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Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\) be a random sample from a Cauchy distribution with median \(\theta\), that is, with pdf $$ f(x ; \underline{\theta})=\frac{1}{\pi} \frac{1}{1+(x-\theta)^{2}}, \quad-\infty

Short Answer

Expert verified
To estimate the median of a Cauchy distribution using maximum likelihood estimation, we first derive the log-likelihood function, which is the sum of the logarithms of the probability density function evaluated at each sample point. We then minimize this function using R, by plotting it against a range of values for the median and finding the value that gives the minimum sum of the log-likelihood. This is the maximum likelihood estimate for the median of the Cauchy distribution.

Step by step solution

01

Derive the log-likelihood function

Given the probability density function (pdf) of a Cauchy distribution, the likelihood function for a sample \((X_{1}, X_{2}, X_{3}, X_{4}, X_{5})\) will be the product of the pdf evaluated at each sample point. That is, \(L(\theta)= \prod_{i=1}^5 f(x_i ; \theta) = \prod_{i=1}^{5} \left[\frac{1}{\pi}\frac{1}{1+(x_{i}-\theta)^{2}}\right]\). The log-likelihood function is obtained by taking the natural logarithm of the likelihood function, causing the product to turn into a sum. Hence, \(l(\theta) = \log\left[L(\theta)\right] = \sum_{i=1}^{5} \log \left[\frac{1}{\pi} \frac{1}{1+(x_{i}-\theta)^{2}}\right]\). We can ignore the constant \(-\frac{1}{\pi}\) in the minimization since it doesn't impact where the minimum occurs. Thus, the function to be minimized becomes \(\sum_{i=1}^{5} \log \left[1+\left(x_{i}-\theta\right)^{2}\right]\).
02

Write the R code to implement the function

In R, we create a sequence of potential theta values ranging from -6 to 6 using the seq() function. Then we create an empty vector to hold the sum of the log likelihood for each theta. This is done inside a for loop that iterates over each potential theta value, calculates the sum of the log likelihood at that theta and adds it to the list.
03

Plot the function in R

We plot the sum of the log likelihood calculated in the previous step against the sequence of potential theta values. The theta value at the minimum point of the plot is the approximate MLE of the median, theta.

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Key Concepts

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

Maximum Likelihood Estimation (MLE)
Maximum Likelihood Estimation (MLE) is a statistical method used to estimate the parameters of a probability distribution by maximizing a likelihood function. The likelihood function measures the probability of observing the given sample data as a function of the parameters. In the context of the Cauchy distribution, the MLE aims to determine the median \(\theta\) that would make the observed sample most likely.

The procedure involves finding the parameter value that maximizes the likelihood function or equivalently, minimizes the negative log-likelihood function (since logarithms are monotonic transformations). In the given exercise, this involves summing up the log probabilities of each observation and finding the value of \(\theta\) that results in the lowest sum, indicating the highest likelihood of observing the given data.
Probability Density Function (PDF)
A probability density function (PDF) describes the likelihood of a random variable taking on a value within a specific interval. It is a fundamental concept in probability theory and statistics used to model the distribution of continuous variables. The PDF for a given distribution provides a function that can be integrated over an interval to yield the probability that the random variable falls within that interval.

In the Cauchy distribution example from the exercise, the PDF is defined as \[ f(x ; \theta)=\frac{1}{\pi} \frac{1}{1+(x-\theta)^{2}},\quad-\infty
Log-likelihood Function
The log-likelihood function is a transformation of the likelihood function and plays a crucial role in MLE. By taking the logarithm of the likelihood, the product of probabilities is converted into a sum. This often simplifies the calculation and optimization, especially with regard to the computational stability and efficiency.

