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?

  • $\begingroup$ $X_i$'s have a Beta distribution, and a Gamma prior on $\theta$ should indeed turn out to be a conjugate prior. Check your expression of posterior density carefully. Also add the self-study tag. $\endgroup$ Oct 19, 2021 at 6:03
  • 2
    $\begingroup$ Hint: $x^\theta = \exp(\theta \log(x))$ $\endgroup$
    – B.Liu
    Oct 19, 2021 at 9:27

1 Answer 1


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.

  • $\begingroup$ How did $\left(\prod\limits_{i=1}^n x_i\right)^{\theta-1}$ turn into $\left(\prod\limits_{i=1}^n x_i\right)^{\theta}$ ? $\endgroup$
    – Henry
    Oct 19, 2021 at 12:07
  • 1
    $\begingroup$ I have a $\propto$ before that. $\endgroup$ Oct 19, 2021 at 12:22
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    $\begingroup$ OK, so you seem to be saying the missing $\prod\limits_{i=1}^n x_i$ normalises out when looking at the posterior distribution for $\theta$ $\endgroup$
    – Henry
    Oct 19, 2021 at 12:26

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