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Chapter 6: Problem 27

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solutions yield the maximum likelihood estimators of \(\alpha\) and \(\beta\). Do you think they can be solved explicitly? b. Show that the mle of \(\mu=\alpha \beta\) is \(\hat{\mu}=\bar{X}\).

Short Answer

Expert verified
Maximum likelihood estimation (MLE) for gamma distribution includes complex equations not easily solved explicitly; \(\hat{\mu} = \bar{X}\) confirms that the estimator for \(\mu\) is the sample mean.

Step by step solution

01

Understand the Gamma Distribution

The probability density function (pdf) of a gamma distribution with parameters \(\alpha\) and \(\beta\) is given by \(f(x; \alpha, \beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}\), where \(x > 0\), \(\alpha > 0\), and \(\beta > 0\). In this context, \(\Gamma(\alpha)\) is the gamma function.
02

Write the Likelihood Function

For a random sample \(X_1, \ldots, X_n\) from this distribution, the likelihood function is the product of the individual pdfs: \[ L(\alpha, \beta) = \prod_{i=1}^n \frac{\beta^{\alpha}}{\Gamma(\alpha)} X_i^{\alpha-1} e^{-\beta X_i} = \frac{\beta^{n\alpha}}{(\Gamma(\alpha))^n} \prod_{i=1}^n X_i^{\alpha-1} e^{-\beta \sum_{i=1}^n X_i} \] .

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Key Concepts

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

Gamma Distribution
The gamma distribution is a two-parameter family of continuous probability distributions. It's mainly used to model waiting times or lifetimes of objects. It's characterized by the parameters \( \alpha \) (shape) and \( \beta \) (rate). The parameters quantify:
  • Shape \( \alpha \): Determines the distribution's skewness and how data is dispersed over different ranges.
  • Rate \( \beta \): Affects the scale, dictating how "spread out" the distribution is.
The randomness of events happening over time is well-represented by this distribution, for instance, modeling the time until the next earthquake or till a light bulb burns out. Understanding these two parameters helps in applying the gamma distribution effectively in real-world scenarios.
Probability Density Function
The probability density function (pdf) is pivotal for continuous random variables. It allows us to understand the likelihood of a variable assuming a specific value. For a gamma distribution, the pdf is expressed as: \[ f(x; \alpha, \beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} \] where:
  • \( x > 0 \) ensures we are working within the valid range of the distribution.
  • \( \Gamma(\alpha) \) is the gamma function, an extension of the factorial function.
  • \( x^{\alpha-1} \) and \( e^{-\beta x} \) dictate how the distribution behaves as the variable \( x \) changes.
The pdf is a window into the probability landscape of gamma-distributed data, shaping our understanding of event likelihoods and occurrences.
Likelihood Function
The likelihood function is fundamental in statistics, especially for parameter estimation through maximum likelihood estimation (MLE). It represents the probability of observing the given data under specific model parameters. When dealing with a gamma distribution, the likelihood function given data \(X_1, \ldots, X_n\) is:\[ L(\alpha, \beta) = \frac{\beta^{n\alpha}}{(\Gamma(\alpha))^n} \prod_{i=1}^n X_i^{\alpha-1} e^{-\beta \sum X_i} \] Key points:
  • It is derived by replacing the parameters in the pdf with sample data and multiplying across all observations.
  • The goal is to find parameter values \( \alpha \) and \( \beta \) that maximize this function, yielding the most likely explanation for the observed data.
Through the likelihood function, we make inference not just about the data, but also about how different parameter configurations impact our observations.
Sample Mean
The sample mean is an essential concept in statistics and represents the average of a set of observations. In the context of the gamma distribution, it holds particular significance as part of the maximum likelihood estimation process. Given observations \(X_1, X_2, \ldots, X_n\), the sample mean \( \bar{X} \) is:\[ \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \] This quantity serves as an estimator for \( \mu = \alpha \beta \), the mean of the gamma distribution:
  • Easily computed from data, providing direct insight into the sample's central tendency.
  • Acts as the maximum likelihood estimator for \( \mu \) under specific conditions.
The sample mean thus bridges empirical data with theoretical parameters, showing how simple calculations can uncover deeper statistical truths.

