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Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solution yields 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
MLE equations can be derived but not solved explicitly; \(\hat{\mu} = \bar{X}\).

Step by step solution

01

Understanding the Gamma Distribution

The probability density function for a gamma distribution with parameters \(\alpha\) and \(\beta\) is given by \( f(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \), where \(x > 0\), \(\alpha > 0\) and \(\beta > 0\). We need to find MLE for \(\alpha\) and \(\beta\).
02

Form the Likelihood Function

The likelihood function for the given sample \(X_1, \ldots, X_n\) is \( L(\alpha, \beta) = \prod_{i=1}^{n} \frac{X_i^{\alpha-1} e^{-X_i/\beta}}{\beta^\alpha \Gamma(\alpha)} \).
03

Log-likelihood Function

Taking natural logarithm of the likelihood function, we get the log-likelihood: \( \ell(\alpha, \beta) = (\alpha - 1) \sum_{i=1}^{n} \ln X_i - \frac{1}{\beta} \sum_{i=1}^{n} X_i - n(\alpha \ln \beta + \ln \Gamma(\alpha)) \).
04

Partial Derivatives and Equations

Finding the partial derivatives \(\frac{\partial \ell}{\partial \alpha}\) and \(\frac{\partial \ell}{\partial \beta}\) and setting them to zero gives the equations that the MLEs of \(\alpha\) and \(\beta\) must satisfy. These equations involve terms such as \(\psi(\alpha)\), the digamma function, hence making them difficult to solve explicitly. Thus, they cannot be solved explicitly and require numerical methods.
05

Parameter Relation: \(\mu = \alpha \beta\)

We recognize that \(\mu = \alpha \beta\) is the mean of the gamma distribution. The sample mean \(\bar{X}\) is an unbiased estimator for \(\mu\).
06

MLE of \(\mu\) is \(\hat{\mu} = \bar{X}\)

Since \(\hat{\mu} = \alpha \beta\) and \(\mu = \alpha \beta\), and knowing that the sample mean is a sufficient statistic for the mean of a distribution, the maximum likelihood estimator for \(\mu\) is \(\bar{X}\).

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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 continuous probability distribution that is commonly used to model the time until an event occurs, such as the wait time for a bus or the lifespan of a piece of machinery. It is characterized by two parameters: \(\alpha\) (shape) and \(\beta\) (scale). These parameters help shape the distribution, determining its skewness and the spread of the data it models.
The probability density function (PDF) of the gamma distribution is given by:
  • \( f(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \)
  • where \( x > 0 \), \( \alpha > 0 \), and \( \beta > 0 \)
This equation defines the likelihood of a random variable \(x\) for given \(\alpha\) and \(\beta\). The function \(\Gamma(\alpha)\) is the gamma function, which extends the factorial function to non-integer values and is essential in calculating the PDF.Understanding the gamma distribution is fundamental when dealing with data that aligns with exponential types of processes, as it helps in accurately modeling and making statistical inferences based on such data.
Probability Density Function
In statistics, the probability density function (PDF) is a vital concept used to specify the probability of a random variable falling within a particular range of values as opposed to taking on any specific value. For continuous random variables, the PDF is essential as it describes the likelihood of the variable appearing at any given point.
For the gamma distribution, its specific PDF is expressed as:
  • \( f(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \)
This formula intricately links the random variable \(x\) with the shape parameter \(\alpha\) and the scale parameter \(\beta\). Here, the term \(x^{\alpha-1}\) indicates how steeply the curve rises initially. The exponential part \(e^{-x/\beta}\) ensures the curve declines, ensuring the overall shape typical of a gamma distribution.
The PDF is used for deriving other important statistical measures, such as moments and maximum likelihood estimators, crucial when parameterizing distributions in practical applications.
Log-likelihood Function
The log-likelihood function is a pivotal tool in maximum likelihood estimation (MLE), a commonly used method for estimating the parameters of a statistical model. The MLE seeks parameter values that maximize the likelihood that the observed data would occur under the model.
For the gamma distribution with sample \(X_1, \ldots, X_n\), the likelihood function \(L(\alpha, \beta)\) is expressed as a product of individual probabilities, which is often cumbersome to work with. By taking the natural logarithm, we transform this product into a summation, simplifying calculations greatly. Thus, the log-likelihood function \(\ell(\alpha, \beta)\) becomes:
  • \( \ell(\alpha, \beta) = (\alpha - 1) \sum_{i=1}^{n} \ln X_i - \frac{1}{\beta} \sum_{i=1}^{n} X_i - n(\alpha \ln \beta + \ln \Gamma(\alpha)) \)
The log-likelihood helps in deriving equations by differentiating with respect to \(\alpha\) and \(\beta\), whose solutions yield the MLEs. Direct solutions may not be available due to the complex nature of these functions but numerical methods complete the estimation process.
Unbiased Estimator
An unbiased estimator is a statistical term used to describe an estimator that, on average, hits the true parameter value it is estimating. In simpler terms, if we repeatedly sample from the population and use our estimator, the expected value of our estimator will equal the actual population parameter.
In the context of the gamma distribution, we focus on the parameter \(\mu = \alpha \beta\), known as the mean. Importantly, the sample mean \(\bar{X}\) serves as an unbiased estimator of \(\mu\). This means that across all possible samples, the average value of \(\bar{X}\) equals \(\mu\).
Unbiased estimators are crucial in statistical estimation as they ensure that we are not systematically over or underestimating parameters, providing accuracy and reliability in our inferential procedures. This property is particularly important for applications where predictions based on estimated parameters can significantly impact decision-making processes.

