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Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate normal distribution with \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}, \rho=\frac{1}{2}\), where \(\mu_{1}, \mu_{2}\), and \(\sigma^{2}>0\) are unknown real numbers. Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0, \sigma^{2}\) unknown against all alternatives. The likelihood ratio \(\Lambda\) is a function of what statistic that has a well- known distribution?

Short Answer

Expert verified
The likelihood ratio \(\Lambda\) is based on the likelihoods under the null and alternative hypotheses respectively. The statistic of this ratio used to compare hypotheses follows a chi-squared distribution with 2 degrees of freedom.

Step by step solution

01

Define the Likelihood Ratio Test (LRT) Statistic

The Likelihood Ratio Test (LRT) is often used for hypothesis testing. It is defined as the ratio of the maximum likelihood under the null hypothesis, to the maximum likelihood under the alternative hypothesis. Mathematically, we write it as follows: \[\Lambda=\frac{\sup_{\theta \in \Theta_{0}} L(\theta; x)}{\sup_{\theta \in \Theta} L(\theta; x)}\]where \(\Theta_{0}\) denotes the parameter space under the null hypothesis, and \(\Theta\) denotes the parameter space under the alternative hypothesis. \(L \) denotes the likelihood function.
02

Write down the likelihood function under the Null and Alternative Hypothesis

The likelihood function of a bivariate normal distribution is given by:\[L(\mu_1, \mu_2, \sigma^2; x) = \prod_{i=1}^{n} \frac{1}{2\pi\sigma^2\sqrt{1-\rho^2}}exp\Big\{-\frac{1} {2\sigma^2(1-\rho^2)} [(x_{1i}-\mu_1)^2+ (x_{2i}-\mu_2)^2 - 2\rho(x_{1i}-\mu_1)(x_{2i}-\mu_2)]\Big\}\]Under \(H_0\): \(\mu_1 = \mu_2 = 0\), so the likelihood function simplifies to:\[L(0,0, \sigma^2; x) = \prod_{i=1}^{n} \frac{1}{2\pi\sigma^2\sqrt{1-\rho^2}}exp\Big\{-\frac{1} {2\sigma^2(1-\rho^2)} [x_{1i}^2+ x_{2i}^2 ]\Big\}\]
03

Compute the Ratio

After we have found the likelihoods under the null and alternative hypotheses, we can now compute the ratio \(\Lambda\). By doing this, we would have completed our calculation for the likelihood ratio.
04

Compute the Statistic Value

The statistic value of the ratio will be a measure of how much the data under the null hypothesis differ from the data under the alternative hypothesis. If \(\Lambda\) is close to 1, there is not much evidence against \(H_0\). If \(\Lambda\) is close to 0, there is strong evidence against \(H_0\). This statistic has a well-known chi-squared distribution with 2 degrees of freedom.

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Most popular questions from this chapter

Let \(X\) have the pmf \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), zero elsewhere. \(W_{\text {e }}\) test the simple hypothesis \(H_{0}: \theta=\frac{1}{4}\) against the alternative composite hypothesis \(H_{1}: \theta<\frac{1}{4}\) by taking a random sample of size 10 and rejecting \(H_{0}: \theta=\frac{1}{4}\) if aulul only if the observed values \(x_{1}, x_{2}, \ldots, x_{10}\) of the sample observations are such that \(\sum_{1}^{10} x_{i} \leq 1 .\) Find the power function \(\gamma(\theta), 0<\theta \leq \frac{1}{4}\), of this test.

Consider a distribution having a pmf of the form \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=\) 0,1, zero elsewhere. Let \(H_{0}: \theta=\frac{1}{20}\) and \(H_{1}: \theta>\frac{1}{20} .\) Use the central limit theorcu? to determine the sample size \(n\) of a random sample so that a uniformly most powerful test of \(H_{0}\) against \(H_{1}\) has a power function \(\gamma(\theta)\), with approximately \(\gamma\left(\frac{1}{20}\right)=0.05\) and \(\gamma\left(\frac{1}{10}\right)=0.90\).

. Let \(X\) and \(Y\) have a joint bivariate normal distribution. An observation \((x, y)\) arises from the joint distribution with parameters equal to either $$\mu_{1}^{\prime}=\mu_{2}^{\prime}=0,\quad\left(\sigma_{1}^{2}\right)^{\prime}=\left(\sigma_{2}^{2}\right)^{\prime}=1, \quad \rho^{\prime}=\frac{1}{2}$$ or $$\mu_{1}^{\prime \prime}=\mu_{2}^{\prime \prime}=1,\left(\sigma_{1}^{2}\right)^{\prime \prime}=4, \quad\left(\sigma_{2}^{2}\right)^{\prime \prime}=9, \rho^{\prime \prime}=\frac{1}{2}$$ Show that the classification rule involves a second-degree polynomial in \(x\) and \(y\).

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample of size 10 from a normal distribution \(N\left(0, \sigma^{2}\right) .\) Find a best critical region of size \(\alpha=0.05\) for testing \(H_{0}: \sigma^{2}=1\) against \(H_{1}: \sigma^{2}=2 .\) Is this a best critical region of size \(0.05\) for testing \(H_{0}: \sigma^{2}=1\) against \(H_{1}: \sigma^{2}=4 ?\) Against \(H_{1}: \sigma^{2}=\sigma_{1}^{2}>1 ?\)

Let the random variable \(X\) have the pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

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