Statistics; density functions and likelihood estimators

Question

I dont get a single bit of this, the question is nothing like things I've done before.

Answer 1

This is a Pareto distribution with a scale factor of $1$ and a shape parameter $\theta$.

The maximum likelihood estimate of the shape parameter (in the case of $1$ for the scale parameter): $$\overline{\theta}=\frac{n}{\sum_{i=1}^n\ln{(X_i)}}.$$ You can substitute your data...

As far as the last question. The inverse of the maximum likelihood estimate is the average of the logarithms of the data: $$\frac1n\sum_{i=1}^n\ln{(X_i)}.$$ In order to see if this is unbiased we have to compute the mean: $$\frac1n\sum_{i=1}^nE[\ln{(X_i)}]=E[\ln(X_1)]=\theta\int_1^{\infty}\ln(x)x^{-(\theta+1)}\ dx.$$ Integrate by part with $u'=x^{-(\theta+1)}$ and $v=\ln(x)$ and you will learn if the average above is biased or not.

Answer 2

Hint:

To help you get started,

Write down the log-likelihood function, $$\ln \left(\prod_{i=1}^n \frac{\theta}{x_i^{\theta+1}}\right) = \sum_{i=1}^n \ln \left( \frac{\theta}{x_i^{\theta+1}}\right)=n\ln \theta-(\theta+1)\sum_{i=1}^n \ln x_i$$

To find $\hat{\theta}_{ML}$, differentiate the previous expression with respect to $\theta$ and equate them to zero and solve for $\theta$.