{d (\ln g)}{d\theta}(\theta) & = \left(\left(\sum x_i\right) - 1\right) \frac{1}{\theta} - \left(n-1-\sum x_i\right) \frac{1}{1-\theta} - \frac{1}{\nu} \log\left(\frac{\theta}{1-\theta}\right) \frac{1-\theta}{\theta} \frac{(1-\theta) + \theta}{(1-\theta)^2} \\
& = \left(\left(\sum x_i\right) - 1\right) \frac{1}{\theta} - \left(n-1-\sum x_i\right) \frac{1}{1-\theta} - \frac{1}{\nu} \log\left(\frac{\theta}{1-\theta}\right) \frac{1}{\theta (1- \theta)}
\end{align*}
For $g(\theta)$ to be log-concave, the derivative of the Log of the posterior, i.e. $\displaystyle \frac{d(\ln g)}{d\theta}$, has to exist, and be non-increasing in $\theta$. This derivative seems to exist, how to show that it is not non-increasing in $\theta$ and hence that $g(\theta)$ is not log-concave?

 

[Valkarie]

We can use the fact that $\theta$ and $\phi = \frac{\theta}{1-\theta}$ have an inverse relationship with each other, i.e. $\theta = \frac{\phi}{1+\phi}$. Thus, the posterior density for $\phi$ is \begin{align*} g(\phi) & = f(\phi | \boldsymbol{x}) \propto f(\boldsymbol{x} | \phi) f(\phi) \\ & = \left(\frac{\phi}{1+\phi}\right)^{(\sum x_i) - 1} \left(1-\frac{\phi}{1+\phi}\right)^{n-1-\sum x_i} \exp\left(-\frac{1}{2\nu} \log\left(\phi\right)^2\right) \end{align*} Thus, we can write the Log of the posterior as \begin{align*} \ln g(\phi) & = \left((\sum x_i) - 1\right) \ln \left(\frac{\phi}{1+\phi}\right) + \left(n-1-\sum x_i\right) \ln \left(1-\frac{\phi}{1+\phi}\right) - \frac{1}{2\nu} \log\left(\phi\right)^2 \end{align*} Taking the derivative of the Log of the posterior with respect to $\phi$, we get \begin{align*} \frac{d (\ln g)}{d\phi}(\phi) & = \left(\left(\sum x_i\right) - 1\right) \frac{1}{\phi} - \left(n-1-\sum x_i\right) \frac{1}{1+\phi} - \frac{1}{\nu} \log\left(\phi\right) \frac{1+\phi}{\phi} \frac{(1+\phi) + \phi}{(1+\phi)^2} \\ & = \left(\left(\sum x_i