Standard error of a rare-event rate

Question

In a portfolio of n = 900 loans, a fraction \hat{p} = 0.10 defaulted. What is the standard error of the sample default rate?

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The default rate is a proportion, so its precision follows the proportion standard-error rule: 	ext{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. With \hat{p} = 0.10 and n = 900: 	ext{SE} = \sqrt{\frac{0.10 	imes 0.90}{900}} = \sqrt{\frac{0.09}{900}} = \sqrt{0.0001} = 0.01. So the 10\% default rate is measured to about 1 percentage point. Rarer events (small \hat{p}) have a smaller outcome spread \hat{p}(1-\hat{p}), which helps keep the standard error down. Takeaway. The proportion standard error \sqrt{\hat{p}(1-\hat{p})/n} shrinks both with a larger sample and with a more lopsided rate: a 10…

Standard error of a rare-event rate

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