Cost-optimal cutoff with a wider spread

Question

A sensor reads X. Under "normal" (H0), X \sim 	ext{Normal}(10, 2) with prior \pi0 = 0.95. Under "fault" (H1), X \sim 	ext{Normal}(18, 2) with prior \pi1 = 0.05. A false alarm costs 2 units; a missed fault costs 50 units. You declare "fault" when X c. Find the cost-minimizing cutoff c.

Accepted Answer

The optimal threshold. c = \frac{\mu0 + \mu1}{2} + \frac{\sigma^2}{\mu1 - \mu0}\,\ln\!\left(\frac{\pi0\, cI}{\pi1\, c{II}}ight). Plug in. (\mu0+\mu1)/2 = 14, \sigma^2/(\mu1-\mu0) = 4/8 = 0.5, and \frac{\pi0 cI}{\pi1 c{II}} = \frac{0.95 	imes 2}{0.05 	imes 50} = \frac{1.9}{2.5} = 0.76. c = 14 + 0.5 	imes \ln(0.76) = 14 + 0.5 	imes (-0.274) = 14 - 0.137 \approx 13.86. Read it. Two forces nearly balance. Faults are rare (\pi1 = 0.05), which pushes the cutoff up; but a miss costs 25 times a false alarm, which pulls it down. The cost asymmetry slightly wins, so the cutoff lands just below th…

Cost-optimal cutoff with a wider spread

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