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Most popular questions from this chapter

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of \(n\) students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of type I and 50 are of type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a 0,1 , or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let \(p\) denote the proportion of honor- code violators (i.e., the probability of a randomly selected student being a violator), and let \(\lambda=P(\) yes response). Then \(\lambda\) and \(p\) are related by \(\lambda=.5 p+(.5)(.3)\). a. Let \(Y\) denote the number of yes responses, so \(Y \sim\) Bin \((n, \lambda)\). Thus \(Y / n\) is an unbiased estimator of \(\lambda\). Derive an estimator for \(p\) based on \(Y\). If \(n=80\) and \(y=20\), what is your estimate? [Hint: Solve \(\lambda=.5 p+.15\) for \(p\) and then substitute \(Y / n\) for \(\lambda .]\) b. Use the fact that \(E(Y / n)=\lambda\) to show that your estimator \(\hat{p}\) is unbiased. c. If there were 70 type I and 30 type II cards, what would be your estimator for \(p\) ?

At time \(t=0,20\) identical components are tested. The lifetime distribution of each is exponential with parameter \(\lambda\). The experimenter then leaves the test facility unmonitored. On his return 24 hours later, the experimenter immediately terminates the test after noticing that \(y=15\) of the 20 components are still in operation (so 5 have failed). Derive the mle of \(\lambda\).

Suppose the true average growth \(\mu\) of one type of plant during a 1-year period is identical to that of a second type, but the variance of growth for the first type is \(\sigma^{2}\), whereas for the second type the variance is \(4 \sigma^{2}\). Let \(X_{1}, \ldots, X_{m}\) be \(m\) independent growth observations on the first type [so \(\left.E\left(X_{i}\right)=\mu, V\left(X_{i}\right)=\sigma^{2}\right]\), and let \(Y_{1}, \ldots, Y_{n}\) be \(n\) independent growth observations on the second type \(\left[E\left(Y_{i}\right)=\mu\right.\), \(\left.V\left(Y_{i}\right)=4 \sigma^{2}\right]\) a. Show that for any \(\delta\) between 0 and 1, the estimator \(\hat{\mu}=\delta \bar{X}+(1-\delta) \bar{Y}\) is unbiased for \(\mu\). b. For fixed \(m\) and \(n\), compute \(V(\hat{\mu})\), and then find the value of \(\delta\) that minimizes \(V(\hat{\mu})\). [Hint: Differentiate \(V(\hat{\mu})\) with respect to \(\delta\).]

Each of \(n\) specimens is to be weighed twice on the same scale. Let \(X_{i}\) and \(Y_{i}\) denote the two observed weights for the \(i\) th specimen. Suppose \(X_{i}\) and \(Y_{i}\) are independent of one another, each normally distributed with mean value \(\mu_{i}\) (the true weight of specimen \(i\) ) and variance \(\sigma^{2}\). a. Show that the maximum likelihood estimator of \(\sigma^{2}\) is \(\hat{\sigma}^{2}=\sum\left(X_{i}-Y_{i}\right)^{2} /(4 n)\). [Hint: If \(\bar{z}=\left(z_{1}+z_{2}\right) / 2\), then \(\left.\Sigma\left(z_{i}-\bar{z}\right)^{2}=\left(z_{1}-z_{2}\right)^{2} / 2 .\right]\) b. Is the mle \(\hat{\sigma}^{2}\) an unbiased estimator of \(\sigma^{2}\) ? Find an unbiased estimator of \(\sigma^{2}\).

The mean squared error of an estimator \(\hat{\theta}\) is \(\operatorname{MSE}(\hat{\theta})=E(\hat{\theta}-\theta)^{2}\). If \(\hat{\theta}\) is unbiased, then \(\operatorname{MSE}(\hat{\theta})=V(\hat{\theta})\), but in general \(\operatorname{MSE}(\hat{\theta})=V(\hat{\theta})+(\text { bias })^{2}\). Consider the estimator \(\hat{\sigma}^{2}=K S^{2}\), where \(S^{2}=\) sample variance. What value of \(K\) minimizes the mean squared error of this estimator when the population distribution is normal? [Hint: It can be shown that $$ E\left[\left(S^{2}\right)^{2}\right]=(n+1) \sigma^{4} /(n-1) $$ In general, it is difficult to find \(\hat{\theta}\) to minimize \(\operatorname{MSE}(\hat{\theta})\), which is why we look only at unbiased estimators and minimize \(V(\hat{\theta}) .]\)
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