Question

Let \(X\) be \(N(0, \theta)\) and, in the notation of this section, let \(\theta^{\prime}=4, \theta^{\prime \prime}=9\), \(\alpha_{a}=0.05\), and \(\beta_{a}=0.10 .\) Show that the sequential probability ratio test can be based upon the statistic \(\sum_{1}^{n} X_{i}^{2} .\) Determine \(c_{0}(n)\) and \(c_{1}(n)\).

Short answer

For the sequential probability ratio test based on the statistic \(\sum_{1}^{n} X_{i}^{2}\), the decision boundaries are \(c_{1}(n)=9.5\) and \(c_{0}(n)=0.05556\). reject \(H_{0}\) if \(L_n > 9.5\), and do not reject \(H_{0}\) if \(L_n<0.05556\). Continue with the test if the value of \(L_n\) lies between these two boundaries.

Step-by-step solution

SPRT Setup

We want to test the null hypothesis \(H_{0}: \theta = \theta^{\prime} = 4\) against the alternative hypothesis \(H_{1}: \theta = \theta^{\prime \prime} = 9\). The likelihood ratio for \(n\) observations \(X_{1},... ,X_{n}\) under \(H_{0}\) and \(H_{1}\) is given by the ratio of the densities, raised to the power \(n\), and multiplied by the exponential of the statistic. In this case, the likelihood ratio \(L_n\) is: \(L_n = \left( \frac{\theta^{\prime}}{\theta^{\prime \prime}} \right)^{\frac{n}{2}} \mathrm{exp}\left( \frac{1}{2} \left( \frac{1}{\theta^{\prime}} - \frac{1}{\theta^{\prime \prime}} \right) \sum_{1}^{n} X_{i}^{2} \right)\)

Decision Boundaries

The decision boundaries \(c_{0}(n)\) and \(c_{1}(n)\) can be computed using the formulae for SPRT: - Reject \(H_{0}\) for \(L_n\) > \(c_{1}(n)\) = \(\frac{1-\alpha_{a}}{\beta_{a}}\). - Do not reject \(H_{0}\) for \(L_n\) < \(c_{0}(n)\) = \(\frac{\alpha_{a}}{1-\beta_{a}}\). Substituting the given values \(\alpha_{a}=0.05, \beta_{a}=0.10\) gives: \(c_{1}(n)=\frac{1-0.05}{0.10}=9.5\), \(c_{0}(n)=\frac{0.05}{1-0.10} = 0.05556\)

Final Statement

For each \(n\), calculate \(L_n\). Proceed to the next \(X\) if \(c_{0}(n) < L_n < c_{1}(n)\), otherwise stop and make a decision: If \(L_n < c_{0}(n)\), fail to reject the null hypothesis \(H_{0}\), and if the \(L_n > c_{1}(n)\), reject the null and accept the alternate hypothesis \(H_{1}\).