The large-sample spread of an estimated proportion

Question

Flips X1, \dots, Xn are i.i.d. Bernoulli(p), and the maximum likelihood estimator of the head probability is the sample proportion \hat p = \bar X. Find the large-sample distribution of \hat p and confirm it is asymptotically efficient. What is its asymptotic variance factor?

Accepted Answer

General result. The maximum likelihood estimator satisfies \sqrt{n}\,(\hat p - p) \xrightarrow{d} N\!\left(0, \frac{1}{I(p)}\right). Fisher information. For one Bernoulli flip, I(p) = \frac{1}{p(1-p)}, \qquad \frac{1}{I(p)} = p(1-p). Hence \sqrt{n}\,(\hat p - p) \xrightarrow{d} N\big(0,\ p(1-p)\big), so the asymptotic variance factor is p(1-p) and the standard error is \sqrt{p(1-p)/n}. Direct check. This also follows straight from the central limit theorem, since \hat p is an average of Bernoulli variables each with variance p(1-p). Here the agreement is exact even in finite samples: \operato…

The large-sample spread of an estimated proportion

Your answer

Discussion

Learn the concepts

Related questions