Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the beta distribution with \(\alpha=\beta=\theta\) and \(\Omega=\\{\theta: \theta=1,2\\} .\) Show that the likelihood ratio test statistic \(\Lambda\) for testing \(H_{0}: \theta=1\) versus \(H_{1}: \theta=2\) is a function of the statistic \(W=\) \(\sum_{i=1}^{n} \log X_{i}+\sum_{i=1}^{n} \log \left(1-X_{i}\right)\)

Short answer

The likelihood ratio test statistic \(\Lambda\) can indeed be expressed as a function of the statistic \(W\). This is done by first finding the likelihood function under the null and alternative hypothesis, then calculating their ratio to get the likelihood ratio test statistic, and finally expressing the likelihood ratio test statistic in terms of \(W\).

Step-by-step solution

Construct the likelihood function under \(H_{0}\) and \(H_{1}\)

The likelihood function under \(H_{0}\) (i.e., when \(θ=1\)) and under \(H_{1}\) (i.e., when \(θ=2\)) can be constructed from the probability density function (PDF) of Beta distribution, which is \((1/B(θ,θ))(x^{θ-1}(1 - x)^{θ-1})\). Now, for \(θ=1\), the likelihood function \(L_{H_{0}}\) would be \(\prod_{i=1}^{n} 1/B(1,1)\) and for \(θ=2\), the likelihood function \(L_{H_{1}}\) would be \(\prod_{i=1}^{n} (1/B(2,2))(x_{i}(1 - x_{i})\)

Compute the likelihood ratio

The likelihood ratio \(\Lambda\) is defined as the ratio of the likelihood under \(H_{0}\) to the likelihood under \(H_{1}\), i.e., \(\Lambda = L_{H_{0}}/L_{H_{1}} = (\prod_{i=1}^{n} 1/B(1,1))/(\prod_{i=1}^{n} (1/B(2,2))(x_{i}(1 - x_{i}))\). This simplifies to \(\Lambda = (B(2,2)/B(1,1))^{n}(\prod_{i=1}^{n}x_{i}(1 - x_{i}))^{-1}\)

Formulate the likelihood ratio as a function of statistic \(W\)

In this step, we need to equate \(\Lambda\) to statistic \(W\). Note that \(W = \sum_{i=1}^{n} \log x_{i} + \sum_{i=1}^{n} \log(1-x_{i}) = -\log(\prod_{i=1}^{n}x_{i}(1 - x_{i})\). We can rewrite \(\Lambda\) in terms of \(W\) by observing that \(W\) appears in the denominator of \(\Lambda\) in log form. Therefore, we have: \(log(\Lambda) = n \log(B(2,2)/B(1,1)) - \log(\prod_{i=1}^{n} x_{i}(1 - x_{i})) = n \log(B(2,2)/B(1,1)) - W\), thus \(\Lambda\) is a function of \(W\)