Random variables $X_1, \ldots, X_n$ are i.i.d given $\vartheta = \theta$ and have the following pdf:
\begin{equation}
p(x|\theta)=\begin{cases}
\theta x^{\theta – 1}, & \text{if $0<x<1$}.\\
0, & \text{otherwise}.
\end{cases}
\end{equation}
$\vartheta \sim Gamma(\alpha, \lambda)$(a) Find posterior $\pi(\theta|x_1, \ldots, x_n)$
(b) Find one-dimensional sufficient statistic
(c) Find $\mathbb{E(\vartheta|x_1, \ldots, x_n)}$
Background: I'm self-learning Bayesian statistics from scratch.
For (a) I have
\begin{equation}
\pi(\theta|x_1, \ldots, x_n) \propto \pi(\theta) \Pi_{i=1}^n p(x_i|\theta) \propto \theta^{n + \alpha – 1} e^{-\lambda \theta}\Pi_{i=1}^n x^{(\theta – 1)}
\end{equation}
However, this doesn't look like the pdf of any distribution I know and I would expect (perhaps mistakenly) $X_i$s to be conjugate with the Gamma distribution. I also can't find information about any distribution with $\theta x^{\theta – 1}$ probability density function in the distribution tables of any textbook I own so I don't know what information to search for online.
Perhaps I've made some basic mistake in calculating the posterior or I'm missing some other fundamental concept?
Best Answer
Density of $\boldsymbol X=(X_1,\ldots,X_n)$ given $\theta$ is
$$p(\boldsymbol x \mid \theta)=\theta^n \left(\prod_{i=1}^n x_i\right)^{\theta-1} \mathbf1_{0<x_1,\ldots,x_n<1}$$
And I assume the following kernel for the Gamma prior:
$$\pi(\theta)\propto \theta^{\alpha-1}e^{-\lambda \theta}\mathbf1_{\theta>0}$$
For $0<x_1,\ldots,x_n<1$, the posterior density of $\theta$ given $\boldsymbol X=\boldsymbol x$ is therefore
\begin{align} \pi(\theta\mid \boldsymbol x) &\propto p(\boldsymbol x \mid \theta)\,\pi(\theta) \\&\propto \theta^n \color{darkblue}{\left(\prod_{i=1}^n x_i\right)^{\theta} }\theta^{\alpha-1} e^{-\lambda \theta}\mathbf1_{\theta>0} \\&=\theta^{\alpha+n-1}\color{darkblue}{\exp\left\{\ln\left(\prod_{i=1}^n x_i\right)^{\theta}\right\}} e^{-\lambda \theta}\mathbf1_{\theta>0} \\&= \theta^{\alpha+n-1}\color{darkblue}{\exp\left\{\theta \sum_{i=1}^n \ln x_i\right\}} e^{-\lambda \theta}\mathbf1_{\theta>0} \end{align}
Simplify and hence conclude.