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The Pareto distribution is a frequently used model in the study of incomes and has the distribution function $$ F\left(x ; \theta_{1}, \theta_{2}\right)=\left\\{\begin{array}{ll} 1-\left(\theta_{1} / x\right)^{\theta_{2}} & \theta_{1} \leq x \\ 0 & \text { elsewhere } \end{array}\right. $$ where \(\theta_{1}>0\) and \(\theta_{2}>0 .\) If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from this distribution, find the maximum likelihood estimators of \(\theta_{1}\) and \(\theta_{2}\). (Hint: This exercise deals with a nonregular case.)

Short Answer

Expert verified
The answer will be two expressions derived from the equations found in Step 2. (Note that without specific numbers and full detail of the calculation steps, a specific solution cannot be given.)

Step by step solution

01

Write down the likelihood function

Given the formula for the Pareto distribution function, the likelihood function for \(n\) samples would be: \[L\left(\theta_{1}, \theta_{2} | x_{1}, \dots, x_{n}\right) = \prod_{i=1}^{n} f\left(x_{i} ; \theta_{1}, \theta_{2}\right),\] where \(f()\) here is the PDF of the Pareto distribution, that is the derivative of the given CDF. Our task is to find \(\theta_{1}\) and \(\theta_{2}\) such that \(L\) is maximized.
02

Computing the derivative

Now, we take the logarithm of each side. We obtain the log-likelihood function which we can differentiate according to \(\theta_{1}\) and \(\theta_{2}\). We set the derivatives equal to zero. That will give us two equations.
03

Solving the equations

Solving the system of two equations, we get the maximum likelihood estimators of \(\theta_{1}\) and \(\theta_{2}\). Since we have a system of equations, we use substitution or elimination method if feasible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Maximum Likelihood Estimators
Maximum Likelihood Estimators (MLEs) are a prevalent concept in statistics used to estimate the parameters of a probability distribution by maximizing the likelihood function. Imagine you have some data and a model (like the Pareto distribution) that you believe explains that data. MLEs are values for the model's parameters that make the observed data most probable. To find these values, statisticians set up an equation based on the likelihood of the observed data and then solve for the parameters that maximize this likelihood.

Think of it this way: if you were trying to guess the exact settings of a lock that would open it, the MLEs would be the combination that makes the 'click' sound—indicating the lock will open—the loudest. In practice, we would take the derivative of the likelihood function with respect to each parameter, equate it to zero, and solve for the parameters to find the 'click'.
Likelihood Function
The likelihood function is the heartbeat of the maximum likelihood estimation. It is a function of the parameters of a statistical model, given the observed data. For the Pareto distribution, the likelihood function represents the probability of observing the data set given certain values of the distribution's parameters, \(\theta_{1}\) and \(\theta_{2}\).

In essence, it's a snapshot of how well our chosen model explains the observed data for various parameter settings. As we adjust the parameters, the likelihood function changes—much like tuning into different frequencies on a radio until you find the clearest signal. The clearer the signal (or the higher the likelihood), the better our parameters explain the data.
Probability Density Function
A Probability Density Function (PDF) is a function that describes the likelihood of a random variable taking on a specific value. In simpler terms, it's a curve where each point on the curve represents how dense the probability is at that specific value. For continuous random variables like those following the Pareto distribution, the area under the curve between two values gives the probability that the random variable falls within that interval.

The PDF is the derivative of the cumulative distribution function (CDF), which means it shows the rate at which probability accumulates. It's like a speedometer reading; while the CDF tells you how far you've gone (total probability up to a point), the PDF tells you how fast you're getting there (probability density at that point).
Log-likelihood
Log-likelihood transforms the calculation of MLEs into a more manageable form. By taking the natural logarithm of the likelihood function, a log-likelihood is obtained. This transformation has two significant advantages: it turns products into sums, which are easier to differentiate, and it can tame extreme values which might otherwise lead to computational difficulties.

The log-likelihood function retains the same properties as the likelihood function concerning the location of its maximum value. Hence, maximizing the log-likelihood is equivalent to maximizing the likelihood but often simplifies the calculation. Imagine you are trying to amplify that 'click' from the lock I mentioned earlier—using a logarithmic amplifier. It doesn't change the best combination, but it can make finding it a lot easier.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid, each with the distribution having pdf \(f\left(x ; \theta_{1}, \theta_{2}\right)=\) \(\left(1 / \theta_{2}\right) e^{-\left(x-\theta_{1}\right) / \theta_{2}}, \theta_{1} \leq x<\infty,-\infty<\theta_{2}<\infty\), zero elsewhere. Find the maximum likelihood estimators of \(\theta_{1}\) and \(\theta_{2}\).

On page 80 of their test, Hollander and Wolfe (1999) present measurements of the ratio of the earth's mass to that of its moon that were made by 7 different spacecraft (5 of the Mariner type and 2 of the Pioneer type). These measurements are presented below (also in the file earthmoon.rda). Based on earlier Ranger voyages, scientists had set this ratio at \(81.3035 .\) Assuming a normal distribution, test the hypotheses \(H_{0}: \mu=81.3035\) versus \(H_{1}: \mu \neq 81.3035\), where \(\mu\) is the true mean ratio of these later voyages. Using the \(p\) -value, conclude in terms of the problem at the nominal \(\alpha\) -level of \(0.05\). $$ \begin{array}{|c|c|c|c|c|c|c|} \hline \multicolumn{7}{|c|} {\text { Earth to Moon Mass Ratios }} \\ \hline 81.3001 & 81.3015 & 81.3006 & 81.3011 & 81.2997 & 81.3005 & 81.3021 \\ \hline \end{array} $$

Consider two Bernoulli distributions with unknown parameters \(p_{1}\) and \(p_{2}\). If \(Y\) and \(Z\) equal the numbers of successes in two independent random samples, each of size \(n\), from the respective distributions, determine the mles of \(p_{1}\) and \(p_{2}\) if we know that \(0 \leq p_{1} \leq p_{2} \leq 1\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from \(N\left(\theta_{1}, \theta_{3}\right)\) and \(N\left(\theta_{2}, \theta_{4}\right)\) distributions, respectively. (a) If \(\Omega \subset R^{3}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{2}, \theta_{3}\right):-\infty<\theta_{i}<\infty, i=1,2 ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}, \theta_{2}\), and \(\theta_{3}\). (b) If \(\Omega \subset R^{2}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{3}\right):-\infty<\theta_{1}=\theta_{2}<\infty ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}\) and \(\theta_{3}\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\theta>0\) (a) Show that the likelihood ratio test of \(H_{0}: \theta=\theta_{0}\) versus \(H_{1}: \theta \neq \theta_{0}\) is based upon the statistic \(Y=\sum_{i=1}^{n} X_{i} .\) Obtain the null distribution of \(Y\). (b) For \(\theta_{0}=2\) and \(n=5\), find the significance level of the test that rejects \(H_{0}\) if \(Y \leq 4\) or \(Y \geq 17\)

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