91Ó°ÊÓ

A diagnostic test for a certain disease is applied to \(n\) individuals known to not have the disease. Let \(X=\) the number among the \(n\) test results that are positive (indicating presence of the disease, so \(X\) is the number of false positives) and \(p=\) the probability that a disease-free individual's test result is positive (i.e., \(p\) is the true proportion of test results from disease-free individuals that are positive). Assume that only \(X\) is available rather than the actual sequence of test results. a. Derive the maximum likelihood estimator of \(p\). If \(n=20\) and \(x=3\), what is the estimate? b. Is the estimator of part (a) unbiased? c. If \(n=20\) and \(x=3\), what is the mle of the probability \((1-p)^{5}\) that none of the next five tests done on disease-free individuals are positive? 21\. Let \(X\) have a Weibull distribution with parameters \(\alpha\) and \(\beta\), so $$ \begin{gathered} E(X)=\beta \cdot \Gamma(1+1 / \alpha) \\ V(X)=\beta^{2}\left[\Gamma(1+2 / \alpha)-[\Gamma(1+1 / \alpha)]^{2}\right] \end{gathered} $$ a. Based on a random sample \(X_{1}, \ldots, X_{n}\), write equations for the method of moments estimators of \(\beta\) and \(\alpha\). Show that, once the estimate of \(\alpha\) has been obtained, the estimate of \(\beta\) can be found from a table of the gamma function and that the estimate of \(\alpha\) is the solution to a complicated equation involving the gamma function. b. If \(n=20, \bar{x}=28.0\), and \(\Sigma x_{i}^{2}=16,500\), compute the estimates. [Hint: \(\left.[\Gamma(1.2)]^{2} / \Gamma(1.4)=.95 .\right]\)

Step by step solution

01

Define the likelihood function

Given that the individuals are disease-free, the problem involves determining the maximum likelihood estimator (MLE) of the probability \(p\) for false positives. The outcome \(X\) follows a binomial distribution: \(X \sim \text{Binomial}(n, p)\). The likelihood function is given by\: \(L(p) = \binom{n}{x} p^x (1-p)^{n-x}\).

02

Derive the log-likelihood function

To simplify the differentiation process, convert the likelihood function into a log-likelihood function:\: \(\log L(p) = \log\binom{n}{x} + x \log(p) + (n-x)\log(1-p)\). Here, \(\log\binom{n}{x}\) is a constant and doesn't affect the differentiation with respect to \(p\).

03

Differentiate the log-likelihood function

Differentiate \(\log L(p)\) with respect to \(p\) to find its maximum:\: \(\frac{d}{dp} [x \log(p) + (n-x)\log(1-p)] = \frac{x}{p} - \frac{n-x}{1-p}\).

04

Solve for p

Set the derivative from Step 3 to zero to find the critical points:\: \(\frac{x}{p} - \frac{n-x}{1-p} = 0\). Solve for \(p\), yielding\: \(\hat{p} = \frac{x}{n}\).

05

Calculate the MLE for a Specific Case

Given \(n = 20\) and \(x = 3\), substitute these values into \(\hat{p} = \frac{x}{n}\) to compute the MLE for \(p\):\: \(\hat{p} = \frac{3}{20} = 0.15\).

06

Check if the estimator is unbiased

An estimator \(\hat{p}\) is unbiased if \(E[\hat{p}] = p\). For the sample proportion \(\hat{p} = \frac{X}{n}\), \(E[\hat{p}] = \frac{E[X]}{n} = \frac{np}{n} = p\), showing that \(\hat{p}\) is indeed unbiased.

07

Compute the probability of 5 consecutive negative tests

The probability that none of the next 5 tests are positive is given by \((1-p)^5\). Substitute the MLE \(\hat{p} = 0.15\) into this expression:\: \((1-0.15)^5 = 0.85^5 \approx 0.444\).

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Key Concepts

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

Binomial Distribution

The binomial distribution is a fundamental concept in probability and statistics, often used to model scenarios where there are exactly two outcomes, like success or failure on a given trial. Suppose you administer a diagnostic test to individuals without a disease and want to determine how many return a false positive result. This scenario can be modeled with a binomial distribution. In the binomial distribution, the random variable, often represented as \(X\), follows the notation \(X \sim \text{Binomial}(n, p)\). Here, \(n\) denotes the number of trials, or in our context, the number of tests, while \(p\) represents the probability of a false positive on any individual test.

Key characteristics of a binomial distribution include:

• Each trial is independent.
• There are only two possible outcomes per trial: success (false positive) or failure (true negative).
• The probability of success \(p\) is constant across all trials.
• The random variable \(X\) counts the number of successes in \(n\) trials.

Thus, knowing the binomial distribution, we can use the maximum likelihood estimation approach to derive estimators for the probability \(p\).

Weibull Distribution

The Weibull distribution is another useful probability distribution, particularly popular in reliability analysis and life data analysis. It is specified by two parameters: \(\alpha\) (the shape parameter) and \(\beta\) (the scale parameter). In a Weibull distribution, different combinations of these parameters result in various shapes of the distribution's curve, which can model diverse types of data. The Weibull distribution is recognized for its flexibility in representing defect lifetimes, failure times, and other phenomena related to duration.

For a random variable \(X\) with a Weibull distribution:

• The expected value or mean is defined as \(E(X) = \beta \cdot \Gamma(1+1/\alpha)\).
• The variance, indicating the spread of the distribution, is \(V(X) = \beta^2 [\Gamma(1+2/\alpha) - [\Gamma(1+1/\alpha)]^2]\).

The function \(\Gamma\) represents the gamma function, closely related to factorials for non-integers, aiding in the computation of these moments.

Method of Moments

The method of moments is a statistical technique used to derive parameter estimates from the moments of a sample. A 'moment' is a quantitative measure related to the shape of a set of points. This method involves equating sample moments (such as mean and variance) to theoretical moments derived from the distribution to estimate parameters.

When estimating parameters \(\alpha\) and \(\beta\) for a Weibull distribution from sample data \(X_1, X_2, \ldots, X_n\), the process follows these steps:

• Calculate sample moments like mean \(\bar{x}\) and variance.
• Set these sample moments equal to the theoretical moments of the Weibull distribution.
• Solve the resulting equations to find estimates for the parameters.

In practice, finding these estimates might involve complicated formulas and numerical solutions, especially when dealing with non-integer gamma functions, but this approach offers a way to derive parameters from empirical data effectively.

Bias of Estimators

The bias of an estimator is an essential concept in statistics, providing insight into the accuracy and reliability of estimation methods. An estimator is said to be unbiased if its expected value equals the true parameter of interest. In other words, an unbiased estimator accurately centers around the true value.

Consider, for example, an estimator \(\hat{p}\) for the probability of a false positive in a binomial distribution. The estimator \(\hat{p}\) was derived via maximum likelihood estimation as \(\hat{p} = \frac{X}{n}\), where \(X\) is the number of false positives, and \(n\) is the total number of tests conducted. For \(\hat{p}\) to be unbiased, the expected value \(E[\hat{p}]\) should equal the true probability \(p\). Through mathematical derivation, it can be shown that \(E[\hat{p}] = \frac{np}{n} = p\), confirming that \(\hat{p}\) is indeed unbiased.

Understanding estimator bias helps in choosing the correct estimation method for analysis, ensuring that the predictions and conclusions drawn are as accurate as possible.