When false alarms are the expensive error

Question

A trading alert reads X. Under "no trade" (H0), X \sim 	ext{Normal}(0, 1); under "trade" (H1), X \sim 	ext{Normal}(2, 1). The two states are equally likely, \pi0 = \pi1 = 0.5. Acting on a false alarm costs 8 units (a bad trade), while missing a real opportunity costs only 1 unit. You declare "trade" 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 = 1, \sigma^2/(\mu1-\mu0) = 1/2, and \frac{\pi0 cI}{\pi1 c{II}} = \frac{0.5 	imes 8}{0.5 	imes 1} = 8. c = 1 + 	frac{1}{2}\ln(8) = 1 + 0.5 	imes 2.079 = 1 + 1.040 \approx 2.04. Read it. This is the mirror image of the costly-miss case. Because a false alarm now costs 8 times a miss, the optimal cutoff rises above the midpoint of 1, so the system demands strong evidence before it declares "trade" and rarely fires by mistake. The price is …

When false alarms are the expensive error

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