The variance floor for a geometric success probability

Question

You run a sequence of independent trials, each succeeding with probability $p$ , and record how many trials it takes to reach the first success. Repeating this $n$ times gives i.i.d. geometric counts $X_1, \dots, X_n$ on $\{1, 2, 3, \dots\}$ , with probability mass $f(x;p) = (1-p)^{x-1}p$ .

Compute the Fisher information for $p$ and state the Cramer-Rao lower bound on the variance of any unbiased estimator of $p$ .

Accepted Answer

Score of one observation. The log mass is \log f = (x-1)\log(1-p) + \log p, so the score is s(x;p) = \frac{\partial}{\partial p}\log f = -\frac{x-1}{1-p} + \frac{1}{p}. Fisher information by the curvature form. Differentiate the score again: \frac{\partial^2}{\partial p^2}\log f = -\frac{x-1}{(1-p)^2} - \frac{1}{p^2}. Take minus its expectation, using \mathbb{E}[X] = 1/p so \mathbb{E}[X-1] = (1-p)/p: I(p) = \frac{\mathbb{E}[X-1]}{(1-p)^2} + \frac{1}{p^2} = \frac{(1-p)/p}{(1-p)^2} + \frac{1}{p^2} = \frac{1}{p(1-p)} + \frac{1}{p^2}. Put over a common denominator: I(p) = \frac{p + (1-p)}{p^2(1-p…

The variance floor for a geometric success probability

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