Above the Threshold

Question

An auction's clearing price is uniformly distributed on [0, 1]. You learn only that the print came in above 0.60. What is the expected clearing price given that information?

Accepted Answer

Condition directly. For u in (0.6, 1), f(u mid U > 0.6) = fracf(u)P(U > 0.6) = frac10.4 = 2.5, a constant. A uniform conditioned to land in a sub-interval is uniform on that sub-interval, so the conditional mean is the midpoint: E[U mid U > 0.6] = frac0.6 + 12 = 0.8. Integral check, the long way: E[U mid U > 0.6] = fracint_0.6^1 u, duP(U > 0.6) = frac(1 - 0.36)/20.4 = frac0.320.4 = 0.8. The trap is forgetting to renormalize: int_0.6^1 u,du = 0.32 is not a conditional mean, and neither is anything involving the unconditional mean 0.5. Conditioning throws away the part of the distribution that is ruled out and rescales what survives. Worth registering the contrast with the exponential: an exponential conditioned on exceeding t keeps its full mean ahead of it (memorylessness), while the uniform has increasing hazard, so the expected overshoot beyond the threshold, here 0.2, shrinks as the threshold climbs. Knowing which way a distribution ages is the fast way to sanity-check any conditional-mean answer. On a desk this is the simplest case of reasoning about censored or filtered prints: average fill given the fill beat a limit, average size given size cleared a reporting threshold. The

Above the Threshold

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