The Cramer-Rao floor and how efficient the mean is

Question

The Cramer-Rao lower bound says no unbiased estimator of a parameter can have variance below $1/[n\,I(\theta)]$ , where $I(\theta)$ is the Fisher information per observation.

For a normal sample with known variance $\sigma^2$ , state the bound for estimating $\mu$ , then compare the efficiency of the sample mean and the sample median.

Accepted Answer

For Xi \sim N(\mu, \sigma^2) the Fisher information about \mu is I(\mu) = 1/\sigma^2 per observation, so the Cramer-Rao lower bound is \operatorname{Var}(\hat\mu) \ge \frac{1}{n\,I(\mu)} = \frac{\sigma^2}{n}. Sample mean. \operatorname{Var}(\bar X) = \sigma^2/n, exactly the bound. The mean is efficient: no unbiased estimator does better. Sample median. Its large-sample variance is \frac{\pi}{2}\cdot\frac{\sigma^2}{n} \approx 1.57\,\frac{\sigma^2}{n}, above the floor. Its efficiency is \frac{\sigma^2/n}{(\pi/2)\sigma^2/n} = \frac{2}{\pi} \approx 0.64, so the median effectively uses only 64\% o…

The Cramer-Rao floor and how efficient the mean is

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