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Most popular questions from this chapter

Assume that the number of defects in a car has a Poisson distribution with parameter \(\lambda\). To estimate \(\lambda\) we obtain the random sample \(X_{1}\), \(X_{2}, \ldots, X_{n}\). a. Find the Fisher information in a single observation using two methods. b. Find the Cramér-Rao lower bound for the variance of an unbiased estimator of \(\lambda\). c. Use the score function to find the mle of \(\lambda\) and show that the mle is an efficient estimator. d. Is the asymptotic distribution of the mle in accord with the second theorem? Explain.

A random sample of \(n\) bike helmets manufactured by a company is selected. Let \(X=\) the number among the \(n\) that are flawed, and let \(p=P\) (flawed). Assume that only \(X\) is observed, rather than the sequence of \(S\) 's and \(F\) 's. 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 helmets examined is flawed?

Let \(X_{1}, \ldots, X_{n}\) be a random sample of component lifetimes from an exponential distribution with parameter \(\lambda\). Use the factorization theorem to show that \(\sum X_{i}\) is a sufficient statistic for \(\lambda\).

Two different computer systems are monitored for a total of \(n\) weeks. Let \(X_{i}\) denote the number of breakdowns of the first system during the \(i\) th week, and suppose the \(X_{i}\) 's are independent and drawn from a Poisson distribution with parameter \(\lambda_{1}\). Similarly, let \(Y_{i}\) denote the number of breakdowns of the second system during the \(i\) th week, and assume independence with each \(Y_{i}\) Poisson with parameter \(\lambda_{2}\). Derive the mle's of \(\lambda_{1}, \lambda_{2}\), and \(\lambda_{1}-\lambda_{2}\). [Hint: Using independence, write the joint \(\mathrm{pmf}\) (likelihood) of the \(X_{i}\) 's and \(Y_{i}\) 's together.]

Let \(X\), the payoff from playing a certain game, have pmf $$ f(x ; \theta)=\left\\{\begin{array}{cc} \theta & x=-1 \\ (1-\theta)^{2} \theta^{x} & x=0,1,2, \ldots \end{array}\right. $$ a. Verify that \(f(x ; \theta)\) is a legitimate pmf, and determine the expected payoff. [Hint: Look back at the properties of a geometric random variable discussed in Chapter 3.] b. Let \(X_{1}, \ldots, X_{n}\) be the payoffs from \(n\) independent games of this type. Determine the mle of \(\theta\). [Hint: Let \(Y\) denote the number of observations among the \(n\) that equal \(-1\) that is, \(Y=\Sigma I\left(Y_{i}=-1\right)\), where \(I(A)=1\) if the event \(A\) occurs and 0 otherwise \(\\}\), and write the likelihood as a single expression in terms of \(\sum x_{i}\) and \(y\).] c. What is the approximate variance of the mle when \(n\) is large?

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