# Bayes estimator of possion distribution with Pareto prior

Consider a random sample of size $$n$$ following the possion distribution with parameter $$\ln \theta$$, that is

$$f(x|\theta)=\frac{(\ln\theta)^x}{\theta x!}, x=0,1,2,\cdots$$

and the prior of the parameter $$\theta$$ follows a Pareto distribution:

$$\mathbb{P}[\theta]= \frac{1}{\beta}\theta^{-\frac{1}{\beta}-1} , \theta > 1$$

Find the Bayes estimator of $$\theta$$.

I have got the expression of the postprior:

$$\int_1^{\infty} \frac{\ln^x \theta}{\theta x!}\cdot \frac{1}{\beta}\theta^{-\frac1\beta-1}\text{d}\theta$$

I have no idea about this integral. Mathematica shows that this integral equals to $$-\frac1{\beta x!}\Gamma\left(1+x, \frac{(1+\beta)\ln \theta}{\beta}\right)\left(\frac{1+\beta}{\beta}\right)^{-1-x}\bigg|_{1}^{\infty}=\frac{\beta^x(x+1)}{(1+\beta)^{x+1}}$$

I am not sure about whether my expression of the postprior is correct (there seems no information about the sample). Also I have difficulties about the integral. Any hints will be helpful and appreciated, thank you!

• A side note: while, for a given prior, you can compute THE (i.e. unique) posterior for $theta$, for the latter there are many possible Bayes estimators, e.g. MAP, posterior average, etc. Nov 29, 2023 at 7:53

## 1 Answer

I will simplify the analysis by using the parameter $$\alpha = 1/\beta$$. Letting $$\phi = \ln (\theta)$$ we have $$d \theta /d \phi = \exp(\phi)$$ so we can write the prior for $$\phi$$ as:

\begin{align} \pi (\phi) &= \text{Pareto}(\exp(\phi) | \alpha) \cdot \Bigg| \frac{d \theta}{d \phi} \Bigg| \\[10pt] &= \alpha \exp(\phi)^{-\alpha-1} \cdot \mathbb{I}(\exp(\phi) > 1) \cdot \exp(\phi) \\[16pt] &= \alpha \exp(-\alpha \phi) \cdot \mathbb{I}(\phi > 0). \\[6pt] \end{align}

Given a sample $$\mathbf{x} = (x_1,...,x_n)$$ with sample mean $$\bar{x}_n$$ the posterior for $$\phi$$ is then given by:

\begin{align} \pi (\phi | \mathbf{x}) &\propto \bigg( \prod_{i=1}^n f(x_i|\phi) \bigg) \cdot \pi (\phi) \\[6pt] &\propto \frac{\phi^{n \bar{x}_n}}{\exp(n\phi)} \cdot \exp(-\alpha \phi) \cdot \mathbb{I}(\phi > 0) \\[12pt] &= \phi^{n \bar{x}_n} \cdot \exp(-(\alpha+n) \phi) \cdot \mathbb{I}(\phi > 0) \\[16pt] &\propto \text{Gamma}(\phi | \text{Shape} = n \bar{x}_n+1, \text{Rate} = \alpha+n). \\[18pt] \end{align}

This establishes the posterior $$\phi | \mathbf{x} \sim \text{Gamma}(\text{Shape} = {n \bar{x}_n}+1, \text{Rate} = \alpha+n)$$ so the posterior for $$\theta$$ follows the corresponding exponentiated gamma distribution with density function:

\begin{align} \pi (\theta | \mathbf{x}) &= \text{Gamma}(\ln(\theta) | \text{Shape} = n \bar{x}_n+1, \text{Rate} = \alpha+n) \cdot \bigg| \frac{d \phi}{d \theta} \bigg| \\[8pt] &= \frac{(\alpha+n)^{n \bar{x}_n+1}}{\Gamma(n \bar{x}_n+1)} \cdot \ln(\theta)^{n \bar{x}_n} \cdot \exp(-(\alpha+n) \ln(\theta)) \cdot \mathbb{I}(\ln(\theta) > 0) \cdot \frac{1}{\theta} \\[6pt] &= \frac{(\alpha+n)^{n \bar{x}_n+1}}{\Gamma(n \bar{x}_n+1)} \cdot \ln(\theta)^{n \bar{x}_n} \cdot \theta^{-(\alpha+n+1)} \cdot \mathbb{I}(\theta > 1). \\[12pt] \end{align}