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Consider the two uniform distributions with respective pdfs $$ f\left(x ; \theta_{i}\right)=\left\\{\begin{array}{ll} \frac{1}{2 \theta_{i}} & -\theta_{i}

Short Answer

Expert verified
In this problem, the likelihood ratio statistic \(\Lambda\) used to test \(H0\) against \(H1\) is found to be 1. The distribution of \(-2 \log \Lambda\) when \(H0\) is true is 0. The degrees of freedom is 2.

Step by step solution

01

Identifying the Likelihood Function

For a random sample, the joint pdf is the product of the individual pdfs. The likelihood function \(L(\theta_1, \theta_2)\) is the joint pdf of the two samples viewed as a function of \(\theta_1\) and \(\theta_2\). Since both samples have the same number \(n_1 = n_2 = n\) for ease of the mathematics, the likelihood for \(H0:\theta_1 = \theta_2 =\theta\) is \(L(\theta) = 1/(2\theta)^{2n}\) if \(-\theta < X_1, Y_1 < \theta \) and 0 otherwise. While under the \(H1:\theta_1\neq \theta_2\), \(L(\theta_1, \theta_2) = 1/(2\theta_1)^n (2\theta_2)^n, -\theta_1 < X_1 < \theta_1, -\theta_2 < Y_1 < \theta_2\) and 0 otherwise.
02

Constructing the Likelihood Ratio

The likelihood ratio \(\Lambda\) is the ratio of maximum of the likelihood under \(H0\) to the maximum of the likelihood under \(H1\). Here, \(\Lambda = L(\theta)/L(\theta_1, \theta_2)\) where \(\theta\) maximizes the likelihood under \(H0\) and \(\theta_1\) and \(\theta_2\) maximize the likelihood under \(H1\). Maximizing the likelihood functions gives the maximum likelihood estimates \(\hat{\theta} = max{X_n, Y_n}\), \(\hat{\theta_1} = X_n, \hat{\theta_2} = Y_n\), substituting these estimates back into \(\Lambda\) gives \(\Lambda = (max{X_n, Y_n}/max{X_n, Y_n})^{2n} = 1.\)
03

Finding the distribution of \(-2 \log \Lambda\)

The distribution of \(-2 \log \Lambda\) when \(H0\) is true, where \(\Lambda = 1\), using the fact that \(\log (1) = 0\) and thus \(-2 \log \Lambda = 0\). This is a well-established result in statistics known as Wilk's theorem.
04

Understanding degrees of freedom.

Having found the likelihood ratio statistics, we need to find the distribution under Null hypothesis. Here, the dimensions of \(\Omega\) and \(\omega\) refers to the number of parameters to be estimated under the alternate and null hypotheses respectively. For this case, under \(H0\) there is 1 parameter, \(\theta\), to be estimated. But under \(H1\), there are 2 independent parameters, \(\theta_1 and \theta_2\), to be estimated. Thus, the number of degrees of freedom is 2 times the difference, which equals \(2*(2 - 1) = 2\).

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Most popular questions from this chapter

Let the table $$ \begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & 6 & 10 & 14 & 13 & 6 & 1 \end{array} $$ represent a summary of a sample of size 50 from a binomial distribution having \(n=5\). Find the mle of \(P(X \geq 3)\). For the data in the table, using the \(\mathrm{R}\) function pbinom determine the realization of the mle.

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?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with pmf \(p(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), where \(0<\theta<1 .\) We wish to test \(H_{0}: \theta=1 / 3\) versus \(H_{1}: \theta \neq 1 / 3\). (a) Find \(\Lambda\) and \(-2 \log \Lambda\). (b) Determine the Wald-type test. (c) What is Rao's score statistic?

Rao (page 368,1973 ) considers a problem in the estimation of linkages in genetics. McLachlan and Krishnan (1997) also discuss this problem and we present their model. For our purposes, it can be described as a multinomial model with the four categories \(C_{1}, C_{2}, C_{3}\), and \(C_{4}\). For a sample of size \(n\), let \(\mathbf{X}=\left(X_{1}, X_{2}, X_{3}, X_{4}\right)^{\prime}\) denote the observed frequencies of the four categories. Hence, \(n=\sum_{i=1}^{4} X_{i} .\) The probability model is $$ \begin{array}{|c|c|c|c|} \hline C_{1} & C_{2} & C_{3} & C_{4} \\ \hline \frac{1}{2}+\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4} \theta \\ \hline \end{array} $$ where the parameter \(\theta\) satisfies \(0 \leq \theta \leq 1\). In this exercise, we obtain the mle of \(\theta\). (a) Show that likelihood function is given by $$ L(\theta \mid \mathbf{x})=\frac{n !}{x_{1} ! x_{2} ! x_{3} ! x_{4} !}\left[\frac{1}{2}+\frac{1}{4} \theta\right]^{x_{1}}\left[\frac{1}{4}-\frac{1}{4} \theta\right]^{x_{2}+x_{3}}\left[\frac{1}{4} \theta\right]^{x_{4}} $$ (b) Show that the log of the likelihood function can be expressed as a constant (not involving parameters) plus the term $$ x_{1} \log [2+\theta]+\left[x_{2}+x_{3}\right] \log [1-\theta]+x_{4} \log \theta $$ (c) Obtain the partial derivative with respect to \(\theta\) of the last expression, set the result to 0 , and solve for the mle. (This will result in a quadratic equation that has one positive and one negative root.)

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 \(\theta_{3}\).

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