Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N\left(\mu_{0}, \sigma^{2}=\theta\right)\) distribution, where \(0<\theta<\infty\) and \(\mu_{0}\) is known. Show that the likelihood ratio test of \(H_{0}: \theta=\theta_{0}\) versus \(H_{1}: \theta \neq \theta_{0}\) can be based upon the statistic \(W=\sum_{i=1}^{n}\left(X_{i}-\mu_{0}\right)^{2} / \theta_{0}\). Determine the null distribution of \(W\) and give, explicitly, the rejection rule for a level \(\alpha\) test.

Short answer

The likelihood ratio test can be based on \(W = \sum_{i=1}^{n}(X_{i}-\mu_{0})^{2}/\theta_{0}\). The null distribution of \(W\) is \(\chi_{2}(n)\). For a level \(\alpha\) test, \(H_{0}\) should be rejected if either \(W > \chi_{2}^{2}(1 - \alpha/2)\) or \(W < \chi_{2}^{2}(\alpha/2)\).

Step-by-step solution

- Proof for the Statistic

Firstly, the likelihood function for this normal distribution with known \(\mu_0\) and unknown \(\sigma^2\) is \(\frac{1}{\sqrt{2\pi \theta}}\exp\left(-\frac{1}{2\theta}\sum_{i=1}^{n}\left(X_{i}-\mu_{0}\right)^{2}\right)\). For the hypothesis \(H_{0}: \theta =\theta_{0}\) and Alternative hypothesis \(H_{1}: \theta \neq \theta_{0}\), the likelihood ratio is the maximum likelihood of \(H_{1}\) divided by the likelihood of \(H_{0}\), which equates to \(\exp\left(-\frac{1}{2}\left(\frac{\sum_{i=1}^{n}(X_{i}-\mu_{0})^{2}}{\hat{\theta}}-\frac{\sum_{i=1}^{n}(X_{i}-\mu_{0})^{2}}{\theta_{0}}\right)\). It's easier to make decisions based on \(W\) rather than the exponentiated version; hence, \(W = \sum_{i=1}^{n}(X_{i}-\mu_{0})^{2}/\theta_{0}\) is derived from this.

- Determining the Null Distribution

Given the null hypothesis \(H_{0}: \theta =\theta_{0}\), the null distribution of \(W\) is then a chi-square distribution with \(n\) degree of freedom; hence \(W \sim \chi^{2}(n)\). This is due to the property of a standard normal distribution that the sum of squares of standard normally distributed variables follows a chi-square distribution.

- Setting Up the Rejection Rule

For a level \(\alpha\) test, the rule is set up as rejecting \(H_{0}\) if either \(W > \chi_{2}^{2}(1 - \alpha/2)\) on the right side or if \(W < \chi_{2}^{2}(\alpha/2)\) on the left side because the likelihood ratio test for this context is a two-tailed test. Here, \(\chi_{2}^{2}(1 - \alpha/2)\) and \(\chi_{2}^{2}(\alpha/2)\) are the \((1 - \alpha/2)\)th and \(\alpha/2\)th percentiles of the chi-square distribution with \(n\) degrees of freedom respectively.