Power of an alarm rule from its threshold

Question

An early-warning system reads a value X. In calm conditions (H0), X \sim 	ext{Normal}(0, 2). In alert conditions (H1), X \sim 	ext{Normal}(5, 2). The rule raises the alarm whenever X 3. What is the power of this rule (the probability it correctly raises the alarm when conditions truly warrant it)?

Accepted Answer

Power is the chance the alarm fires when H1 is true, that is, X 3 under the alert distribution. Compute power. Under H1, X \sim 	ext{Normal}(5, 2), so standardize with \sigma = 2: 	ext{power} = P(X 3 \mid H1) = 1 - \Phi\!\left(\frac{3 - 5}{2}ight) = 1 - \Phi(-1) = \Phi(1) \approx 0.841. So the alarm correctly fires about 84% of the time; the type II error (missed alert) is the remaining 1 - 0.841 \approx 0.159. For context. The false-alarm rate is \alpha = 1 - \Phi((3-0)/2) = 1 - \Phi(1.5) \approx 0.067. Takeaway. Power from a fixed cutoff is the signal distribution's mass on the "declare…

Power of an alarm rule from its threshold

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