The variance floor for estimating signal power

Question

Readings X1, \dots, Xn are i.i.d. Normal with unknown mean \mu and known variance \sigma^2. You want the signal power g(\mu) = \mu^2. State the Cramer-Rao lower bound for an unbiased estimator of \mu^2, and comment on the naive plug-in \bar X^2.

Accepted Answer

Fisher information for the mean. For a normal with known variance, the score for one reading is (x-\mu)/\sigma^2, so I(\mu) = \frac{1}{\sigma^2}. Transform to \mu^2. With g(\mu) = \mu^2, g'(\mu) = 2\mu, \qquad [g'(\mu)]^2 = 4\mu^2. So 	ext{CRLB} = \frac{[g'(\mu)]^2}{n\,I(\mu)} = \frac{4\mu^2}{n/\sigma^2} = \frac{4\mu^2\sigma^2}{n}. The naive plug-in is biased. The obvious estimator \bar X^2 is not unbiased, because \mathbb{E}[\bar X^2] = \operatorname{Var}(\bar X) + (\mathbb{E}\bar X)^2 = \frac{\sigma^2}{n} + \mu^2, so it overstates \mu^2 by \sigma^2/n. The bias-corrected version \widehat{\m…

The variance floor for estimating signal power

Your answer

Discussion

Learn the concepts

Related questions