The variance floor for a Bernoulli variance

Question

From n flips of a coin with head probability p, you want to estimate the variance of a single flip, v = g(p) = p(1-p). State the Cramer-Rao lower bound for an unbiased estimator of v, and interpret what happens at p = 1/2.

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The transformed bound. For target g(p), \operatorname{Var}(T) \ge \frac{[g'(p)]^2}{n\,I(p)}, \qquad I(p) = \frac{1}{p(1-p)}. Derivative. With g(p) = p(1-p) = p - p^2, g'(p) = 1 - 2p, \qquad [g'(p)]^2 = (1 - 2p)^2. Assemble. 	ext{CRLB} = \frac{(1-2p)^2}{n/[p(1-p)]} = \frac{(1-2p)^2\,p(1-p)}{n}. The vanishing floor at p = 1/2. At p = 1/2 the derivative g'(1/2) = 0, so the floor is exactly 0. The reason is that v = p(1-p) is at its maximum there, the function is flat, so first-order errors in p do not move v at all. The bound going to zero does not mean you can estimate v with no error; it mean…

The variance floor for a Bernoulli variance

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