Solution to Problem 6.

(6) [15 points] The Pareto($\theta$) family of distributions has pdf given by

$$f(x\vert\theta) \; = \; \theta x^{-(\theta+1)} I_{(1,\infty)} (x) ,$$

where $\theta > 0$. (a) Verify that this is a legitimate pdf.

Solution:

$$\begin{eqnarray*} \int_1^{\infty} \; \theta x^{-(\theta+1)} \; dx & \; = \; & ... ...]_{x=1}^{\infty} \\ & \; = \; & 0 - (-1) \\ & \; = \; & 1 . \end{eqnarray*}$$

Clearly $f(x\vert\theta) \ge 0$, so it is a legitimate pdf.


(b) Is this an exponential family?

Solution:

$$f(x\vert\theta) \; = \; \theta \exp[ - \theta \log x ] x^{-1} I_{(1,\infty)} (x)$$

which is an exponential family with

$$\begin{eqnarray*} c(\theta) & \; = \; & \theta \\ w(\theta) & \; = \; & - \th... ...; = \; & \log x \\ h(x) & \; = \; & x^{-1} I_{(1,\infty)} (x) \end{eqnarray*}$$

(c) Does this distribution have a moment generating function that is finite in a neighborhood of the origin?

Solution: Note that

$$\int_1^{\infty} \, x^{\theta + 1} f(x\vert\theta) \, dx \; = \; \theta \int_1^{\infty} \, 1 \, dx \; = \; \infty .$$

That is, the $\theta + 1$ moment is not finite. If a distribution has a mgf that is finite in a neighborhood of $0$, then it has moments of all orders. Since this distribution does not have moments of all orders, it cannot have a mgf that is finite in a neighborhood of $0$.


(d) Suppose we have $n$ i.i.d. observations from the Pareto($\theta$) family with $\theta$ unknown. Find the maximum likelihood estimator of $\theta$.

Solution: The log likelihood is

$$\log L(\theta) \; = \; n \log \theta \, - \, \theta \sum_{i=1}^n \log X_i$$

except for the irrelevant terms involving the $h(X_i)$. Since this is an exponential family, we can take derivatives and set to $0$ to find the mle (otherwise, we would have to check that it gives a maximum):

$$n/\theta \, - \, \sum_{i=1}^n \log X_i \; = \; 0 \; \Long... ...= \; \left[ \frac{1}{n} \sum_{i=1}^n \log X_i \right]^{-1}.$$

(d) A Bayesian wants to make inferences about $\theta$. He uses a gamma($\alpha$,$\beta$) prior for $\theta$. Show that the posterior for $\theta$ is also a gamma distribution and find its parameters. Also, find the posterior mean of $\alpha$.

Solution: The posterior is

$$f(\theta\vert x_1,\ldots,x_n) \; \propto \; \theta^{\alpha-1+n} \exp[-\theta(1/\beta + \sum_{i=1}^n \log x_i) ] .$$

This is a

$$\mbox{gamma} \left(\alpha+n, \left[ 1/\beta + \sum_{i=1}^n \log x_i \right]^{-1} \right)$$

pdf. The posterior mean of $\theta$ is

$$\frac{\alpha +n}{1/\beta + \sum_{i=1}^n \log x_i} .$$