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Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with pdf \(f(x ; \theta)=\theta(1-x)^{\theta-1}, 00\) (a) Find the form of the uniformly most powerful test of \(H_{0}: \theta=1\) against \(H_{1}: \theta>1\) (b) What is the likelihood ratio \(\Lambda\) for testing \(H_{0}: \theta=1\) against \(H_{1}: \theta \neq 1 ?\)

Short Answer

Expert verified
For part (a), the rule will have the form \(Reject H_0: \theta =1 if \frac{1-n}{s} > k\), where \(s\) is the sample mean. The likelihood ratio for part (b) would be \(\Lambda = \frac{1}{(\theta(1-x)^{\theta-1})}\).

Step by step solution

01

Writing Down the Likelihood Function

The likelihood function can be given as \(L(\theta; x) = \prod_{i=1}^{n} \theta(1-x_i)^{\theta-1}\). Notice that likelihood function is a function of \(\theta\) given the observed sample.
02

Deriving the Uniformly Most Powerful Test

The Neyman-Pearson lemma states that the most powerful test for testing \(H_0: \theta = \theta_0\) against \(H_1: \theta = \theta_1\) has a critical region defined by \{x : \frac{f(x|\theta_1)}{f(x|\theta_0)} > k \} where \(k\) a constant depending on the size of the test. Applying this to our hypotheses, the rejection rule becomes \(\frac{(1-x)^0}{(1-x)} > k\). Solve for \(k\), this gives us the uniformly most powerful test.
03

Calculating the Likelihood Ratio

For the likelihood ratio test, we need to compute the ratio \( \Lambda = \frac{L(\theta_0)}{L(\theta_1)} \). This can be computed by substituting \(\theta = 1\) and \(\theta \neq 1\) in the derived likelihood function.

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Most popular questions from this chapter

Illustrative Example \(8.2 .1\) of this section dealt with a random sample of size \(n=2\) from a gamma distribution with \(\alpha=1, \beta=\theta .\) Thus the mgf of the distribution is \((1-\theta t)^{-1}, t<1 / \theta, \theta \geq 2 .\) Let \(Z=X_{1}+X_{2}\). Show that \(Z\) has a gamma distribution with \(\alpha=2, \beta=\theta .\) Express the power function \(\gamma(\theta)\) of Example \(8.2 .1\) in terms of a single integral. Generalize this for a random sample of size \(n .\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution \(N(\theta, 16)\). Find the sample size \(n\) and a uniformly most powerful test of \(H_{0}: \theta=25\) against \(H_{1}: \theta<25\) with power function \(\gamma(\theta)\) so that approximately \(\gamma(25)=0.10\) and \(\gamma(23)=0.90\).

Let \(X_{1}, X_{2}, \ldots, X_{25}\) denote a random sample of size 25 from a normal distribution \(N(\theta, 100)\). Find a uniformly most powerful critical region of size \(\alpha=0.10\) for testing \(H_{0}: \theta=75\) against \(H_{1}: \theta>75\)

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.

Let the independent random variables \(Y\) and \(Z\) be \(N\left(\mu_{1}, 1\right)\) and \(N\left(\mu_{2}, 1\right)\), respectively. Let \(\theta=\mu_{1}-\mu_{2}\). Let us observe independent observations from each distribution, say \(Y_{1}, Y_{2}, \ldots\) and \(Z_{1}, Z_{2}, \ldots\) To test sequentially the hypothesis \(H_{0}: \theta=0\) against \(H_{1}: \theta=\frac{1}{2}\), use the sequence \(X_{i}=Y_{i}-Z_{i}, i=1,2, \ldots\) If \(\alpha_{a}=\beta_{a}=0.05\), show that the test can be based upon \(\bar{X}=\bar{Y}-\bar{Z} .\) Find \(c_{0}(n)\) and \(c_{1}(n)\)

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