In the Cauchy distribution problem, the log-likelihood function \( l(\theta) \) was derived from the PDF by converting the product of densities into a sum, leading to \[ l(\theta) = \sum_{i=1}^{5} \log \left[\frac{1}{\pi} \frac{1}{1+(x_{i}-\theta)^{2}}\right] \]. The derivation of the log-likelihood is crucial since directly working with the likelihood itself would be less practical due to the smaller scale of raw probability values, which can lead to numerical underflow in computers.
R Programming
R is a programming language widely used for statistical computing and graphics, favored for its vast array of packages tailored for data analysis. The R code provided in the exercise demonstrates how to implement the process of MLE by codifying the log-likelihood function for the Cauchy distribution and using numerical methods to find \(\theta\) that minimizes the function, hence approximating the MLE of the median.

The code includes the creation of a sequence of potential \(\theta\) values and the computation of the cumulative log-likelihood for these values. It then visualizes the results with a plot, helping to identify the value of \(\theta\) that minimizes the log-likelihood, thus employing both the computational and graphical capabilities of R.
Statistical Inference
Statistical inference is the process of drawing conclusions about a population's characteristics based on information contained in a sample. MLE is a method used in statistical inference to estimate population parameters, such as the median \(\theta\) in a Cauchy distribution. Through inference, we aim to understand the underlying distribution of data and to make predictions or decisions based on data analysis.

The Cauchy example illustrates how MLE is used for parameter estimation within the broader context of statistical inference. By estimating the optimal value of \(\theta\), we're able to infer the most likely parameter that defines the central tendency of the population from which the sample was drawn.

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Most popular questions from this chapter

Let \(n\) independent trials of an experiment be such that \(x_{1}, x_{2}, \ldots, x_{k}\) are the respective numbers of times that the experiment ends in the mutually exclusive and exhaustive events \(C_{1}, C_{2}, \ldots, C_{k} .\) If \(p_{i}=P\left(C_{i}\right)\) is constant throughout the \(n\) trials, then the probability of that particular sequence of trials is \(L=p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{k}^{x_{k}}\). (a) Recalling that \(p_{1}+p_{2}+\cdots+p_{k}=1\), show that the likelihood ratio for testing \(H_{0}: p_{i}=p_{i 0}>0, i=1,2, \ldots, k\), against all alternatives is given by $$ \Lambda=\prod_{i=1}^{k}\left(\frac{\left(p_{i 0}\right)^{x_{i}}}{\left(x_{i} / n\right)^{x_{i}}}\right) $$ (b) Show that $$ -2 \log \Lambda=\sum_{i=1}^{k} \frac{x_{i}\left(x_{i}-n p_{i 0}\right)^{2}}{\left(n p_{i}^{\prime}\right)^{2}} $$ where \(p_{i}^{\prime}\) is between \(p_{i 0}\) and \(x_{i} / n\). Hint: Expand \(\log p_{i 0}\) in a Taylor's series with the remainder in the term involving \(\left(p_{i 0}-x_{i} / n\right)^{2}\). (c) For large \(n\), argue that \(x_{i} /\left(n p_{i}^{\prime}\right)^{2}\) is approximated by \(1 /\left(n p_{i 0}\right)\) and hence \(-2 \log \Lambda \approx \sum_{i=1}^{k} \frac{\left(x_{i}-n p_{i 0}\right)^{2}}{n p_{i 0}}\) when \(H_{0}\) is true. Theorem \(6.5 .1\) says that the right-hand member of this last equation defines a statistic that has an approximate chi-square distribution with \(k-1\) degrees of freedom. Note that dimension of \(\Omega-\) dimension of \(\omega=(k-1)-0=k-1\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with one of two pdfs. If \(\theta=1\), then \(f(x ; \theta=1)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2},-\infty

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 \(0.05\). $$ \begin{array}{|c|c|c|c|c|c|c|} \hline \multicolumn{7}{|c|} {\text { Earth to Moon Mass Ratios }} \\ \hline 81.3001 & 81.3015 & 81.3006 & 81.3011 & 81.2997 & 81.3005 & 81.3021 \\ \hline \end{array} $$

Let \(X\) and \(Y\) be two independent random variables with respective pdfs $$ f\left(x ; \theta_{i}\right)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta_{i}}\right) e^{-x / \theta_{i}} & 0

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 \(0 \leq p_{1} \leq p_{2} \leq 1\)

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