Is the plug-in estimate of the odds consistent?

Question

An ad is shown n times and clicked with unknown probability p (with 0 < p < 1). You estimate the odds of a click, \omega = p/(1-p), by plugging in the sample click rate \hat p = k/n: \hat\omega = \frac{\hat p}{1 - \hat p}. Is \hat\omega a consistent estimator of the true odds \omega?

Accepted Answer

Step 1: the click rate converges. The clicks are i.i.d. Bernoulli(p), so by the law of large numbers the sample proportion converges in probability to the true probability: \hat p = \frac{k}{n} \xrightarrow{p} p. Step 2: push through the odds map. The function g(t) = t/(1-t) is continuous everywhere except at t = 1. Since the true value p is strictly below 1, g is continuous at p, and the continuous mapping theorem gives \hat\omega = g(\hat p) \xrightarrow{p} g(p) = \frac{p}{1-p} = \omega. So the plug-in odds estimator is consistent. A caution on bias. Consistency does not mean unbiasedness. …

Is the plug-in estimate of the odds consistent?

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