Rao-Blackwell builds an unbiased estimate of p squared

Question

You flip a coin with unknown head probability $p$ , collecting i.i.d. outcomes $X_1, \dots, X_n$ (each $1$ for heads, $0$ for tails). You want to estimate $p^2$ , the probability that two specific flips both come up heads. A crude unbiased estimator is $T = X_1 X_2$ , since $\mathbb{E}[X_1 X_2] = p\cdot p = p^2$ by independence.

Rao-Blackwellize $T$ by conditioning on the sufficient statistic $S = \sum_i X_i$ , and give the resulting unbiased estimator of $p^2$ .

Accepted Answer

Setup. The total number of heads S = \sumi Xi is sufficient for p. Rao-Blackwell says T^\star = \mathbb{E}[X1 X2 \mid S] is unbiased for p^2 and has variance no larger than the crude indicator. Condition on the total. Given S = s heads spread across n flips, all \binom{n}{s} arrangements are equally likely. The product X1 X2 equals 1 exactly when both flip 1 and flip 2 are heads. Choosing which flips are heads without replacement, \mathbb{E}[X1 X2 \mid S = s] = P(	ext{flip 1 and flip 2 both heads} \mid s	ext{ heads}) = \frac{s}{n}\cdot\frac{s-1}{n-1}. So the improved estimator is T^\star = …

Rao-Blackwell builds an unbiased estimate of p squared

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