Question

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

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

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.

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.

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.