# Posterior distribution of $\theta x^{\theta - 1}$ with $Gamma(\alpha, \lambda)$ prior

Random variables $$X_1, \ldots, X_n$$ are i.i.d given $$\vartheta = \theta$$ and have the following pdf: $$$$p(x|\theta)=\begin{cases} \theta x^{\theta - 1}, & \text{if 0 $$\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

$$$$\pi(\theta|x_1, \ldots, x_n) \propto \pi(\theta) \Pi_{i=1}^n p(x_i|\theta) \propto \theta^{n + \alpha - 1} e^{-\lambda \theta}\Pi_{i=1}^n x^{(\theta - 1)}$$$$

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?

• $X_i$'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. Oct 19, 2021 at 6:03
• Hint: $x^\theta = \exp(\theta \log(x))$ Oct 19, 2021 at 9:27

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

And I assume the following kernel for the Gamma prior:

$$\pi(\theta)\propto \theta^{\alpha-1}e^{-\lambda \theta}\mathbf1_{\theta>0}$$

For $$0, the posterior density of $$\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.

• How did $\left(\prod\limits_{i=1}^n x_i\right)^{\theta-1}$ turn into $\left(\prod\limits_{i=1}^n x_i\right)^{\theta}$ ? Oct 19, 2021 at 12:07
• I have a $\propto$ before that. Oct 19, 2021 at 12:22
• 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 $\theta$ Oct 19, 2021 at 12:26