Averaging then inverting is not inverting then averaging

Question

You have an unbiased estimate \hat{R} of some positive quantity R (say an average fill rate or a price), so \mathbb{E}[\hat{R}] = R. You then report 1/\hat{R} as your estimate of 1/R (an average time-per-fill, say). Is 1/\hat{R} an unbiased estimate of 1/R? If not, which way is it biased and by roughly how much?

Accepted Answer

Direction, from Jensen. For x 0, g(x) = 1/x has g''(x) = 2/x^3 0, so it is strictly convex. Jensen's inequality says for convex g, \mathbb{E}[g(X)] \ge g(\mathbb{E}[X]), strictly unless X is constant. Hence \mathbb{E}\!\left[\frac{1}{\hat{R}}ight] \ \ \frac{1}{\mathbb{E}[\hat{R}]} = \frac{1}{R}. The reciprocal of an unbiased estimate is biased upward: inverting a noisy average systematically overestimates the true reciprocal. Size, from the delta method. Expand g(\hat{R}) = 1/\hat{R} to second order around R: \mathbb{E}[g(\hat{R})] \approx g(R) + 	frac12 g''(R)\,\mathrm{Var}(\hat{R}) = \fr…

Averaging then inverting is not inverting then averaging

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