The square of an unbiased mean is not unbiased

Question

You draw n independent observations with mean \mu and variance \sigma^2 and compute the sample mean \bar{X}, which is unbiased: \mathbb{E}[\bar{X}] = \mu. A colleague reports \bar{X}^2 as an estimate of \mu^2. Is \bar{X}^2 an unbiased estimate of \mu^2? If not, which way is it biased, by how much, and how would you correct it?

Accepted Answer

Direction, from Jensen. The map g(x) = x^2 is strictly convex (g'' = 2 0), so Jensen gives \mathbb{E}[g(\bar X)] \ge g(\mathbb{E}[\bar X]), i.e. \mathbb{E}[\bar X^2] \ge \mu^2. The square of an unbiased mean is biased upward. Exact size. Here there is no need to approximate; the variance identity gives it exactly: \mathbb{E}[\bar{X}^2] = \mathrm{Var}(\bar{X}) + \big(\mathbb{E}[\bar{X}]\big)^2 = \frac{\sigma^2}{n} + \mu^2. So \bar{X}^2 overestimates \mu^2 by exactly \sigma^2/n, the variance of the sample mean. (This matches the delta-method formula 	frac12 g''(\mu)\,\mathrm{Var}(\bar X) = 	f…

The square of an unbiased mean is not unbiased

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