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Chapter 6: Problem 13

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?

Short Answer

Expert verified
To achieve the tasks, one would load the data and plot a histogram to visualize the data. The MLE for the parameter of the beta distribution would be calculated, which estimates the parameter from the given data. After that, a 95% confidence interval would be computed to provide a range of likely values for the parameter. The result would then be assessed to determine whether the confidence interval is successful or not.

Step by step solution

01

Loading Data and Obtaining Histogram

Load the data set 'beta30.rda'. Using either a built-in function or custom function, obtain a histogram of the data, with frequency density on the y-axis.This can be done with the hist() function in R, with pr=T argument to get the density plot. Overlay the pdf of a beta (4,1) distribution using lines() function with the density function dbeta() to show the beta distribution.
02

Computing Maximum Likelihood Estimate (MLE)

Using the results of exercise 6.2 .12, calculate the maximum likelihood estimate. MLE is a method of estimating the parameters of a distribution by maximizing a likelihood function. The formula used for maximum likelihood estimation for a beta distribution will be \(\theta=n( \frac{1}{ \bar{x}}-1)\), where n is the number of observations and \(\bar{x}\) is the mean of observations.
03

Computing Confidence Interval

Calculate the 95% confidence interval for \(\theta\). Confidence intervals give a range in which the parameter will lie with a certain probability. It gives an estimated range of values which is likely to include an unknown population parameter. The formula from Exercise 6.2.12 will be used for this calculation, and determining whether the interval contains the true value of the parameter \(\theta =4\). If it does, then it is successful.

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Most popular questions from this chapter

Consider a location model $$ X_{i}=\theta+e_{i}, \quad i=1, \ldots, n $$ where \(e_{1}, e_{2}, \ldots, e_{n}\) are iid with pdf \(f(z)\). There is a nice geometric interpretation for estimating \(\theta\). Let \(\mathbf{X}=\left(X_{1}, \ldots, X_{n}\right)^{\prime}\) and \(\mathbf{e}=\left(e_{1}, \ldots, e_{n}\right)^{\prime}\) be the vectors of observations and random error, respectively, and let \(\boldsymbol{\mu}=\theta \mathbf{1}\), where \(\mathbf{1}\) is a vector with all components equal to 1 . Let \(V\) be the subspace of vectors of the form \(\mu\); i.e., \(V=\\{\mathbf{v}: \mathbf{v}=a \mathbf{1}\), for some \(a \in R\\}\). Then in vector notation we can write the model as $$ \mathbf{X}=\boldsymbol{\mu}+\mathbf{e}, \quad \boldsymbol{\mu} \in V $$ Then we can summarize the model by saying, "Except for the random error vector e, X would reside in \(V\)." Hence, it makes sense intuitively to estimate \(\boldsymbol{\mu}\) by a vector in \(V\) that is "closest" to \(\mathbf{X}\). That is, given a norm \(\|\cdot\|\) in \(R^{n}\), choose $$ \widehat{\boldsymbol{\mu}}=\operatorname{Argmin}\|\mathbf{X}-\mathbf{v}\|, \quad \mathbf{v} \in V $$ (a) If the error pdf is the Laplace, \((2.2 .4)\), show that the minimization in \((6.3 .27)\) is equivalent to maximizing the likelihood when the norm is the \(l_{1}\) norm given by $$ \|\mathbf{v}\|_{1}=\sum_{i=1}^{n}\left|v_{i}\right| $$ (b) If the error pdf is the \(N(0,1)\), show that the minimization in \((6.3 .27)\) is equivalent to maximizing the likelihood when the norm is given by the square of the \(l_{2}\) norm $$ \|\mathbf{v}\|_{2}^{2}=\sum_{i=1}^{n} v_{i}^{2} $$

Suppose the pdf of \(X\) is of a location and scale family as defined in Example 6.4.4. Show that if \(f(z)=f(-z)\), then the entry \(I_{12}\) of the information matrix is 0 . Then argue that in this case the mles of \(a\) and \(b\) are asymptotically independent.

Given \(f(x ; \theta)=1 / \theta, 00\), formally compute the reciprocal of $$ n E\left\\{\left[\frac{\partial \log f(X: \theta)}{\partial \theta}\right]^{2}\right\\} $$ Compare this with the variance of \((n+1) Y_{n} / n\), where \(Y_{n}\) is the largest observation of a random sample of size \(n\) from this distribution. Comment.

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 \(\widehat{\theta}=\max \\{0, \bar{X}\\}\).

Let \(X\) be \(N(0, \theta), 0<\theta<\infty\). (a) Find the Fisher information \(I(\theta)\). (b) If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from this distribution, show that the mle of \(\theta\) is an efficient estimator of \(\theta\). (c) What is the asymptotic distribution of \(\sqrt{n}(\widehat{\theta}-\theta) ?\)
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