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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N\left(\theta, \sigma^{2}\right)\) distribution, where \(\sigma^{2}\) is fixed but \(-\infty<\theta<\infty\) (a) Show that the mle of \(\theta\) is \(\bar{X}\). (b) If \(\theta\) is restricted by \(0 \leq \theta<\infty\), show that the mle of \(\theta\) is \(\widehat{\theta}=\max \\{0, \bar{X}\\}\).

Short Answer

Expert verified
For unrestricted \(\theta\), the MLE is \(\bar{X}\). When \(\theta\) is restricted to non-negative values, the MLE is \(\max\{0, \bar{X}\}\).

Step by step solution

01

Write the Likelihood Function

The likelihood function for a set of observations \(X_{1}, X_{2}, \ldots, X_{n}\) from a normal distribution \(N(\theta, \sigma^2)\) is given by: \[ L(\theta) = \prod _{i=1} ^n \frac{1}{\sqrt{2\pi}\sigma} e^{-(X_{i}-\theta)^2 / (2\sigma^2)} \]
02

Logarithm of the Likelihood Function

Taking the natural logarithm of the likelihood function simplifies the task of finding the maximum of the function. The log-likelihood function is given by: \[ l(\theta) = \ln(L(\theta)) = -\frac{n}{2} \ln(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^{n} (X_i - \theta)^2 \]
03

Derive the MLE for Unrestricted θ

Take the derivative of \(l(\theta)\) with respect to \(\theta\) and set equal to zero, to locate the maximum of the function. Solve for θ: \[ \frac{d}{d\theta} l(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^{n} (X_i - \theta) = 0 \] Solving for \(\theta\) gives the MLE for \(\theta\) as \(\bar{X}\), the sample mean.
04

Derive the MLE for Restricted θ

When \(\theta\) is restricted to non-negative values, the MLE is found by maximizing the likelihood over the interval [0, ∞). Therefore, the MLE will be the maximum of 0 and \(\bar{X}\) - i.e., if \(\bar{X} < 0\), then the MLE of \(\theta\) will be 0, otherwise it is \(\bar{X}\). Therefore, \(\widehat{\theta} = \max\{0, \bar{X} \}\).

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Key Concepts

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

Likelihood Function
The likelihood function is a fundamental concept in statistical inference, especially in the context of maximum likelihood estimation (MLE). It represents the probability of observing the sample data given specific values of the parameters of the model. In a more technical sense, for a set of independent and identically distributed observations, the likelihood function is the product of the probability density functions (PDFs) for each observed value.

When dealing with continuous distributions such as the normal distribution, the likelihood function for observing a particular sample \(X_1, X_2, ..., X_n\) given a parameter \(\theta\) is expressed as \(L(\theta) = \prod_{i=1}^{n} f(X_i|\theta)\), where \(f(x|\theta)\) is the PDF of the distribution. To compute the MLE, one needs to find the parameter value \(\theta\) that maximizes this likelihood function. The act of maximizing the likelihood function indirectly maximizes the chances of observing our sample, which makes the estimated parameter more plausible.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. It is characterized by its mean \(\mu\) and variance \(\sigma^2\), and its PDF is defined as \(f(x|\mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-(x-\mu)^2 / (2\sigma^2)}\).

The normal distribution is widely used in statistics due to the Central Limit Theorem, which states that the sum of a large number of independent and identically distributed random variables will be approximately normally distributed, regardless of the original distribution of the variables. This property is particularly useful when dealing with sample means and inference about population parameters.
Sample Mean
The sample mean, denoted as \(\bar{X}\), is the arithmetic average of a set of sample values. It is computed by summing up all the observed values in a sample and then dividing by the number of observations \(n\): \(\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i\).

The sample mean is of paramount importance when it comes to estimating parameters of a population, especially the population mean. In the context of MLE, when the distribution under consideration is the normal distribution, the sample mean becomes the MLE of the population mean \(\mu\) under certain conditions, particularly when the variance \(\sigma^2\) is known and the mean \(\mu\) is the parameter to be estimated. The sample mean serves as a central concept in both descriptive and inferential statistics due to its property of being the best unbiased estimate of the population mean when the samples are randomly drawn.
Log-Likelihood Function
The log-likelihood function is simply the natural logarithm transformation of the likelihood function and is widely used for ease of computation and better numerical stability. For the normal distribution, the logarithm transformation turns the product of exponentials into a sum, which makes differentiation with respect to the parameter \(\theta\) simpler and more manageable.

The general form of the log-likelihood function for a sample from a normal distribution is given by \(l(\theta) = \ln(L(\theta))\), which, after applying the logarithm, becomes a sum of the logged PDFs. When deriving the MLE, we take the derivative of this function with respect to \(\theta\) and set it equal to zero to find the maximum. This process is known as maximization of the log-likelihood, and it often yields the same estimates as the original likelihood function, with the added benefits of simpler algebraic manipulation and a reduced risk of computational underflow or overflow.

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Most popular questions from this chapter

A survey is taken of the citizens in a city as to whether or not they support the zoning plan that the city council is considering. The responses are: Yes, No, Indifferent, and Otherwise. Let \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\) denote the respective true probabilities of these responses. The results of the survey are: $$ \begin{array}{|c|c|c|c|} \hline \text { Yes } & \text { No } & \text { Indifferent } & \text { Otherwise } \\ \hline 60 & 45 & 70 & 25 \\ \hline \end{array} $$ (a) Obtain the mles of \(p_{i}, i=1, \ldots, 4\). (b) Obtain \(95 \%\) confidence intervals, \((4.2 .7)\), for \(p_{i}, i=1, \ldots, 4\).

Given the pdf $$ f(x ; \theta)=\frac{1}{\pi\left[1+(x-\theta)^{2}\right]}, \quad-\infty

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from \(N\left(\theta_{1}, \theta_{3}\right)\) and \(N\left(\theta_{2}, \theta_{4}\right)\) distributions, respectively. (a) If \(\Omega \subset R^{3}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{2}, \theta_{3}\right):-\infty<\theta_{i}<\infty, i=1,2 ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}, \theta_{2}\), and \(\theta_{3}\). (b) If \(\Omega \subset R^{2}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{3}\right):-\infty<\theta_{1}=\theta_{2}<\infty ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}\) and \(\theta_{3}\).

The data file beta30. rda contains 30 observations generated from a beta \((\theta, 1)\) distribution, where \(\theta=4\). The file can be downloaded at the site discussed in the Preface. (a) Obtain a histogram of the data using the argument \(\mathrm{pr}=\mathrm{T}\). Overlay the pdf of a \(\beta(4,1)\) pdf. Comment. (b) Using the results of Exercise \(6.2 .12\), compute the maximum likelihood estimate based on the data. (c) Using the confidence interval found in Part (c) of Exercise 6.2.12, compute the \(95 \%\) confidence interval for \(\theta\) based on the data. Is the confidence interval successful?

Prove that \(\bar{X}\), the mean of a random sample of size \(n\) from a distribution that is \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\), is, for every known \(\sigma^{2}>0\), an efficient estimator of \(\theta\).

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