Question

Recall that \(\widehat{\theta}=-n / \sum_{i=1}^{n} \log X_{i}\) is the mle of \(\theta\) for a beta \((\theta, 1)\) distribution. Also, \(W=-\sum_{i=1}^{n} \log X_{i}\) has the gamma distribution \(\Gamma(n, 1 / \theta)\). (a) Show that \(2 \theta W\) has a \(\chi^{2}(2 n)\) distribution. (b) Using part (a), find \(c_{1}\) and \(c_{2}\) so that $$ P\left(c_{1}<\frac{2 \theta n}{\hat{\theta}}

Short answer

In part (a), it was shown that \(2 \theta W\) follows a \(\chi^{2}(2 n)\) distribution. For part (b), we found the values for \(c_{1}\) and \(c_{2}\) that give the \(100(1-\alpha)\%\) confidence interval for \(\theta\) as \(\left(\frac{n}{c_{2}}, \frac{n}{c_{1}}\right)\). And finally in part (c), we compare the length of this interval with the one found in Example 6.2.6 when \(\alpha = 0.05\) and \(n = 10\).

Step-by-step solution

Proving the Expression Follows a Chi-square Distribution

We need to show that \(2 \theta W\) follows a \(\chi^{2}(2 n)\) distribution. We know that \(W\) follows the gamma distribution, which is \(W \sim \Gamma(n, 1 / \theta)\). The chi-square distribution is a special case of the gamma distribution with shape parameter \(n\) and scale parameter \(2\), that is, \(X \sim \Gamma(n/2, 2)\). So to make \(W\) follow a chi-square distribution, we would multiply by \(2\theta\) to get \(2\theta W \sim \chi^{2}(2 n)\).

Finding the Constants \(c_{1}\) and \(c_{2}\)

We now want to find \(c_{1}\) and \(c_{2}\) such that the probability \(P(c_{1} < \frac{2 \theta n}{\hat{\theta}} < c_{2}) = 1 - \alpha\). Since \(2\theta W = \frac{2\theta n}{\hat{\theta}}\) follows a \(\chi^{2}(2n)\) distribution, we can use chi-square quantiles to find \(c_{1}\) and \(c_{2}\). Using the properties of the chi-square distribution, we find that \(c_{1} = \chi^{2}(2n, \alpha / 2)\) and \(c_{2} = \chi^{2}(2n, 1 - \alpha / 2)\).

Construct the Confidence Interval

Using \(c_{1}\) and \(c_{2}\), we can construct a \(100(1-\alpha)\%\) confidence interval for \(\theta\) as \(\left(\frac{n}{c_{2}}, \frac{n}{c_{1}}\right)\).

Compare the Lengths of Different Confidence Intervals

For \(\alpha = 0.05\) and \(n = 10\), we find the length of this interval and then compare it with the length of the interval found in Example 6.2.6. The length of an interval \((a, b)\) is \(b - a\), and we use it to compare the two intervals as requested.

Key concepts

Maximum Likelihood Estimation (MLE)

Understanding Maximum Likelihood Estimation (MLE) is fundamental when it comes to estimating parameters in statistical models. It's a method that determines the parameter values of a statistical model by maximizing the likelihood function. The likelihood function measures how well the model with a particular set of parameters explains the observed data.

When applying MLE to a beta distribution, you must find the parameter value that makes the observed data most probable. In the given exercise, the MLE is calculated for a Beta \(\theta, 1\) distribution, and the value of \(\widehat{\theta}\) is given by \(\widehat{\theta}=-n / \sum_{i=1}^{n} \log X_{i}\), which maximizes the likelihood function for the observed data set \(X_1, X_2, ..., X_n\).

Beta Distribution

The Beta Distribution is a continuous probability distribution characterized by two shape parameters, \(\alpha\) and \(\beta\), which determine its shape. Its support is on the interval \(0, 1\). The beta distribution is often used in Bayesian statistics, project management, and order statistics.

In our example, we work with a specific case where the beta distribution's second parameter is fixed at 1, i.e., Beta \(\theta, 1\). The MLE of the first parameter \(\theta\) is sought, which can be understood as adjusting the 'concentration' of the distribution towards 0 while holding the other parameter constant.

Gamma Distribution

The Gamma Distribution is another continuous distribution that is specified by two parameters, shape \(k\) and scale \(\theta\). Often used to model waiting times and life data, it generalizes the exponential and chi-squared distributions.

In the context of the exercise, \(W\) is defined to follow a gamma distribution with parameters \(n\) and \(1/\theta\), written as Gamma \(\Gamma(n, 1 / \theta)\). The gamma distribution's connection to the chi-squared distribution is utilized to transition from the gamma \(\Gamma(n, 1 / \theta)\) to the chi-square \(\chi^{2}(2 n)\) distribution through a transformation of the variable \(W\).

Chi-square Distribution

A Chi-square Distribution, typically denoted as \(\chi^2(d)\), is a special case of the gamma distribution with its shape \(k = d/2\) and scale \(\theta = 2\), where \(d\) is the degrees of freedom. It's often used in hypothesis testing and confidence interval estimation.

Through multiplication by \(2\theta\), we can transform the gamma-distributed variable \(W\) into a chi-square distributed variable with degrees of freedom \(2n\), as shown in part (a) of the exercise. This elegant statistical property allows us to use the well-tabulated chi-square distribution to find critical values for constructing confidence intervals.

Confidence Interval

Constructing a Confidence Interval gives us a range of plausible values for an unknown parameter based on the data from a sampled population. A confidence interval with a 1-\(\alpha\) confidence level means that if we were to take many samples and build a confidence interval from each sample, then approximately 100(1-\(\alpha\))% of those intervals would contain the true parameter value.

In step 2 and step 3 of the example, \(c_1\) and \(c_2\) are found using the chi-square distribution, which are then used to construct the confidence interval for \(\theta\). This interval gives us a statistical range within which we can say, with the desired level of confidence, that the true value of \(\theta\) lies. For \(\alpha=0.05\) and \(n=10\), the confidence interval's length is an indicator of precision – the shorter the interval, the more precise the estimation is considered.