Estimating the Fleet From Four Serial Numbers

Question

A venue stamps every order with a sequential id starting at 1. You observe four orders with ids 22, 7, 41, and 15. Estimate how many orders the venue has processed in total, and justify why your estimator beats simply answering 41.

Accepted Answer

Why the maximum is biased Let N be the total and let m be the largest of n observed ids. Since m le N always, using m as the estimate is systematically too low. The question is by how much. Think of the n observations as cutting 1, dots, N into n+1 gaps. By symmetry each gap has the same expected size, so each has expected width fracNn+1, including the gap above the largest observation. So on average E[m] = N - fracNn+1 = Nfracnn+1. The unbiased estimator Invert that relation and replace the expectation by the observation: hatN = mfracn+1n = m + fracmn - 1 quadtext(discrete form) qquad hatN = m + fracmn - 1. With m = 41 and n = 4: hatN = 41 + frac414 - 1 = 41 + 10.25 - 1 = 50.25 approx 50. Sanity checks worth saying out loud The estimate must be at least m, and it is. It approaches m as n grows, which is right: with many samples the largest one is nearly the true top. It adds a large correction when n is small, which is also right: four samples tell you little, so you should assume you have not seen near the top. This is the classic wartime estimator applied to serial-numbered anything, and interviewers like it because the naive answer (41) is defensible-sounding and wrong, while t

Estimating the Fleet From Four Serial Numbers

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