Is the squared sample proportion unbiased for p squared?

Question

The sample click rate \hat p = k/n is unbiased for the true click probability p. To estimate p^2, the probability that two independent impressions both get clicked, you square it: \hat p^2. Is \hat p^2 an unbiased estimator of p^2? If not, which way is it biased, and by exactly how much?

Accepted Answer

Squaring breaks unbiasedness. We have \mathbb{E}[\hat p] = p, but squaring is nonlinear, so \mathbb{E}[\hat p^2] 
e (\mathbb{E}[\hat p])^2 in general. Compute it exactly. Use the identity \mathbb{E}[\hat p^2] = \operatorname{Var}(\hat p) + (\mathbb{E}[\hat p])^2. Since \operatorname{Var}(\hat p) = p(1-p)/n, \mathbb{E}[\hat p^2] = \frac{p(1-p)}{n} + p^2 = p^2 + \frac{p(1-p)}{n}. So \hat p^2 is biased high by exactly \frac{p(1-p)}{n}. (This also matches Jensen: g(x) = x^2 is convex, so \mathbb{E}[\hat p^2] p^2.) How to fix it. The bias is the variance term, so subtract an unbiased estimate of …

Is the squared sample proportion unbiased for p squared?

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