The variance floor for estimating the log-odds

Question

From n flips of a coin with head probability p, you want to estimate the log-odds (the logit), \eta = g(p) = \log\dfrac{p}{1-p}, the natural scale in logistic regression. State the Cramer-Rao lower bound for an unbiased estimator of \eta.

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The transformed bound. For a smooth target g(p), \operatorname{Var}(T) \ge \frac{[g'(p)]^2}{n\,I(p)}, \qquad I(p) = \frac{1}{p(1-p)}. Derivative of the log-odds. Write g(p) = \log p - \log(1-p). Then g'(p) = \frac{1}{p} + \frac{1}{1-p} = \frac{1}{p(1-p)}, \qquad [g'(p)]^2 = \frac{1}{p^2(1-p)^2}. Assemble. 	ext{CRLB} = \frac{1/[p^2(1-p)^2]}{n/[p(1-p)]} = \frac{p(1-p)}{n\,p^2(1-p)^2} = \frac{1}{n\,p(1-p)}. A neat observation. The log-odds has the reciprocal of the Fisher information as its floor, 1/[np(1-p)], and this is smallest at p = 1/2 (where p(1-p) is largest). So the log-odds is easiest…

The variance floor for estimating the log-odds

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