Is one over an unbiased mean an unbiased rate?

Question

You estimate the average time per task with the sample mean $\bar X$ , which is unbiased for the true mean $\mu > 0$ . You then report the throughput rate $1/\mu$ using the plug-in $1/\bar X$ .

Is $1/\bar X$ an unbiased estimator of $1/\mu$ ? If not, which way is it biased?

Accepted Answer

Unbiasedness does not survive nonlinear maps. Being unbiased means \mathbb{E}[\bar X] = \mu. But expectation does not pass through a nonlinear function: in general \mathbb{E}[g(\bar X)] 
e g(\mathbb{E}[\bar X]). So there is no reason to expect \mathbb{E}[1/\bar X] = 1/\mu, and indeed it fails. Which way it breaks, by Jensen. The function g(x) = 1/x is strictly convex for x 0. Jensen's inequality says that for a convex function, \mathbb{E}\!\left[\frac{1}{\bar X}ight] \frac{1}{\mathbb{E}[\bar X]} = \frac{1}{\mu}. So 1/\bar X is biased high: it overestimates the rate on average. The strict i…

Is one over an unbiased mean an unbiased rate?

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